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Calculating the effective impedance of a current source

  1. Jun 9, 2009 #1
    I'd like to calculate the gain of a common emitter transistor circuit I'm looking at that has a a constant current source as the emitter load. Now I know the gain is going to be -gm(Re||RL) but how do I calculate what the effective resistance of the current source is? The current source I'm using is one that uses a Zener diode to keep a constant voltage across base to emitter junction, and thus a constant current through the collector of and the emitter to collector of the input transistor.

    A tangentially related question is, if I were using a Darlington pair as the device connected to the constant current source instead of a transistor, how would I go about calculating the transconductance of the Darlington? Thanks for any assistance!
     
  2. jcsd
  3. Jun 9, 2009 #2
    Here is a constant current source I have used in the past, For this circuit, when changing the pullup resistor R3 from 1 ohm to 50 ohms, the collector current changed from 176.569 mA to 176.566 milliamps in simulation. R1 can be changed to change the output current. For improved temperature compensation of the Q2 base-emitter junction, put a diode in series with R1.
     

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  4. Jun 9, 2009 #3

    vk6kro

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    Could you post your actual circuit please?

    The gain of a common emitter amplifier with an unbypassed emitter resistor approximates (RL/RE) provided the RE is large compared with the R of the base emitter junction.

    This is a very handy way of getting a predictable gain.
     
  5. Jun 10, 2009 #4
    I was suggesting in my post that you use the Q1-Q2-R1 circuit in my post and thumbnail as your emitter-to-ground connection instead of just an emitter resistor for a very stable constant current sink. The reason this connection is so stable is that the Q2 collector-to-emitter voltage is about 2 diode drops, so the Q2 gain is pretty good. Using a zener from emitter to base would give you no current adjustability and very poor thermal stability.
    -
     
  6. Jun 10, 2009 #5

    vk6kro

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    Hi Bob,

    My request was to the original poster to post his circuit.


    Yours is fairly standard although a bipolar transistor which was fully conducting would be quite a shunt across the input impedance of the main transistor.
    One advantage of the unbypassed emitter resistor is that you get an increase in input impedance. It is approximately (Hfe times RE) which is quite a useful lift in impedance for a bipolar transistor.
     
  7. Jun 11, 2009 #6
    I was trying to point out that the effective impedance of the collector of Q1 in my circuit is very high and has a good frequency (flat) response, and if used as the emitter load in your circuit would give you a very high RE (emitter resistance). Just substitute your circuit for R3, and select R! to adjust the Q1 collector current.
     
  8. Jun 14, 2009 #7
    Sorry for the delay in my reply. The tutorial online at: http://www.ecircuitcenter.com/Circuits_Audio_Amp/Active_Load_Gain_Stage/Active_Load_Gain_Stage.htm says about the impedance of the active load "And rumor has it - its quite high." Well, how high is it? That's what I hate about some of these tutorials, they'll lead you 90% of the way there and leave you hanging. If I knew the impedance of the active load I could figure out the gain of the stage - is the impedance looking into the emitter of the active load just 1/gm? If that's the case though and the emitter resistance of the transistor the active load is loading is also 1/gm then finding the gain through Rl/Re will give me unity! How's that supposed to give the big gain that current source loads are supposed to provide? Figuring this out has been gnawing at me for days now - any pointers as to where I'm going wrong would help so much!
     
  9. Jun 14, 2009 #8

    vk6kro

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    Thanks for the circuit. Makes it much clearer.

    A current source has infinite impedance theoretically.
    However a real transistor has output admittance which is like a resistor in parallel with the output.
    This is typically 10 k but can be as low as 1 K. So, this would be your load from the current source Q4, and it appears in parallel with the output admittance of the gain transistor Q3.

    So, worst case, you might really have a load of 500 ohms. More likely it would be about 5000 ohms. This limits your final gain.

    The driver circuit apparently includes another current source but details of this are not shown. The input of Q3 will be quite low impedance (maybe 1000 ohms) and the voltage output of the driver would be the current source output times this resistance.
    So,
    V at input of Q3 = gm * v(1-2) * R (input of Q3).

    .
     
    Last edited: Jun 14, 2009
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