JFET amplifier problem

Discussion in 'Homework Help' started by PsySc0rpi0n, Jul 6, 2015.

  1. PsySc0rpi0n

    Thread Starter Well-Known Member

    Mar 4, 2014
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    I have this new problem to solve.

    The attached circuit is an amplifier with a JFET!

    We are asked to find Av, Ri and Ro for that circuit with gm = 10mA/V and Rd = ∞ (I think Rd is the same as 1/hoe).

    First step is to identify the type of amplifier! I would say that this is a common-source amplifier but I'm not sure!
     
  2. kyka

    New Member

    Jun 7, 2015
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    No, it is not common-source, it is common-drain (the input is being applied in the gate and the output is being taken from the source).

    With some fast approximate calculations, it turns out that

    Av=0.94, because it is a common-drain topology and there is a resistor divider at the input
    Ri=180 kOhm, because it is a JFet
    Ro=500 Ohm because the source resistance is much smaller than the 10k resistance. In absence, of the 10k resistor the Ro would be 1/gm.
     
  3. PsySc0rpi0n

    Thread Starter Well-Known Member

    Mar 4, 2014
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    kyka, and can you tell me about if my equivalent small AC circuit and hybrid model are correct? And also how to identify common-drain/source/gate amplifiers?
     
    Last edited: Jul 7, 2015
  4. PsySc0rpi0n

    Thread Starter Well-Known Member

    Mar 4, 2014
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    With the hybrid model I have that I hope is correct (please anyone confirm it with the screens attached at 1st post), I was trying to calculate the voltage gain but I'm not quite sure how to do it:

    Av = Vout/Vin

    Vout = gm*Rs//Rload <-- Is this correct?

    For Vin, has Ig = 0 I would say that would be:
    Vin = (Rg1//Rg2*Vsig)/(Rsig + Rg1//Rg2)

    But I don't know Vsig so I can't calculate Av.
     
  5. ericgibbs

    AAC Fanatic!

    Jan 29, 2010
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    hi,
    Read thru this PDF.
    E
     
  6. PsySc0rpi0n

    Thread Starter Well-Known Member

    Mar 4, 2014
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    After reading and looking for those schematics in that link, I can't understand what is done for Vout.

    For Vout I was multiplying the current gm*Vgs*(Rs//Rload). Wouldn't this be the voltage drop at that parallel that would also be the voltage drop at R_load?
     
  7. ericgibbs

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    Jan 29, 2010
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    Calculate the Output impedance for the Common Drain, then calculate Av [voltage gain]., it will be less than 1

    You are given Vin [Vsig], calculate actual Vgate signal level.

    You now know the actual Vin' and Av so calculate Vout.

    Is this what you are asking.?
     
  8. PsySc0rpi0n

    Thread Starter Well-Known Member

    Mar 4, 2014
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    Kind of...

    From the other link I didn't understood what was being done to calculate Vout.


    By your reply, do I have (mandatory) to calculate output impedance prior to calculating Av?
     
  9. ericgibbs

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    How could you calculate Av without knowing the total Load resistance ie Rs in || with R_Load, as Zout [Rs] is required in the equation for Av.
     
  10. PsySc0rpi0n

    Thread Starter Well-Known Member

    Mar 4, 2014
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    I usually try to calculate Av by means of Vout/Vin. And for Vout I say it would be:

    Vout = gm*Vgs*(Rs//R_Load)

    and for Vin I would say maybe:

    Vin = Vgs but I think this one is wrong.
     
  11. ericgibbs

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    Using the PDF equations and your equations, please post the calculations for Vout , using both methods.

    E
     
  12. PsySc0rpi0n

    Thread Starter Well-Known Member

    Mar 4, 2014
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    I'm using now the PDF equation for Vout but I will ignore Rds because if this Rds is what I think, 1/hoe, we were learned to ignore it as it is usually too high that will not affect calcs!

    Edited;

    Attached are my results using both PDF equations and gif equations for the common-drain.


    Edited1;
    From my class notebook I have that for a common drain setup:

    Av = (Rs*gm*Vgs)/(Vgs+gm*Vgs*Rs) = gm*Rs/(1+gn*Rs)

    which matches the one from the gif but I don't understand why Vin = Vgs+gm*Vgs*Rs.
     
    Last edited: Jul 7, 2015
  13. Jony130

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    From kvl we have Vi = Vgs + Id*Rs and Id = Vgs*gm
    So we have this Vi = Vgs + Vgs*gm*Rs.
    Vout = Id*Rs = Vgs*gm*Rs

    Vout/Vin = (Vgs*gm*Rs)/(Vgs+Vgs*gm*Rs) = Rs/(1/gm + Rs)
     
    Last edited: Jul 8, 2015
  14. PsySc0rpi0n

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    Mar 4, 2014
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    Ok, thanks Jony1030.

    For Ri, I'm thinking about the paths that current can take to ground, so I think Ri = Rg1//Rg2 = 1meg*220k/(1.220meg) = 180.33kΩ.

    For Rout I think it's the current flowing through Rs because this Rout is from the R_Load point of view, so Rout = Vout'/Iout.

    Vout' = gm*Vgs*Rs
    Iout = Id = gm*Vgs

    Rout = Vout'/Iout = gm*Vgs*Rs/gm*Vgs = Rs

    Gathering all info:

    Av = (gm*Vgs*(Rs//R_Load))/(1+gm*Vgs*(Rs//R_Load)) = 10*500/(1+10*500) = 0.999

    Ri = Rg1//Rg2 = 1meg*220k/(1.220meg) = 180.33kΩ

    Rout = Rs = 1kΩ
     
  15. Jony130

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  16. PsySc0rpi0n

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    Mar 4, 2014
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    Ok, sure about Ro. The path for the current generated by imaginary Vs to GND flows through the parallel of gm impedance which is 1/gm and Rs.
    So:
    Rout = (1/gm)//Rs = 1000*1000/(2000) = 500Ω

    But Ri is correct, right?
     
  17. Jony130

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    Yes your Rin looks good.
     
  18. PsySc0rpi0n

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    Mar 4, 2014
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    After noting a small units error at Av, I recalculated it and got:

    Av = 0.833

    The error was that gm = 10mA/V. So I should use 10*10^(-3) and I used simply 10 which was wrong!

    So:

    Av = 0.833
    Ri = 180.33kΩ
    Ro = 90.91Ω

    Also Ro was not correct!
     
  19. Jony130

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    Well for gm = 10mS (milli siemens ) and RL = 1kΩ the JFET gain is
    Av =(Rs||RL)/(1/gm + Rs||RL) = 500Ω/(100Ω + 500Ω) = 0.833V/V
    But the overall gain is less than 0.833 because Rsig together with Rin form a voltage divider.
    Aover = 180.33kΩ/(180.33kΩ + 10kΩ)*0.833 = 0.7892V/V

    Rin = 180.33kΩ and Rout = 1k||100Ω = 90.90Ω
     
  20. The Electrician

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    Oct 9, 2007
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    You've chosen Av to be defined as Vsource/Vgate. Are you sure you shouldn't take into account the loading effect of the input impedance on the source impedance, so that Av would be Vsource/Vsig?
     
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