Talk:Current source

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Looking at the information shown here there is nothing showing a circuit that has a jfet in it. See http://www2.eng.cam.ac.uk/~dmh/ptialcd/csink/csink.htm for an example. I am willing to draw up such a circuit. Chaney44145 (talk) 18:50, 28 December 2007 (UTC)[reply]

Heh. That's from my engineering course 1st year (currently in second year). Anyway, my actual point is related... the symbol we've been taught to use for a current source is different to those used here (we use two overlapping circles to represent a current source)... unless my university is completely out of touch, there must be a lot of people who are used to seeing this - could it be listed as an alternative symbol scheme? Whose standard is the one adopted by wikipedia? Andipi (talk) 08:50, 20 May 2009 (UTC)[reply]

Against: Galvanostat article is more technical, this keeps more general. Tswsl1989 12:06, 8 May 2007 (UTC)[reply]

-) Honestly,(Light current) I am getting somewhat stressed since you arrived. I understand your motives and you have a lot of good knowledge to contribute, but we obviously clash on what's appropriate info to include, and I feel like I have to closely follow all your edits to make sure you're not breaking anything. (And your enthusiasm and spending lots of time on here is great, and I'm glad you're contributing so much, but now I'm spending a lot more time on here than I normally would trying to keep up.)  :-)Omegatron 17:51, August 9, 2005 (UTC)

We need diagrams for the current mirror type source to make it more understandable.Light current 15:05, 3 August 2005 (UTC)[reply]

The current mirror article already covers this. - Omegatron 15:39, August 3, 2005 (UTC)

Doesn't belong in article. Not a current source. Same argument as Talk:Voltage source#Potential divider "source"?? - Omegatron 14:53, August 8, 2005 (UTC)

Wrong again. look here http://sound.whsites.net/ism.htm#intro.

How is that any different from your previous arguments? - Omegatron 17:22, August 12, 2005 (UTC)

Its not, I'm just reinforcing them by using a source which you appear to respect. Also I have put this link right at the end of the para on resistor curent sources to reaassure any (Doubting Thomases). (New Testament-- disciple of Jesus who would not believe that he had risen from the dead). Light current 17:52, 12 August 2005 (UTC)[reply]

Anything that tends to provide a current can be called acurent source. Anything that tends to provide a voltage can be called a voltage source. Can you explain more fully what you see as the difference between these simple sources and those using semiconductors(aprat from the fact that their performanne is genaerally worse - but hey -- no ones perfect!!). Just repeating that you feel its not right does not advance the argument.

Also would you consider an inductor to be a current source or not? Light current 15:48, 8 August 2005 (UTC)[reply]

Take look at one of WPs star articles electronic amplifier under Class E and tell me if you agree with them describing an inductor as a source of constant current. Light current 16:02, 8 August 2005 (UTC)[reply]

I'm not just repeating what I feel. I said the same arguments apply that I made over at Talk:Voltage source: They don't have any immunity to changes in loading and therefore don't approximate a voltage or current source at all.

No, an inductor's not a current source, either. It smooths out variations in the supply, but if the load changes and stays changed, the current changes, too. - Omegatron 17:27, August 8, 2005 (UTC)

Most people seem to want to limit the discussion of current sources to those of which they have heard and are not prepared to look at the wider view of the concept.

If people want to limit this page's content to current source circuits based on semiconductors then that is fine by me. But then it should be called 'current sources made from semiconsuctors' or such like and we put the other sorts of current sources in a different article. Exactly the same argument I apply to te voltage source page where people are trying to limit content to what they have personal experience of. Surely Wikipedia is not intended to be narrow minded like that is it?? Comments please from other interested parties Light current 16:22, 8 August 2005 (UTC)[reply]

Unfortunatley ther seems to be a little confusion over the terms describing current sources in this article. Some of the terms that have already been used and applied (too) loosely are:

Ideal

Non_ideal

Independent

Non-independent (or controlled)(CCS)

Practical

Theoretical

Voltage controlled (VCCS)

Current controlled (CCCS)

(Real world)??

In order to avoid gross confusion and endless arguments it is important for editors to understand the proper meaning of these terms before attempting to use them in this article. Any good book on electrical or electronic engingeering should define these terms adequately.Light current 16:47, 8 August 2005 (UTC)[reply]

Certainly a good idea. Some words, like "practical" already have obvious meanings, but we should define what we mean by "independent current source" the first time it is used. - Omegatron 17:22, August 8, 2005 (UTC)

Now speaking from another universe but we still have the internet here:-)).

Just a thought to try to resolve this issue. If we we put all the types of current (voltage) sources that 'O' and others like under heading called Active Device Sources and put the others under another heading (in the same article )would that be acceptable to everyone. Please discuss.Answers in no more than 50 words pls Dave Bowman Light current 20:03, 8 August 2005 (UTC)[reply]

How about this for terminology, this may seem obvious but shouldn't the simple transistor constant current source be called a constant current sink? It is not sourcing current, but rather sinking current from the load...Just a thought.

I agree with this last comment - all the circuits described are current sinks rather than sources. Mike. 80.195.172.5 (talk) 10:05, 7 March 2012 (UTC)[reply]

• Note that the same circuit (e.g. a JFET with gate-source tied) coul be used as a current source or a current sink, and the term used really depends on your perspective. One could say that a current sink has to go to ground while a source is a supply that drives a load that is connected to ground; that makes some sense but it is simply not the definition (think of circuits inside opamps with balanced power supplies and the current doesn't really involve earth much at all). Or if a source is one where that uses a PNP transistor, or comes from a positive supply instead of negative supply... all these aren't reliabile definitions of the difference (and also ask whether we are talking electron flow or conventional current... and this generation normally assumes negative earth but when positive-earth was more common in the 1960's did all the textbooks flip their definitions of source and sink?). I think the answer is that current source circuits are current sink circuits and vica versa, and writers typically settle on one name (often "source") simply to avoid writing the long phrase: "current source/sink (depending on your perspective)". Whether it is sinking or sourcing current depends on the circumstances - what is around it and what "sense" the current flow has in one person's analysis of it (it is quite valid, for example, to say something sources a negative current). Also think about some push-pull output stages with (intentionally) very high output impedances that are essentially current sources, but do we call then current sinks on the negative half of each cycle? Maitchy (talk) 01:01, 16 August 2014 (UTC)[reply]

I have some questions on the bootstrap insertion.

1) It says the bootstrap cct is used to bias the class A stage. But this is the Class A stage so how is the transistor biased by the this cct?

2) How is the collector current of Q determined by R1 or R2 exactly?

3) Could you explain your last paragraph in more detail

THanks Light current 21:36, 9 August 2005 (UTC)[reply]

More questions:

When I studied electronics at university (many years ago now), we were taught that bootstrapping is a form of positive feeback used to increase the input impudance of a stage and improve the dynamic range of large signal circuits. Your circuit does not have any feedback so can you say how your circuit achieves these two aims? Or has the definition of bootstrapping in the high end audio world been changed without my knowledge?Light current 21:49, 9 August 2005 (UTC)[reply]

I don't understand this circuit. I think the major problem is that half the circuit is missing. Here's an explanation: http://sound.whsites.net/ism.htm#bootstrap - Omegatron 00:11, August 10, 2005 (UTC)

Yes I agree. This is where the circuit was probably sourced. The ESP article is very good and I have downloaded it for further study. We must be careful not to get into copyright problems here though!. I suggest the full cct is drawn independently. I also agree with the description Rod Elliot has written about bootstrap operation. I also agree that it does provide a constant current drive to the output stage (sort of)and therfore should be left in in its modded form.So what now? Light current 00:24, 10 August 2005 (UTC)[reply]

See "Temp home for Bootstrap para" below for calculation. Rohitbd 08:29, August 10, 2005 (UTC)

Added references section, the page pointed out by omegatron is the source for the boot-strap section here. Now, should this section be moved back to the article or is it best omitted...? Rohitbd 08:55, August 10, 2005 (UTC)

If you copied that schematic from the Westhost site, it is missing a transistor. I will redraw it if you'd like.

Copyright doesn't cover information; only presentation of information. If we provide more than one reference and put it in our own words and draw our own images we are fine. - Omegatron 07:51, August 11, 2005 (UTC)

I did not copy the schematic from that website. I have drawn it, and added the text from my own understanding after reading the article. The second transistor in the source article is an emitter follower stage (unity voltage gain), and hence (IMHO) can be omitted for the purpose of describing the boot-strap CCS operation. I have already simulated the schematic as I drew it here on my PC and it works exactly as I described it here. If you wish, I can send the results (wave-forms) of the simulated analysis to you. However, if you would like to make the schematic more complete, please go ahead. I also request you to go through the analysis that I have given below and correct it as may be required, although I am quite confident that it is correct in essence - may be it needs to be re-worded. Rohitbd 08:44, August 11, 2005 (UTC)

Although not as good as the active constant current sources depicted above, the boot-strap current source is commonly found in many audio power amplifiers to bias the class-A voltage gain stages in them. These offer a cheap alternative to active sources where the output impedance of the CCS is of secondary concern. As can be seen in the image on the left, a boot-strap CCS is formed by resistors R1, R2 and capacitor CB. Transistor Q is wired as a class-A voltage gain stage. The circuit works as follows:

At no signal (i.e., DC conditions), CB acts as an open circuit and the collector current is determined simply by (R1 + R2).

