benzo
  • benzo
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?
MIT 6.002 Circuits and Electronics, Spring 2007
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benzo
  • benzo
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?
MIT 6.002 Circuits and Electronics, Spring 2007
jamiebookeater
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anonymous
  • anonymous
what are ur values given.............give the cumplete question
anonymous
  • anonymous
dude answer something between 2 to 7
anonymous
  • anonymous
try it

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anonymous
  • anonymous
i guess you got it
benzo
  • benzo
not yet
anonymous
  • anonymous
vdd=5.5. just try it
benzo
  • benzo
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
anonymous
  • anonymous
is you check. i though it must be corect
anonymous
  • anonymous
vdd is 5.5 for me. other vin is diferent
benzo
  • benzo
ok right for me....thanks.....H5P1.....whats the step by step solution 1, 2, 3
anonymous
  • anonymous
-.5,8000,0.093 is the answere
benzo
  • benzo
step by step solution
anonymous
  • anonymous
, 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
anonymous
  • anonymous
@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=
anonymous
  • anonymous
vOUT=iDS*RS
benzo
  • benzo
thanks lab and hw 100%
anonymous
  • anonymous
i get the vout for sorce flower small signal but its doesn't work what you vout for source flower small signal
anonymous
  • anonymous
vi=0.0009V what is the incremental output vo? also Gain
anonymous
  • anonymous
gain=0.9498
anonymous
  • anonymous
and vo?
anonymous
  • anonymous
yes vo=0.9498
anonymous
  • anonymous
Wont that conclude a gain of 1055?
anonymous
  • anonymous
Bro Gain is .94 vo=?
anonymous
  • anonymous
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.
anonymous
  • anonymous
vo= 0.0009498 gives X (wrong)
anonymous
  • anonymous
for each value each answer
anonymous
  • anonymous
for VIN = 6.4, 6.9 and 5.9?

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