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I have a function generator SFG-2110 from GW Instek. I set it to generate a square wave with max/min of aroudn +- 1 V. And I checked with an oscilloscope.

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However, when I connected outputs of the function generator to an op-amp comparator with TL972IP from Texas Instrument, voltage significantly drops, even though I DID NOT make any change on the function generator.

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I think there's something wrong with circuit impedance. But I've only heard about that and have no idea how to adjust impedance of a circuit.

Why does an oscilloscope show a different voltage from a signal generator? says some generators have High Z mode. But, I can't find such a thing on my one.

Currently, my circuit looks like:

schematic

simulate this circuit – Schematic created using CircuitLab

Will this problem be solved if I put a tiny resistor between (+) and (-) of the function generator?

If I am thinking wrong that impedance is a source of the problem, please point it out.

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    \$\begingroup\$ opamp inputs is out of common mode value \$\endgroup\$
    – User323693
    Jul 9, 2015 at 16:37
  • \$\begingroup\$ @Umar, could you explain in detail a little bit more? \$\endgroup\$
    – Jeon
    Jul 9, 2015 at 16:40
  • \$\begingroup\$ One thing you are doing is putting a -1V to +1V signal on an op-amp with only a ground connection if your drawing represents the real-world. That's going to not be appreciated much by any op-amp. \$\endgroup\$
    – Asmyldof
    Jul 9, 2015 at 16:40
  • \$\begingroup\$ The input common-mode voltage range will be about a few 100 mV from the negative supply range. if the input is kept above that, there wont be issues \$\endgroup\$
    – User323693
    Jul 9, 2015 at 16:59

2 Answers 2

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That op-amp (although it's not shown on the 'functional block diagram') almost surely has a network that looks something similar to back-to-back diodes across the inputs.

enter image description here

Hence the absolute maximum input voltage of +/-1V.

Also, look at this:

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Putting 0V on the input with respect to the Vcc- does not damage the op-amp but it causes a whole whack of current to flow out of the input terminal.

You need to respect the common mode range of the amplifier- it is rail-to-rail on the output, not on the input, and when you go outside the input CM range or apply significant differential voltage then substantial currents can flow.

This is why it's often better not to try to use an op-amp as a comparator. If you reduce the input voltage to a few hundred mV and offset it so that it's within the input CM range (or give it a small negative supply) it should work okay. Note that if you are applying -1V you need a negative supply Vcc- of perhaps 2.5V.

Even without the quirkiness of this particular amplifier this would trip you up- even with a comparator that allows input voltages down to ground (or possibly a bit below), you should not apply voltages less than Vcc-. In the case of this particular part, you should not exceed the supply voltages- you should not even get closer than about 1.5V.

As Scott says this information is in the datasheet explicitly here:

enter image description here

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  • \$\begingroup\$ Thanks @Spero Pefhany. You mean that it is not because of the impedance, but because a signal goes below Vcc- - 0.3 V written in datasheet? \$\endgroup\$
    – Jeon
    Jul 9, 2015 at 16:58
  • \$\begingroup\$ No, it's because either differential voltage approaches 1V or the input goes below Vcc- + 1.15V. The latter is kind of hidden in the datasheet because they show the input common mode range for a +/-2.5V supply as +/-1.35V at 25°C. Subtract -1.35 from -2.5 and you get -1.15V So you should really have both inputs above maybe Vcc- + 1.5V at all times (to account for temperature) and not apply a differential voltage exceeding perhaps 500mV to be safe. Or use a more appropriate chip- this is not 'single supply' or rail to rail input and it's not designed to be used as a comparator. \$\endgroup\$ Jul 9, 2015 at 17:12
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Sphero's answer is correct, but references the wrong table

enter image description here

Common mode input is (V+ + V-)/2. (-1V + 0V)/2 = -0.5V. Vcc- + 1.15 = 1.15V, so you're using non-allowed input voltage range. (Note that even your 1V input is below the recommended range)

Absolute maximum ranges refer to that range of conditions for which if you exceed them, the IC is never guaranteed to function correctly again. You're exceeding that as well, but not exceeding absolute maxima does not necessarily mean that the IC will function as normal -- it just means you won't break it.

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  • \$\begingroup\$ Thanks! I had already added the same numbers to the bottom of my answer derived from the specs. Even within the common mode range the differential voltage applied to this particular op-amp should not be more than a few hundred mV. \$\endgroup\$ Jul 9, 2015 at 18:48

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