NO ITS NOT I asked this question of you before; you seem to have missed it.! The collector current is determined by the base current. Please revise your transistor theory! Light current 15:35, 10 August 2005 (UTC)[reply]

Excuse me? how does the base current determine the collector current?? simply by the transitor's Hfe?? how?? Rohitbd 18:53, August 10, 2005 (UTC)

Yes, ${\displaystyle I_{C}=\beta I_{B}}$, to a first approximation. For the common emitter circuit without emitter resistor (which is what this is, no?), a better approximation is ${\displaystyle I_{C}=\beta {r_{o} \over r_{o}+R_{C}}\cdot I_{B}}$ [1], which takes into account the "Early effect". r_o is the transistor's internal output resistance (>100 kΩ) and R_C is the collector resistance (10 kΩ here). The collector current is then affected by the collector resistor as well. So LC's statements that the collector current is determined solely by the base current aren't totally true. This would have a pretty small effect, though, less than 10% difference, so I'm not sure if it's relevant to this circuit. More info here (section 3): [2] [3] [4] These approximations are for the active mode only, and you can keep adding more and more terms to the equations to get better and better approximations, taking into account internal capacitances of the transistor, saturation and cutoff, etc. It can get tremendously complicated: [5] - Omegatron 22:28, August 14, 2005 (UTC)

The equation you have quoted seems correct but it does not include the Early effect as you said. The Early effect modification/addition has been quoted (correctly I believe) by Alfred Centuari lower down this page. Anyway, if anybody looks at the characteristic curves of a transistor, it is blindingly obvious that the collector current is mainly controlled by the base current and that only a small variation is cause by the collector voltage (indicated by the current curves not being completely horizontal). In the bootstrap circuit this effect is NOT used to bias the transistor. It may move the Q point a bit but nowhere near enough to worry about. This is a Minor , minor effect (unless you are working somewhere near the breakdown voltage of the transistor) and in practice is never taken into account in design calculations for lowish voltage stages Light current 22:56, 14 August 2005 (UTC)[reply]

Ah. Didn't see his reply. Yes, they're the same equation. ${\displaystyle r_{o}={V_{A} \over I_{C}}}$, where I_C is the collector current corresponding to a given V_CE

While the dependence on V_CE is not usually included in DC bias design parameters, the value of the equivalent r_o can have a significant effect on the gain of the transistor (for 10 kΩ and 100 kΩ, as I showed, the effect is to reduce the gain by ~9%). I don't understand this circuit (probably because half of it is missing), so I'm not sure whether this is relevant. - Omegatron 23:31, August 14, 2005 (UTC)

This is just a nitpick so I'm prepared to be flamed but ${\displaystyle r_{o}={\frac {V_{A}+V_{CE}}{I_{C}}}}$. Of course, ${\displaystyle V_{A}}$ is usually an order of magnitude greater than ${\displaystyle V_{CE}}$ so the equation you gave above is commonly used. However, I would like to point out that ${\displaystyle r_{o}}$ is a small-signal parameter and the equations I gave are large signal equations where the concept of ${\displaystyle r_{o}}$ is invalid. For the record, I use the following notation convention:

${\displaystyle I_{C}}$ - quiescent (DC) collector current

${\displaystyle i_{c}}$ - small-signal collector current

${\displaystyle i_{C}}$ - total (DC + small-signal) collector current

By the way, a very good summary of BJT equations is located here: http://users.ece.gatech.edu/~mleach/ece3050/notes/bjt/bjtsumm.pdf. I'm hopeful that you two are familiar with professor Leach or at least have heard of the 'Leach Amp'. Alfred Centauri 02:26, 15 August 2005 (UTC)[reply]

"This is just a nitpick so I'm prepared to be flamed but ${\displaystyle r_{o}={\frac {V_{A}+V_{CE}}{I_{C}}}}$."

Approximations on approximations. Let's just say ${\displaystyle I_{C}\propto I_{B}}$ and be done with it. :-)

That summary is very helpful. No, I have not heard of the Leach amp. - Omegatron 03:14, August 15, 2005 (UTC)

Assume that R1 and R2 are chosen such that the DC bias current (collector current) is much larger than the load connected to the collector. When a signal is applied to the transistor's base, the collector voltage will change. This change in voltage is coupled back to R2 by CB in such a way that the voltage across R2 remains relatively constant. Constant voltage across R2 means constant current through R2 which is also the collector current, and hence R1, R2 and CB form a CCS. This CCS is a passive current source, and the transistor Q forms the load for it.

The cap just shorts out R2 at ac. Therefore, as it stands, this circuit is NOT a current source as you desribe it. THere is NO pfb. Its output impedance is roughly R1 -- no higher. Please re read the article where you got the cct.Light current 15:30, 10 August 2005 (UTC)[reply]

That's exactly why the DC voltage across R2 remains constant. Instead of criticising, it would be far more helpful if you provide some objective explanation. Besides if YOU read the source article, you will see it agrees with what I said. Rohitbd 18:53, August 10, 2005 (UTC)

Calculation of R1 & R2: the collector voltage (V_C) is assumed to be at half the total supply (V_supply) - in this case since the emitter is shown floating, it could be either connected to an equal −Vcc or ground. In either case we assume V_C equals half the total supply.

THe rest of this argument must be discounted as it is based on 2 suppositions above that I have shown to be patently untrue. I suggest you delete your latest submission from this talk page to save yourself further embarassment. Light current 15:45, 10 August 2005 (UTC)[reply]

1. Relax, please.
2. Embarassment should be embraced, but please don't use it as an attack, or defensive behavior will result instead of belief restructuring.
3. No personal attacks - Omegatron 16:23, August 10, 2005 (UTC)

The reason for assuming that V_C is half the total supply is that this particlar configuration appears in a VFB design - similar to that of an op-amp in discrete form with a differential pair input stage and DC coupling internally. When negative feedback is applied around the ckt, the o/p (the collector of the transistor, in this case) always sits mid-way of the total supply - i.e., if we have a split ±V_CC, then the o/p will be at 0V. If a single supply (V_CC) is used then the o/p will be at V_CC/2.

If the next stage requires say, Y mA of current, then this transistor must be biased at atleast 1.2 to 2 times this for class-A operation. We take twice that (2Y mA) for calculation, call it X mA. Now, ${\displaystyle (R1+R2)={\frac {V_{supply}-V_{c}}{X}}}$ KΩ. Usually R1 = R2. Therefore ${\displaystyle R1=R2={\frac {V_{supply}-V_{C}}{2X}}}$ KΩ. Since V_C = V_supply / 2, therefore, ${\displaystyle R1=R2={\frac {V_{supply}}{4X}}}$ KΩ. The boot-strap capacitor is calculated simply by assuming a lowest frequency (f) of operation (say 1 Hz) and applying the formula ${\displaystyle CB={\frac {1}{2.\pi .f.R_{P}}}}$ Farads, where R_P = R1 || R2. Since most power amps have very little gain below about 10 Hz by design, it means that at low frequencies (as 1 or 2 Hz), R1 and R2 by themselves are sufficient to ensure adequate bias. For R1 = R2 = 2.2KΩ, and f = 1Hz, CB turns out to be about 114.7µF and most amps use the standard 100µF with no degradation in performance. Rohitbd 09:05, August 10, 2005 (UTC)

Light current, please back up your counter arguments with some objective formulas/explanations. Simply saying that the collector current is determined by the base current does not mean anything. In fact if base current solely determines the collector current, we simply wouldn't need any collector biasing - the collector resistor does determine the DC collector current. All the current sources depicted can be used as collector loads and their function is precisely to bias the collector in such applications. You have only made a lot of criticism and noise without actually trying to provide an alternative or better explanation. Atleast I tried to give an analysis of the ckt...and what did you do?? simply fired a statement or two, called upon me not to embarass myself. Stop this childishness please. This website is not your sole ownership and nobody is answerable to you. So you please mind your language!! Rohitbd 18:53, August 10, 2005 (UTC)

Please, please go look up some good books on electronics saying how transistors work then we will not need to have this pointless argument. Light current 01:34, 11 August 2005 (UTC)[reply]

Saying "go read some books, you don't know what you're talking about" isn't a good argument. Try explaining yourself. - Omegatron 01:59, August 11, 2005 (UTC)

Anyone who doesnt understand the basics of transistor operation should not be trying to edit this page. WP is not a place for editors to learn, it is a place for people with some knowledge to contribute that knowledge to the 'pedia. Any way if you have so much time to watch everyone else, why dont you try to explain transistor operation to Rotbild?? Light current 02:17, 11 August 2005 (UTC)[reply]

Why don't you try? - Omegatron 07:48, August 11, 2005 (UTC)

I have tried, my God how I ve tried. It just seems that R does not seem to wish to modify his erroneous view of how a transistor operates. I am banging my head against a brick wall here, and its starting to hurt! Light current 13:52, 12 August 2005 (UTC)[reply]

Any way if you have so much time to watch everyone else, why dont you try to explain transistor operation to Rotbild??

