I have a two stage op amp. I was able to find the TF if it was just one stage, which would be: $$\frac{V_o}{V_i}=\frac{R_2C_2s}{(R_1C_1s+1)(R_2C_2s+1)}$$ Can I apply the same procedure to the second half (after the second node) then end up with the final TF as the one right here, or do they change when the circuits are cascaded? $$TF=TF_1*TF_2$$ Hope someone can give me some insight!
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\$\begingroup\$ Cascaded circuits multiply transfer function, not divide. \$\endgroup\$– HearthNov 27, 2020 at 22:48
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\$\begingroup\$ Thank you! Will fix now \$\endgroup\$– DapperNov 27, 2020 at 22:53
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For perfect op-amps, the answer is absolutely yes, because due to feedback, and the fact that a perfect op-amp has infinite gain, the output impedance of op-amp stages like you show here is zero.
Assuming that the ratio of \$C_1/C_2\$ and \$C_3/C_4\$ are low enough that stability is maintained*, then almost yes, for the same reason (because the op-amp gain is really big, at least if the op-amp gain-bandwidth product is really big in relation to the circuit's design bandwidth).
Whether or not that almost is good enough for your circuit depends on your requirements, of course.
answered Nov 27, 2020 at 23:50
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\$\begingroup\$ Thank you, Tim! Very informative. I'm going with the assumption that it is a perfect op-amp so no worries there. As for my transfer function for the one stage, does it look correct to you? \$\endgroup\$– DapperNov 28, 2020 at 0:00
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\$\begingroup\$ That's a separate enough question that you should ask it as a separate question. And be prepared to get hints, not answers -- it looks a lot like homework. \$\endgroup\$ Nov 28, 2020 at 0:14