proper use of PNP transistor

Proper use of PNP transistor

Hi folks,

I'm actually curious about this, (still a beginner, self-studying, not a genius, doesn't have 1 year experience on electronics, doesn't have solid foundation on basics, doesn't have a mentor, always feels dizzy w/ books but always trying hard to understand this fun hobby ) I was in the process of looking for Inverted NPN darlington array transistors (at least 10) to replace my 10 PNP transistors w/o at least using inverter IC because I want it to be less spacious and few components. Then i found this picture.



I thought transistors are mostly used as simple switches (NPN = P means should input (+) to connect N and N connections; PNP = N means should input (- or gnd) to connect P and P connections). That's why I have used PNP transistors this way on the outputs of open-drain IC w/pull up resistor to drive relays or LEDs.



the above works fine together with the open-drain IC I used for a long time and nothing is wrong. Am I using PNP wrong and how it should be used? can I still use PNP on this way since it works? I hope you could please explain it to me in layman's term or as basic as possible :bang:

The PNP transistor makes little sense in this circuit because it's inverted and has no current gain. You can either omit R4 and Q1, driving the load directly from open drain.

Or, if the open drain output current capability is insufficient to drive the load, flip Q1 emitter and collector and use higher resistance value for R4.

Hi FvM,

no current gain? then I can still use it like this? current gain is for amplification right but i just wanted to use it as a driving switch since that LED will be replaced like an SSR, would that be uhm - ethical in electronics?

I am using the MAX7317 i/o expander. It is said that I can only sink only few micro amps on the open-drain output of the IC so the R4 there is really around 100k to 220k. I just want to add the transistor because i dont want to directly sink the led load (SSR, etc) to the IC. Since I am currently using 100K as R4 already, then only flipping E and C won't change anything or affect my circuit?

no current gain? then I can still use it like this? current gain is for amplification right but i just wanted to use it as a driving switch since that LED will be replaced like an SSR, would that be uhm - ethical in electronics?

Click to expand...

Operating the transistor in its low gain region is senseless because 1) you do not have gain and 2) you have a voltage drop higher than when operated as a switch.
All in all, there is no reason of its use in your specific application.

Hi,

There are so many documents, internet sites and even tutorial videos about basic transistor operation, basic connection schemes and part value calculation.
This are very basic things, thus a simple internet search will answer all your questions. Did you ever try to use this huge source of information?

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One problem for us (or at least me) is, that you give very vage and confusing informations.
Please give us a schematic that exactly shows your situation....with voltages...and input circuit and output circuit.

Depending on your load requirements the MAXxxxx IC may be able to sink 500...5000uA ... this is more than "a few".

Btw: even a cheap standard logic shift register (like 74xx595) may sink more current.

Klaus