Rotbild....was that a typo?? I think not!! Omegatron/wiki administrators, this user "light current" is crossing his limits now. I take strong offence to the mangling of my username by light current and request the administrators of wikipedia to take some action. Rohitbd 07:55, August 11, 2005 (UTC)

Dont be so silly R Light current 13:52, 12 August 2005 (UTC)[reply]

Moved from my talk page Rohitbd 08:04, August 11, 2005 (UTC)

I'll probably regret sticking my nose into this debate but I can't resist - when a BJT is in the active mode, the collector current is a primarily an exponential function of the base-emitter voltage and secondarily a linear function of the collector-emitter voltage:

${\displaystyle i_{C}=I_{S0}e^{\frac {v_{BE}}{V_{T}}}\left(1+{\frac {v_{CE}}{V_{A}}}\right)}$

the base current is given by:

${\displaystyle i_{B}={\frac {1}{\beta }}I_{S0}e^{\frac {v_{BE}}{V_{T}}}}$

so we have:

${\displaystyle i_{C}=\beta i_{B}\left(1+{\frac {v_{CE}}{V_{A}}}\right)}$

Since the collector resistance affects ${\displaystyle v_{CE}}$, this resistance does have some effect on the collector current where the magnitude of this effect is a function of the Early voltage, ${\displaystyle V_{A}}$.

Secondly, the circuit as drawn above is incomplete so it is difficult to say what the heck it is. However, on thing is clear to me, the combination of R1, R2, and CB create a pole and zero in the transfer function of this circuit. The pole frequency is 1 / (R2 * CB) while the zero frequency is 1 / (R1||R2 * CB). In other words, at low frequencies, the collector 'sees' R1 in series with R2, at high frequencies, the collector 'sees' R1 only as CB shunts R2. I do not see any feedback at all here. Alfred Centauri 19:28, 13 August 2005 (UTC)[reply]

Collector voltage will have some effect (I think its called the Early efect or base width modulation) but not enough to bias the transistor. If it did have a large effect then none of our transistor circuits would work at all. You are correct that there is no feedback in this cct.(but there was before half the cct was stripped away by Rohitbd (follow the link) http://sound.whsites.net/ism.htm#bootstrap Light current 19:50, 13 August 2005 (UTC)[reply]

That's correct, the Early effect is typically considered in small-signal modeling of the BJT. However, for PNP transistors especially, this effect should be taken into account when solving the bias equations. For example, the Early voltage, ${\displaystyle V_{A}}$, might be of the order of 50V for a low-power PNP transistor. If the desired quiescent collector-emitter voltage is say, 10V, the quiescent collector current will by 20% higher than a 'back-of-the envelope' calculation that doesn't take the Early effect into account. By the way, we are in agreement that, to "zero'th" order, the quiescent collector current is determined by solving the bias equation for the base-emitter circuit and the quiescent collector voltage is then set by the external collector resistance Alfred Centauri 18:13, 14 August 2005 (UTC)[reply]

Agreed! Good point about PNPs . I try to avoid them whenever possible 18:54, 14 August 2005 (UTC)

I assume you wrote the 'bootstrap' piece in current sources.

THe trouble is that, as O pointed out, half the diagram is missing. I dont think it can be copied 'en bloc'. because of copyright. It needs redrawing fully before the text surrounding it can be discussed. Are you any good at drawing ccts??

It does seem to act as a current source drive and so maybe should be included. But maybe it should be included in an audio amplifier page. THis is where I think these are most used.:-)Light current 15:04, 10 August 2005 (UTC)[reply]

No comments. Rohitbd 08:59, August 11, 2005 (UTC)

Ic = BIb Where B is the gain of the transistor. This is the only way the collector current can be modulated. The collector has a very high resistance so its current is not affected by the voltage on it (much). Please try to get hold of some books on transistors. I think you will find them quite interesting and informative. :-) Light current 00:48, 11 August 2005 (UTC)[reply]

Please refrain from sending private messages to me. If you have anything to say, say it here. Rohitbd 08:04, August 11, 2005 (UTC)

I believe that R has unfortunately failed to realise the importance of the emitter follower in the bootstarp cct. THis is essential to the operation as it provides unity voltage (positve) feedback to the top of R2 making R2 look like a higher resistance than it really is. In the cct as R has drawn it, there is no feedback at all and R2 is just shorted out by C at signal frequencies. Does this make sense?

BTW Ic = BIb + Iceo.( where B is the gain of the transistor and Iceo is the leakage current). This is the basic transistor equation developed by Ebers + Moll. Collectors are not biased as you thought, it is the base emmitter junction that has to be biased to control the collector currrent in accordance with the above equation. THe collector terminal is a high impedance terminal that does not respond to applied voltages(much). THat is why the collector of a transistor is sometimes used as a current source (because it is a high impedance point) Do you understand it now? Light current 13:13, 11 August 2005 (UTC)[reply]

Light current, what is the meaning of ...provides unity voltage (positve) feedback to the top of R2 making R2 look like a higher resistance...? Could you please give a more qualitative explanation...a few equations would help, both with the emitter follower and without. Prove to me that the boot-strap ckt will not work without the emitter follower and also how you would determine R1, R2 and CB. Of course you are not obliged to do so, but since you have been so vehement in your debunking of my views, I think it is fair to ask you for some mathematical analyses of the boot-strap CCS. Rohitbd 13:56, August 11, 2005 (UTC)

If you dont know how a transistor operates, who are you to tell me about current sources??Light current 01:37, 11 August 2005 (UTC)[reply]

No one invited you to see it or comment on it. YOU saw the article. YOU took it personally that someone has written something that either YOU do not understand or does not agree with what YOUR textbook says. YOU think YOU know all of electronics. Well, IT'S YOUR PROBLEM!! Rohitbd 08:04, August 11, 2005 (UTC)

Please do not shout!. No I dont have any problem with my current knowledge of electronics. It is other people who have a problem in believing me. I am used to it. As I said before, I am very concerned that WP does not mislead readers. That is why I roam the pages to try to correct any errors. If I am not certain that they are in error, I either a) look it up in a text book, or b) leave it alone for someone with superior knowledge to fix. Its not what I THINK that's important here, it is what is CORRECT. Anyway a policy of WP is to supply proper references to the articles we write. Do you have a problem with that? Light current 17:00, 11 August 2005 (UTC)[reply]

It is other people who have a problem in believing me. That says it all. I don't want to argue with you any further...too bad I didn't see this earlier... Rohitbd 21:22, August 11, 2005 (UTC)

Rohitbd could look here *Bipolar junction transistor for info on transistors 212.74.96.201

It would be nice if 212.74.96.201 would identify him/herself. All said, it is true that Ic = Ib + Ie and that Hfe = Ic/Ib. But that is only as far as the transistor's DC current gain is concerned. While using a transistor in the active region in common emitter mode, we have to bias it - and biasing is done by setting the collector current and the Vce, with the base-emitter junction forward biased.

There is a very good easy to understand article here [http://www.faqs.org/docs/electric/Semi/SEMI_4.html on transistors. If Rohitbd scolls down to the characteristcs graph , he will see that the collector current curves are horizontal thus indicating that the collector current is not really affected by the collector voltage much, but it is affected by the base current. This is why the transistor can't be biased by altering things at the collector. Hope this help you Rohitbd! 212.74.96.201 14:19, 11 August 2005 (UTC)[reply]

Thank you for that link. Yes, the base current determines the collector current, but it cannot be used for biasing the transistor. My argument was about biasing a transistor. We cannot rely on solely the base current to bias a transistor. All the CCS (including bootstrap) can (and are) used instead of a plain collector resistor precisely to set the collector current at a fixed level - i.e. BIASING. A collector resistor or a CCS is used to drive a current (called the quiescient current or bias current denoted by Icq) through the collector - in the case of a CCS instead of a resistor, it acts to give a very large (open-loop) voltage gain (in approximate form, |Av| = hfe(Rc/Re), if Rc = infinite, |Av| = infinte. Now would you please identify yourself (is that you, Light current??). Rohitbd 15:15, August 11, 2005 (UTC)

Assume the bootstrap idea applied to a simple emitter follower with the emitter resistor being called R3. At the junction of the bias chain resistors(R1,R2)(that would normally be connected directly to the base) and the transistor base a resistor R4 is inserted. THe input to the emitter follower is applied to the transistor base as ususal. NOW, you connect a capacitor from the emitter to the bias end of R1 ( ie the end not connected to the base.) Let us assume that the capactitor is very large. THis is a simple BOOTSTRAP CIRCUIT

THe effective emitter load is R1||R2||R3. Let us call this R' For an input Vat the base, the voltage across R4 is V-KV wher k is the the voltage gain of the transistor and is approx 1. From the poin of view of input signal (ac) the input of the stage appears as follows: a simple emitter follower with a resitor R4 connected to the base. THe input to the stage is still directly at the base as normal, but the other end of R4 now as avoltage on it (from the emitteer) of KV (Yes?). So, the current thro this resistor R4 is: I =V(1-K)/R4. So the total input resistance as seen by the input signal is now the original input resistance in parallel with Reff. Where Reff= V/i= R4/(1-K) NOW, as K approaches 1, THe new input resistance approaches Rin which is roughly (1+hfe)R'. So the input resistance of the stage has been increased by the bootstrap and the effects of the bias chain are now negligible. THe important point is that an extra source of energy has to be used in a bootstrap circuit and this is provided by the current supplied to the capacitor from the transistor emitter. This is why your circuit does not work-- there is no extra source of energy invoked. Hope this resolves your difficulty. Light current 14:57, 11 August 2005 (UTC)[reply]

Well, you could said this before instead of resorting to mud-slinging. Care to depict your point of view with a schematic...? Rohitbd 15:15, August 11, 2005 (UTC)

Moved from my talk page Rohitbd 15:30, August 11, 2005 (UTC)

I have replied on Talk:current source. I hope you will find the argument convincing. BTW Keep Calm -I'm only interested in the accuracy of WP :-) Light current 15:22, 11 August 2005 (UTC)[reply]

