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There are some op amps that allow nulling of input offset voltage by provding decidcated inputs where, typically, a potentiometer is connected for the offset correction. My question is that if a op amp, does not provide such inputs, then can the offset voltage still be corrected? How can this be done?

For example, How would one offset input voltage in Current-to-voltage configuration?

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  • \$\begingroup\$ A modern laser-trimmed op-amp has input offset in the dozens of uV (microVolts) range. It's highly unlikely that a potentiometer will be able to improve on that, as pretty much any external device you can add would add more noise/inaccuracy. \$\endgroup\$ – Zuofu Mar 25 '15 at 4:21
  • \$\begingroup\$ @Zuofu Thanks for your response. It makes sense to me; however I am using AD712 which states the following " The AD712 is a high speed, precision, monolithic operational amplifier offering high performance at very modest prices. Its very low offset voltage and offset voltage drift are the results of advanced laser wafer trimming technology." and then into more details, it says that the "Input Offset Voltage" is 0.3mV (which is Large for my case). Link to datasheet:www.analog.com/static/imported-files/data_sheets/ad712.pdf \$\endgroup\$ – Anuj Purohit Mar 25 '15 at 4:56
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Okay, so you've got a transimpedance amplifier (inverting) so there's a virtual ground at the inverting input. That virtual ground might be a few mV one way or the other from actual ground, depending on the Vos. If it's a true current input you'll just see the same offset at the output. You can subtract the offset after the amplifier or do this:

You'll need two suitably(do the math) stable references (plus and minus). Say you have +/-2.5V and you want to be able to adjust the offset by +/-5mV. You can connect a pot across the references (so +/- 2.5V appears at the wiper) then divide that down appropriately say with 100K and 200 ohms (100K should be much less than the pot element resistance) , then apply that voltage to the non-inverting input. If the offset is entirely due to the internal offset of the op-amp the virtual ground will then be exactly 0v when the output is adjusted to 0mV (assuming negligible bias current).

By the way, the offset adjust terminals on an op-amp (where available) should not be used to compensate for offsets that are external to the op-amp, as the are not derived from a reference, the voltage will tend to be something like proportional to absolute temperature (not very stable).

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  • \$\begingroup\$ I dont get this "the offset adjust terminals on an op-amp (where available) should not be used to compensate for offsets that are external to the op-amp". As I understand, I need to ground the inputs of opamp and set a known gain (say 10) and then measure the output voltage to know the offset voltage. This will require 2 resistors (like 1 Meg and 100K) to introduce the gain and be able to measure the offset voltage (amplified by 10). Once this is done, I will need to offset that by using one of the techniques (like a pot). After offset, I may use the amp. Is this not true? \$\endgroup\$ – Anuj Purohit Mar 25 '15 at 5:22
  • \$\begingroup\$ No.. Perhaps you should ignore that paragraph for now as it does not apply to your situation (no adjust terminals). Also, that's not how to measure offset voltage- usually you make a high gain amplifier with 0V input and divide the measured output offset by the gain to get the inout-referred offset. Finally, you don't normally need to measure the input offset voltage to compensate for it, you simply trim the input until the output is what it should be. \$\endgroup\$ – Spehro Pefhany Mar 25 '15 at 5:28
  • \$\begingroup\$ I think you said the same thing i.e. inputs to be zero with a known gain and then see the output voltage. We know that the inputs are at zero therefore any reading on output voltage is because of input offset voltage(ignore the bias current for a moment). Trim the pot to see a zero and connect the real inputs now. This is the way I was learned in college. \$\endgroup\$ – Anuj Purohit Mar 25 '15 at 6:17
  • \$\begingroup\$ Inputs to entire amplifier zero, not the op-amp inputs, yes. \$\endgroup\$ – Spehro Pefhany Mar 25 '15 at 6:47
  • \$\begingroup\$ @Arjun,what you are telling is probably the best way to do it. I also learned the same thing in college. If there are no dedicated pins on the IC to deal with the offset voltage , then offset can be adjusted, with considerable accuracy, by setting the input voltage zero and adjust the pot to see a zero voltage at the output(which, of course, is due to offset only!) \$\endgroup\$ – Abhishek Tyagi Jul 27 '15 at 5:01

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