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# what is the role of these two VCVS?

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#### wjxcom

##### Full Member level 5
Hi, all: I design a full differential opa, and I process the ac analysis.

Now, I have a simulation circuit that can do the ac analysis, but I do not know what the role of these two VCVS, i.e. E1 and E2.

at the same time, capacitance C1 and C3 were added to the input port of the opa, so I think they are short circuit for AC source, i.e. V1.[/list]

The two VCVS are put there to preserve the symmetry.
V1 is the differential voltage.

E1 is used to +0.5 * V1 on the + node.
E2 is used to put -0.5 * V1 on the - node.
Then the differential voltage appearing accross the + and the - terminal
will be V1 .

is the circuit hooked up correctly?

it seems theres -0.5 * V1 on the + node (Vinp)
and -0.5 * V1 on the - node. (Vinn)

The caps C3, C4 are there to provide an AC ground. It also allows the input gates
in the + and - nodes to be biased by the output voltage from vop and von, which is probably close to vcm (2.5) in this case.

Most likely, C1, C2 effects the location of the dominant pole in this circuit. Making C1, C2 bigger will shift the dominant pole to a lower frequency.
And thus making C1, C2 bigger will increase the phase margin of this circuit,
making it more stable. At the same time, if you make C1, C2 bigger, it will reduce your open loop bandwidth.

Hi, eecs4ever: you said "it seems theres -0.5 * V1 on the + node (Vinp) and -0.5 * V1 on the - node (Vinn) ", do you mean that -0.5 * V1 add to the both +node (Vinp) and - node. (Vinn) ?

wjxcom said:
Hi, eecs4ever: you said "it seems theres -0.5 * V1 on the + node (Vinp) and -0.5 * V1 on the - node (Vinn) ", do you mean that -0.5 * V1 add to the both +node (Vinp) and - node. (Vinn) ?

yes. the AC voltage at Vinp is -0.5 * V1 . and the AC votlage at Vinn is also -0.5 * V1.

maybe your circuit is setup to measure common mode gain?

if you want to measure the differential gain of this circuit, you will want to
change egain=0.5 to egain = -0.5 for one of the VCVS.

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