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I'm new in electrical engineering (since I'm a chemical engineer) but I've been working on system dynamics through simulation of a DC circuit that involves Op Amps. One of my main variables to watch is supply voltage. Now, I'm struggling with relating the open-loop gain to supply voltage, especially at low values of the latter. Data handling from simulations of a class of Op Amp (LMC6482) in NI Multisim showed that open-loop gain drops drastically when a certain supply voltage is reached (from high to low voltages) as is expected, but it continues decreasing, reaching almost zero when supply voltage is not zero yet. Is it physically possible to get open-loop gain values that low? Can saturation voltage levels play a part in this situation by decreasing too much while open-loop gain remains at low but not-less-than-unity value? Is there a way of relating other Op Amp parameters to obtain the open loop gain as a function of supply voltage?

I attached a plot showing the numerical data obtained in NI Multisim.enter image description here

If simulation data is not trustworthy at supply voltage lower than recommended operation conditions, what could be the best way to approach to transition between the Op Amp operating from active to passive?

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    \$\begingroup\$ If your design has dependence on the open loop gain being a particular value, it's a bad design. Use a part that has "more than plenty" of open loop gain at the voltages you will feed it, or arrange to provide adequate voltage to the part you choose. \$\endgroup\$
    – Ecnerwal
    Jul 14, 2017 at 0:52
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    \$\begingroup\$ Yes its possible to get open-loop gains that low at small supply voltages. Equivalently its like asking how much current passes through a capacitor at DC. \$\endgroup\$
    – sstobbe
    Jul 14, 2017 at 1:48
  • \$\begingroup\$ If you take a class on analog circuit design you'll learn that it takes a great deal of cleverness to make the op-amp gain flat over any range of power supply voltages at all. \$\endgroup\$
    – The Photon
    Jul 14, 2017 at 2:05

2 Answers 2

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First, your question is misguided for at least two reasons:

  1. Gain of an op-amp is not well controlled. It might be dramatically different from device to device. Simulators won't tell you about this variation, they'll just give you the performance of a "typical" device.

    Well-designed op-amp circuits will give good results as long as the gain is "high enough" and won't depend on the exact value of the gain.

  2. You should not operate your op-amp with a lower supply voltage than its datasheet specifies. There is nothing guaranteed about its performance under these conditions.

    You also shouldn't expect the simulation model to be accurate when the operating parameters (like supply voltage) are outside the specified ranges. So you can't trust that your simulation results are reflecting the real behavior of these op-amps.

Now, what is going on in your simulation?

Data handling from simulations of a class of Op Amp (LMC6482) in NI Multisim showed that open-loop gain drops drastically when a certain supply voltage is reached (from high to low voltages) as is expected, but it continues decreasing, reaching almost zero when supply voltage is not zero yet.

Probably, as the supply voltage drops, the input common mode range changes. At some point, the bias point you chose for your simulation went out of the common mode range for that supply voltage, and the amplifier gain dropped.

It's also possible that at some point the bias point of some internal nodes in the circuit get out of whack, breaking the connection from one stage to another. Again this could easily lead to a drop in gain.

As mentioned before, you shouldn't trust the simulation model to be accurate under these conditions --- the simulator is meant to tell you how the circuit will behave under typical operating conditions. It's very unlikely that the engineers (or interns) who developed the model optimized it or verified its accuracy for simulating operating conditions outside the datasheet limits.

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  • \$\begingroup\$ Thanks for your answer. I'd like to comment that the Op Amps involved in the circuit work as Negative Resistor Converters and the mathematical model developed is sensitive to open-loop gain, and I care about the Op Amp transition from "active" to "passive" (when supply voltage is not enough), what do you think could be the best way to approach to it without taking into account the simulation data? \$\endgroup\$
    – Sergio Munguía
    Jul 14, 2017 at 2:14
  • \$\begingroup\$ You should power your op-amp with voltages that are in the range recommended in the data sheet. \$\endgroup\$
    – The Photon
    Jul 14, 2017 at 2:16
  • \$\begingroup\$ The problem here is that battery discharge is being modeled, so there exists a point where voltage supply is not enough, and that transition plays a big part. \$\endgroup\$
    – Sergio Munguía
    Jul 14, 2017 at 2:24
  • \$\begingroup\$ You could replace the batteries before they get so low they aren't able to operate your circuit. Or you can power the op-amp from a different source than those batteries. \$\endgroup\$
    – The Photon
    Jul 14, 2017 at 2:26
  • \$\begingroup\$ Bigger picture: Only you know the constraints of the problem you're trying to solve. If you want our help, give us enough information to help you. Post a new question asking about the problem you're actually trying to solve. \$\endgroup\$
    – The Photon
    Jul 14, 2017 at 2:26
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It's very simple. Don't run your power supplies below the recommended operating conditions for the op-amp: -

enter image description here

Supply voltage range for the LMC6482 is 3 to 15.5 volts and you are simulating it down to about 0.5 volts. Would you try and run your petrol lawnmower on water?

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  • \$\begingroup\$ What happens when you work with an Op Amp supplying less voltage? Doesn't it just act like a resistor? \$\endgroup\$
    – Sergio Munguía
    Jul 14, 2017 at 14:37
  • \$\begingroup\$ Op-amps receive a supply from a power supply, they don't supply voltage other than to their output pin. And no, any complex interconnected group of transistors is very unlikely to act as a resistor. \$\endgroup\$
    – Andy aka
    Jul 14, 2017 at 15:18
  • \$\begingroup\$ Even if Op Amps are connected as negative resistance converters? \$\endgroup\$
    – Sergio Munguía
    Jul 14, 2017 at 16:22
  • \$\begingroup\$ @SergioMunguía please understand this - the supply voltage for the op-amp in question has a minimum value of 3 volts - if you feed it with less than 3 volts then it might not work properly and all bets are off. You need to study op-amps because asking about negative resistance converters is totally irrelevant. Go study. \$\endgroup\$
    – Andy aka
    Jul 14, 2017 at 16:32
  • \$\begingroup\$ Can you tell me where I can look for what happens with Op Amps when they're supplied with lower voltage? I understand your point and there's no need to be rude. Not only the analysis I'm carrying out cares about the op amps working at the supply voltage needed, but also at low voltages. I wouldn't mind if op amps don't work as they should, but I need to know how they do. Since op amps are connected as negative resistance converters for which there are mathematical expressions of its "equivalent resistance", I thought there would probably be one for lower voltages (or even the same expression) \$\endgroup\$
    – Sergio Munguía
    Jul 16, 2017 at 19:38

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