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Doubts about current flows in this op-amp

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Full Member level 1
Jul 4, 2007
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So i am wondering the current flows into Mb1 equals the current coming out of M9 since Mb1 and M9 forms a current mirror. however, the book says that when designing this op-amp, allows 330ua flows into Mb1 and 3mA flows into M9. my question is how can you have 330 ua flows into Mb1 while having 3mA flows into M9. The input current of the current mirror does not equal to output current of the current mirror. Please explain it to me. I know it sounds kinda stupid, but I really dont get it....

Re: Op-Amp

oh yeah that's true. I forgot to look at the sizing of the trasistors.
thax for mentioning that to me.

Re: Op-Amp

The output impedance of this opamp is (gm3ro3ro1 || gm5ro5ro7), i believe. correct me if that is wrong. Assume that UnCox = 60u UpCox = 30u lapta n = 0.1 lapta p = 0.2, how to calculate the output impedance?
The drain current in M9 is 3mA. M1 = 1.5mA M2 = 1.5mA.
ro = 1 / (lapta * Id)

I have calculated the otuput impedance of the op-amp which comes out to be 93k ohms. However, the book says it is 266k ohms. I used the equation,
Id = (UnCox/2) (W/L) (Vgs-Vt)^2
to figure out (Vgs-Vt)
Then I used (Vgs-Vt) to calculate gm. gm = (UnCox/2) (W/L) (Vgs-Vt)
I dont know where I made mistake. Checked my work twice. Please show me
a way to calculate the output impedance. thx

Re: Op-Amp


I got where the problem is :

Please pay attention to that the 266k output impedence is derived by increase the length of PMOS transistors M5--M8,and then the lapta P is reduced to 0.1, not 0.2 before, that's the difference.

Re: Op-Amp

yeah i finally figured it out. thank you so much for pointing my mistake and helping me to solve this problem.

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