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.



  • \$\begingroup\$ 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. \$\endgroup\$
    – Darko
    Commented Mar 22, 2016 at 12:43
  • \$\begingroup\$ 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). \$\endgroup\$
    – Darko
    Commented Mar 22, 2016 at 12:49
  • \$\begingroup\$ 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. \$\endgroup\$ Commented Mar 22, 2016 at 15:41
  • \$\begingroup\$ Haha yes, I knew there had to be something wrong with it. electroschematics.com/9010/… \$\endgroup\$
    – Brian C.
    Commented Mar 23, 2016 at 13:21

1 Answer 1


I've simulated the circuit you've posted and it doesn't seem to work at all.

The problem, I'm sure, is at least partly due to the emitter resistance killing the gain of the circuit.

I've redesigned it and made the first stage NPN common emitter with a pot to adjust the low voltage switching point, and added the second stage to invert the output of the first stage, isolate the LED load, and snap the LED ON quickly at the switching point.

I've shown a white LED because that's what I ran the simulation with, but any garden-variety LED will work by adjusting its forward current via R3.

In the graphic, below, the red trace is the battery voltage decaying from 12 volts to zero volts, and the green trace shows the LED popping ON at about 8.9 volts and then getting dimmer and dimmer as the battery voltage decays.

enter image description here

  • \$\begingroup\$ It is inverted :(. LED should be on when battery is low, not the other way around. \$\endgroup\$
    – ilkhd
    Commented Mar 22, 2016 at 21:34
  • \$\begingroup\$ @ilkhd: Oops... you're right! Sorry about that, and thanks for the reality check. I should have been paying more attention. I think it's fixed now. :-) \$\endgroup\$
    – EM Fields
    Commented Mar 23, 2016 at 11:10

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