## ankit3177 Group Title H5P3 Source Follower Small Signal 2 years ago 2 years ago

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1. Srikanth_Gangula

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!!

2. summum

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

3. summum

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

4. Srikanth_Gangula

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

5. summum

6. Srikanth_Gangula

16.229... Are you facing trouble in submitting formulae

7. summum

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.

8. summum

:( Not good

9. summum

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

10. Srikanth_Gangula

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

11. summum

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

12. Srikanth_Gangula

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

13. summum

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

14. summum

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

15. summum

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

16. summum

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. :/

17. Srikanth_Gangula

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

18. Srikanth_Gangula

Let me what you got ...

19. summum

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!

20. Srikanth_Gangula

Welcome Bro...!!!

21. summum

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

22. Srikanth_Gangula

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

23. Coolamigo

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

24. yekanth

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

25. ali110

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

26. ali110

:)))

27. Coolamigo

:((((

28. kashyap

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