I do not find your argument convincing. And it is you who needs to be calm. By your language, it appears you are more interested in forcing across your views rather than in accuracy. If you really cared about accuracy, you would have agreed to both Omegatron's and my view that a "potential divider" is not a voltage source. I am sorry to say this, but by your rather insulting language you have turned this discussion into a mud-slinging match. Rohitbd 15:30, August 11, 2005 (UTC)

Whilst the voltage on the collector is determined by the collector current and the collector resistance, this is not usually called biasing the transistor. Is that what you mean by biasing( ie setting the value of the collector voltage?). As I siad before, biasing of the transistor is usually done at the base. That is why the 2 resistors shown connected to the base in some circuits is called the bias network (or chain). If you dont believe me ask from another source, but I thnk you'll find I'm right. Light current 16:18, 11 August 2005 (UTC)[reply]

Unfortunately I am not able to draw diagrams on my monitor. That is why I explained the circuit in detail, so that you could draw it on paper from my description and then look at what I wrote about it. The fundamental point about a bootstrap circuit is that it mus have psitve feedback. Sometimes this is applied round 2 stages as in the esp article. In the example I have given you is applied just round one stage: the emitter follower which is the essential part in your circuit that you have missed out. Is this becoming any clearer?. I'm not trying to sling mud, I'm just suggesting that maybe you could do with looking for some more info on transistors and transistor circuitry. I'm trying to point you in that direction, because I think you may have a bit of trouble understanding my arguments without some of the fundamentals -- thats all. No offence intended.;-)Light current 16:36, 11 August 2005 (UTC)[reply]

Biasing involves both a voltage and a current. Base current controls the collector current, but the base is not used to set (fix, bias) the quiescient collector current. The base-bias network is there to ensure that the transistor base-emitter junction is forward biased and can supply enough base current - not to set the collector bias. If the base current were to control the collector bias (Q-point/quiescient current) current, we would have to change the ckt components every time we change the transistor. My arguments were related to any CE amplifier biased using a CCS (including bootstrap) as the collector load. Once a CCS has been used, the transistor can no longer be used in open-loop as a very small base current will cause the output voltage to sit at the supply rails, the reason is the high open-loop voltage gain. Even a voltage-divider base bias won't help much. That's why in my analysis of the bootstrap CCS I mentioned The reason for assuming that V_C is half the total supply is that this particlar configuration appears in a VFB design - similar to that of an op-amp in discrete form with a differential pair input stage and DC coupling internally. When negative feedback is applied around the ckt, the o/p (the collector of the transistor, in this case) always sits mid-way of the total supply - i.e., if we have a split ±VCC, then the o/p will be at 0V. Rohitbd 16:44, 11 August 2005 (UTC)[reply]

You seem to be contradicting yourself here. First you say 'Base current controls the collector current',(CORRECT). Then you say 'the base is not used to set (fix, bias) the quiescient collector current. (INCORRECT) . (ie what is the difference between the collector current and the quiescent collector current in your opinion? I cant see any difference except one is the steady state current, and the general term applies to the instantaneous current which is ALSO controlled by the base current. Do you see that? Light current 17:11, 11 August 2005 (UTC)[reply]

Just had another look at the esp article and AHA! I think I can see wher Rohitbd get his (erroneous) idea from. In the article describing the 2 stage bootstrap cct, I says" R1 + R2 supply a current of 5mA". THis statement is misleading. THey only supply this current because the transistor is telling them to. If Rohitbd looks at the left of the diagram, (s)he will see a little word 'bias' near the input. This denotes that the bias voltage is fed in here to program the collector current. I think this now clears up the misunderstanding. ( but again I may be wrong!) THis now concludes (I hope) the case for the prosecution(defence?)Light current 17:46, 11 August 2005 (UTC)[reply]

The base current controls the collector current but is not used to set the collector bias. Of course, change in base current causes changes in collector current, but this doesn't mean that the base current is fixing the quiescient collector current. Clear enough? The QUIESCIENT collector current is fixed by the collector resistor or CCS. I have verified this in the context of the bootstrap ckt with my SPICE simulator. Base current is used for the AC behaviour where it does control the collector current (rather changes in collector current from the quiescient). The DC bias is not fixed using the base drive. That's what I am trying to say. The 'bias on the left' that you mentioned is meant for the base - to ensure that the transistor's base-emitter junction is forward biased NOT for the collector quiescient current. Whenever we design a CE stage with self-bias, don't we choose the value of the voltage divider resistors at the base such that the current through them is atleast 2 times that of the base current (chosen based on Ic/Hfe(min) )? That's the base "bias" and it is so that transistors of the same type but with different Hfe's (due to device tolerances) can be replaced without having to change anything else - meaning we do not have to readjust the transistor's biasing. If base current were used to set the collector bias current (quiescient current, rather), don't you think changing the transistor would lead to change the quiescient collector current, since Ic = Hfe.Ib and different transistors of the same lot have different Hfe's? Rohitbd 20:59, 11 August 2005 (UTC)[reply]

I'm afraid I cannot explain the circuit in any simpler terms. Perhaps someone else can try to illuminate Rohitbd based on what we have already discussed or form a judgement as to whose view is more correct. Light current 22:41, 11 August 2005 (UTC)[reply]

The DC bias is not fixed using the base drive.

This is just dead wrong. Sorry to have to put it that way and certainly don't take just my word for it. Just look at the large signal model for the BJT used by SPICE and you will see that what have just stated is incorrect. If that's not good enough, take a look at these:

http://people.deas.harvard.edu/~jones/es154/lectures/lecture_3/bjt_models/ebers_moll/ebers_moll.html

http://users.ece.gatech.edu/~mleach/ece3050/su03notes/CEAMP.pdf (see page 2 - DC Bias Solution)

To reiterate, when the BJT is in the active region of operation, the total (DC and small-signal) collector current is related to the total base current as follows:

${\displaystyle i_{C}=\beta _{0}i_{B}\left(1+{\frac {v_{CE}}{V_{A}}}\right)}$

That is, ${\displaystyle I_{C}}$ is determined entirely by ${\displaystyle I_{B}}$ when ${\displaystyle V_{CE}}$ is small, and is adjusted by ${\displaystyle V_{CE}}$ as ${\displaystyle V_{CE}}$ gets 'large'. Thus, the external collector resistance ${\displaystyle R_{C}}$ primarily sets ${\displaystyle V_{C}}$ and secondarily alters ${\displaystyle I_{C}}$ due to the Early effect. Alfred Centauri 18:57, 14 August 2005 (UTC)[reply]

Thanks for the article. Yes, the calculations in the article are correct and I agree with them - but there is a caveat: That article is about finding the bias conditions given the values of all the circuit components and to prove that it is indeed operating in class-A (i.e. the transistor is in the active region). It is about analysing a circuit which has already been designed. My argument is designing for a given bias condition (to find the values of the ckt. components) - such that the bias point is relatively independent of the device used. If a particular design is supposed to give the final ckt. independence from device variations (i.e variations in β), how can the base current be used to set the collector quiescient current? ${\displaystyle i_{C}=\beta _{0}i_{B}\left(1+{\frac {v_{CE}}{V_{A}}}\right)}$ Doesn't this equation apply to changes in I_C and V_CE rather than DC conditions...? Rohitbd 08:06, 15 August 2005 (UTC)[reply]

Hmmm... I must have missed the memo ;<) Try this: create a simple SPICE deck or schematic where the base-emitter circuit is driven by a DC current source and the collector-emitter circuit is driven by a DC voltage source. Now, perform a DC analysis while sweeping the current source. Plot the collector and base currents or better, plot the ratio of the collector current to the base current. For even more fun, step the DC current source in an outside loop and sweep the DC voltage source in the inside loop. Plot the collector current against the collector-emitter voltage. The plot should look like a curve tracer display. You should see that for a given ${\displaystyle V_{CE}}$, the collector current is proportional to the base current where the proportionality constant is ${\displaystyle \beta }$. You should also see that for constant base current, the collector current is affected (but not completely determined) by ${\displaystyle V_{CE}}$ where the magnitude of this effect is inversely proportional to ${\displaystyle V_{A}}$. Enjoy! Alfred Centauri 13:12, 15 August 2005 (UTC)[reply]

Regardless of whether this is right or not, you really shouldn't be posting a circuit that you designed. We have a policy of no original research. You can definitely write about the one included on the Westhost site, though. - Omegatron 23:50, August 11, 2005 (UTC)

Its a bit late to say this now 'O', after I've spent all this time trying to help young Rohitdb. I mean, could you not have come to this decision earlier? Light current 01:44, 12 August 2005 (UTC)[reply]

Omegatron, I think there's a misunderstanding. I mentioned the circuit I designed because I made it for simulation as "proof of concept" to actually verify my statements about the bootstrap ckt, given the stiff "opposition" :-). It is not my intention to put my work here. I understand that this is an encyclopedia and as such must not delve deeper into topics - but whatever I posted in this discussion was only to explain my point of view and to give some analysis of the bootstrap source - all because you as well as Light Current expressed a need for greater clarity in the description of the ckt. All said, I would suggest that the bootstrap section be re-instated with the original description (without the analysis, if it be) but with an updated schematic. Rohitbd 07:28, 12 August 2005 (UTC)[reply]

Please read this article: http://www.eecs.utoledo.edu/~rking/elecdesign/EDmanualB.pdf

The bootstrapping technique is described just before the middle of the article. Although this schematic also has an emitter-follower, according to it, R2A + R2B (R1 + R2 as per my description & schematic) determines the Q-point current of the transistor - A point that I made, and was opposed by Light Current. Also the article has no mention of the emitter follower being required (or otherwise), so I guess my assumption for the sake of describing it is neither right nor wrong. The effect of the emitter follower is simply that the voltage at the output end of the bootstrap capacitor is just less than that at the collector - in my case it would simply be equal. In short, the resulting constant current is given by (K.Vx + Vb)/R2B as per the article. In the PDF article, K < 1 (due to emitter-follower Vbe drop), Vx is the collector voltage and Vb is the voltage at the emitter of the emitter-follower which is 0.65V less than Vx. In my case, looking at the schematic, K = 1 and Vb = 0 since no emitter-follower is there, so as per the equation it still works. It becomes Vx/R2 in my schematic.

