pnptransistors
I have seen a few examples of a low voltage warning LED using two NPN transistors, like this;
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;
If the output voltage is high enough you could consider an emitter follower which has a high input impedance β*Zl and therefore hardly loads the jack.
This doesn't satisfy everything that Dwayne speaks of, but for a simple tester it should serve your purposes.
simulate this circuit – Schematic created using CircuitLab
Q1 can be either an NPN or a PNP. This is really just testing the polarity of the base terminal by assuming that current will flow out of the base in a pnp and into the base for an npn.
This will dimly light the LEDs depending on the base current the transistor can handle. If you need more brightness, I would go ahead and use an op-amp or a comparator to detect current flow between the voltage divider and Q1. That might look like this:
simulate this circuit
Note that this circuit doesn't detect if there's a short in your transistor or not so it's still a rudimentary transistor checker.
Best Answer
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.