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anonymous
 3 years ago
H5P1 third question and H5P2 third question
anonymous
 3 years ago
H5P1 third question and H5P2 third question

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Can you post the exercise text and figure?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0As 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
 3 years ago
Best ResponseYou've already chosen the best response.0Sorry, 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.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ok. Thanks. anyway: mosfet cames into sattriode 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.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0please solve these problem

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0what is the solution of ids?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0the answer for H5P1 third ques cud b the formula in S9E3: MOSFET AMPLIFIER 2 ... but still im not getting the answer !!

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0H5P1 third question .. these derivations i have found which which give the answer algebraically .. but still im unable to get the value

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0H5P1 third question is easy.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0For Nurali: question on ids of source follower. It is almost simalar to that of teh zerooffset 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
 3 years ago
Best ResponseYou've already chosen the best response.0H5P2 (RS*K*(vINVT)+1sqrt((K*(vINVT)*RS+1)^2K^2*RS^2*(vINVT)^2))/(K*RS^2)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0For 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?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0?? 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).

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0please solve these anwer

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0please solve these problem

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0if 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*(VGSVT)*VD^2 Then, observe circuit and... apply kvl to the VsRVDS loop.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0answer is wrong please solve complete expression

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Derive 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<VT≤vIN≤VS. Write your expression in the space provided below. Remember that algebraic expressions are case sensitive.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Please 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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