PowerBoost Low Battery Output to Safely Shutdown Raspberry Pi

Question

I've got an Adafruit PowerBoost 1000C and I'd like to connect the LBO (Low Battery Output) to a Raspberry Pi GPIO pin to detect low battery and shut down safely. I'm using a 2500 mAh battery.

What I think I want is this:

PowerBoost LBO -> diode -> resistor -> GPIO

I have a few questions about this:

I'm using the diode to prevent current flow back into LBO, which I've read can cause the low battery LED to light up at the wrong time. I have some 1N914 / 1N4148 switching diodes. Are these appropriate for the task?
What resistor value do I want to use? I think Ohm's law is relevant here, but I'm not totally sure what values to plug in and why.
Is it better to have the diode before the resistor like above, or should I reverse those? Does it matter?

I saw a recommendation not to source or sink more than 0.5mA into an input pin. In that case, it seems like a 10Kohm resistor would be appropriate (4.2V / 10000Ω = 0.00042A) since 4.2 is the voltage of a fully-charged LiPo and 0.42mA is close to 0.5mA. Am I thinking about this correctly?

shabaz · Accepted Answer

Hi Andy, You're right, some resistor based circuit is needed to get that 4.2V down to a usable value. You can do it with a potential divider circuit, e.g. have a 10k or 12k resistor, and put a 33k resistor in series. Connect one end of the 10k or 12k resistor to the LBO pin (which is being pulled high as you say, up to 4.2V), and connect the other end of the 33k resistor to ground. The junction between the two resistors can be directly connected to the Pi (first confirm with a multimeter that you see a voltage of around 3.3V at the junction). No need for any other resistors/diodes etc hopefully. If you have a multimeter, you can check that the junction voltage drops to close to 0V when the battery voltage reduces. Provided it goes to less than 0.8V, the Pi will detect that as a valid logic '0'. The exact voltage levels I'm unsure of, but definitely by being under 0.8V, the Pi should see the pin as a logic '0'. One thing to bear in mind is that some small amount of current will flow through the resistors even when the system is off, so the battery will slowly drain - but it could be low enough not to worry, depending on your use-case. There are ways around that, but they involve more circuitry (a MOSFET) or tapping off the connection elsewhere.But hopefully it will work well enough to get you going for now. There are some input/output interface circuit ideas by the way for the Pi here: Raspberry Pi GPIO Explained EDIT: I just checked the IC datasheet, and the datasheet for the BJT on that board (I should have done that first), and looks like the internal resistor is about 47k. So, the values above are too high. Please ignore it all. Here is an easier approach that will work and needs just one resistor. Connect a 100k resistor between LBO and ground. Then connect LBO to the Pi. You're all set.

cmelement14 · Answer

I agree with Shabaz. Potential divider is the way to go. No need of diode.

shabaz · Answer

Just to clarify, it is still a potential divider with the 100k resistor, just that we're relying on the other resistor that is part of that transistor that is already on the board.

andywest · Answer

Thanks for the great explanation, shabaz ! That sounds really simple. Regarding the current flow when the system is off, I've connected EN and GND with a switch. I have confirmed that it toggles the power LED when I press it. Will the current still flow, even when I turn off the power this way? (The idea is that the Pi will be safely shut down in software (either manually or by monitoring LBO pin), and then the "hard" shutdown with the switch will cut the power to the Pi. Then the Pi can be started again by using the switch.) This conversation has inspired me to seek out some introductory electronics books so I can start answering these questions for myself.

shabaz · Answer

Hi Andy, That's perfect, then you're ok, if you're using the EN pin according to the datasheet http://www.ti.com/lit/ds/symlink/tps61090.pdf it should disconnect the battery (some leakage will be present), so you'll get long life without it discharging significantly. jw0752 has a list of good books by the way, it is here: I Want to Do Electronics! Where do I start?

andywest · Answer

Thanks, Charles. I thought a diode might be necessary because someone on another forum had an issue where the low battery LED was lighting as soon as they connected LBO pin to GPIO, regardless of whether the battery was low. The diode (in addition to a 10k resistor) fixed it for them.

