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FET common source analysis with source resistor 
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#1
Sep113, 10:57 PM

#2
Sep213, 02:46 AM

P: 404

Can you tell me what is a rd resistance. rd should not be connected between drain and source (parallel to the gm*Vgs current source) ?
http://forum.allaboutcircuits.com/sh...154#post208154 


#3
Sep213, 04:30 AM

P: 14

rd= 1/gm 


#4
Sep213, 05:14 AM

P: 404

FET common source analysis with source resistor
So you need to use this smallsignal model to find output impedance. Zo = Vx/Ix 


#5
Sep213, 06:00 AM

P: 14

as far as my understanding says, gm= Id/Vgs and if the slope of characterstic curve is constant the gm can be approximated as rd. Please correct me if i am worng.
In the image attached by you the current source should be open because at the time of calculating output impedence we remove input voltage source. since Vgs=0, current source is open circuit. so if we remove the current source then the output impedece will be, Zo=(ro+Rs)Rd. Is this the way to do it? 


#6
Sep213, 04:43 PM

P: 404

And now the voltage gain is equal to Av = Rd/(1/gm + Rs) Also notice that ro = rds = ΔVds/ΔId represent channellength modulation similar to the Early effect in bipolar devices. And to represent channellength modulation, i.e., variation of Id with Vds we add a resistor ro to small signal model. So we connect the FET gate to ground. But we also replace the load resistance with a Vx test voltage source. So we can find Zo = Vx/Ix So if you look at the circuit diagram You will see that Vgs is not equal 0V (only gate is connect to GND) not the source. Vgs = Is * Rs Test voltage provide a current for Rs resistor so there will be a voltage drop across Rs resistor, so the gm*Vgs current source is not a open circuit. There will be additional current flow through Rs resistor. Is = Iro + (gm*Vgs). But if we remove ro resistance from our smallsignal model, then Vgs is indeed equal to Vgs = 0V, current source is open circuit. The output impedance will be, Zo = Rd. 


#7
Sep413, 05:27 AM

P: 14

thanks a lot for telling me the difference in between gm and rd. I had a missconfussion regarding this concept. thanks for sharing this with me.
so Vo=Vds+Vs where, Vds = (IogmVgs)*(roRd): here Vgs=Vs Vds=(Io+gmVs)*(roRd) Vs=Rs*(Io) sum of current source and ro current will be the source current. Vo=Vds+Vs Vo=(Io+gmVs)*(roRd)+(Rs*Io) or Vo=((Io+gm(Rs*Io))*(roRd))+(Rs*Io) Vo=Io*((1+gmRs))*(roRd))+(Rs) Vo/Io=((1+gmRs))*(roRd))+(Rs) right...? but text books says the zo=Rd/(1+gmRs). please clear this. 


#8
Sep413, 06:27 AM

P: 404

Try to find Zo if you remove Rd from the circuit.
Vx  Iro•ro  Ix•Rs = 0 (1) Ix = gm•Vgs + Iro (2) Vgs = (Ix)•Rs (3) And now I substituting equation (2) and (3) to (1) Vx  Iro•ro  (gm•(Ix) •Rs + Iro)•Rs = 0 Now solve for Iro Iro = (gm• Ix• Rsē + Vx) / (Rs + ro) (4) And substituting (4) to (1) and solve for Ix Ix = Vx / (Rs + ro + gm•Rs•ro) And finally Vx/Ix = = Re + ro + gm•Re•ro = ro • (1+gm•RE + RE/ro) Or we can write this equation Ix•Rs + Iro•ro = Vx (1) Iro = (Ix  Id) (2) Id = (gm•(Ix)•Rs) (3) so after substitution Ix•Rs + ( Ix  (gm•(Ix)•Rs))*ro = Vx next solve for I2 Ix = Vx / ( Rs + ro + gm*Rs*ro ) Vx/Ix = Rs + ro + gm*Rs*ro so now if we add Rd resistor we have this 'exact' value Rout = Rd(ro+Rs+gm•ro•Rs) = Rd( ro •(1+gm•Rs + Rs/ro) ) This expression, Rout = RD  (ro * (1 + gm * Rs)), is a good approximation provided that ro >> 1/gm. Can you tell me the title of the book? 


#9
Sep913, 10:56 PM

P: 14

Hey thanks a lot. I get it where I was wrong. I thought that ro is directly connected to ground but it's connected to Rs. That's why I am taking Rdro.
Thanks a lot again. 


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