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I would like to know if that is correct schematic1

Ideal Operational amplifier

$$Z_{\text{i}}=\infty$$

$$Z_{\text{o}}=Z_{\text{c}} \parallel R$$

741 Operational amplifier

$$Z_{\text{i}}=2\text{ M}\Omega+R+Z_{\text{c}}$$

$$Z_{\text{o}}=Z_{\text{c}} \parallel (R+75 \Omega)\text{ (at DC)}$$

schematic2

Ideal Operational amplifier

$$Z_{\text{i}}=R+Z_{\text{c}}$$

$$Z_{\text{o}}=0$$

741 Operational amplifier

$$Z_{\text{i}}=R+Z_{\text{c}}$$

$$Z_{\text{o}}=75\Omega + (R \parallel Z_{\text{c}})\text{ (at DC)}$$

I have used plot of output resistance from LM741 datasheet.

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1 Answer 1

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Because you are using the op-amp in a closed loop configuration, any output impedance at DC is virtually zero (see my previous answer). So, for the first circuit, when using the 741 op-amp the output impedance Zo=Zc||R.

For the input impedance, the wrong answer is Zi=2Mohm+R+Zc because the op-amp buffers the output resistor and capacitor from having any influence at the input and Zin is what the data sheet says about the 741 (typically 2 Mohm).

For your 2nd circuit, the output impedance is virtually zero ohms at DC for the reasons given in my previous answer. As frequency rises the output impedance rises.

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  • \$\begingroup\$ Hi Andy I promise this is the last question, thanks in advance! electronics.stackexchange.com/questions/321860/… \$\endgroup\$
    – Fran
    Aug 2, 2017 at 12:27
  • \$\begingroup\$ That is a subtley much more difficult question to answer because of the feedback capacitor. unfortunately I'll be on a tiny and puny android for the next three days so I can't do it justice. \$\endgroup\$
    – Andy aka
    Aug 2, 2017 at 12:53

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