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From Sedra-Smith:

Another limitation on the operation of op-amps is that their output current is limited to a specified maximum. [...] This, of course, has to include both the current in the feedback circuit as well as the current supplied to a load resistor. If the circuit requires a larger current, the op-amp output voltage will saturate at the level corresponding to the maximum allowed output current.

The meaning of the paragraph is clear, but I still don't get why op-amps have such a limitation on the output current. How can you deduce that from the internal circuitry of an amplifier? For example, if we consider the two-stage CMOS op-amp below, is there a way to calculate the maximum output current?

enter image description here

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    \$\begingroup\$ In the circuit you show, suppose a current is flowing out of the output into a load to ground. What device in the circuit delivers this current? Can this device deliver infinite amounts of current or not? I see that this device is part of a very common circuit structure. If Q8 has the same size as the device then I already know that the maximum current is equal to \$I_{REF}\$. Do you understand why that is so? \$\endgroup\$
    – Bimpelrekkie
    Commented Apr 20, 2020 at 13:24
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    \$\begingroup\$ @Bimpelrekkie assuming the current mirror has a unity-gain (i.e. when Q7 is the same size as Q8) then the current in Q7 should be $I_{REF}$. Considering that the output current is the difference between the current in Q7 and the one in Q6, then the maximum current is reached when the current in Q6 is minimum, that is when it is zero. Is this correct? \$\endgroup\$
    – Adrian
    Commented Apr 20, 2020 at 13:41
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    \$\begingroup\$ That is correct but you forgot that Q7 needs to be in saturation mode as well. If Q7 has a low Vds then it will be in linear mode and even less current can flow. \$\endgroup\$
    – Bimpelrekkie
    Commented Apr 20, 2020 at 13:59

4 Answers 4

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The output current that your op amp is able to source is determined by IREF and the size ratio of Q7 and Q8. The maximum output current is sourced when Q4 is in triode and Q6 is in cutoff.

The output current that the op amp is able to sink is more complicated, since it will depend on the common mode input voltage, but it happens when the gate of Q6 is pulled high by driving the non inverting input below the inverting input.

Usually that is an op amp you will find inside an IC and if it has to source current, that current has to come from the bias current. Usually commercially available op amps have an output stage, usually class B or AB, which buffers the high impedance output of the previous stage and is able to supply mA of current.

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  • \$\begingroup\$ Q1 through Q4 circuit deserves a separate analysis as to how and why exactly it would behave. At first glance, I thought they would limit the sinking current, but then I realized there is more to it. Your answer is the most correct here. \$\endgroup\$
    – Edin Fifić
    Commented Sep 5, 2022 at 2:24
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The output current depends on the reference current IREF through Q8, but, in order to know the output current, we also need to know the output MOSFETs' (Q6 and Q7) channel widths relative to the Q8 channel width, whose ratio determines the ratio of output current versus the reference current. If they are all equal in size or channel width,the maximum output current would be the same as the reference current. However, their relative sizes could have up to 1:10 ratio in a typical IC op-amp, so the output current could be up to 10 times the reference current.

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    \$\begingroup\$ Thanks for the insightful comment. What you say is certainly true, however, the text says that the amplifier will saturate when the current limit is exceeded. While there is certainly a current level which the transistors physically cannot reach, is it possible that there is another (presumably lower) current value which instead will make the transistors saturate? \$\endgroup\$
    – Adrian
    Commented Apr 20, 2020 at 21:35
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    \$\begingroup\$ He's asking why opamp short circuit current is typically well below the current required to destroy the part. I don't think this answer addresses that. \$\endgroup\$
    – Navin
    Commented Aug 13, 2022 at 1:02
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    \$\begingroup\$ @Nicol : (sorry for this much delay :D ) > The reference current and the MOSFET (channel) size ratios determine the maximum output current. I have updated my answer above. \$\endgroup\$
    – Edin Fifić
    Commented Sep 5, 2022 at 2:05
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    \$\begingroup\$ @Navin : You're 100% right. I have updated my answer to reflect that, though after a long time, as I had been very busy. \$\endgroup\$
    – Edin Fifić
    Commented Sep 5, 2022 at 2:07
  • \$\begingroup\$ @EdinFifić better late than never. Thank you so much! \$\endgroup\$
    – Adrian
    Commented Sep 11, 2022 at 23:16
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Steady-state current to the output has to come from Q6 or Q7 as those are the only DC paths to the output.

If we have a drive voltage Vgs on a MOSFET (at most it will be at one of the supply rails), consider what the Ids vs. Vds curve of a MOSFET looks like, and what that implies as to the maximum output current.

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  • \$\begingroup\$ The current in Q7 should be I_REF, hence the output current is I_o = I_REF - I_6, which reaches its maximum value when the current in Q6 is minimum. But the minimum current in Q6 is zero, i.e. when the transistor is cut-off. Therefore the maximum output current is I_REF, is this right? \$\endgroup\$
    – Adrian
    Commented Apr 20, 2020 at 13:37
  • \$\begingroup\$ Why? Q7 is part of a current mirror with Q8, so its current is Iref. \$\endgroup\$
    – Adrian
    Commented Apr 20, 2020 at 14:13
  • \$\begingroup\$ Ah, okay you're right! Q7 maximum current is Iref. Q6 has no such limitation. \$\endgroup\$
    – Spehro 'speff' Pefhany
    Commented Apr 20, 2020 at 14:41
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Bipolar devices, used in the early silicon opamps (UA702 from Fairchild, also UA709), and in SiGe today, often were in emitter-follower output configuration that for slewrate performance have

enormous base drive currents

which for survival, when the engineer's scope probe tip might slip or when the C_load accidentally becomes 1uf instead of the intended 1nanofarad, needs current limiting.

Also FETs at least have a self-protective behavior as temperatures rise, whereas bipolars definitely are not self-protective.

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