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Srikanth_Gangula Group TitleBest ResponseYou've already chosen the best response.2
Incremental 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!!
 2 years ago

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

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

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

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

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

summum Group TitleBest ResponseYou've already chosen the best response.0
Yes, 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.
 2 years ago

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

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

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

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

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

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

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

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

Srikanth_Gangula Group TitleBest ResponseYou've already chosen the best response.2
So 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
 2 years ago

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

summum Group TitleBest ResponseYou've already chosen the best response.0
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!
 2 years ago

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

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

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

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

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

ali110 Group TitleBest ResponseYou've already chosen the best response.0
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*(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
 2 years ago

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