I'm working with ICL7650S(online datasheet). Today, when I was working with the Op-Amp, I came across to an odd thing. this is my circuit:


simulate this circuit – Schematic created using CircuitLab

Note that 22uF is for reducing the supply osilation(this question might help you that why I put this cap. also at first I used 100nF and it was getting better and when I changed it to 22uF, the output just 10mv increased(a bit better than earlier)).

The problem is that the Gain isn't correct and especially when I reduce the input, the Gain is going to be reduced(due to the input). for example look at this table:


This table is an example of my measurements(from above circuit). the gain for this circuit should be (357/1.033)+1=346.59

The question is Why the Gain is reducing due to the input?

P.S. A not-completely relevant question. How can I smooth the output of my switching supply as possible as?

  • \$\begingroup\$ did you measure the gain resistors or are these nominal values? \$\endgroup\$ May 4, 2015 at 7:41
  • \$\begingroup\$ Where did you get the circuit from - the configuration of R1/R2 is strange? It's possible for R2 to short the 5V supply to ground; also note that R1 loads R2, you cannot simply multiply the two potentiometer gains (even if you reconfigure the potentiometer connections) \$\endgroup\$
    – Chu
    May 4, 2015 at 7:51
  • \$\begingroup\$ @VladimirCravero No, I measured them. because I would get real amount as possible as. \$\endgroup\$
    – Roh
    May 4, 2015 at 7:54
  • \$\begingroup\$ @Chu If you look at the circuit, you will see "milivolt" lable. I measure the input from there. don't worry about it. \$\endgroup\$
    – Roh
    May 4, 2015 at 7:58
  • \$\begingroup\$ @Roh I did look at the circuit. Did you check the voltage on P11? \$\endgroup\$
    – Chu
    May 4, 2015 at 8:11

2 Answers 2


You're trying to operate this amplifier very close to its negative supply rail, and it really isn't optimized for that. If you want accurate results near 0V input and output, you need to connect the -V supply pin to a source of -5V.

  • \$\begingroup\$ I'm not sure this is true, although the specs don't address the issue directly. At +/- 5v supplies, the minimum common mode voltage is -5, with a typical value of -5.2. So there's no obvious reason why it wouldn't work as shown. \$\endgroup\$ May 4, 2015 at 19:02
  • \$\begingroup\$ @WhatRoughBeast: It isn't just a question of the input range, it's also a question of how close the output can drive to the rail. Lightly loaded, it can get to within 300 mV guaranteed, but even then you're already in a region of reduced overall gain. \$\endgroup\$
    – Dave Tweed
    May 4, 2015 at 20:43
  • \$\begingroup\$ True enough, but his measurements go no lower than 600 mV. \$\endgroup\$ May 4, 2015 at 22:13

I suggest that your multimeter is not as good as you think. Then again, the circuit is not as you think either.

When measuring gains, you've fallen prey to the error of not accounting for offsets. Your 4 individual measurements are OK (well, not really, but keep reading) but your conclusions are wrong. Instead of computing gain from individual input/output combinations, compare separate readings. So, for instance, between the first and last readings, the input changes by (9.6 - 2.1) mV, or 7.5 mV. For the same measurements, the output changes by (3.09 - .578) volts, or 2.512 volts. So the gain is 2.512 / .0075, or 324. Your nominal gain is 1 + (357/1.033), or 346.6.

Is this a problem? Probably not. If you have the manual for your multimeter, take a look at the accuracy section, and pay attention. The first thing you'll see is, for voltage measurements, a number such as 0.1% +/- 1 digit. Since your input measurements were all taken at the same range, you can assume the scale factor remains constant, but the 1-digit uncertainty remains. This means that 9.6 mV could be 9.5 or 9.7, and the 2.1 could be 2.0 or 2.1. As a result, your input difference could be anywhere in the range of 7.3 to 7.7 mV. The output readings are worse. The change in resolution (from x.xx to .xxx) suggests that the meter has internally changed ranges. In doing so, the scale factors have changed. The two readings will have an accuracy which is different between them. This is (or should be) part of the manual. Let's assume that the scale factors are good to 0.1%. Additionally, the fixed 3 digits means that the 3.09 volt measurement could vary by 10 mV, and you'd never know it. And the 1-digit uncertainty remains. So take it in order. 3.09 could be 3.08 to 3.10. In turn, this could be a range of 3.075 to 3.105 when you fill in the missing digit. Finally, multiplying by the scale factor of 1 +/- .001 gives a range of 3.072 to 3.109.

Combining the two, the actual gain could be anywhere in the range of (3.075 - .577)/ .0077) to (3.109 - .576) / .0073), or 324 to 345.

Aha! you cry. 345 is still too low! Yup. Now we come to your resistance measurements. You're using a 3 1/2 digit meter, and as with the voltage measurements, it's working on different scales for the two values. In each case, the measurement references an internal resistor, and the resistors are not perfectly accurate. In your case, let's be charitable and assume 0.5% accuracy. Then your 1.033 k could be in the range of 1.028 k to 1.038 k, while the 357 k could be in the range of 355.2 k to 358.8 k. Plugging these in gives a predicted gain in the range of 343 to 350.

Your measured gain range (324 to 346.6) overlaps your predicted range (343 to 350). There is no reason to think that your circuit is behaving other than expected. Actually, I've used what I consider quite conservative assumptions about voltage and resistance accuracy, particularly the resistance accuracy. For a cheap DMM, I would not be surprised if the numbers are worse, and the ranges shown above are larger.


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