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How can I estimate the output impedance of a unity-gain opamp buffer circuit?

With an ideal opamp the impedance would be \$0 \Omega\$, but we don't have ideal parts. E.g. the datasheet of the AD8226 instrumentation amplifier recommends a source impedance below of \$2 \Omega\$ as voltage reference. The example uses the OP1177 opamp to realize a buffer circuit. How do I know that the choosen opamp fulfills the design constraint or how would I choose a suitable opamp?

example from AD8226 datasheet

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You don't need to know the output impedance of the op amp -- You need to know the output impedance of the noninverting amplifier as a block. This can be complicated because the output is in the feedback loop.

It can be demonstrated (http://masteringelectronicsdesign.com/the-non-inverting-amplifier-output-resistance/) that for an op-amp with output resistance \$R_o\$, that the effective output resistance of the non-inverting amplifier, \$R_oout\$, is \$R_{out} = R_o * \frac{ A_{CL}} {A_{OL}}\$,

where CL and OL refer to open loop and closed loop gains. With a closed loop gain of 1 and an open loop gain in the hundreds of thousands, the effective output impedance is quite small (regardless of the op amp!!)

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  • \$\begingroup\$ How does Figure 14/Figure 34. Output Impedance vs. Frequency relate to this? At the Mhz region the output impedance is above \$2\Omega\$. \$\endgroup\$ – someonr Mar 23 '16 at 18:50
  • \$\begingroup\$ @someonr - that's the output impdance \$R_o\$ of the op amp. With an open loop gain in the hundreds of thousands, you're still fine to use the buffer as your source. \$\endgroup\$ – Scott Seidman Mar 23 '16 at 19:23
  • \$\begingroup\$ thx for the clarification. \$\endgroup\$ – someonr Mar 23 '16 at 19:27
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The output impedance \$r_\text{out}\$ of a circuit with negative feedback can be calculated by dividing the output resistance \$r_\text{o}\$ (without feedback) by the loop gain. For a unity-gain amplifier we have a feedback factor \$\beta=1\$ and the loop gain is identical the open-loop gain \$A_\text{ol}\$ of the opamp. This results in a very very small value which can be always neglected. (set to zero).

Therefore: \$r_\text{out} = \frac{r_\text{o}}{A_\text{ol}}\$

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  • \$\begingroup\$ i cannot decode that r,out=r,o/Aol. statement. can you replace it with some \$\LaTeX\$ and define your terms so that it is unambiguous? \$\endgroup\$ – robert bristow-johnson Mar 23 '16 at 19:51
  • \$\begingroup\$ Is it ambigious? I can repeat the wording from my answer:The output impedance r,out of a circuit with negative feedback can be calculated by dividing the output resistance r,o (without feedback) by the loop gain. This is the result from negative feedback analyses. To be exact: We have to divide by (1+loop gain). It is the same factor which is relevant for the closed-loop gain also: Acl=Aol/(1+loop gain).) \$\endgroup\$ – LvW Mar 23 '16 at 21:16
  • \$\begingroup\$ please use math. \$\endgroup\$ – robert bristow-johnson Mar 23 '16 at 23:24
  • \$\begingroup\$ Sorry - I do not understand your recommendation. \$\endgroup\$ – LvW Mar 24 '16 at 9:32
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    \$\begingroup\$ so i edited you answer to show you how \$\LaTeX\$ is used. the final equations is a little off, it should be \$R_\text{out}=\frac{R_\text{o}}{1+\beta A_\text{ol}}\$ with \$\beta=1\$, if i understand your usage of \$\beta\$ correctly. the usage of commas in math notation is usually in the argument of a multivariable function like \$f(x,y)\$. even using "ASCII math" (which we need not do, since this is not USENET), your usage of commas has no identifiable convention. that makes it your own language and not one of common information exchange. \$\endgroup\$ – robert bristow-johnson Mar 24 '16 at 18:57

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