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Finding a high frequency differential amplifier Transfer Function

Discussion in 'Electronics Homework Help' started by Mobin, Apr 5, 2018.

  1. Mobin

    Mobin

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    0
    May 1, 2017
    hi friends,
    I have been assigned to analyze this circuit below(it's an improvement to a differential amplifier in order to have better BandWidth):
    Capture22.PNG
    like a differential amplifier , i tried to find the half-circuit and i came up with this:
    Capture22 - Copy.PNG
    in order to find the transfer function i used hybrid-pi model but now I'm stuck with a problem: that there is no connection between the two transistor and the second transistor seems non-functionalo_O so how can i find the transfer function? have i done any mistake in a part of solution? :(
    Capture2 - Copy (2) - Copy.png
     
  2. Mobin

    Mobin

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    0
    May 1, 2017
    can somebody help me, please...?
     
  3. Ratch

    Ratch

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    302
    Mar 10, 2013
    What do you want us to do? Do you want the equations for the voltages V1, V2, Vout/2 which you marked in red? Obviously V2 is grounded. A transfer function needs to be defined before is can be calculated (Vout/Vin,Iout,Iin,Vout,Iin,Iout,Vout). Which is it? I have never done double differential amps before, or any differential amps for that matter. But if you give me more info, I will take a swipe at it.

    Ratch
     
  4. LvW

    LvW

    604
    143
    Apr 12, 2014
    Why do you ignore the current source at the common emitter node of Q1 and Q2.
    Yes - in this case, the "second transistor is not-functional".
     
  5. Mobin

    Mobin

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    0
    May 1, 2017
    in ac analysis dc current source would act like open circuit
    but i shouldn't make that point grounded, that way the problem could be solved
    now i have another problem, with biasing it .
    when i write down dc and ac load lines,in order to get maximum swing, it comes up that there is more variables than equations, and also that Q2 bias current is zero!!!!

    is that possible?

    Capture22 - Copy.PNG
     
  6. Ratch

    Ratch

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    302
    Mar 10, 2013
    Why don't you answer the question I asked in post #3?
     
  7. Mobin

    Mobin

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    May 1, 2017
    sorry i missed that
    the transfer function i'm looking for is Vout/Vd
     
  8. LvW

    LvW

    604
    143
    Apr 12, 2014
    For ac analyses, the current source may be replaced by an open circuit - that`s correct.
    However, if you want to analyze bias conditions...what do you think: Is the DC current source important or not?
     
    Mobin likes this.
  9. Ratch

    Ratch

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    302
    Mar 10, 2013
    Very interesting circuit. Differential amps with a current source have a very small turn on/off range. As you can see from the plot below, ±4 Vt , where Vt = 25.4 mV, is all it takes to switch the paired transistors from ON-OFF to OFF-ON. To stay in the linear region, Vd should be within ±Vt. During operation, the voltages of the transistors from left to right are HIGH-LOW-HIGH-LOW or LOW-HIGH-LOW-HIGH. The two inner transistors from where the voltages are taken switch between HIGH-LOW and LOW-HIGH. The advantage of this double differential configuration is that the output voltage can vary between ±v. A single differential configuration could only vary between 0 to v. That is double the voltage range. The transfer function is 2v/Vt. The higher the supply voltage, the greater the voltage amplification.
    Mobin.JPG
    Ratch
     
    Mobin likes this.
  10. Mobin

    Mobin

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    May 1, 2017
    that was really useful thanks a lot
     
  11. Mobin

    Mobin

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    May 1, 2017
    of course that is important, and it shows the bias current of both transistors. but the problem is that i can't recognize the relation between bias current of the two transistors. are they biased independently?
     
    Last edited: Apr 12, 2018
  12. Mobin

    Mobin

    13
    0
    May 1, 2017
    is there anybody who could help me with biasing this
    Capture22.PNG
     
  13. Ratch

    Ratch

    1,023
    302
    Mar 10, 2013
    What's the problem? If Vd is zero, all the transistors will be conducting with 1/4 of the current source in each transistor. All the emitter voltages will be at -0.7 volts and all the E-B junctions will be forward biased. If Vcc is greater than 0.7 volts, the the C-B junction will be reversed bias and all the transistors will be operating in the active region.

    Ratch
     
    Mobin likes this.
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