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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
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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thoppaBest ResponseYou've already chosen the best response.0
what are ur values given.............give the cumplete question
 one year ago

thoppaBest ResponseYou've already chosen the best response.0
dude answer something between 2 to 7
 one year ago

muhammedvaseem007Best ResponseYou've already chosen the best response.2
vdd=5.5. just try it
 one year ago

benzoBest ResponseYou've already chosen the best response.0
i got it...VINVOUT greater or equall ti VT...VT=2.0 6.13.6=2.5 5.42.9=2.5 6.84.2=2.6 VINVT 6.82=4.8 VDD=4.8
 one year ago

muhammedvaseem007Best ResponseYou've already chosen the best response.2
is you check. i though it must be corect
 one year ago

muhammedvaseem007Best ResponseYou've already chosen the best response.2
vdd is 5.5 for me. other vin is diferent
 one year ago

benzoBest ResponseYou've already chosen the best response.0
ok right for me....thanks.....H5P1.....whats the step by step solution 1, 2, 3
 one year ago

muhammedvaseem007Best ResponseYou've already chosen the best response.2
.5,8000,0.093 is the answere
 one year ago

muhammedvaseem007Best ResponseYou've already chosen the best response.2
, we have to find max Vin ; to find Vin let us apply kvll to the input loop Vin  Vgs+1 =0; ie, Vin = Vgs1 ...............................(eqn1) now the unknown parameter is Vgs, hence to find it lets apply KVL to external loop : 1Ids*RVds+1=0 , further substituting the value of Ids = k/2 *(VgsVt)^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 = VgsVt ie Vgs = Vds+Vt therefore, Vgs = 0.593+0.5=1.093 and substituting the above valuw in eqn1 we have Vin=1.0931=0.093
 one year ago

akash97Best ResponseYou've already chosen the best response.0
@benzo : In saturation: iDS=K/2*(vGSVT)2, and since here vGS is vINvOUT, then iDS=K/2*(vINvOUTVT)2 Q2 vOUT=
 one year ago

muhammedvaseem007Best ResponseYou've already chosen the best response.2
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

CoolamigoBest ResponseYou've already chosen the best response.0
vi=0.0009V what is the incremental output vo? also Gain
 one year ago

muhammedvaseem007Best ResponseYou've already chosen the best response.2
yes vo=0.9498
 one year ago

CoolamigoBest ResponseYou've already chosen the best response.0
Wont that conclude a gain of 1055?
 one year ago

CoolamigoBest ResponseYou've already chosen the best response.0
Bro Gain is .94 vo=?
 one year ago

ali110Best ResponseYou've already chosen the best response.0
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*(VINVT))) Simplifying the expression above will give: vout/vin = (11/d) à vout=vin*(11/sqrt(1+2K*RS*(VINVT)))) 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 = (11/d) à vout=0.001*(11/19.925)=0.001*0.9498 à vout = 0.0009498 V Q3 We've already found voltage gain solving for Q2. vout/vin = (11/d) = 11/19.925 = 0.9498 You can see that source follower has a voltage gain slightly less than 1.
 one year ago

CoolamigoBest ResponseYou've already chosen the best response.0
vo= 0.0009498 gives X (wrong)
 one year ago

muhammedvaseem007Best ResponseYou've already chosen the best response.2
for each value each answer
 one year ago

brunofornariBest ResponseYou've already chosen the best response.0
for VIN = 6.4, 6.9 and 5.9?
 one year ago
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