# Help me understand calculating poles for op-amp

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#### firsttimedesigning

##### Full Member level 1 So I have finished designing this op-amp.(please see the file Op-Amp)
There are few things that I dont understand though.
First of all, I think there are two poles in this circuit because there two stages ( stage 1 is the differential amplifier, stage 2 is the common source with current source load)

Second, when calculating the first pole,
C1 = Cgs7 + Cgd7(1+ abs(A2)) + Cdb4 + Cgd4 + Cdb2 + Cgd2
abs( ) = absolute value of
(please see file pole1)

what i dont understand is that why is Cgd7 multiplied by (1+ abs(A2))?

when calculating the second pole,
C2 = Cgd7(1+ abs(1/A2)) + Cdb7 + Cgd8 + Cdb8
(please see file pole2)

why is Cgd7 multiplied by (1+ abs(1/A2))?

#### kgl_13gr

##### Member level 5 Re: Op-Amp

Search for Miller theorem.
In brief when a capacitance is connected to the input and the output of a gain stage it can be approximated by equivalent capacitances connected accross the input and the output terminals. These capacitances are given by the Miller theorem (is also quite easy to prove).

### firsttimedesigning

Points: 2

#### dasong

##### Member level 2 Op-Amp

i think you will get the answer fater you read the chapter 10 of Raviva's book

### firsttimedesigning

Points: 2

#### firsttimedesigning

##### Full Member level 1 Re: Op-Amp

thank you for respodning..
i actually got this from another book and i am currently reading chapter 9.
almost done though, i guess i will read chapter 10 first.
anyway, thank you for the help.

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