I hope this clarifies my point of view a bit. Rohitbd 15:12, 12 August 2005 (UTC)[reply]

The statement shown on p56 of of the article http://www.eecs.utoledo.edu/~rking/elecdesign/EDmanualB.pdf

is incorrect. R2a and R2b do NOT determine the Q point current of Q2. What they do do is determine the VOLTAGE at the collector of Q2 (in conjuction with the bias network (which is probably an amplified diode voltage source)). The rest of the page does appear to describe bootstrap operation correctly although I have not checked every line. This article apears to have been written by a student and therfore can not be given much credence. Any one agree??Light current 15:27, 12 August 2005 (UTC)[reply]

ON second thoughts, this article looks to me like a draft form of a book judging by the page numbering etc. But it looks as if the author(who is probably a lecturer not a student (as I had erroneously first assumed)) has not yet picked up his own error about the Q point of Q2. Light current 09:32, 13 August 2005 (UTC)[reply]

Hey 'O' ,in a spirit of cooperation, if you were to do the bootstrap diagram ( I havent found how to do diags yet - too busy editing), I would be prepared to write some text to go with it. Then we could both look at it and agree a compromise edit on it. How does that sound to you? Light current 15:54, 12 August 2005 (UTC)[reply]

Actually, the emitter follower is crucial to the operation of the circuit. The circuit drawn above without the follower has no feedback but it does have a pole and zero in the transfer function due to CB. Inserting the emitter follower amounts to the insertion of a current amplifier. NOW, there is feedback. The output of the current amp is 'fedback' to the input of the current amp via CB and R2. It is the current amplification by the emitter follower that produces the additional base drive current that is the heart of the operation of a true bootstrap circuit. Alfred Centauri 15:00, 15 August 2005 (UTC)[reply]

Exactly!! Light current 16:11, 15 August 2005 (UTC)[reply]

Moved from my talk page Rohitbd 07:35, August 12, 2005 (UTC)

As it says on each and every editing page:

If you do not want your writing to be edited mercilessly and redistributed at will, do not submit it

Now please dont be a sore loser. We all have to learn. Its just that you have chosen the hard way to do it. Light current 01:13, 12 August 2005 (UTC)[reply]

Look who's talking...? Rohitbd 07:35, August 12, 2005 (UTC)

This discussion has been going on for very long now. A couple of points...

1. Yes, during the analysis of a CE ckt., where the components are already known, the base current controls the collector current. This is evident in the equations. BUT IMHO, equations used to analyse a ckt. from known components are not necessarily applicable going the other way round.
2. While designing (i.e., calculating the components with a given goal in mind), we do not use the base current to obtain the QUIESCIENT current - it is always fixed by a collector resistor or a CCS and the base voltage is adjusted to achieve operation in the active region. I reiterate once more, if the designed value of the quiescient collector current is based on the base current, it will be too device dependent to be of any use.

In my explanation of the circuit, I described it from the point of view of designing and not analysing the circuit - although I termed it as "analysis" - for want of a better word (sorry about that). Also, I do not feel that the ckt requires in-depth analysis or design steps, it can be described simply in a few words. If needed, the schematic can be updated to include an emitter-follower, but IMHO it makes no difference to the operating principle. Rohitbd 08:56, August 15, 2005 (UTC)

It's hard for me to believe that you are claiming that device and design equations are valid only after the circuit is designed. Further, please reconsider your claim that ${\displaystyle R_{C}}$ sets the quiescent current. Such claims represent fundamental conceptual errors. If one of my students made such claims on a quiz, I would mark it an 'F'. If one of my engineers made such claims, well... Sorry to be so blunt, but I'm only trying to help.

By the way, I do agree that setting the operating point on a true CE (where the emitter is connected to common) circuit is hardly repeatable. This is why the emitter degeneration resistor is almost always used. The addition of ${\displaystyle R_{E}}$ provides local negative feedback that stabilizes the operating point over a fairly wide range of device variability. Alfred Centauri 13:50, 15 August 2005 (UTC)[reply]

Re-reading your post again, I've come to the conclusion that what you are describing is actually the following process. (1) Pick a quiescent collector current. (2) Pick ${\displaystyle R_{C}}$ such that for the desired collector current, the desired ${\displaystyle V_{C}}$ is obtained. (3) Play around with the base bias voltage divider resistors in SPICE until you get ${\displaystyle V_{C}}$ where you want it. Is my interpretation correct? Alfred Centauri 14:00, 15 August 2005 (UTC)[reply]

We are drifting away from the topic. I am not talking about self-biased circuits here, although everyone seems to have assumed self-biased circuits (I mentioned VFB/op-amp type circuits in my "analysis" of the boot-strap CCS. These are not self-biased). I am not claiming that device and design equations are valid only after the circuit is designed - only that the design process need not be as detailed. I say this after having designed and built audio power amplifiers ranging from 20W to 100W with reliability and repeatability (if that matters at all). Yes I do use SPICE - in order to verify the design before actually building it and minimise wastages. Of course, some tweaking is also done to reduce distortion and improve frequency response. Your interpretation is correct except the last part which is valid only for self-bias circuits - the base voltage is fixed by calculating the voltage divider using the drop across the emitter resistor and the Vbe drop, and the required divider current - I should not have used the word "adjusted" here. Besides, as far as a CCS in the collector is concerned (and is the topic of this article including boot-strap CCS), the base cannot be used to bias the transistor all by itself: global negative feedback is always required as in direct-coupled VFB circuits. Look for "The reason for assuming that VC is half the total supply is that this particlar configuration appears in a VFB design - similar to that of an op-amp in discrete form with a differential pair input stage and DC coupling internally" above.

Rohitbd 15:36, August 15, 2005 (UTC)

I partially agree with your comment about active loads. It is certainly true that global negative feedback is required for the simple reason that the ${\displaystyle V_{CE}}$ of an active load at a particular current is virtually unpredictable as it depends heavily on process variations, temperature, etc. However, let me state once more for the record that the desired quiescent collector current is set primarily by the base bias circuit and the quiescent collector voltage is then given by: ${\displaystyle V_{CC}-I_{C}R_{C}}$. If ${\displaystyle R_{C}}$ happens to be the collector of an active load, ${\displaystyle V_{C}}$ is virtually unpredictable and must be set by other means such as global negative feedback. If we fundamentally disagree on this, then there no sense in dicussing this further. Your view would then be at odds with all the electrical engineering literature. Alfred Centauri 16:29, 15 August 2005 (UTC)[reply]

Alfred, my view is assuming CCS loads and global -ve feedback circuits. All along I had a CCS instead of a collector load in mind while discussing the bootstrap principle. I don't know, maybe repetitive work with VFB amps has perhaps put me into the habit of always using a CCS to bias the individual stages of the amp. Nevertheless, as defence of my argument that the collector resistor can be used to bias the transistor, let me give an example, I just designed and simulated this CE amplifier as per my method — These are the known parameters and design goals:

• Vo(pk) = 2V
• Vin(pk) = 200mV
• Io(pk) = 500µA (here I am replacing the load on this CE stage with the input impedance of a buffer stage represented by a 4K resistor for simulation and hence the particular value for the Io(pk) which is the load on the CE stage)
• Vcc = 12V

Step 1: Assumptions

• Hfe(min) = 100
• Collector bias current = 10-50 times the peak load current, I chose 20 which makes Icq = 10mA
• |Av| = Hfe(min) * (Rc / Re) (using approximate model)

Step 2: Calculate Rc & Re

• Assuming the collector to be at half supply, i.e., 6V, Rc becomes 6 / 10mA = 600Ω, higher standard: 680Ω
• Av = Vo/Vin = 10 and hence Re = Rc / 10. Thus, Re = 60Ω lower standard: 56Ω
• The reason for using higher std. for Rc and lower std. for Re is so that the unloaded gain is higher than designed and will be closer to the required value when loaded.

Step 3: Calculate base voltage divider

• Ib(max) = Icq/Hfe(min) = 100µA, choose the current through R1 & R2 (Ibb) to be 5 times this. Thus Ibb = 500µA
• Lower resistor R2 = Vr2 / 500µA = (Vbe + Vre) / 500µA = (0.65 + (10mA * 60)) / 500µA = 2.5K, nearest standard 2.7K
• Upper resistor R1 = (Vcc - Vr2) / 500µA = 21.5K, nearest standard 22K

After calculations, the actual voltage gain into a 4K load is about 9.6 as simulated (instead of the intended 10), practically we can expect about 9. In terms of decibels, this is a difference of about 0.35db (1db practically), and the voltage gain is stable across following transistors: BC547, BC337, 2N2222 and BD139. This can be improved by using a CCS instead of Rc and using global negative feedback. Rohitbd 18:03, August 15, 2005 (UTC)

This is a textbook example of CE amplifier design. Well done! My question to you is this: at what point did you design the quiescent collector current? From my perspective, it was in step 3 where you designed the base current to be 100µA. Now, as you have pointed out earlier, the actual collecter current may or may not be the desired value of 10mA for a number of reasons including but not limited to differences in β. On the other hand, assuming the transistor stays in the active region of operation, the quiescent collector current is quite insensitive to changes in Rc.

