I'm comfortable with Ohm's law and analysing series/parallel circuits with resistors, but I'm struggling when it comes to analysing some simple circuits with transistors.

The attached circuit is from an entry level electronics book, but I'm not sure what the purpose of the 100k and 47k resistors is. It's a continuity tester, so the probes may be shorted (no resistance). The idea is that if current flows between the emitter and the base, the LED will light up. So, what's the purpose of the 100k resistor? If it was omitted, current would still flow through the base and turn on the LED wouldn't it?

And is the purpose of the 47k resistor to divide the voltage between the emitter and ground like in the discussion about the common collector amplifier in vol. III of All About Circuits? Ie, the voltage drop between E and B is 0.7 volts, and the drop over the 47k resistor is therefore 5.3V?? And the reason for that would be to prevent the transistor getting the whole current and burning out? But this is a common emitter arrangement isn't it, not a common collector?

What am I missing?
Thanks.

 

You are correct. The PNP transistor is connected in a common-emitter configuration.

Without the 100K and the 47K resistor there would be no certainty the transistor would be turned off when the two probe lines were open circuited. The 47K is there to limit the current to the base of the transistor with the two probe terminals are shorted together.

hgmjr

 

Thanks Hgmjr.

I can see why you would want to limit current through the base with the 47k resistor, but I'm still not totally clear on the purpose of the 100k one. Say the 100k resistor wasn't there and the probes are open circuited, wouldn't it be impossible for current to flow through the base? Ie., we could be certain the transistor is turned off even without the 100k resistor?

Is it there as a voltage divider to ensure there's at least a 0.7V difference between base and emitter? If the probes are shorted, then there's a full 6V drop across the 100k resistor. But I don't see what this would achieve because the probes would be at ground anyway.

 

What you are overlooking is that there is a small amount of leakage current internal to the transistor that makes its way into the base and can cause the transistor to begin to turn on. It will not turn the transistor on significantly but the transistor may not be fully off.

The 100K plus the 47K in series are there to overpower that leakage current and insure that the transistor is in cutoff.

hgmjr

 

Thanks, I wasn't aware of the leakage current - it makes sense to me now.

 

Typically, modern silicon transistors will not have enough c-b leakage to turn the transistor on, certainly not enough to turn on an LED, unless the transistor gets really hot (leakage current doubles every 10 deg C). I always use a resistor, because I'm generally concerned about turn-off time. The resistor helps discharge base capacitance (or clean out excess base charge, if you want to get technical). In design for production, it's good engineering practice, because high temperature is generally a possibility. In your case, you could probably get along without it.
Back in the bad old days, before silicon trannies, we had germanium, which was really leaky. The base-emitter resistor was mandatory in a switching circuit.