H5P1 third question and H5P2 third question

- Nurali

H5P1 third question and H5P2 third question

- schrodinger

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- anonymous

Can you post the exercise text and figure?

- 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

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

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

thanks

- Nurali

please solve these problem

- Nurali

what is the solution of ids?

- anonymous

the answer for H5P1 third ques cud b the formula in S9E3: MOSFET AMPLIFIER 2 ... but still im not getting the answer !!

- 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

H5P1 third question is easy.

- 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

H5P2
(RS*K*(vIN-VT)+1-sqrt((K*(vIN-VT)*RS+1)^2-K^2*RS^2*(vIN-VT)^2))/(K*RS^2)

- Nurali

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- Nurali

Stefo please help

- 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

iC=CdvC(t)dt

- 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

thanks

- anonymous

you're welcome :)

- Nurali

please solve these anwer

- Nurali

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- Nurali

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- Nurali

please solve these problem

- 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

answer is wrong please solve complete expression

- 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

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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