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powerbank voltage LED

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winnetouch

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Hy. I am planing on doing a simple raspberry pi project. I will power the chip with a 5v powerbank. The problem is that the raspberry pi zero (or the powerbank itself) does not have any means of indicating if the battery has enough charge to power it. I don't want the battery just to cut out when it's empty I want to power down the pi safely.

I want to make a simple circuit that will light up an LED when the voltage of the battery gets to low. The raspberry pi needs at least 3.3 volts to run so if the led would light up at about 3.5 volts that would be awesome. But most circuits I found online are overcomplicated for what I need with multiple LEDs or connecting to the raspberry pi GPIOs that I find unnecessary (not to mention I have them all populated anyway).

So finally my question. Is there a simple circuit I can connect to a 5v battery that will light up an LED when the voltage drops to 3.5v? I am no electronics wiz... This project is basically how I want to learn some basics. So please be patient if I don't understand everything.

Thank you :)
 

crutschow

Well-Known Member
Most Helpful Member
The simplest would be a low-power IC dedicated to the task such as this or these.
 

DirtyLude

Well-Known Member
The 5V power bank will output 5v until the very end. It doesn't drop in voltage. You would need to measure the battery voltage inside it before the boost converter.
 

winnetouch

New Member
The 5V power bank will output 5v until the very end. It doesn't drop in voltage. You would need to measure the battery voltage inside it before the boost converter.
That shouldn't be a problem. I plan on dissasembling it beforehand anyway. If I do use a STM1061 what would the circuit look like?

I was originally thinking of using something like this even if the green LED is on there. So if I use this before the power boost to 5v this should work? It's for 3.7v battery.

 
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Flyback

Well-Known Member
I would just use a comparator to turn on the led via a switch mode led driver, -and obviously a small low power switch mode led driver, so it wasn’t consuming much at all until the battery actually goes below 3.5v

Have some hysteresis with the comparator, and when the comparator trips, the led gets turned on
 

audioguru

Well-Known Member
Most Helpful Member
A 3.7V lithium battery is 3.2V when its is almost dead and its load should be disconnected and it is 4.2V when fully charged. Therefore its average discharging voltage is 3.7V.

Your simple circuit has many variables:
1) What battery voltage causes the powerbank to drop its output voltage or stop working?
2) Temperature changes the base-emitter voltage of the transistor and the forward voltage of the diode and the green LED.
3) Is the green LED a 2.2V dim old one or a 3.2V bright modern one?
4) Are the diode voltage and the base-emitter voltages 0.6V, 0.65V or 0.7V?

I don't think your simple circuit will work with such a low voltage anyway:
 

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winnetouch

New Member
Well you got some components wrong. My bad for not writting them up. On my tests it works. I built the circuit with some parts laying arround (I used a different diode though). It's a bit messy because I didn't use a pcb. On the simulation it works. The actual build also works on a full battery. The green LED isn't fully lit but it's enough. I had no way of testing it if the charge isn't full be cause I don't have other power sources than the actuall battery and all other bateries I have have too low a voltage to even light up the red LED.

All I really need is the red one anyway. I'll try a solution with a comparator when I get one but I would still want to see what can be done with the circuit I posted.

Here are the simulations:
Screen Shot 2016-06-05 at 19.56.39.png Screen Shot 2016-06-05 at 19.58.40.png
 
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