# Confused about battery voltage warning circuit with PNP transistor and Zener diode

Below is a circuit where an LED will light on when the battery voltage (Vcc) is below approximately 8.8V. However, I'm confused as to how it works. My understanding is that when the battery voltage is less than 8.8V (6.8V Zener + 2.0V LED), the PNP will be in saturation mode, and thus the LED will turn on. However, when the battery voltage is greater than 8.8V, why does the LED turn off? Wouldn't the PNP be in active mode?

Please correct me if anything I've said is wrong. I apologise if this is a trivial question, but I've searched everywhere and asked a few older peers already. It also isn't very helpful when most resources on BJT transistors talk about NPN.

Thanks!

• To me it looks like constant current driver for LED. If voltage increases it will still provide around 12mA to the led which will kip it lit. But I may be wrong. Commented Mar 22, 2016 at 12:43
• My understanding is that voltage on the base of transistor would be Vcc-6.8V. Transistor "will maintain" ~0.6V Vbe, so on resistor R2 would be ~6.6V. Current is than 6.6V / 510 Ohm = ~0.012A. Small part of that current is Ib (trough resistor R4) and majority would go through collector (Ic) to the LED (Ib x hfe = Ic). Commented Mar 22, 2016 at 12:49
• Please provide a link to your source. As shown, the circuit will not work as a low-voltage indicator. With increasing voltage above about 1 volt the LED turns on, with brightness increasing linearly with voltage to about 9 volts, and then stays on at that brightness for higher voltages. Short answer is, you are confused because it doesn't work the way you have been told. Commented Mar 22, 2016 at 15:41
• Haha yes, I knew there had to be something wrong with it. electroschematics.com/9010/… Commented Mar 23, 2016 at 13:21