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Thread: P-N-P used as a switch - a problem

  1. #1 P-N-P used as a switch - a problem 
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    I have a query about a simple switch circuit using a p-n-p transistor (2N3906) - I ought to know better, but it doesn't work as I expected.
    There are two transistors in parallel, i.e. bases bonded together and collectors bonded together.
    Each emitter is connected to +12v via a 33k resistor and these feed directly into independent 555's (pin 2/4 inputs) running in bistable mode. The collectors are direct to 0v. The bases are connected to +12v via a common 8k2 resistor. The bases can be taken to 0v by a reed switch (i.e both bases can be either +12v or 0v depending on the reed switch) - this part of the circuit works fine and does the job; the reed switch taking the bases to 0v thus turning 'on' the transistors and putting 0v to both 555's from the respective emitters. So far, so good.
    The part of the circuit I'm having the problem with is this - essentially, the removal of the +12v supply on one emitter:
    A second reed switch is connected between the emitter of the first transistor and 0v thereby - when closed - it takes the emitter and the input to the first 555 down to 0v - by effectively removing the supply to that emitter. In effect, this does what it needs to do.....however.....
    My problem is....
    This second reed switch on the emitter of the first transistor is also affecting the second transistor by taking the base voltages to 0v when the base should remain at +12v due to the first reed switch (in the base circuit) being open circuit.
    Likewise, I also have another reed switch in the emitter circuit of the second transistor (same config as previous) and this behaves in the same manner, i.e. taking the bases to 0v and both outputs go to at 0v - not as the doctor ordered!
    The reed switch in the base circuit should take both outputs to 0v - which it does. But the second and third reed switches in the emitter circuits should only affect the one (relevant) output - NOT both as is happening.
    I always thought that so long as the emitter of a pnp was >0.7v higher than the base it would be turned off - yet the bases have been taken low (to 0v) for some unknown reason when only one emitter is forced low - but because the bases are taken low, so BOTH emitters are also low.
    Any pointers would be appreciated as to how this can be rectified so that the second reed relay on the emitter will only affect the input to the first 555 (on the emitter of the first transistor) leaving the other transistor and its output unaffected - and vice versa.
    What is it I'm missing here?
    Maybe I'm kidding myself it should work, but I can't see why it doesn't.
    Hell! I've been in electronics for more years than I care to recall, but something is bypassing me here.


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  3. #2  
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    Quote Originally Posted by DaveC View Post
    I have a query about a simple switch circuit using a p-n-p transistor (2N3906) - I ought to know better, but it doesn't work as I expected.
    There are two transistors in parallel, i.e. bases bonded together and collectors bonded together.
    Each emitter is connected to +12v via a 33k resistor and these feed directly into independent 555's (pin 2/4 inputs) running in bistable mode. The collectors are direct to 0v. The bases are connected to +12v via a common 8k2 resistor. The bases can be taken to 0v by a reed switch (i.e both bases can be either +12v or 0v depending on the reed switch) - this part of the circuit works fine and does the job; the reed switch taking the bases to 0v thus turning 'on' the transistors and putting 0v to both 555's from the respective emitters. So far, so good.
    The part of the circuit I'm having the problem with is this - essentially, the removal of the +12v supply on one emitter:
    A second reed switch is connected between the emitter of the first transistor and 0v thereby - when closed - it takes the emitter and the input to the first 555 down to 0v - by effectively removing the supply to that emitter. In effect, this does what it needs to do.....however.....
    My problem is....
    This second reed switch on the emitter of the first transistor is also affecting the second transistor by taking the base voltages to 0v when the base should remain at +12v due to the first reed switch (in the base circuit) being open circuit.
    Likewise, I also have another reed switch in the emitter circuit of the second transistor (same config as previous) and this behaves in the same manner, i.e. taking the bases to 0v and both outputs go to at 0v - not as the doctor ordered!
    The reed switch in the base circuit should take both outputs to 0v - which it does. But the second and third reed switches in the emitter circuits should only affect the one (relevant) output - NOT both as is happening.
    I always thought that so long as the emitter of a pnp was >0.7v higher than the base it would be turned off - yet the bases have been taken low (to 0v) for some unknown reason when only one emitter is forced low - but because the bases are taken low, so BOTH emitters are also low.
    Any pointers would be appreciated as to how this can be rectified so that the second reed relay on the emitter will only affect the input to the first 555 (on the emitter of the first transistor) leaving the other transistor and its output unaffected - and vice versa.
    What is it I'm missing here?
    Maybe I'm kidding myself it should work, but I can't see why it doesn't.
    Hell! I've been in electronics for more years than I care to recall, but something is bypassing me here.
    You are pulling the emitter too far below the base, causing the base-emitter junction to go into reverse ("zener") breakdown at about 8 volts or so (typ for general-purpose devices, give or take a volt). That causes the base to follow the emitter with an offset of that amount.

