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benzo
 4 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?
benzo
 4 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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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0what are ur values given.............give the cumplete question

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dude answer something between 2 to 7

benzo
 4 years ago
Best ResponseYou've already chosen the best response.0i 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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0is you check. i though it must be corect

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0vdd is 5.5 for me. other vin is diferent

benzo
 4 years ago
Best ResponseYou've already chosen the best response.0ok right for me....thanks.....H5P1.....whats the step by step solution 1, 2, 3

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0.5,8000,0.093 is the answere

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0, 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

anonymous
 4 years ago
Best 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=

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i get the vout for sorce flower small signal but its doesn't work what you vout for source flower small signal

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0vi=0.0009V what is the incremental output vo? also Gain

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Wont that conclude a gain of 1055?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Q1 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.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0vo= 0.0009498 gives X (wrong)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0for each value each answer

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0for VIN = 6.4, 6.9 and 5.9?
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