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benzo

  • 2 years ago

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?

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  1. thoppa
    • 2 years ago
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    what are ur values given.............give the cumplete question

  2. thoppa
    • 2 years ago
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    dude answer something between 2 to 7

  3. thoppa
    • 2 years ago
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    try it

  4. thoppa
    • 2 years ago
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    i guess you got it

  5. benzo
    • 2 years ago
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    not yet

  6. muhammedvaseem007
    • 2 years ago
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    vdd=5.5. just try it

  7. benzo
    • 2 years ago
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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

  8. muhammedvaseem007
    • 2 years ago
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    is you check. i though it must be corect

  9. muhammedvaseem007
    • 2 years ago
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    vdd is 5.5 for me. other vin is diferent

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

  11. muhammedvaseem007
    • 2 years ago
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    -.5,8000,0.093 is the answere

  12. benzo
    • 2 years ago
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    step by step solution

  13. muhammedvaseem007
    • 2 years ago
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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

  14. akash97
    • 2 years ago
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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=

  15. akash97
    • 2 years ago
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    vOUT=iDS*RS

  16. benzo
    • 2 years ago
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    thanks lab and hw 100%

  17. muhammedvaseem007
    • 2 years ago
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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

  18. Coolamigo
    • 2 years ago
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    vi=0.0009V what is the incremental output vo? also Gain

  19. muhammedvaseem007
    • 2 years ago
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    gain=0.9498

  20. Coolamigo
    • 2 years ago
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    and vo?

  21. muhammedvaseem007
    • 2 years ago
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    yes vo=0.9498

  22. Coolamigo
    • 2 years ago
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    Wont that conclude a gain of 1055?

  23. Coolamigo
    • 2 years ago
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    Bro Gain is .94 vo=?

  24. ali110
    • 2 years ago
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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.

  25. Coolamigo
    • 2 years ago
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    vo= 0.0009498 gives X (wrong)

  26. muhammedvaseem007
    • 2 years ago
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    for each value each answer

  27. brunofornari
    • 2 years ago
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    for VIN = 6.4, 6.9 and 5.9?

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