AM modulator circuit help plz ..

Discussion in 'Wireless & RF Design' started by Walidj, Jun 10, 2011.

  1. Walidj

    Thread Starter New Member

    Jun 10, 2011
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    http://www.zen22142.zen.co.uk/spice/ammod.htm

    in this circuit can anyone plz explain how does changing Ie through the audio source affect the AC gain of the transistor ?

    The AC gain is Av = - Rc / re , (re = rπ/β) , it does not depend on Ie so how would changing Ie affect the AC gain ?

    thnx,
     
  2. steveb

    Senior Member

    Jul 3, 2008
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    It's not clear what your equation for "re" exactly is, but it should depend on transistor current. What is "r" and what is "π"? Probably beta is the current gain, but we can't be sure of any of it, unless you define it more clearly.
     
  3. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    Presumably re is the small signal dynamic emitter resistance.

    The value of re is dependent upon Ie. Typically re=26/Ie(mA) at room temperature. Modulating Ie at audio frequency will modulate the circuit gain for the carrier through gain.

    It produces a pretty awful AM signal!
     
    Last edited: Jun 10, 2011
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  4. Walidj

    Thread Starter New Member

    Jun 10, 2011
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    As t n k said : re is the small signal dynamic emitter resistance , rπ (r pi) is sometimes written as (hie) the small signal dynamic resistance in B-E junction , β is hfe (the current gain).
     
  5. steveb

    Senior Member

    Jul 3, 2008
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    Yes, I already knew that. As I said, re "should depend on the transistor current".

    But, I was trying to see if you knew that. If you know that, then you know it depends on transistor current. Yet, you claimed that it does not depend on the current. By making you explain what re is (or, more particularly, what rπ is), I was hoping you would be able to answer the question yourself. In general, spitting out an equation without defining the variables and parameters is bad practice. rπ is actually β/gm, where gm is the junction transconductance, and gm=Ic/Vt, where Vt is the thermal voltage KT/q, where K is Boltzmann's constant, T is temperature in Kelvin and q is the electron charge. That's where the 26 mV comes from. By writing all this out, you can see that there is dependence not only on current, but also on temperature. (strictly, beta also depends on current and temperature, but that is a second order effect that is often ignored)

    If re depends on the current, then Av=-Rc/re must depend on the current, and there you have your answer.
     
    Last edited: Jun 11, 2011
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  6. Walidj

    Thread Starter New Member

    Jun 10, 2011
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    Thanks ,

    One more question , what is the role of the capacitor C3 here ? What happens if it's omitted from the circuit ? does this affect the modulation ?
     
  7. steveb

    Senior Member

    Jul 3, 2008
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    The capacitor C3 is the typical emitter bypass capacitor. It is usually used to get higher gain by eliminating (or sometimes only reducing) the emitter resistor (that 4.7K resistor) from the AC gain equation. Normally, the 4.7 K emitter resistor would dominate over re and the gain would be about -Rc/Re which is only -2 in this case.

    Elimination of Re from the gain increases the gain and also makes it sensitive to collector current. Normally, this means higher harmonic distortion which is bad, but in this case, you want nonlinearity to allow for modulation. As t_n_k implied, it's not the best circuit, but it is intended to illustrate the effect.
     
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