Nurali
  • Nurali
H5P1 third question and H5P2 third question
MIT 6.002 Circuits and Electronics, Spring 2007
schrodinger
  • schrodinger
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anonymous
  • anonymous
Can you post the exercise text and figure?
Nurali
  • Nurali
As VIN is increased, the output voltage VOUT decreases and the MOSFET goes out of saturation. For the value or RL found above, what is the maximum input voltage VIN in volts such that the MOSFET will remain in the saturation region?
anonymous
  • anonymous
Sorry, I don't fully understand the exercise: "...For the value or RL found above..." refers to a question before that one u reported? Moreover: if Vin = VT, VGS = 1.5 V > VT and nMOS is saturation region and IDS is not 0.

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anonymous
  • anonymous
ok. Thanks. anyway: mosfet cames into sat-triode border region if VGD = VT. To find the IDS at which such a situation is entered are: Vin - Vo = VT Vo = VS - ID x RL ID = 0.5 x K x (VGS - VT)^2 VGS = Vin - (-VS) You'll have to solve a quadratic equation in Vin. Only one root has a sense (I found Vin = 0.093 V). The other one is negative and has not a physical sense.
Nurali
  • Nurali
thanks
Nurali
  • Nurali
please solve these problem
Nurali
  • Nurali
what is the solution of ids?
anonymous
  • anonymous
the answer for H5P1 third ques cud b the formula in S9E3: MOSFET AMPLIFIER 2 ... but still im not getting the answer !!
anonymous
  • anonymous
H5P1 third question .. these derivations i have found which which give the answer algebraically .. but still im unable to get the value
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Nurali
  • Nurali
H5P1 third question is easy.
anonymous
  • anonymous
For Nurali: question on ids of source follower. It is almost simalar to that of teh zero-offset amplifier. You correctly find: ids = 0.5 * K * (VIN - VOUT - VT)^2 and : VOUT = RS * Ids. So u can write the (implicit) expression for ids: Ids = 0.5*K*(VIN - RS*Ids - VT)^2 then find the equation of Ids: a * Ids^2 + b * Ids + c = 0 which then gives: ids = [ -b +/- sqrt(b^2 - 4ac) ] / (2a). If I didn't make errors you'll probably have: Ids = (1/(K*Rs^2) ) * [ 1 + K*Rs*(Vin - VT) - sqrt( 1 + 2*K*Rs*(Vin - VT) ) ] Please note: only the "negative" choice is correct because Ids = 0 @ Vin = VT has a physical meaning. I'll control my result, but you can make the culcula as indicated before.
anonymous
  • anonymous
H5P2 (RS*K*(vIN-VT)+1-sqrt((K*(vIN-VT)*RS+1)^2-K^2*RS^2*(vIN-VT)^2))/(K*RS^2)
Nurali
  • Nurali
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Nurali
  • Nurali
Stefo please help
anonymous
  • anonymous
For an RC circuit, the time constant tau = Req x C where Req is the resistance seen by the capacitor. What does the capacitor see in circuit A and B, respectively?
Nurali
  • Nurali
iC=CdvC(t)dt
anonymous
  • anonymous
?? well... it is true the formula for the Ic. However, to find a time constant you can simply use Kirchhoff and Ohm. Req seen by the capacitor can be found substituing C with an "Ohmmeter": For example, for circiut A, you put a voltage source Vx instead of C. Then the current the generator sources is Ix = Vx/(R1+R2). So: Req = Vx / Ix = R1+R2 (Indeed C sees the series R1+R2). The time constant is (R1+R2)*C and the capacitor discharges from Vo (at t=0) following the law: Vc(t) = Vo * exp[ -t / (Req*C)]. For circuit B you apply the same method. (I suggest to use a current source and to find the Vx).
Nurali
  • Nurali
thanks
anonymous
  • anonymous
you're welcome :)
Nurali
  • Nurali
please solve these anwer
Nurali
  • Nurali
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Nurali
  • Nurali
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Nurali
  • Nurali
please solve these problem
anonymous
  • anonymous
if u observe circuit of fig. 2 u can write: VGS = Vin VDS = Vo Having Vin > VT newfet works in the active region and IDS = K*(VGS-VT)*VD^2 Then, observe circuit and... apply kvl to the Vs-R-VDS loop.
Nurali
  • Nurali
answer is wrong please solve complete expression
Nurali
  • Nurali
Derive an expression of vOUT as a function of vIN in terms of the power supply voltage VS, the resistance R, and the NewFET parameters, K and VT. Do so for the NewFET biased into the active region 0
anonymous
  • anonymous
Please Nurali. I don't understand if u need help to understand the method to solve the problem or simply the exact solution. In the latter, I think I'm not so helpful to you...

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