andywest · Answer

It just occurred to me that 4.2V is also higher than the maximum safe voltage of 3.3V. So now I'm just confused. Is it possible to limit current to 0.5mA and voltage to 3.3V with a single resistor, or do I need to introduce another component somewhere? As you can see, my basic electronics knowledge is lacking. Any help is appreciated. As a side note, I'm not positive that 0.5mA is really the limit for GPIO pins on the Pi. I've also seen 16mA in places, but I can't find the official spec. Thanks!

shabaz · Answer

Hi Andy, I just did a rough calculation, knowing that the Pi input supports a maximum of 3.3V, and that is two-thirds of 5V, which is close enough to the 4.2V, so if the top resistor in the potential divider is about 47k (and it will vary a fair amount, it is not accurately defined because it is implemented in silicon), then a ballpark of 100k for the lower half of the potential divider should be sufficient. You could reduce it further (say down to 68k) and it should still be fine, the Pi has a bit of input protection. If you're above 2V then that is a logic '1' in a lot of device datasheets, but not all - some devices use different logic levels and that might not be sufficient. Below 0.8V is a logic '0' in many device datasheets. Regarding the input current, the Pi will just take what it needs (it has an input resistance) which I assumed will be of the order of a few microamps (because it will be a MOSFET gate at the input inside the Pi's chip, and they don't consume a lot of power unless switching many times a second, because the gate looks like a small capacitance), which the potential divider should support. The 0.5mA specification I think it only applies if the voltage is taken above 3.3V, but I'm not sure. In any case, it won't apply when you're using normal logic level voltages, because then the Pi's GPIO pin will just consume what a logic gate input may consume, i.e. a value in microamps.

comperem · Answer

Perfect. Great sleuth work @shabaz. The 100kOhm resistor is the other part of the voltage divider with the built-in 47kOhm resistance in the LBO circuit. Nice. Simple.

andywest · Answer

I'm trying to understand the "why" behind your answer. I found the equation Vout = Vin * (R2 / (R1 + R2)). I plugged in the known values. So for LBO pin voltages of 3.2V - 4.2V, Vout is between 2.18V and 2.86V. This is less than the maximum 3.3V but hopefully high enough to register as logic '1'. The one thing I'm still not sure about is current. Supposedly a Pi GPIO pin set to input can only handle a certain amount of current. I'm thinking if V=IR, and V = 4.2, and R = 47kΩ between LBO pin and GPIO pin, then I = 0.09mA? That is well below 0.5mA and should therefore be safe. Is any of what I said above remotely correct? Thanks for bearing with me while I figure this out.

larsks · Answer

I tried Shabaz's idea of putting a 100K resistor between LBO and ground in order to connect the LBO output to a Pi.

Prior to adding the resistor, I read about 5V at the LBO pin. The docs say "By default it is pulled high to BAT", but since BAT reads 4.2V I'm a little confused. I see that on the schematic Vbat is not actually the battery voltage (that's Vlipo), so maybe this is just some confusion between the PCB labels and the schematic (so the pin labelled BAT on the PCB is actually Vlipo, and LBO actually tracks Vbat, which is exposed on the pin labelled Vs).

After adding a 100K resistor between the LBO pin and ground, I see:

- The LOW LED goes on immediately.

- I read about 4.2V at the LBO pin.

That doesn't seem to match the discussion here. Does this seem unexpected, or am I just misreading this thread?

Thanks!

shabaz · Answer

Hi Lars,

Whoever drew the schematic at Adafruit took no pride in their work, it is a mess but looks like it should only be possible to go to the battery voltage level, not the boosted 5V level.

Anyway, just in case the schematic has changed and is inconsistent with the physical circuit, please try this (i.e. instead of 100k to ground, try this circuit instead):

Before you connect to the Pi, please confirm with a multimeter that the voltage level never exceeds 3.3V.