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Srikanth_Gangula
 2 years ago
Best ResponseYou've already chosen the best response.2Incremental ouput voltage ::::\[vout=(11\div \sqrt{1+2*RS*K(VINVT)}) *vin\]For calculating Gain::::\[Gain=vout \div vin \] You got all values to substitute in..Go head and try it out .... Cheers!!

summum
 2 years ago
Best ResponseYou've already chosen the best response.0Hey Sri, I am not getting Vout with Vin=6.3V and vi=0.002? What am I doing wrong?

summum
 2 years ago
Best ResponseYou've already chosen the best response.0What did you get? I get 0.00187, which is not right :/

Srikanth_Gangula
 2 years ago
Best ResponseYou've already chosen the best response.2In my Problem set VIN=6.1 and vin=0.003 i got vout=2.815 m volts and gain=0.938

summum
 2 years ago
Best ResponseYou've already chosen the best response.0What is your sqrt(1+2K*RS*(VINVT)

Srikanth_Gangula
 2 years ago
Best ResponseYou've already chosen the best response.216.229... Are you facing trouble in submitting formulae

summum
 2 years ago
Best ResponseYou've already chosen the best response.0Yes, I got d= 16.61 and then when I substitute in the formula using my values : 0.002*(11/16.61), I get 0.00187, which it doesnt accept.

summum
 2 years ago
Best ResponseYou've already chosen the best response.0And sir, what do you get for H5P2 SOURCE FOLLOWER LARGE SIGNAL, Vdd, = ?

Srikanth_Gangula
 2 years ago
Best ResponseYou've already chosen the best response.25.1 my values are max Vin=7.1 and Min Vin = 5.7

summum
 2 years ago
Best ResponseYou've already chosen the best response.0Yes, I was trying to put in 5.5. 5.1 works!! Thanks Sri! Your help is greatly appreciated!

Srikanth_Gangula
 2 years ago
Best ResponseYou've already chosen the best response.2oh summum, i thought you just miss places Sri with Sir.... hmm ....Need any further help regarding problems on H5?

summum
 2 years ago
Best ResponseYou've already chosen the best response.0No, Sri, I did address you as Sir and Sri both. :)

summum
 2 years ago
Best ResponseYou've already chosen the best response.0Not much of a difference between the two though. This is almost philosophical but I'll rest my case here

summum
 2 years ago
Best ResponseYou've already chosen the best response.0Sri,I am not getting small signal source follower, last two answers

summum
 2 years ago
Best ResponseYou've already chosen the best response.0I 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. :/

Srikanth_Gangula
 2 years ago
Best ResponseYou've already chosen the best response.2So if it is not accepting the exprresion .... Jst do this ....Copy the below expression as it is and paste it ...and hit check (11/sqrt(1+2*RS*K*(VINVT)))*vin

Srikanth_Gangula
 2 years ago
Best ResponseYou've already chosen the best response.2Let me what you got ...

summum
 2 years ago
Best ResponseYou've already chosen the best response.0I 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!

Srikanth_Gangula
 2 years ago
Best ResponseYou've already chosen the best response.2Welcome Bro...!!!

summum
 2 years ago
Best ResponseYou've already chosen the best response.0I hope I could be your help someday. See you around Sri!

Srikanth_Gangula
 2 years ago
Best ResponseYou've already chosen the best response.2So long !!!!!!!!!!!!!

Coolamigo
 2 years ago
Best ResponseYou've already chosen the best response.0I am getting it wrong when I put my value for vo. What is the right value?

yekanth
 2 years ago
Best ResponseYou've already chosen the best response.0i am not getting voltage eqns of source follower large signal########$$$$$$$$$ help me!!!!!!!!!1

ali110
 2 years ago
Best ResponseYou've already chosen the best response.0Parameters 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*(VGSVT)2, where VGS=VIRS*ID Therefore ID=K/2*(VIID*RSVT)2, which is similar to the quadratic form we already solved in H5P1.Q3. Thus: ID = (VIVT)/RS + (1/K*RS 2)*(1sqrt(1+2K*RS*(VIVT))) Plugging in the given values will result in ID=0.185 A. Q3 VO=VDDRL*ID=9082*0.185=74.83 V Q4 vo/vi = ϑvO/ϑvI = RL* ϑiD/ϑvI Taking the derivative, defining d=sqrt(1+2K*RS*(VIVT))), and simplifying the terms will give: vo/vi= RL/RS * [11/d] Using the values given in this case, vo/vi = 82/22*(11/19.925) = 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 * [11/d] Q6 When K approaches infinity (i.e. very high current transistor), d will become infinite and thus vo/vi = RL/RS * [11/d] = RL/RS

kashyap
 2 years ago
Best ResponseYou've already chosen the best response.1(vin*(11/sqrt(1+2*K*RS*(VINVT))))
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