By the way, there are many ways to set up and solve the base bias problem. Here's a way that I find quite illustrative. Apply KVL around the loop that includes R2, the base-emitter junction, and ${\displaystyle R_{E}}$. With a little algebra, you get the following:

${\displaystyle V_{BE}=\left(I_{BB}-I_{B}\right)R2-I_{B}\left(\beta +1\right)R_{E}}$

Note the ${\displaystyle \left(\beta +1\right)}$ term multiplying ${\displaystyle R_{E}}$. If ${\displaystyle \beta }$ is greater than the designed for value, ${\displaystyle V_{BE}}$ is decreased thus decreasing the collector current and vica versa. That is, the presence of (a large enough) ${\displaystyle R_{E}}$ stabilizes the operating point against variations in ${\displaystyle \beta }$. Also, this form of the equation clearly shows why ${\displaystyle I_{BB}}$ should be much larger than ${\displaystyle I_{B}}$ - it is so that the first term on the right can be decoupled from the left side of the equation (${\displaystyle I_{B}}$ is a function of ${\displaystyle V_{BE}}$). Alfred Centauri 20:13, 15 August 2005 (UTC)[reply]

Well, from my perspective, I set the Icq in step 2. Step 3 is only about ensuring that the base-emitter junction is forward biased. Now, of course this is called biasing the base, IMHO it still does not set Icq. All along I have held that the collector bias current is not set using the base current - only the change in collector current is controlled by the changes in base current.

<insert>

And here is our fundamental disagreement... Your statements above are inconsistent not only with the fundamental BJT device equations, but are also inconsistent with basic circuit theory. Further, I am convinced that you are convinced that your statements are correct. Thus, there is really no need to discuss this further. Let me close with the following suggestion. Prepare a paper supporting your statements above and submit this paper to the IEEE Journal of Solid-State Circuits. Should your paper be published, it will come as quite a shock to the electrical engineering community at large to learn that the established literature on BJT device and circuit theory is incorrect. Alfred Centauri 23:28, 15 August 2005 (UTC)[reply]

Well it is neither my intention nor desire to go against any accepted norms. My understanding of the circuit works for me - you and I both agreed on the overall CE biasing method above and IMHO that is what matters as far as designing goes. This whole discussion has become much longer than needed and has taken a rather ugly form due to certain personal remarks against me simply because of this very point. It was neither worth it, nor warranted. When I mentioned in the bootstrap CCS section that R1 + R2 set the collector bias, I was required to give an explanation as to why. I did and it led to even more personal attacks. Now, if I was indeed so wrong (and others so confident about their knowledge) then why didn't anyone simply make the appropriate change to the article and leave it at that instead of choosing to just keep arguing this over and over again - changes have been made to stuff that I wrote and I have not questioned them. I am willing to hold back my views (not necessarily change them) if the majority says otherwise - that's ok and is in the spirit of wikipedia. IMHO, writing something which is not 100% accurate will not bring down the sky - over a period of time it will be corrected. Rohitbd 07:52, 16 August 2005 (UTC)[reply]

</insert>

The very reason behind making Ibb very large compared to Ib is that we want Icq (NOT i_c) to be independent of Ib. To clarify what I think is meant by "the base current fixing the collector current", to me it means that the quiescient collector current will depend on the base current and not the voltage drop across Rc. And this is what I don't agree to. Changing Rc will change the Q-point due to the presence of Re. If Re were absent, then Rc necessarily has to be a CCS - the base bias cannot be used to set the Q-point and global -ve feedback is a must. Also, note that in the kind of audio amp design that I have done, biasing means setting the DC operating current of one stage so that it can supply the current required by the next stage without itself running out of juice (a.k.a. bias). So may be we have a terminology conflict here and hence this rather long "war" :-).

Now, as you have pointed out earlier, the actual collecter current may or may not be the desired value of 10mA for a number of reasons including but not limited to differences in β.

I do not recollect saying the above. I only said that the quiescient collector current cannot be determined using the base current due to device tolerance or variation in Hfe... Rohitbd 21:21, August 15, 2005 (UTC)

Follow-up question: What can be improved by using an active load with global feedback? Alfred Centauri 20:35, 15 August 2005 (UTC)[reply]

Well, with a simple resistor as the collector load, the voltage gain is determined by the ratio Rc/Re (using the approximate model) as long as β is large enough. This works fine as long as the output current swing is such that the collector or output voltage swing is small compared to the supply. With large signal excursions towards Vcc, as the collector current drops, the effect of the transistor's inner r_e becomes significant. r_e is non-linear, which means that it changes with the collector (or emitter) current. The result is non-linear voltage gain and in turn results in distortion. This is also one of the reasons why an external Re is connected - if |Av| = hfe(Rc/(Re + re)) and Re >> re, we can ensure that the gain is linear and essentially a function of Re. There is a limit to Re - to get useful gain, Rc will have to be made so large that the CE stage simply won't be able to source any current to the load connected to it. When Rc is replaced by a CCS - which theoretically has infinite impedance, we can see from the approximate equation, that both Re & r_e become redundant, and we end up with a very huge voltage gain (which again is of no use). The global negative feedback is now used to reduce the gain to a usable level and also to stabilise the output voltage. Note that when a CCS is used in place of Rc, the base bias network alone cannot be used to set the Vc - this requires output to input negative feedback and possibly an extra error amplifier stage before the CE stage. This also rules out capacitive coupling - the individual stages have to be direct-coupled (except for the signal source and the load). Rohitbd 21:21, August 15, 2005 (UTC)

In the last section of this article, there is the statement that an ideal voltage source cannot be connected to an ideal short circuit. Well, here is a thought experiment to consider. Connect a resistor to a voltage source. The voltage across the resistor is the voltage of the source and the current is just the ratio of the source voltage and the resistance of the resistor. Now, let the resistance tend to zero... At all times, the voltage across the resistor is the voltage across the source and the current increases without bound. In the limit, we have that the voltage across the resistor of zero ohms is still the source voltage. How can this be? It is because the current through the short circuit has passed to infinity and recall that infinity multiplied by zero is an indeterminate form. In order to determine the value, a limit process must be used just as we did above. A similar argument can be made for an ideal current source without an external circuit connected (infinite voltage multiplied by zero conductance). Enjoy! Alfred Centauri 02:09, 16 August 2005 (UTC)[reply]

Yeah, but that's taking a limit. That's what we used for the section that tells the power dissipated in such a system. But in reality, an ideal voltage source has a non-zero voltage across it, always, and a short circuit has zero voltage across it, always, so you're saying 5 = 0 if you actually connect them together. - Omegatron 03:23, August 16, 2005 (UTC)

Not so fast! I submit that the circuit element that has zero volts across it, always, is the ideal voltage source with source voltage equal to zero. On the other hand, the circuit element that we call a short circuit is the ideal model of a wire - a resistor with resistance of zero ohms. A resistance of zero ohms certainly has zero volts across it for any finite current but that voltage is indeterminate for an infinite current. If you connect a short circuit across a 5 volt voltage source, the resulting equation is not 5 = 0, but is instead, 5 = I*R = (infinity)*0. Of course, all of this amounts to argueing how many angels can fit on the head of a pin - over a few beers. Alfred Centauri 12:42, 16 August 2005 (UTC)[reply]

OK - if you didn't bite on that one you probably won't bite on this one but I think this example is irrefutable! Consider an ideal voltage source - that is, there is 0 source resistance or equivalently, the Thevenin resistance is 0 ohms. The Norton dual of this is a current source in parallel with the Thevenin resistance, 0 ohms. The open circuit voltage of this Norton equivalent circuit must be equal to the (assumed non-zero) voltage of the voltage source, right? Hmmm... An ideal voltage source is equivalent to a ideal current source supplying an infinite current into a short circuit such that not only is there a voltage across the short circuit, there is an unlimited amount of current 'left over' for any load connected to this circuit! Ain't infinity fun? Alfred Centauri 17:48, 30 August 2005 (UTC)[reply]

...But in reality, an ideal voltage source has a non-zero voltage across it...

