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A Newbie Question on PNP Transistors

Discussion in 'General Electronics Discussion' started by haroldjclements, Oct 17, 2013.

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  1. haroldjclements

    haroldjclements

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    1
    Oct 17, 2013
    I have a question on PNP transistors. Everywhere I read it tells me that a PNP transistor needs a negative voltage to the base in order for the transistor it turn on. However, I have created a Micky Mouse circuit (attached) and do not understand how it is working.

    I am happy with the NPN part when the switch is open. A am also happy that when the switch is closed, the base of the NPN drops to 0v causing it to turn off. However, this is where I get confused. The PNP transistor turn on, but the voltage to base is high (not negative as I am led to believe should turn the PNP transistor on). I have used a meter to vitrify this.

    Now the current going to the base of PNP transistor is high when the switch is open and is at 0mA when the switch is close. Could this be what is turning the PNP transistor on. If so, why dose all the literature that I have read say that its negative voltage?

    Thank you for your help,
    Harold Clements
     

    Attached Files:

  2. duke37

    duke37

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    769
    Jan 9, 2011
    There seems to be little current limiting in your circuit. Presumably the battery will be running flat out.

    I prefer circuits to be drawn with the positive at the top. I find this easier to understand. The austalians and new zealanders may prefer it the other way up.

    When the switch is open, Q1 is turned on by a current through R3.
    When the switch is closed Q2 is turned on with base current through R1.

    The positive and negative confusion is that you should measure voltages relative to the emitter.

    I do not see the pupose of D3.
     
  3. haroldjclements

    haroldjclements

    24
    1
    Oct 17, 2013
    Hello duke37, thank you for your reply.

    I take your point on the way I drew the circuit. As I said, it was just a "throw away" example to illustrate my question.

    As for D3 (remembering that I am a complete novice and trying to teach myself electronics), the idea was that I would preventing any current flowing to the base of NPN transistor when the switch was closed. (Obviously not a good idea)

    Now, I am still a little confused and appreciate your time and effort. I understand your explanation that "When the switch is closed Q2 is turned on with base current through R1". However, I still do not understand how the transistor turns on despite the lack of a negative voltage. You say that I "should measure voltages relative to the emitter". Could you elaborate, as emitter of Q2 is always high?

    As I say, I am a novice trying to educate myself and appreciate your help.

    Kind Regards,
    Harold Clements
     
  4. haroldjclements

    haroldjclements

    24
    1
    Oct 17, 2013
    Hello,

    Just following up my question. I have been reading the transistor tutorials on this site.

    "No current through the base of the transistor, shuts it off like an open switch and prevents current through the collector. A base current, turns the transistor on like a closed switch and allows a proportional amount of current through the collector. Collector current is primarily limited by the base current, regardless of the amount of voltage available to push it."

    This is how the Q2 is working on my circuit. So, does a PNP transistor need a non-positive signal on the base for it turn on (as I have keep reading else where), or does it just need a small base current?

    Thanks in advance,
    Harold Clements
     
  5. duke37

    duke37

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    769
    Jan 9, 2011
    This is how I would have drawn the circuit with the positive at the top.

    Taking Q1 first.
    When the switch is open, R3 goes to the 9V supply and provides a base current to turn on the transistor.
    When the switch is closed, R3 current is shorted to ground so there is no current into the base and the transistor is off. Things are complicated a little by the voltage drop across the led.

    Now Q2
    Q2 is turned on by current through R1. When the switch is open, there will be no current through R1 so the transistor will be off.
    When the switch is closed, there will be current through R1 so the transistor will be turned on.

    Note that Q1 is npn so the base current has to come from a positive supply relative to the emitter.
    Q2 is pnp so the base current has to come from a source negative to the emitter (which is at 9V).
    If the base is provided with a positive voltage relative to the emitter, then negligible base current will fow and the transistor will be off. If this positive voltage is too high, perhaps 6V or so, then reverse current will flow and the transistor can be damaged, this cannot occur in this circuit.

    The currents passed by the transistors will depend on their gain which will depend on the transistor type and often has a very wide tolerance. A resistor in series with each led can define the current.
     

    Attached Files:

  6. BobK

    BobK

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    Jan 5, 2010
    You seem to be missing the point that voltages are always a difference between two points. The base of a PNP has to be negative with respect to the emitter to turn on. Here the emitter is at 9V and the base connected to 0V through a resistor. Which is the same as saying that the base is connected through a resistor to -9V with respect to the emitter. So there is your negative voltage.

    Bob
     
  7. haroldjclements

    haroldjclements

    24
    1
    Oct 17, 2013
    Thank you very much duke37 and Bob...

    I think I have it now. As long as the base of the PNP transistor is in some way pinned to ground (0v rail) then it should satisfy the "switching on" criteria regardless of the positive voltage that I am seeing at the base.

    Again, I would like to thank both of you for your time and effort.

    Kind Regards,
    Harold Clements
     
  8. BobK

    BobK

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    Jan 5, 2010
    Actually, the correct learning is that in order to turn on a PNP transistor, the voltage on the base must be less then voltage on the emitter.

    Bob
     
  9. boylesg

    boylesg

    2
    0
    Oct 19, 2013
    PNP transistors

    If no current flows from the emitter to the base of a PNP, i.e. if the voltage level applied to the base is equal or greater than the voltage level applied to the emitter, then the PNP transistor is open and current will flow from the emitter to the collector.

    If you reduce the voltage level applied to the base of a PNP transistor, resulting in current starting to flow from the emitter to the base, then the transistor starts to turn off.

    In this way they are opposite to NPN transistors.
     
  10. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    25,411
    2,779
    Jan 21, 2010
    I think you mean "will not flow".

    toy mean "on".

    I remember it this way:

    for an NPN or PNP transistor, firstly the arrow on the emitter points to N (inward for PNP, outward for NPN). To turn them on, the base must be more positive (or negative) than the emitter. The middle letter (NPN, PNP) tells you if it is more negative or more positive.

    So for a PNP transistor, the base must be more negative then the emitter.

    If you look at the arrow on the emitter, it points to N(egative) and the direction it points must be negative with respect to the direction it comes from. It tells you which way around the diode junction is that must conduct.

    NPN and PNP are opposite of each other in that they behave EXACTLY the same way if you replace NPN for PNP (or vice versa) and then swap +ve for -ve in the circuit.

    Because we normally draw +ve at the top, this also generally results in us drawing the circuit upside-down with respect to the other, but that's just convention not a real difference.

    More negative == less positive.
    More positive == less negative.
     
  11. boylesg

    boylesg

    2
    0
    Oct 19, 2013
    Damn, I should have picked up that I had it wrong because I am building an 8x8 RGB cube that is controlled by BC327s at present.
     
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