I'd recommend reducing R7 by an order of magnitude. If it's too large, there'll be enough voltage to turn on Q2, which will activate the red LED as well. The voltage drop across R5 is therefore small, which means Vce of Q1 will be somewhat large. When the battery is still good, that resistor is keeping the current into the base of Q1 very low, which allows very little current through the green LED. My answer is based on your 'Version 2' circuit.įirst, the value for R7 is a bit high. I'll try to address them and answer your specific questions. There's a couple issues with your circuit. I updated the circuit to reflect that of the answer provided by Attaching DC sweep simulation. Second, is it safe to assume the regulator can run from the output of the source of Q4 and D4? My estimates suggest that I will, when drawing current, have a 0.2V drop ( V_ds) across Q4, and another 0.3V drop across D4.Įdit: Adding updated circuit as per earlier suggestions. I'm convinced I'm overlooking V_ce on Q2 though. If the red LED is on, there's a 0.7V drop across it, then a moderate drop across R6 (ie: estimated current, (7.0V-0.7V) / 330R = 19mA, so a 6.27V drop across the resistor roughly), and since this will make the Q4 bias less than that connected to Q4's drain, Q4 is turned off, cutting off power to the buck-boost converter. I have two questions with respect to this circuit:įirst, does the cutoff logic make sense? My assumption is that if the red LED is off, no current is going through D3, R6, and Q2, so the voltage at the node between R6 and the collector of Q2 would be the same as the power rail connected at the top to D1, D2, D3, and Q4's drain. The circuit should also cut off the current coming out of the source of Q4. I already have a working 9V low-battery indicator circuit (shown below), which turns on the green LED if the battery is supplying a voltage greater than 7.0V, and if the voltage dips below 7.0V, the red LED is enabled. The buck-boost regulator does not have an enable pin, and I am working out a way to cut the power to the regulator. One important part of this external circuitry is a 12V buck-boost switching power supply which is powered off of a 9V battery.

I have a Raspberry PI board connected to some external circuitry, and am using Microchip MCP23008 I2C GPIO expander chips to control external LEDs and motor circuitry.

9v low battery indicator circuit