Not to play with words but are there any ideal voltage sources in reality? :-)

IMHO, we are reaching the limits of physics when we talk of an ideal voltage source connected across an ideal short ckt...there may be a solution though - it may be possible for a voltage source to supply infinite current only and only to the load (i.e. short) connected to it - it will be a voltage source to the load (short), but not to the external world. It becomes a sort of closed system (?) - so that any attempt to measure the voltage will yeild zero volts. So in all, we can say that an ideal voltage source will maintain a constant voltage across any non-zero load impedance, irrespective of the impedance. So we have a limiting condition here. Similarly, trying to measure the voltage across a current source feeding an infinite impedance (open ckt) load will simply yeild the voltage across the measuring instrument - not the load, since the measuring instrument necessarily has to draw some current in order to perform the measurement. Rohitbd 08:32, August 16, 2005 (UTC)

I think, as someone pointed out elsewhere, these degenerate cases simply depend on how you define components before taking the limit. If you define a short circuit as a section of wire with 0 V across it you have a different situation than defining it as a section of wire with 0 resistance. — Omegatron 18:11, August 30, 2005 (UTC)

Ok, now that I have received enough brick-bats, would anybody care to present an analyis of the bootstrap CCS? I tried and was opposed vehemently (not to mention personal attacks)...so I guess it is only fair of me to ask my detractors to give an analysis themselves. Of course, there are no obligations - but for a change, why don't you try and then include it in the article? Rohitbd 13:19, August 16, 2005 (UTC)

Well, I have a suggestion. Consider describing only the AC operation of the circuit. At DC, there is no bootstrap effect anyhow so the whole DC / bias debate going on here is actually irrelavent to the central operating principle of this circuit.

Here's another suggestion. Whenever there is feedback, the analysis of the circuit is made far simpler if the feedback path is replaced with the Thevenin equivalent circuit(s). In this case, the analysis is trivial. The emitter follower output 'looks' like a voltage amp with a gain of nearly one in series with a relatively small resistance. The input of this amp is the voltage on one side of R2 and the output of the amp is (nearly) connected to the other side of R2. Thus, the AC voltage across R2 is nearly the same so that the AC current through R2 is nearly zero so that the collector of our CE amplifier 'sees' a nearly open circuit as its AC load. In other words, the positive feedback has the effect of increasing the AC resistance seen by the collector of the CE amp thereby increasing the gain of this stage.

And yes, the emitter follower is necessary for correct operation of this circuit. Alfred Centauri 19:00, 16 August 2005 (UTC)[reply]

Follow-up: What do you guys use to draw your schematics? I'd be happy to give this thing a try... Alfred Centauri 19:06, 16 August 2005 (UTC)[reply]

OK, I propose putting the following text as the description; and updating the schematic to include an emitter follower —

Although not as good as the active constant current sources depicted above, the boot-strap current source is commonly found in many audio power amplifiers to bias the class-A voltage gain stages in them. These offer a cheap alternative to active sources where the output impedance of the CCS is of secondary concern. As can be seen in the image on the left, a boot-strap CCS is formed by resistors R1, R2 and capacitor CB. Transistor Q1 is wired as a class-A voltage gain stage. Operation of the circuit is as follows: R1 + R2 supply the nominal collector current at no signal (i.e., DC). When the output voltage changes due to an input signal, this change is coupled back to R2 such that the voltage across R2 remains relatively constant. Constant voltage across R2 means constant current through it and thus the circuit operates as a constant current source.

PS: If needed I can update the schematic...I used MS-paintbrush to create the original image. Rohitbd 19:21, August 16, 2005 (UTC)

My only suggestion is that the 'commonly found' be changed to 'was commonly found'. Now this is just my opinion, but I'm pretty sure that bootstrapping is much less common today than it was before symmetrical amplifier circuits became the rage where the NPN and PNP CE stages act as active loads for each other. Otherwise, it sounds good to me.

Well, IMHO boot-strapped CCS are still widely used in modern amps - but only for discrete designs. Older power op-amp ICs like the STK series used to have an external boot-strapped CCS, but most modern power op-amp ICs do not use the boot-strap CCS. Besides "is" or "was" hardly makes any difference. Rohitbd 19:39, August 16, 2005 (UTC)

This is what I've been doing for drawing circuits: User:Omegatron#Electronics_diagrams It's not the best solution in the world, but it works. Examples here: Commons:User:Omegatron/Gallery I'd like to improve the program, but I don't know much about coding and I'm not the best artist. It would be nice if a good, free, program existed already... - Omegatron 20:26, August 16, 2005 (UTC)

(see also Current and Voltage Source Suggestion)

Hi everybody! Two months ago, I placed a link on constant current source page pointing to my story about this subject. In this way, I tried to resume the discussion about constant current source and the dual voltage source. Browsing through these and talk pages I found a lot of brilliant thoughts. Only, the pages look quite cluttered; there isn't good structure and hierarchy in their arrangements. So, I suggest rearranging the materials according to the principles below:

1. At every page, expound the subject step-by-step by moving from simple to complex (imperfect to perfect, passive to active, transistor- to op-amp versions etc.)

2. At every step, first reveal the basic circuit idea; then show the concrete circuit solution.

In order to illustrate my suggestion, I have made a web page containing two versions of the contents - a short and a long one. Here is the short version:

Current Source

1. Theoretical current source.
1.1. Ideal current source: definition, examples of natural current sources.
1.2. Comparison between current and voltage sources.
1.3. Real current source: imperfections.

2. Practical current sources.
2.1. Ohmic resistor current source. Imperfections.
2.2. Dynamic resistor current sources.
2.3. Current sources with compensating voltage.
2.4. Current sources with compensating current.
2.5. Current sources using negative feedback.

Voltage Source

1. Theoretical voltage source.
1.1. Ideal voltage source: definition, examples of natural voltage sources.
1.2. Comparison between voltage and current sources.
1.3. Real voltage source: imperfections.

2. Practical voltage sources.
2.1. Voltage divider source. Imperfections.
2.2. Dynamic resistor voltage sources.
2.3. Voltage sources with compensating negative resistance.
2.4. Voltage sources using negative feedback.

I realize that I have proposed major changes in the structure of these pages. So, I would like first to coordinate them with you, especially with those who are created these pages; then, I will try to modify gradually (in small steps) the pages. But yet, don't remember that I am a Wiki newbie; so, don't bite me desperately:) Circuit-fantasist 16:24, 29 June 2006 (UTC)[reply]

You say "imperfect to perfect", but I think it's better to teach ideal, perfect sources first ("an ideal voltage source does whatever it takes to keep the voltage across it the same, no matter what"), and then teach the imperfect sources ("a real source can't do this because of internal resistances and reactances"). — Omegatron 14:48, 8 July 2006 (UTC)[reply]

Omegatron, thank you for your response (it is the first reaction to my intervention in the domain of electronics in Wikipedia). If you look at the first parts of the contents above, you will see that I have still followed your advice. Both of the contents begin with info about ideal sources. I have suggested exposing the practical circuits of constant current sources in consecutive logically connected steps moving from simple to complex and explaining the basic ideas at every stage. Usually, the simple is imperfect and the complex is (more) perfect; that is why I have suggested moving "from imperfect to perfect". IMO, it is so much natural as to write from left to right and from top to bottom.

Trying to explain the unexplained things in electronics to my students, I have been noting that the large part of the truth about circuit phenomena is hidden in this movement, rather in the final perfect circuit solutions. That is why, I ever show the evolution of electronic circuits to my students by building and reinventing them.

You have suggested "to teach ideal, perfect sources first". Only, don't you think that there are not so many things to say about these bare components? Believe me, no matter what you say about them in the introductory parts, readers will understand what they actually are only when they meet the practical electronic circuits. You have said "an ideal voltage source does whatever it takes to keep the voltage across it the same, no matter what". However, can readers really understand what it is means, if they do not know what negative feedback is? Also, apply these considerations especially for even more abstract constant current source.

Let's some dull info about ideal voltage and current source (traditional for classical books about electricity) stays in the introductory parts of the pages. Only, keep in mind that it should serve mainly as a final point (an ideal goal) of the next circuit exposition.

Sorry for my incompetence; what do you mean when write «54 words changed»? Also, what should I fill in the field "Edit summary" when reply?

Finally, what do you think about the philosphy of negative resistance? An interesting discussion is going there. --Circuit-fantasist 19:01, 10 July 2006 (UTC)[reply]

I have reordered a bit the galvanostat voice and have written a brief comment on its talk page Talk:Galvanostat: please give it a look :). Daniele.tampieri 14:54, 24 December 2006 (UTC)[reply]

Rsduhamel, you have to connect a resistor in series with the zener diode on the circuit diagram of an op-amp current source. Circuit-fantasist 08:49, 26 January 2007 (UTC)[reply]

Circuit-fantasist, I presume that you are referring to Figure 7, where the Zener diode is connected between the op-amp's non-inverting (+) input and ground. No resistor is needed in this portion of the circuit. Current limiting for the Zener diode is provided by resistor between the op-amp's non-inverting input and the power supply. In fact, including a resistor in the same branch as the Zener diode will prevent the voltage across the sense resistor from tracking the voltage across the Zener diode. That would be prevent the circuit from working properly as a constant current source. The circuit is designed so that if the voltage across the sense resistor should increase (indicating an increased current flow), the op-amp will reduce the output voltage, thereby causing a reduction in the current through the load and sense resistor. If the voltage across the sense resistor should decrease (indicating a reduced current flow), the opposite happens. For the constant current in the load to be stable, the voltage at the op-amp's non-inverting input needs to be stable and the resistance of the sense resistor needs to be stable for the expected currents. Thus, no additional resistor is needed. OhioFred (talk) 22:53, 17 September 2014 (UTC)[reply]

Circuit-fantasist, perhaps you meant to remark on the fact that the op-amp inputs are biased near the Zener diode voltage and the op-amp output is biased near the power supply voltage. See the comment below (in section Mistake in Figure 7: Typical op-amp current source) regarding the placement of the load resistance. The solution recommended there would reduce the bias differences. (Don't forget that the op-amp positive and negative supply voltages may differ from the circuit voltages. For best circuit performance, the op-amp should operate half-way between its positive and negative supply voltages.) OhioFred (talk) 02:35, 18 September 2014 (UTC)[reply]

OhioFred, thanks for the thorough analysis. My explanation is very simple - I wrote this comment 7 years ago when Fig. 7 looked like this:

https://upload.wikimedia.org/wikipedia/commons/archive/9/97/20120221031928%21Op-amp_current_source_with_pass_transistor.png

Regards, Circuit dreamer (talk, contribs, email) 09:01, 18 September 2014 (UTC)[reply]

Alfred Centauri, I highly appreciate your last edits. Now, the section about the so-called "resistor current source" is where it has to be - at the beginning of the page.