    A better way to turn off that transistor is to pull the base up to +12. That's a lot gentler than what you're doing now.


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    Thanks tk421, but the base IS at 12v - the first reed switch on the base is 'open' pulling the base high to 12v at the base via the 8k2 resistor and thereby turning it off. Which is good and wanted.
    What I'm also wanting is the second reed switch to take the output low by 'killing' the emitter voltage. Which it does.
    So it's the emitter that is going to 0v by the action of the second reed switch on the emitter being closed to the 0v rail. In the process this also takes the base to 0v also. But this in turn is also taking the second transistor base to 0v (bases strapped together) and turning it 'on' also - and this is not wanted on the second transistor.
    I can accept your comment about a zener breakdown effect (not aware of such an effect) between the base and the collector by me taking the emitter to 0v while leaving the base (supposedly) at 12v (collector is at 0v).
    Guess I might have to think it out again and go for a two transistor OR arrangement on each input to the 555's.
    Shame - I wanted to keep it simple.
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    Quote Originally Posted by DaveC View Post
    Thanks tk421, but the base IS at 12v - the first reed switch on the base is 'open' pulling the base high to 12v at the base via the 8k2 resistor and thereby turning it off. Which is good and wanted.
    Nope. The base is at 12V when the emitter is left alone, but NOT when you ground the emitter; that's the whole point of what I wrote! Instead of assuming that the base is at 12V when you ground the emitter, MEASURE the base voltage. When you ground the emitter of a PNP transistor, the base-emitter junction goes into reverse bias, as I said. Since 12V exceeds the breakdown voltage of that junction (not the collector-base junction -- that has a much higher breakdown voltage), the base will be pulled down in voltage. It will be, in fact, at the breakdown voltage of that junction (e.g., 8V).(The reason for a low breakdown voltage for the base-emitter junction is that the emitter is purposely very heavily doped to improve beta. A side-effect of heavy doping, though, is a reduction in the junction depletion layer width, which results in a higher field for a given voltage, which means that you produce a critical breakdown field at a lower voltage.)
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    Thanks again, I kinda misunderstood your earlier comments.
    I always believe in the KISS principle but maybe I took it too far this time by not understanding the reverse breakdown of the BE junction.
    At least I know what is going on now.
    Ah well, back to the drawing board - I thought it would be too simple to work!
    Thanks again.
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  7. #6  
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    Quote Originally Posted by DaveC View Post
    Thanks again, I kinda misunderstood your earlier comments.I always believe in the KISS principle but maybe I took it too far this time by not understanding the reverse breakdown of the BE junction.At least I know what is going on now.Ah well, back to the drawing board - I thought it would be too simple to work!Thanks again.
    To quote one of my former students who faced a similar type of disappointment in trying to get his experiment to work, "Nature is mean." He's now a hardcore theoretical physicist. He's happier now.


    ETA: You won't see this problem if you simulate the circuit in SPICE with the transistor models that typically come with SPICE libraries. You have to manually add a zener diode across the base-emitter junction. Then you will see what happens.
    Last edited by tk421; February 24th, 2014 at 10:27 PM.
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    Did I miss something here? How can you have a voltage on a grounded emitter?
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    Quote Originally Posted by Mayflow View Post
    Did I miss something here? How can you have a voltage on a grounded emitter?
    Huh? I don't know what your question means in the context of anything in the thread.

    The problem is caused by a "zener" (more correctly, avalanche) breakdown of the emitter-base junction. Instead of the base voltage staying at 12V, it gets pulled down when the emitter gets pulled down.
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