Im looking to drive an LED, but am not sure if i should use an op-amp or bjt configured as a switch. Is it possible for an op-amp to source roughly 10mA; the current needed to drive an LED in series with a resistor?

If this depends on the type of op-amp, then how would I go about figuring out the maximum current an op-amp could source?

  • \$\begingroup\$ Are you trying to avoid damaging a part (by taking too much current), or are you looking for a part which will provide a certain minimum current, or something else? \$\endgroup\$ – gbulmer Aug 13 '14 at 18:40
  • \$\begingroup\$ I was mainly wondering if i could use the output of a comparator op-amp to drive an LED in series with a resistor? Or, is it necessary to use a transistor configured as a switch? The question came about because I was not sure how to tell if a given op-amp would be able to source the current necessary to drive the LED \$\endgroup\$ – mayfield512 Aug 13 '14 at 21:44
  • 1
    \$\begingroup\$ It might be worth adding that explanation to your question as an "Edit"; you might then get even more insight in the answers. \$\endgroup\$ – gbulmer Aug 13 '14 at 21:55

Usually we're concerned with how much loading can be put on the output and still have it function properly. The output swing specification will be at a specific load resistance, usually load is between the output and ground on both dual and single-supply op-amps. For example, the dual-supply OP-07:


You can see that over the whole temperature range with +/-15 supplies it is guaranteed to have a swing of +/-11V with a 2K load, meaning that it can supply +/-5.5mA. If the supplies are different then the current may be different.

If you're primarily interested in short-circuit maximum current, that is not always specified, though some op-amps do have numbers. For example, the ubiquitous LM324:


As you can see, it guarantees (at 25°C) the output short circuit current won't exceed 60mA, but it's only guaranteed to source or sink 5 or 10mA. Also note that the LM324 has a (nominal) 50uA current sink on the output so if you ask it to sink more than some tens of uA the output will no longer get that close to the negative rail. If the load is a resistor to the negative rail, the issue does not arise.

In general the minimum guaranteed current (for it to work properly) is going to be much smaller than the maximum guaranteed short-circuit current (if they even give a number for the latter).

enter image description here


It's usually called Output Current. Parameter "Io" on this TI part, page 4.

ADI calls it Maximum Output Current or "Iout", page 3.

Occasionally, it is called Short Circuit Current or "Isc" like in this Linear part which is 5-9mA depending on operating conditions, pages 4 and 6. It is a slightly roundabout way of saying even if you short the output with 0 ohms, the op-amp can only supply 9mA so that is the max output current.

If you're still having trouble finding it post the datasheet or part number.


Most of the time if you need much power beyond what the op amp can do, you can add a little boost to the circuit. TI has a nice paper on that here: AN-272 Op Amp Booster Designs. It looks scarier than it really is. Definitely worth learning.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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