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I have a current source connected to a load and I want to measure the voltage at the output of the source, and therefore I would like to buffer this voltage.

In this configuration, I am under the impression that the opamp will always draw the same amount of current from the current source regardless of what current it is set to output. (Eg an opamp with a bias current of 1uA buffering the output of a current source producing 1000uA will result in 999uA going to the load. If the current source produces 2000uA, 1999uA goes to the load.)

If this is the case, it seems like I don't need to necessarily seek out an opamp with a low bias current, but I could simply offset the current source by the bias current to ensure that my load gets the desired current.

Is this thought process correct? Or would using a very low bias current opamp still be of significant benefit here?

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  • \$\begingroup\$ Could you tell us something about the signal that you're trying to buffer? Amplitude and speed (bandwidth)? How much noise is okay? It may turn out that an op-amp with FET input is okay - or, on the contrary, that you indeed need to bite the bullet and trade off lower noise for some BJT bias current... \$\endgroup\$ Commented May 24, 2020 at 4:58
  • \$\begingroup\$ 20kHz current square wave up to +/-12v. I guess worst case it will be driving a purely resistive load and have a somewhat square voltage output. I am not quite sure how to determine an acceptable noise window, but I would be using a 12bit ADC if that gives some indication. It seems like trading bias current for lower noise and cost is an easy choice in this scenario? \$\endgroup\$ Commented May 24, 2020 at 5:20
  • \$\begingroup\$ At a first look, I went "eek, 24V square wave at 20 kHz? Slew rate limit, check the datasheet." Looking better at your spec, did you actually mean some current square wave, at 20 kHz, super-imposed on a slow moving, common mode DC offset that's anywhere between + and - 12 V against the reference GND? Still, precision doesn't mix with high frequency and high input impedance. The fastest op-amps are nominally current-coupled. Is your load just a shunt resistor by any chance? Shunt or not, what's the nominal resistance/impedance of the load? \$\endgroup\$ Commented May 24, 2020 at 16:59
  • \$\begingroup\$ You have mentioned an ADC - what's going to be your sampling rate? Is it going to be synchronized to the 20 kHz clock of your signal? Is there a chance to sync your S&H to "just before another edge in the signal" = for optimal settling? \$\endgroup\$ Commented May 24, 2020 at 17:01

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The input bias current may be relatively constant, but it will tend to change with temperature, input voltage and supply voltage, and will generally not be well specified to begin with.

Take an example, such as the MC33078. At 25°C input bias current is typically 300nA but may be as high as 750nA (no minimum is given). Over temperature, it could be as high as 800nA. Pretty wide range- but you could measure it easily at room temperature anyway and calibrate it out. Remember that's only with one particular set of power supply voltages, one temperature, and one common mode and output voltage.

Now, how does it behave with the other parameters.. (typically, no worst case figures are offered):

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Changes with supply voltage, but if you have regulated supply voltage it's not too bad.

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Not great, changes about 50% with temperature.

enter image description here

Okay here the current changes with voltage, meaning it appears like a resistance in parallel with your current source output (degrading it). It typically changes 100nA between -5V and 5V, which means it acts like a 100M\$\Omega\$ resistor. Of course it could be better or worse than that.

So it really depends on your accuracy requirements, and there is no substitute for a thorough analysis if accuracy is important.

JFET or MOSFET op-amps tend to have much lower bias currents, so even if they vary a lot it won't make much difference to a relatively large current like 1uA.

That all is at DC. At relatively high frequencies like your 20kHz, other (possibly much larger) error effects can come into play.

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The two input bias currents are not equal to each other.

The data sheet input bias current specification is actually the average of the two input bias currents.

The input offset current specification is the difference between the two input bias currents.

Input bias currents can flow into or out of the opamp's inputs depending on type of input transistor. NPN transistor bias current flows in whereas for PNP input transistors bias current flows out.

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In my point of view, you would benefit greatly from using an instrumentation amplifier here, because your thought process is not valid until you use one.

James already highlighted some of the problems in calculating the input bias current of your measurement device in general. The comments and the answers to your question so far, clearly indicate that this is an unpractical way. To make your thought process mentioned in the question applicable, the current through your measurement device must be predictable.

How is this achieved? If you use a voltage divider consisting of e. g. 4M7 - 1M - 4M7 resistors in series and then measure the voltage across the 1M resistor with an instrumentation amplifier, you have what you originally wanted. The current through the resistors can be calculated easily (12V/10,4MOhm = 1,15µA). Therefore you can correct your measurements or increase the current of the source to compensate for the current through the measurement resistors, as your thought process in your question suggested.

But the instrumentation amplifier still has an unknown and varying input bias current, that adds to the current through the resistors, you might object. Yes, of course, but this bias current is only one thousandth parts of the current through the resistor divider (taking an AD8421 as an example). Does this really care in your application? If you want to compensate for this, then model the instrumentation amplifier with its input resistors according to the data sheet and then calculate the current through the resulting resistor network including the voltage divider as mentioned above. The result is a single resistance value in parallel with your load that you can calculate with as described in your question. From a practical standpoint, in my point of view, this is the only feasible way to get to a valid result.

Good luck, all the best and remember to give us feedback, how you managed to solve the issue.

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  • \$\begingroup\$ Everything you have said makes sense. For my application a bias current in the low nanoamps range wouldn't cause a deviation in output and measurement that would be noticeable. So selecting a low bias current opamp is probably adequate. \$\endgroup\$ Commented May 24, 2020 at 9:22

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