H5P3 Source Follower Small Signal

- anonymous

H5P3 Source Follower Small Signal

- jamiebookeater

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- anonymous

Incremental ouput voltage ::::\[vout=(1-1\div \sqrt{1+2*RS*K(VIN-VT)}) *vin\]For calculating Gain::::\[Gain=vout \div vin \]
You got all values to substitute in..Go head and try it out ....
Cheers!!

- anonymous

Hey Sri, I am not getting Vout with Vin=6.3V and vi=0.002? What am I doing wrong?

- anonymous

What did you get? I get 0.00187, which is not right :/

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## More answers

- anonymous

In my Problem set VIN=6.1 and vin=0.003
i got vout=2.815 m volts
and gain=0.938

- anonymous

What is your sqrt(1+2K*RS*(VIN-VT)

- anonymous

16.229...
Are you facing trouble in submitting formulae

- anonymous

Yes, I got d= 16.61 and then when I substitute in the formula using my values : 0.002*(1-1/16.61), I get 0.00187, which it doesnt accept.

- anonymous

:( Not good

- anonymous

And sir, what do you get for H5P2 SOURCE FOLLOWER LARGE SIGNAL, Vdd, = ?

- anonymous

5.1 my values are max Vin=7.1 and Min Vin = 5.7

- anonymous

Yes, I was trying to put in 5.5. 5.1 works!! Thanks Sri! Your help is greatly appreciated!

- anonymous

oh summum, i thought you just miss places Sri with Sir.... hmm ....Need any further help regarding problems on H5?

- anonymous

No, Sri, I did address you as Sir and Sri both. :)

- anonymous

Not much of a difference between the two though. This is almost philosophical but I'll rest my case here

- anonymous

Sri,I am not getting small signal source follower, last two answers

- anonymous

I do, but it is strangely accepting either the expression or the second answer. Not both at the same time. Life is hard. and then I am not able to find the gain because of life being hard. otherwise I would have aced this. :/

- anonymous

So if it is not accepting the exprresion ....
Jst do this ....Copy the below expression as it is and paste it ...and hit check
(1-1/sqrt(1+2*RS*K*(VIN-VT)))*vin

- anonymous

Let me what you got ...

- anonymous

I got it! Yeeesss! Than you! Thank you! So much Sri ji! If I had I could send you Rosogullas online, this would be it! Thanks brother!

- anonymous

Welcome Bro...!!!

- anonymous

I hope I could be your help someday. See you around Sri!

- anonymous

So long !!!!!!!!!!!!!

- anonymous

I am getting it wrong when I put my value for vo. What is the right value?

- anonymous

i am not getting voltage eqns
of source follower large signal########$$$$$$$$$ help me!!!!!!!!!1

- anonymous

Parameters used here are: VDD=90V, RS=22Ω, RL=82Ω, VI=6.5V
Q1
If VI < VT à Transistor is off à IDS=0 à VO=VDD=90V
Q2
ID=K/2*(VGS-VT)2, where VGS=VI-RS*ID
Therefore ID=K/2*(VI-ID*RS-VT)2, which is similar to the quadratic form we already solved
in H5P1.Q3. Thus:
ID = (VI-VT)/RS + (1/K*RS
2)*(1-sqrt(1+2K*RS*(VI-VT)))
Plugging in the given values will result in ID=0.185 A.
Q3
VO=VDD-RL*ID=90-82*0.185=74.83 V
Q4
vo/vi = ϑvO/ϑvI = -RL* ϑiD/ϑvI
Taking the derivative, defining d=sqrt(1+2K*RS*(VI-VT))), and simplifying the terms will
give:
vo/vi= -RL/RS * [1-1/d]
Using the values given in this case, vo/vi = -82/22*(1-1/19.925) = -82/22*0.949= -3.54
Note: In this problem the gain should be found using the large signal approach. In the coming
week, we will cover small signal approach, which makes the solution to this problem much
simpler removing the need to take tedious derivatives.
Q5
Just like how we solved Q4, vo/vi = ϑvO/ϑvI = -RL* ϑiD/ϑvI = -RL/RS * [1-1/d]
Q6
When K approaches infinity (i.e. very high current transistor), d will become infinite and thus
vo/vi = -RL/RS * [1-1/d] = -RL/RS

- anonymous

:)))

- anonymous

:((((

- anonymous

(vin*(1-1/sqrt(1+2*K*RS*(VIN-VT))))

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