MTaylor

Product PRT-14411 about 7 years ago

FYI, the babysitter in the post uses high to disable rather than low to disable.

Product PRT-14411 about 7 years ago

This is a pretty basic circuit, I've found that projects I make have a variety of needs so I have to get a bit creative depending on how I want the circuit to operate.

If you want the load to automatically be disconnected, use a small N type mosfet to pull the enable pin low with the gate connected to the USB in rail through a resistor.

If you require the load to stay active (and can tolerate a 0.5v drop) do the above but also put a pair of schottky diodes ORing the USB rail and the output rail to your load. That way the system will redirect power through the supply while pulling the booster offline. If there's too much of a gap in power, build an RC filter on the gate to delay the disconnection.

One project I built has a button wired to turn on the system (microcontroller) when held, and the microcontroller can then power itself off. So in this instance I simply shut the device down before charging. See this Enginursday.

Thanks for the bump

Product PRT-14411 about 7 years ago

Chris is correct. Pulling the enable pin low disconnects the load from the battery -- I can't remember now but it draws something in the order of microamps during disable.

Alternately, I've found that minimal loads (<10mA) will still allow the battery cycle to complete. So if you can put your device to sleep or disable all of its peripherals, that's probably good enough.

Product PRT-14411 about 7 years ago

Sounds great! Sorry if it was confusing.

As for that inductor, it shouldn't be chipped like that but it probably won't effect operation -- if the whole top comes off then maybe. It's part of the boost circuit anyway. If it fails you'll have poor (or no) current output capacity but the charger will still work.

Good luck with your project!

Product PRT-14411 about 7 years ago

Good question. While the PAM has current limiting capacity through that resistor, the current limiting factor here is actually the inductor (as well as the source's current capacity). Let's say the thing is 100% efficient, 5W out = 5W in. If the output is delivering 1A at 5V, that's 5W power output. If the input is at 2.5V, and 5W is required going in, it needs 2A of input current. The winding of the inductor has resistance which can get in the way of the operation at high currents. See more including some graphs in the hookup guide connecting a load section. All boosters have this property-- the lower the input voltage, the higher the input current.

I designed this such that it can deliver 1A over the entire input voltage range of a LiPo. I actually took my measurements up to 1.3A on the output, but I'd rather provide something with a healthy margin that to call it a 1.3A booster.

News - Enginursday: Using a Moto… about 7 years ago

That is fun! Thanks for posting. I'm curious, do you know when that was?

News - Enginursday: Using a Moto… about 7 years ago

Hey Casey! I'm glad you enjoyed. I'm pretty sure it scales waaay up, to the limit of DC power supply and mosfet cost. It's a similar technique to one some robot builder I know is investigating.

Product PRT-14411 about 7 years ago

That's kind of an interesting problem, you want to continue to boost but not charge, all while leaving the charger connected. There's no shutdown pin on the charger IC. I would investigate using a mosfet to switch the charger in and out of the circuit.

Product PRT-14411 about 7 years ago

Thanks!

My main concern when designing this was to make sure the voltage was very noise-free on the output. I can't stand supplies that whine! With that in mind, output capacitance required is a based on the transient load characteristics, which I don't have any control over. a single 22uF may have been fine, but two is better here! And yes, I was considering that less specific components on the board means less work to build, even if it's only a single reel in a machine.

The diodes allow the boards to be used in series -- see the hookup guide for more info. If one goes dead in a chain, the current will be passed by.

Product PRT-14411 about 7 years ago

The booster is regulating to 0.6V on the feedback pin (FB). Adjust the voltage divider made by R1 and R4 using 0.6 and your desired output voltage as constraints, and it should be OK. I solved R4=600k (with R1=100k) for 4.2V output... maybe there's something else going on here.

The inductor value is not critical to regulation voltage.

Also, R4 is stuffed with 750K from the factory so I'm not sure what's up. I'll let you know if I discover anything.

Keep in mind you can't buck down from a larger battery voltage! The PAM2401 only boosts.