NPN vs PNP, a fundamental or just a mental difference
November 23, 2008
Like many here, I were in need of a bit of amplification of a weak signal. This signal had "low impedance", I suppose. The supplied circuit diagram shows two different ways of meeting my need.
Update: labels corrected as Oddbot suggested.
It works. When I shine a laser directly into the phototransistor, my Picaxe picks up a "high" on the digital input. But I have a couple of questions about my choice for the PNP based amplifier.
- These two circuits are equivalent, aren't they?
- I used them as a digital input, but I suppose they could also serve an analogue input. Correct? Maybe tweak the values a bit?
- The big question: does it matter which circuit I choose? Is there a fundamental difference between the two transistor circuits above? Or is it just a difference in fabrication/price/practice?
- Perhaps the fact that my photo transistor is itself NPN makes the two circuits substantially different? What if that was not a concern? It could be a high value potentiometer as voltage divider. Does it still (not) matter?
Just to be clear: I am not looking for help with my laser detection (not today anyway). I just want to learn a bit more about this NPN vs PNP difference.
Update 28 nov:
I thought I needed two more circuits to make this picture complete. Two NPN circuits plus Two PNP circuits. One of each inverting, the other one not.
Boy, was I deceived! I ended up (before I gave up) with EIGHT permutations. (eagle file attached)
But while I was drawing all of these, it dawned on me that I do not need to include the photo transistor. The whole NPN vs PNP debate can be pictured with a single transistor. So I drew this one. NPN version only.
This demonstrates the two basic methods of using a transistor, I think. The left circuit has the output on the collector of the transistor. An NPN version of this left circuit will invert the signal (that's what the line above the words means).
The right hand circuit has the output on the emitter of the NPN and will give a normal output.
If you replace the NPN in any of these by a PNP, the function will reverse.
As I understand now, the emitter-bound output will give slightly better amplification because the resistor is receiving the B-E current in addition to the C-E current.
Any comments, thoughts or aditional questions are still welcomed!
|eight permutations: eagle schematic (posing as .dir fil)||63.9 KB|