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. One important part of this external circuitry is a 12V buck-boost switching power supply which is powered off of a 9V battery. The buck-boost regulator does not have an enable pin, and I am working out a way to cut the power to the regulator.

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 circuit should also cut off the current coming out of the source of **Q4**. 

---

**[Circuit - Version 1][1]**

![Device Under Test (DUT)][2]

---

I have two questions with respect to this circuit:

First, 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. 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'm convinced I'm overlooking **V_ce** on **Q2** though.

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**. 


  [1]: https://i.sstatic.net/esKDI.png
  [2]: https://i.sstatic.net/esKDI.png