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benzo Group Title

h5p2.....im lost on this....give me step by step how you do this please....What is the minimum value of VDD (in Volts) that we must supply to keep the transistor in the region of saturated operation?

  • one year ago
  • one year ago

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  1. thoppa Group Title
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    what are ur values given.............give the cumplete question

    • one year ago
  2. thoppa Group Title
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    dude answer something between 2 to 7

    • one year ago
  3. thoppa Group Title
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    try it

    • one year ago
  4. thoppa Group Title
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    i guess you got it

    • one year ago
  5. benzo Group Title
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    not yet

    • one year ago
  6. muhammedvaseem007 Group Title
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    vdd=5.5. just try it

    • one year ago
  7. benzo Group Title
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    i got it...VIN-VOUT greater or equall ti VT...VT=2.0 6.1-3.6=2.5 5.4-2.9=2.5 6.8-4.2=2.6 VIN-VT 6.8-2=4.8 VDD=4.8

    • one year ago
  8. muhammedvaseem007 Group Title
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    is you check. i though it must be corect

    • one year ago
  9. muhammedvaseem007 Group Title
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    vdd is 5.5 for me. other vin is diferent

    • one year ago
  10. benzo Group Title
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    ok right for me....thanks.....H5P1.....whats the step by step solution 1, 2, 3

    • one year ago
  11. muhammedvaseem007 Group Title
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    -.5,8000,0.093 is the answere

    • one year ago
  12. benzo Group Title
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    step by step solution

    • one year ago
  13. muhammedvaseem007 Group Title
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    , we have to find max Vin ; to find Vin let us apply kvll to the input loop Vin - Vgs+1 =0; ie, Vin = Vgs-1 ...............................(eqn1) now the unknown parameter is Vgs, hence to find it lets apply KVL to external loop : 1-Ids*R-Vds+1=0 , further substituting the value of Ids = k/2 *(Vgs-Vt)^2 in the above equation and solving the quadratic equation we will get, Vds = 0.593, -0.84 : as Vds needs tobe +ve hence neglecting he -ve term we have Vds = 0.593 and we know Vds = Vgs-Vt ie Vgs = Vds+Vt therefore, Vgs = 0.593+0.5=1.093 and substituting the above valuw in eqn1 we have Vin=1.093-1=0.093

    • one year ago
  14. akash97 Group Title
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    @benzo : In saturation: iDS=K/2*(vGS-VT)2, and since here vGS is vIN-vOUT, then iDS=K/2*(vIN-vOUT-VT)2 Q2 vOUT=

    • one year ago
  15. akash97 Group Title
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    vOUT=iDS*RS

    • one year ago
  16. benzo Group Title
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    thanks lab and hw 100%

    • one year ago
  17. muhammedvaseem007 Group Title
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    i get the vout for sorce flower small signal but its doesn't work what you vout for source flower small signal

    • one year ago
  18. Coolamigo Group Title
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    vi=0.0009V what is the incremental output vo? also Gain

    • one year ago
  19. muhammedvaseem007 Group Title
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    gain=0.9498

    • one year ago
  20. Coolamigo Group Title
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    and vo?

    • one year ago
  21. muhammedvaseem007 Group Title
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    yes vo=0.9498

    • one year ago
  22. Coolamigo Group Title
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    Wont that conclude a gain of 1055?

    • one year ago
  23. Coolamigo Group Title
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    Bro Gain is .94 vo=?

    • one year ago
  24. ali110 Group Title
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    Q1 vOUT = iDS*RS vout/vin = ϑvOUT/ϑvIN @ VIN = RS * ϑiDS/ϑvIN = RS * (1/RS + 1/(K*RS 2)*(-2K*RS/(2d)) where d=sqrt(1+2K*RS*(VIN-VT))) Simplifying the expression above will give: vout/vin = (1-1/d) à vout=vin*(1-1/sqrt(1+2K*RS*(VIN-VT)))) Q2 For K=2 [A/V2], VT=2V, RS=16Ω, VIN=6.5V and vin=0.001 V, d=19.925 and thus vout/vin = (1-1/d) à vout=0.001*(1-1/19.925)=0.001*0.9498 à vout = 0.0009498 V Q3 We've already found voltage gain solving for Q2. vout/vin = (1-1/d) = 1-1/19.925 = 0.9498 You can see that source follower has a voltage gain slightly less than 1.

    • one year ago
  25. Coolamigo Group Title
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    vo= 0.0009498 gives X (wrong)

    • one year ago
  26. muhammedvaseem007 Group Title
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    for each value each answer

    • one year ago
  27. brunofornari Group Title
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    for VIN = 6.4, 6.9 and 5.9?

    • one year ago
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