4
\$\begingroup\$

I'm building a piece of automatic test equipment to be used in-house and have a need to measure relatively small currents (e.g. 3 µA). So I'm planning on measuring the voltage across a 1% 1K resistor (3 mV) and multiplying it by 100 or so (in two 10x stages) and then feeding it into an 12-bit ADC with a 4.096 V reference (so one bit = 1 mV).

I picked what I thought was a suitable op amp for this task (ALD1722), and hooked it up in a non-inverting amplifier configuration (Rf = 9.09K, Rg = 1K) with a +5V supply, and discovered for input voltages in the tens of millivolts I got a value close to what I expected but as I decreased the input down to a few mV or so the output went to 0.

Puzzled, I hooked up the same op amp in a unity gain configuration, and discovered that the output was lagging the input by about 10 mV; i.e. an input of 50 mv results in an output of 40 mv, and any input below 10 mV results in no output.

I'm a mostly digital guy, so op amps have always been a little mysterious. (I never studied IC op amps in class, as the 741 wasn't even released until my senior year in college.) I picked this op amp because the datasheet says it has a typical Input Offset Voltage of 25 µV, so I wasn't expecting this kind of error, but maybe I don't understand what offset voltage is.

Is there some sort of biasing circuit I can add to correct for this, so I can feed mV or smaller voltages into the op amp and get a 10x output?

\$\endgroup\$
4
\$\begingroup\$

I suspect the following note from the datasheet may describe what you are seeing:

  1. The ALD1722/ALD1722G has complementary p-channel and nchannel input differential stages connected in parallel to accomplish rail to rail input common mode voltage range. This means that with the ranges of common mode input voltage close to the power supplies, one of the two differential stages is switched off internally. To maintain compatibility with other operational amplifiers, this switching point has been selected to be about 1.5V above the negative supply voltage. Since offset voltage trimming on the ALD1722/ALD1722G is made when the input voltage is symmetrical to the supply voltages, this internal switching does not affect a large variety of applications such as an inverting amplifier or non-inverting amplifier with a gain larger than 2.5 (5V operation), where the common mode voltage does not make excursions below this switching point. The user should however, be aware that this switching does take place if the operational amplifier is connected as a unity gain buffer and should make provision in his design to allow for input offset voltage variations

Note the last paragraph. Below 1.5V the offset voltage will vary due to one of the differential stages switching off.
I would try either running from a +/-2.5V supply so the input is at the centre of common mode range, or biasing the input to 2.5V.
Also make sure you are not loading the output too much, e.g. < 5k or so (I assumed not)

\$\endgroup\$
3
  • \$\begingroup\$ Thanks, I'll try using a split supply and see if the problem goes away. I'm usually pretty good about reading the fine print, but totally missed this. I assume I can use +/-5V as well, as that would be more convenient in this case. \$\endgroup\$
    – tcrosley
    Feb 8 '12 at 5:09
  • \$\begingroup\$ Yes, +/- 5V will actually give you more range on the input. \$\endgroup\$
    – Oli Glaser
    Feb 8 '12 at 6:31
  • \$\begingroup\$ The split +5V / -5V solved my problem perfectly. I can now measure voltages below 1 mV. Thanks! \$\endgroup\$
    – tcrosley
    Feb 8 '12 at 19:10
1
\$\begingroup\$

The best amplifier for current measure is the instrumentation (differential) amplifier, or InAmp, such as the AD620; they have a good offset compensation, a (almost) fully differential gain that allows you to amplify the signal at the first stage and make it more robust for the second stage.

\$\endgroup\$

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.