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I was working a mathematical model of a non ideal op-amp circuit and I saw a slide in the following presentation:

enter image description here

The question I have is the following: Why does the close-loop gain \$\frac{\text{V}_\text{o}}{\text{V}_\text{i}}\$ not contain de resistor \$\text{R}_\text{o}\$?

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  • \$\begingroup\$ Simply because they don't include Ro resistor in their equations. \$\endgroup\$ – G36 Feb 7 '18 at 16:13
  • \$\begingroup\$ The closed-loop gain look like this $$ \frac{Vo}{Vi} = \frac{-\frac{R_F}{R_a}}{1 + (\frac{R_F+R_O}{AR_F - R_O})\cdot (1+ \frac{R_F}{R_a} +\frac{R_F}{R_{\pi}} )}$$ \$\endgroup\$ – G36 Feb 7 '18 at 16:23
  • \$\begingroup\$ @G36 How did you come up with that equation? \$\endgroup\$ – klopiu Feb 7 '18 at 17:31
  • \$\begingroup\$ I simply solve this equations $$\frac{V_{\pi} }{R_{\pi}} + \frac{V_{\pi} - V_i}{R_a}+\frac{V_{\pi} - V_o}{R_F}=0 ; \frac{V_o - V_{\pi}}{R_F} +\frac{V_O - (-AV_{\pi}))}{R_O}=0$$ \$\endgroup\$ – G36 Feb 7 '18 at 17:46
  • \$\begingroup\$ @G36 And where did you find those equations? \$\endgroup\$ – klopiu Feb 7 '18 at 17:50
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Why does the close-loop gain Vo/Vi not contain the resistor Ro?

For this reason: -

enter image description here

The only problem Rout gives is the ability for the op-amp to deliver current to a connected load. An op-amp might internally have 100 ohm resistance built into its output stage in order to protect it from extracting too much current and overheating it yet, we do not need to consider that internal resistance.

What is the difference here - the open loop gain is still determined by R2 and R1 and even under light/moderate loads it will be the same.

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  • \$\begingroup\$ But In the comments @G36 gives an equation with \$\text{R}_\text{o}\$ \$\endgroup\$ – klopiu Feb 7 '18 at 17:31
  • \$\begingroup\$ @klopiu But as long as A >> (R2/R1) the Ro will have a very little effect on output voltage (Vo). \$\endgroup\$ – G36 Feb 7 '18 at 17:50
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Why does the close-loop gain Vo/Vi not contain the resistor Ro?

Because Ro is embedded in a feedback loop who's gain is controlled by Ra/Rf. For open loop gain Ro is part of the gain equation, in series with Vo.

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    \$\begingroup\$ Well I do not see \$\text{R}_\text{o}\$ in the last equation in the sheet, and I do not see why it is not in it?! \$\endgroup\$ – klopiu Feb 7 '18 at 15:37
  • \$\begingroup\$ Why put Ro in the equation if it has no effect? \$\endgroup\$ – Sparky256 Feb 7 '18 at 15:44
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In the open loop configuration, they say \$ V_o = AV_\pi\$. They have calculated the open loop gain A, including the effect of the output resistance Ro. So in the closed loop gain calculation, it need not be considered again. Only the feedback network resistances should be considered.

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  • \$\begingroup\$ In which step they include Ro? \$\endgroup\$ – G36 Feb 7 '18 at 16:14
  • \$\begingroup\$ Otherwise they should have included Ro in the equation. As Andy Aka pointed out, In open loop, Vout is calculated including the series Ro. Hence the open loop gain Vout/Vin = A is calculated including Ro too. \$\endgroup\$ – Mitu Raj Feb 7 '18 at 16:22

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