0
\$\begingroup\$

In the image below, for an ideal op-amp, the output resistance is zero.

But when, we add the load, how does it not affect op-amp output resistance? I can't understand intuitively. I guess I am confusing ideal and real op-amp concepts, as I understand output load affects the gain.

enter image description here

\$\endgroup\$
4
  • \$\begingroup\$ What voltage do you get when you open circuit an ideal current source or, should my question be regarded as missing the point? \$\endgroup\$
    – Andy aka
    Commented Feb 1, 2021 at 17:13
  • \$\begingroup\$ Re-draw the circuit with opamps having an internal resistor that sits in series with the output. If this resistor has a non-zero value, the gain would be affected. Now what if that resistance was actually zero? Spoiler: an ideal opamp is considered to have an output resistance of zero. \$\endgroup\$
    – Bimpelrekkie
    Commented Feb 1, 2021 at 17:21
  • 2
    \$\begingroup\$ Does this answer your question? Resistor Load Effect on inverting op-amp \$\endgroup\$
    – Elliot Alderson
    Commented Feb 1, 2021 at 19:17
  • \$\begingroup\$ I’m voting to close this question because its homework without an attempt at a solution \$\endgroup\$
    – Voltage Spike ♦
    Commented Feb 3, 2021 at 5:06

3 Answers 3

Reset to default
2
\$\begingroup\$

In the image below, for ideal op-amp, output resistance is zero.

You've said it in your opening statement. Due to the feedback the op-amp will adjust the output to maintain the desired output voltage.

schematic

simulate this circuit – Schematic created using CircuitLab

Figure 1. The equivalent circuit.

The output impedance and the load form a voltage divider and we can apply the formula

$$ V_{OUT} = V_1 \frac {R_2}{R_1 + R_2} $$

Since the output impedance is zero this becomes

$$ V_{OUT} = V_1 \frac {R_2}{0 + R_2} = V_1 $$

so

$$ V_{OUT} = V_1 $$

Since R2 doesn't appear in the equation it has no effect on the output.

\$\endgroup\$
1
  • \$\begingroup\$ thanks, for clearly explaining it. \$\endgroup\$
    – negative_feedback
    Commented Feb 1, 2021 at 17:28
1
\$\begingroup\$

Well simply because an ideal opamp can provide infinite current. So the load resistor has no effect on the output whatsoever. The load current could be 1 million amps for all the opamp cares.

All an opamp does is compare its inputs and try to make them the same. It uses its output to do this. Is my input too high? make the output low. Is my input low? make the output high. The load resistor has zero effect on its operation. Feedback is simply used to manipulate this opamp behaviour to set the output to a certain level.

In non-ideal opamps, the output load can cause the output voltage to sink, opamp oscillate if complex components such as capacitors or inductors are on the output. Lots of effects. Again, the opamp output will do whatever it wants in an ideal opamp, the load does nothing because ideal opamp is infinitely fast, with infinite power rails and output current.

\$\endgroup\$
0
\$\begingroup\$

The load does not affect the op-amp output resistance in any way. An ideal op-amp is ideal, it has no output impedance and it can provide any amount of voltage and current into any load.

\$\endgroup\$

Not the answer you're looking for? Browse other questions tagged or ask your own question.