How can I make the current of the transient to be constant?
I don't understand this question. What is "the current of the transient" ?
you shoud try a long time in transient simulation, because the parasitic capancitance may make a long delay while DC analyses ignores parasitic capacitance.
hi,
from your description the problem seems to be loss of open-loop gain, this is most likely caused by operating the real opamp outside its input range.
please try simulating the open-loop gain in circuit not standalone, ask if you don't know how
allennlowaton, what about the power supply for the opamps? Single oder double?
thank you dgnani...
yes, I don't know where to break the circuit and insert the AC signal as well as the point to be measured. Please teach me.
...
I didn't understand this and I think that's the reason I'm having a difficulty on plotting the data. I did include these codes the .print vdb(n5) vdb(n3). AOL(dB) = N5(dB) - N3(dB). Is this correct? I inserted the "vn1 vn1 gnd dc 0V ac 1V" at the positive input of the OP.....and the name for the net split by adding the AC source is kept on the opamp side
First, I suggest not measuring the last current branch for Ibias. In fact, the circuit can work without MP6, MP7, MnN, MN5. The last branch has two current comparing. The current will not reflect correct current in the second current branch (Mp5).
yes as I mentioned you break either one or both the input wires to the opamp (if you break both just make sure you are creating a net differential signal for the opamp), then use the equation above for Aol
Aol = V(N5) / (V(N1)-V(N3))
since you are looking at open loop gain the output is N5, and the name for the net split by adding the AC source is kept on the opamp side
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