Talk:Capacitance multiplier

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"the output voltage is about 0.65V less than the base voltage, which in turn is about 2 to 3V less than Vs." - Why is the base voltage about 2 to 3V less than Vs? Surely this will depend on the load impedance. If there's no load then the base voltage will gradually rise to Vs.

When loaded, the voltage across the capacitor which is also the base voltage will drop depending on the charging current through the resistor - this current causes a voltage to be developed across the resistor. The drop (and hence the base voltage) is not constant DC but a ripple superimposed on a DC. User:Rohitbd

As far as I can see this topic has nothing to do with power supplies. Perhaps an editor noticed that some PSUs need large caps, and here's a cap multiplier circuit. This circuit can not be used to increase reservoir capacitance on psus. I'm going to remove the power supplies category. Tabby (talk) 02:27, 9 May 2010 (UTC)[reply]

i dont understand why, but the first paragraph says, that it can be used to create a PSU with a better stabilization with a smaller footprint (e. g. in "class a amplifiers" whatever that might be...)... --Homer Landskirty (talk) 07:34, 9 May 2010 (UTC)[reply]

Tabby: If you follow the link to the first reference (ESP), you will see an example of the circuit (in its transistor version) applied to filtering in a DC power supply. Homer: Amplifier class (like "A") is explained in the Class A section of the Electronic amplifier article.--Theodore Kloba (talk) 16:42, 28 July 2010 (UTC)[reply]

here u can simulate it urself: http://www.falstad.com/circuit/index.html with this:

$ 1 5.0E-6 135.0639324902207 16 5.0 50
v 192 32 192 128 0 1 50.0 15.0 0.0 0.0 0.5
d 256 80 304 32 1 0.805904783
d 256 80 304 128 1 0.805904783
d 304 128 352 80 1 0.805904783
d 304 32 352 80 1 0.805904783
w 192 32 304 32 0
w 192 128 304 128 0
w 256 80 256 192 0
w 352 80 352 192 0
c 256 192 352 192 0 0.0047 -13.207051567646102
t 384 256 416 256 0 1 -1.0377366965067019 0.6941953627951065 100.0
r 352 192 352 256 0 1000.0
c 352 256 256 256 0 0.0010 12.1693148711394
w 352 256 384 256 0
w 352 192 416 192 0
w 416 192 416 240 0
w 416 272 416 336 0
w 416 336 352 336 0
w 256 256 256 192 0
w 256 256 256 336 0
g 256 192 240 192 0
r 256 416 352 416 0 100.0
w 256 336 256 416 0
w 352 336 352 416 0
O 352 80 400 80 0
O 352 416 400 416 0
p 464 336 464 192 0
w 416 336 464 336 0
R 464 192 464 160 0 0 40.0 11.4755 0.0 0.0 0.5
c 352 336 256 336 0 0.0010 11.475119508344294
o 24 64 0 290 40.0 9.765625E-5 0 -1
o 25 64 0 290 40.0 9.765625E-5 0 -1
o 26 64 0 290 0.0048828125 9.765625E-5 1 -1

(u can use it via "File->Import"...) --Homer Landskirty (talk) 08:31, 9 May 2010 (UTC)[reply]

This isn't unconnected to power supplies... but it's not exactly a power supply in the "power" sense. This circuit is invariably used at the lowest-level input circuit.

This circuit in the real world is most typically used as an "active decoupler" in the very low level input circuits, to prevent power supply fluctuations from getting in and especially causing feedback.

Decoupling is important to prevent e.g. "motorboating". Note that the current Wikipedia article about motorboating has very little to do with reality (typically motorboating is used to describe poor power supply decoupling.... the current wikipedia article is talking about some completely different issue.) —Preceding unsigned comment added by 170.121.14.11 (talk) 14:19, 16 February 2011 (UTC)[reply]

Why the capacitance is multiplied by the beta(transistor's current gain)? I don't think so... Actually I think the time constant of the circuit is still R1*C1, with or without Q in place. With Q, however, the resistance of R1 could be set quite large, so that the R1*C1's time constant can be increased large and not effecting the output impedance looking into terminal Vo. C1 is multiplied by R1 rather than Q's beta, in my point of view. Anybody agree with me here?--Linan0827 (talk) 03:30, 28 November 2012 (UTC)[reply]

sounds good... but this source: [1] seems to say the opposite... i dont understand why... --Homer Landskirty (talk) 06:21, 28 November 2012 (UTC)[reply]

But if you look into the op-amp circuit at node Vi, it does look like a capacitor of value C = C1 * R1 / R2 at low frequencies, until you get to high frequencies where the resistance of R2 dominates. The impedance looking into Vo is just 0. — Preceding unsigned comment added by 71.167.63.118 (talk) 17:56, 20 March 2016 (UTC)[reply]

I was looking at precision capacitors and read on the NPL web site that the can calibrate the General Radio 1417 series of capacitors. When I looked a bit more deeply at this, it is an expensive precision device that allows large values (in this case 1 F) capacitors to be achieved with the use of step up transformers and smaller value more stable capacitors. This seems inportant enough to have in the article, but I know very little about it, so what I added is very incomplete. — Preceding unsigned comment added by 217.42.240.189 (talk) 12:30, 18 March 2015 (UTC)[reply]