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What does the "Output Voltage Swing vs. Load Resistance" signify? Lets say I want to get a gain of 400 from an op-amp. I can use any one of the following combination

Gain = -(R2/R1)

Case 1: R2 = 400 Ohms; R1 = 1 Ohms -> Gain = 400 Case 2: R2 = 40000 (40K) ohms; R1 = 100 Ohms -> Gain = 400.

From op-Amp perspective, does the above scenario do something different?

Picture of "Output Voltage Swing vs. Load Resistance" for AD712 is attached below Output Voltage Swing vs. Load Resistance

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When load resistance decreases, current increases (assuming voltage is fixed). By the looks of it, the opamp you are dealing with can only provide 25mA of current.

In order to ensure that the load doesn't try and source or sink more than 25mA, the graph you provided just shows what the suitable voltage levels are for your resistance.

For instance, lets say your load is 100 ohms. The maximum current you can draw is 25mA. So

$$ V = IR $$ $$ V = 25mA * 100 ohms $$ $$ V = 2.5V $$

Which what you're graph is showing.

So instead of doing this calculation for every single load, they have done it for you.

All this plot is telling you is not to go over 25mA.

Added Found this in the datasheet enter image description here

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  • \$\begingroup\$ Great! The spec. states maximum current as 25mA. You are spot on!. Doe this mean that if I want to amplify a signal to 15 volt peak-to-peak, then my feedback resistance should align with the 15v line? or does it apply only when a resistive load is being drive by op amp? \$\endgroup\$ – Anuj Purohit Mar 11 '15 at 9:51
  • \$\begingroup\$ @AnujPurohit With your current feedback resistors of 40.1k, and if your output is +/- 15V, then the current through the feedback would be 375uA. Which is nothing. The 25mA is the what the opamp can spit out, this includes all paths including feedback and load. So with these resistors, your total load current is 24.625mA which again is pretty much 25mA. So looks like you're ok. \$\endgroup\$ – efox29 Mar 11 '15 at 10:10
  • \$\begingroup\$ thanks for your input. However why did you add 100 ohms resistance with 40K? the 100 Ohm R is on the inverting input and the feedback R is 40K. How does Op Amp sees the resistance at the input? \$\endgroup\$ – Anuj Purohit Mar 11 '15 at 10:17
  • \$\begingroup\$ @AnujPurohit Yes my mistake. The 100 ohm wouldn't really matter since the inverting terminal is made to be near ground potential. So from the opamps perspective, any possible paths that lead to "ground" are what it considers to be a load. Because your feedback resistance is so high (40k) is so high, its negligible. \$\endgroup\$ – efox29 Mar 11 '15 at 10:37
  • \$\begingroup\$ Thanks. Can you please give a look at circuit at electronics.stackexchange.com/questions/159062/… and see if it makes sense? \$\endgroup\$ – Anuj Purohit Mar 11 '15 at 10:47
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What does the "Output Voltage Swing vs. Load Resistance" signify?

It tells you that if you have a 10 ohm load the maximum peak-to-peak voltage you can faithfully produce at the output is a bit less than 1Vp-p. For a 100 ohm load this rises to about 3Vp-p. For an open circuit load this becomes about 28Vp-p. This assumes power rails are +/-15V.

With a feedback resistor of 400 ohms, this is likely the load seen at the output due to the inverting input being a virtual earth so expect an output voltage no greater than 17Vp-p.

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  • \$\begingroup\$ Thanks for your reply. Does this not mean that using feedback resistor >10K is a safe bet? I think I am starting to understand it. Basically the Op Amp's output impedance is high enough to not produce a voltage for a 100 ohm? (kind of a voltage divider?) \$\endgroup\$ – Anuj Purohit Mar 11 '15 at 9:44
  • \$\begingroup\$ A higher resistor is better but it's not output impedance that is at fault directly. It's the output not able to supply the current on low value loads when the volts need to be at their peaks. The output current limits. \$\endgroup\$ – Andy aka Mar 11 '15 at 10:23
  • \$\begingroup\$ Thanks! This question is related to a circuit which I am currently trying to design. It is posted at electronics.stackexchange.com/questions/159062/… Can you please look at it once and see if it makes sense? \$\endgroup\$ – Anuj Purohit Mar 11 '15 at 10:26
  • \$\begingroup\$ No, I'm not analyzing your circuit. It's badly drawn and doesn't name op-amps and is too much to expect of anyone on this site in reality. \$\endgroup\$ – Andy aka Mar 11 '15 at 11:56

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