Basis Transistor Question

Discussion in 'General Electronics Chat' started by DanRilley, Oct 2, 2009.

  1. DanRilley

    Thread Starter Senior Member

    Jan 13, 2008
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    0
    Hi I posted earlier on this but now I have a specific question just about transistors in general. I have read a lot on the internet about the functionality of these but am still not getting it.


    I have a load that measures between 80 and 1000 ohms depending on the configuration.

    I am trying to switch 9V (6 AA) to the load using a PNP 2N3906 transistor.

    I have the transistor base at 0V (which is what I read is necessary for it to be open).
    I have the emitter connected to 9V through a 5K resistor ( I tried calculating this but maybe I'm way off!).
    I then have the collector going through my load to ground.

    Now I don't know if I'm dealing with fried transistors or an incorrect setup but I am reading 9V on all the transistor pins with my meter. Is this the correct setup?
     
  2. Audioguru

    New Member

    Dec 20, 2007
    9,411
    896
    You used the transistor as an emitter-follower (common-collector type). Then the emitter was at about +0.7V (the voltage from base to emitter) so the resistor did not get the full 9V.

    If you change the circuit so that the transistor is a common-emitter type (the resistor from the collector to 0V) and add a series base resistor then the load resistor will get almost the full 9V.
     
  3. hgmjr

    Moderator

    Jan 28, 2005
    9,030
    214
    Rather than trying to describe your circuit in words, how about attaching a schematic to this post. Be sure to clearly label the components.

    hgmjr
     
  4. DanRilley

    Thread Starter Senior Member

    Jan 13, 2008
    107
    0
    OK Here ya go.

    [​IMG]

    I have no idea about transistors when people explain them to me it's always using different terminology so things never make sense.

    I will explain how I understand them and then maybe you can use how I understand them to explain how they actually work.

    From this example I thought the NPN would allow the current to flow from ground to +9V through the heat wire. But no, it works sometimes other times it doesn't, it fries and bad things happen.

    I realize I need to protect the transistor somehow but not sure how. The base needs to be under 5V which it is.

    The maximum current rating I believe is 200ma. So maybe my load is drawing too much current but I don't know how to figure this out. I believe its 9V/1000ohm = 9ma, but I'm sure thats wrong.

    I just need a real basic explanation
    a) Is this even the correct setup for the transistor (in terms of base collector, emitter)
    b) If so, where do I need to resistors to protect the transistor, yet still allow enough current to run through the heat wires to generate heat
     
  5. Audioguru

    New Member

    Dec 20, 2007
    9,411
    896
    Everything is different in your new circuit.

    Ohm's Law is used to calculate current. The current is the voltage across a resistance divided by the resistance. Your load is a minimum of 100 ohms and if the 9V battery is brand new then the current is 9V/100 ohms= 90mA. If the load is 1000 ohms then the current in it is 9mA.

    The transistor needs a resistor in series with its base to limit the current from the PIC (its max allowed output current is 25mA) and limit the base current in the transistor. With a 90mA load then the base current should be 9mA. The base voltage will be about 0.7V when the transistor is turned on and the output from the PIC will be about 4.5V when it has a 9mA load so the base resistor has (4.5V - 0.7V)= 3.8V across it. Ohm's Law calculates the base resistor value to be 3.8V/9mA= 422 ohms. Use 390 ohms which is the nearest standard value.
    If the load is always only 9mA then the base resistor can be 3.9k ohms.
     
  6. DanRilley

    Thread Starter Senior Member

    Jan 13, 2008
    107
    0
    Thank you for the very thorough explanation that was perfect.
     
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