• Only, I do not agree with the first sentence: "By Norton's theorem, a voltage source in series with a resistor is equivalent to a current source in parallel with the same resistor". Norton's theorem (or maybe better, Norton's equivalent circuit) might be mentioned in an eventual section about current source applications as it is made in current-to-voltage applications or in the page of voltage source (if there is any need of this at all). The reason of that is that this arrangement shows how to build a simple voltage source by using a current source. Instead, the section of resistor current source is about how to build a simple current source (and not how to use it). So, I suggest using a phrase similar to "The simplest current source consists of a voltage source in series with a resistor".

• Also, I suggest removing the insertion "(ideally infinite)". If you keep it, you have to explain the phenomenon that an infinite voltage divided by an infinite resistance gives a finite current. In order to do that, I suggest to my students to build consecutively 1mA current source first by using 10V voltage source and 10K resistor, then - 100V and 100K, after - 1000V and 1000K, ...and finally - infinite V and infinite K. In addition, I show that it is a well-known paradoxical trick from our humane routine - to depreciate a harmful quantity by another many times bigger harmful quantity. As a result, they understand the phenomenon.

• From this viewpoint, the assertion below (see Article about current sources on ESP) is just not true

"A perfect current source can be constructed very simply, requiring only an infinite voltage and an infinite resistance. Unfortunately, this will also provide an infinite current (which will be infinitely constant), but is (infinitely?) excessive for an infinite number of applications."

• By the way, the next chain of thoughts in this source (Article about current sources on ESP) is very not true:

"A commonly held belief is that local emitter feedback reduces the output impedance - this is simply not true - the output impedance of a common emitter amplifier is equal to the collector resistance, regardless of emitter degeneration. To reduce the output impedance, it is necessary to apply feedback around the entire circuit - from the output back to the input. The proof of this is outside the scope of this article, but it is nonetheless a fact."

Actually, the collector-emitter part of the transistor of a common-emitter amplifier acting as a current source behaves as an imperfect dynamic resistor (because of the Early effect). The emitter follower with a constant emitter resistor acts as a quite better current source since it is a true circuit with 100% series negative feedback applied exactly around the entire circuit - from the output current back to the input voltage. What more do we want from this bare circuit?

Circuit-fantasist 17:04, 9 April 2007 (UTC)[reply]

The ESP link is a 'leftover' from the original text I edited. Had I bothered to read it, I would have removed it myself. Good catch.

On another note, please do keep in mind that this is an encylopedia, not a text book. That is, Wikipedia is a reference. To keep the articles concise, there must be some assumption of familiarity with the subject material.

I've looked at some of your contributions (voltage-to-current converter and others) and, while I'm impressed with the skills you display for presenting this material for learning and the depth of the information you have put into these articles, it is my opinion that Wikipedia is not the proper venue for this type of presentation. Have you considered making contributions to Wikibooks? The EE sections there could certainly use your help.

Best regards, Alfred Centauri 00:53, 10 April 2007 (UTC)[reply]

Alfred, you are absolutely right! Thank you for the advices about my Wiki contributions; I will consider them. As you can see, the true is quite simple - I have a philosophy in teaching electronic circuits and want to share it with a large audience. Maybe, I will really begin creating Wikibooks and will compress the existing Wiki pages where I have contributed. I would be happy, if you help me to simplify and clarify these pages. Thank you again. Circuit-fantasist 17:22, 10 April 2007 (UTC)[reply]

Solar cells are generally regarded as a current source in parallel with a diode and a shunt resistance with an extra resistor connected in series. My background is in Physics as opposed to electronics but are there any objections to my adding this to the page? —The preceding unsigned comment was added by 155.198.186.155 (talk) 17:30, 16 May 2007 (UTC).[reply]

The simplest way to drive an LED from line-voltage AC is with a large series resistor, which functions as a current source. Very inefficient, due to power loss in the resistor.

The article implicitly only covers DC current sources. AC is a related but more complex subject. But some simple cases are important. We could also drop line voltage to a low voltage to drive an LED with an AC capacitor. We can calculate power in the capacitor by V x A, but the result is not real watts - the answer is ideally all "imaginary" VA? In the real world, how do we calculate power dissipation in real capacitors used thus: how hot they would get, what would be safe design, etc? For example, if two LEDs are anti-parallel, and you want the total average current through the pair to be 20 mA, from a series capacitor, at 120 VAC, what capacitance is needed, what capacitor types would be appropriate, and would power issues in the capacitor be a concern?-69.87.204.197 13:20, 28 May 2007 (UTC)[reply]

91.75.70.167 (talk) 04:08, 3 February 2008 (UTC) A batterie's function is to maintain a constant potential difference about itself/ in the ckt. Does a current producing source, work on the same basis? ie. at every instant of time, its job is to provide a constant supply of current? If yes, then how will we balance the produced current?[reply]

"The current then flowing is the IDSS of the FET."

Please list typical values for IDSS. What is the physical reason for the limitations on this value? It seems from CLD datasheets that this is only a solution for currents below about 15 mA. See [6], for instance.

If more than one CLD can be placed in parallel on a PCB, why can't they manufacture multiple CLDs in parallel on the chip? Wouldn't that be exactly the same as paralleling a bunch of transistors on the same chip to make a power transistor? — Omegatron 18:47, 21 March 2008 (UTC)[reply]

Where is the inductor on the LM317 circuit in the section "Inductor type current source"? Section title typo? I think it should be "Linear type current source". — Preceding unsigned comment added by 213.63.100.45 (talk • contribs) 02:03, 11 December 2010

A section title says "Current sources with series negative feedback" and then the first part "Simple transistor current source" doesn't even mention the word feedback. Later there is a negative series feedback circuit presented; however it is not mentioned that the first bjt circuit already in itself has a negative series feedback.

If temperature doesn't change, then Re emitter resistor is not necessary (in theory only!) as the Ic collector current is set by the Vbe base-emitter voltage. But in this case we can't make the assumption that Vbe=0.65V or so, because a very little change in Vbe results in big changes in Ic. (See http://diranieh.com/Electrenicas/Figures/BJT_EbersMollGraph.jpg for example, don't know why the bjt article doesn't have these curves). And of course that's one reason we use the RE resistor, as we can't guarantee a constant base potential (though the zener diode is a good try). The Re resistor kind of forces the current onto the transistor, as small changes in Vbe will not change the emitter resistor's current much. But still it is possible to change the output current not only by Re but by changing the base voltage (if using a potmeter instead of a Zener).

Now for the feedback: the voltage across emitter resistor Re is proportional to the output current. If the temperature rises, then Ic collector current increases - but then the Ve emitter potential will also increase. As the Vb base potential is fixed by the diode or resistor divider, and Vbe=Vb-Ve so Vbe will decrease, which acts towards decresing the collector current. This is a series negative feedback. If you simulate the circuit without the emitter resistor you get a much worse temperature dependency. Of course Re is only good so far, but can be useful in many cases.

As for earlier comments: in the output curve, the collector current does have a very little dependancy on Vce, which can be a problem for a very precise circuit and this is also compensated by the feedbacks. (Also the finite value of h22e is also due to this). Hoemaco (talk) 11:35, 21 April 2011 (UTC)[reply]

Your thoughts are valuable. For the first time I see an appeal to emphasize the role of negative feedback in these circuits. BTW, I have problems when using the terms "series negative feedback" and "parallel negative feedback"; as a rule, people do not understand what I mean when saying them. Circuit dreamer (talk, contribs, email) 21:58, 21 April 2011 (UTC)[reply]

I think the load resistance in this circuit diagram should be located above the transistor. The reason for this is that if most of the voltage drops at the load (as intended), the voltage at the emitter of the transistor can easily become too big for it to conduct (since the base voltage needs to be at least 0.7V above the emitter voltage). For an PNP transistor the depicted way would be okay, but this is an NPN transistor. A detailed discussion of this kind of circuit is given at http://www.mikrocontroller.net/articles/Konstantstromquelle#Konstantstromquelle_mit_Operationsverst.C3.A4rker_und_Transistor (sorry, in German).

I already produced a corrected version of the image but can not replace the file because it's imported from wikimedia. — Preceding unsigned comment added by 5.158.164.237 (talk) 01:09, 12 May 2013 (UTC)[reply]

This paragraph is misleading. A Van de Graaff generator is essentially a current source and must be regulated to maintain a constant voltage. The high voltage capacitor is charged by a belt on which charge is placed at low voltage and mechanically moved to a region of high voltage. When used a particle accelerator, the load current is determined by an ion source, which generates a constant current and does not resemble a resistor.

I think we need to include the "interlinked circles" symbol for a current source. See link. This is very commonly used in UK engineering practice and in text books/app notes/papers, and is what was used throughout my university course. I much prefer this symbol because it can be quickly and unambiguously interpreted as a current source. Amj4321 (talk) 22:31, 22 November 2015 (UTC)[reply]

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