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anonymous
 4 years ago
In H5 the parser don't "like" my formulas. What's wrong with them:
vin*(11/sqrt(1+2K*RS*(VINVT))))
and:
(vINVT)/RS + (1/K*RS^2)*(1sqrt(1+2K*RS*(vINVT)))
anonymous
 4 years ago
In H5 the parser don't "like" my formulas. What's wrong with them: vin*(11/sqrt(1+2K*RS*(VINVT)))) and: (vINVT)/RS + (1/K*RS^2)*(1sqrt(1+2K*RS*(vINVT)))

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Worked, only one left :)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0This one left  (vINVT)/RS + (1/K*RS^2)*(1sqrt(1+2K*RS*(vINVT))), now there isn't a problem with parsing, but seem the formula isn't correct.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[(vin*(11/\sqrt(1+2*K*RS*(VINVT))))\] does not work and i cannot see why

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0that is the th equation i wrote

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0These formulas are for H5P3 and it's correct. I need another one: Q3 for H5P2 SOURCE FOLLOWER LARGE SIGNAL

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I got it!!! Actually in finding valid Operating ranges \[V0 = [1 + \sqrt{1+2*K*RL*Vs} ] \div KRL\] In the above formula Vs= applied voltage across RL and Drain to Source( i.e Vds).. \[Vs=ids∗RL+Vds(or)Vs=ids∗RL+V0\] But in given problem , by applying KVL to output loop we get \[Vds+idsRL=Vs+(−Vs−)\] \[Vds+ids∗RL=1−(−1)=2\] So in finding V0 substitute\[Vs=2 \] (NOT 1) And we know Valid maximum input range is \[Vgs = V0 + Vthershold\] Applying KVL to input loop \[Vin=Vgs+Vs−\] Substitue the value ...You will get it !!!!!! Cheers!!!!

benzo
 4 years ago
Best ResponseYou've already chosen the best response.0help please h5p2 question# 7....What is the minimum value of VDD (in Volts) that we must supply to keep the transistor in the region of saturated operation?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Conditions for Saturation region Opereation are.... \[VGS \ge VT\]\[VDS \ge VGSVT\]For the given Circuit \[VDS = VDDVOUT\]\[VGS=VINVOUT\] By using all conditions we get final Condition ....\[VDD \ge VINVT\]Here Substitue Maximum Possible \[VIN\] (As given in your problem.i.e maximum positive swing) Now you have \[VIN\] and \[VT\] So subtitue in the Final Condition for \[VDD \] ,You will get the Solution... Cheers!!!!

benzo
 4 years ago
Best ResponseYou've already chosen the best response.0help please h5p2 question# 7....What is the minimum value of VDD (in Volts) that we must supply to keep the transistor in the region of saturated operation?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@benzo I gave an clear explanation above to find VDD ....are you interested in knowing answer or the process of getting answer ?... Anyway for the numbers given in my problem set Max \[VIN=7.1\] and Minimun \[VIN=5.7\] So I got Minimum Possible \[VDD=5.1\]

benzo
 4 years ago
Best ResponseYou've already chosen the best response.0Srikanth_Gangula i got it thanks lab and home work 100%

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0You are Welcome Benzo!!!!
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