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ankit3177 Group Title

H5P3 Source Follower Small Signal

  • one year ago
  • one year ago

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  1. Srikanth_Gangula Group Title
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    Incremental ouput voltage ::::\[vout=(1-1\div \sqrt{1+2*RS*K(VIN-VT)}) *vin\]For calculating Gain::::\[Gain=vout \div vin \] You got all values to substitute in..Go head and try it out .... Cheers!!

    • one year ago
  2. summum Group Title
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    Hey Sri, I am not getting Vout with Vin=6.3V and vi=0.002? What am I doing wrong?

    • one year ago
  3. summum Group Title
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    What did you get? I get 0.00187, which is not right :/

    • one year ago
  4. Srikanth_Gangula Group Title
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    In my Problem set VIN=6.1 and vin=0.003 i got vout=2.815 m volts and gain=0.938

    • one year ago
  5. summum Group Title
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    What is your sqrt(1+2K*RS*(VIN-VT)

    • one year ago
  6. Srikanth_Gangula Group Title
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    16.229... Are you facing trouble in submitting formulae

    • one year ago
  7. summum Group Title
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    Yes, I got d= 16.61 and then when I substitute in the formula using my values : 0.002*(1-1/16.61), I get 0.00187, which it doesnt accept.

    • one year ago
  8. summum Group Title
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    :( Not good

    • one year ago
  9. summum Group Title
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    And sir, what do you get for H5P2 SOURCE FOLLOWER LARGE SIGNAL, Vdd, = ?

    • one year ago
  10. Srikanth_Gangula Group Title
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    5.1 my values are max Vin=7.1 and Min Vin = 5.7

    • one year ago
  11. summum Group Title
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    Yes, I was trying to put in 5.5. 5.1 works!! Thanks Sri! Your help is greatly appreciated!

    • one year ago
  12. Srikanth_Gangula Group Title
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    oh summum, i thought you just miss places Sri with Sir.... hmm ....Need any further help regarding problems on H5?

    • one year ago
  13. summum Group Title
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    No, Sri, I did address you as Sir and Sri both. :)

    • one year ago
  14. summum Group Title
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    Not much of a difference between the two though. This is almost philosophical but I'll rest my case here

    • one year ago
  15. summum Group Title
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    Sri,I am not getting small signal source follower, last two answers

    • one year ago
  16. summum Group Title
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    I do, but it is strangely accepting either the expression or the second answer. Not both at the same time. Life is hard. and then I am not able to find the gain because of life being hard. otherwise I would have aced this. :/

    • one year ago
  17. Srikanth_Gangula Group Title
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    So if it is not accepting the exprresion .... Jst do this ....Copy the below expression as it is and paste it ...and hit check (1-1/sqrt(1+2*RS*K*(VIN-VT)))*vin

    • one year ago
  18. Srikanth_Gangula Group Title
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    Let me what you got ...

    • one year ago
  19. summum Group Title
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    I got it! Yeeesss! Than you! Thank you! So much Sri ji! If I had I could send you Rosogullas online, this would be it! Thanks brother!

    • one year ago
  20. Srikanth_Gangula Group Title
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    Welcome Bro...!!!

    • one year ago
  21. summum Group Title
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    I hope I could be your help someday. See you around Sri!

    • one year ago
  22. Srikanth_Gangula Group Title
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    So long !!!!!!!!!!!!!

    • one year ago
  23. Coolamigo Group Title
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    I am getting it wrong when I put my value for vo. What is the right value?

    • one year ago
  24. yekanth Group Title
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    i am not getting voltage eqns of source follower large signal########$$$$$$$$$ help me!!!!!!!!!1

    • one year ago
  25. ali110 Group Title
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    Parameters used here are: VDD=90V, RS=22Ω, RL=82Ω, VI=6.5V Q1 If VI < VT à Transistor is off à IDS=0 à VO=VDD=90V Q2 ID=K/2*(VGS-VT)2, where VGS=VI-RS*ID Therefore ID=K/2*(VI-ID*RS-VT)2, which is similar to the quadratic form we already solved in H5P1.Q3. Thus: ID = (VI-VT)/RS + (1/K*RS 2)*(1-sqrt(1+2K*RS*(VI-VT))) Plugging in the given values will result in ID=0.185 A. Q3 VO=VDD-RL*ID=90-82*0.185=74.83 V Q4 vo/vi = ϑvO/ϑvI = -RL* ϑiD/ϑvI Taking the derivative, defining d=sqrt(1+2K*RS*(VI-VT))), and simplifying the terms will give: vo/vi= -RL/RS * [1-1/d] Using the values given in this case, vo/vi = -82/22*(1-1/19.925) = -82/22*0.949= -3.54 Note: In this problem the gain should be found using the large signal approach. In the coming week, we will cover small signal approach, which makes the solution to this problem much simpler removing the need to take tedious derivatives. Q5 Just like how we solved Q4, vo/vi = ϑvO/ϑvI = -RL* ϑiD/ϑvI = -RL/RS * [1-1/d] Q6 When K approaches infinity (i.e. very high current transistor), d will become infinite and thus vo/vi = -RL/RS * [1-1/d] = -RL/RS

    • one year ago
  26. ali110 Group Title
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    :)))

    • one year ago
  27. Coolamigo Group Title
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    :((((

    • one year ago
  28. kashyap Group Title
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    (vin*(1-1/sqrt(1+2*K*RS*(VIN-VT))))

    • one year ago
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