1
\$\begingroup\$

I have seen a few examples of a low voltage warning LED using two NPN transistors, like this;

enter image description here

My question is, can this be simplified further using only one PNP transistor, if R2 were a value so that the PNP base current was low enough the transistor would be 'on' below a set voltage? Something like this;

enter image description here

\$\endgroup\$
1
  • \$\begingroup\$ The original one is a self-purposing circuit - If you build it, you'll definitely need a low-battery warning because it draws up to 3.5 mA using the stated input range! \$\endgroup\$
    – pipe
    Commented Jul 27, 2016 at 6:43

2 Answers 2

Reset to default
1
\$\begingroup\$

No. For the PNP transistor to turn on, it's Base Voltage needs to be it's VB-E Forward Drop. For a typical Silicon Transistor, that's 0.6~0.7V. In your proposed circuit, the Base is held high through R2, and at no point will ever go lower than it. So it will not turn on.

\$\endgroup\$
2
  • \$\begingroup\$ The advantage of the second circuit, however, is that is has much loew stand-by power consumption. \$\endgroup\$
    – Curd
    Commented Jul 27, 2016 at 7:20
  • \$\begingroup\$ Ok I understand why this is a stupid question as it is... Can this be done using a PNP and resistors in any configuration? \$\endgroup\$
    – JoshNZ
    Commented Jul 27, 2016 at 9:02
1
\$\begingroup\$

Current needs to flow into the emitter and out of the base to make a PNP turn on. This can never happen in your proposed circuit. With no current flowing through R2, the base and emitter will always be at the same voltage and the transistor off.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.