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So with my one of the part of my circuit below, I only wanted to see the effect on the op amp if I changed the input voltage (knowing that the inverting input is 0). My first thought was that even if I changed the input voltage, the output would changed accordingly so that the amplifier gain stays the same (since it's the same op amp).

So I first set my input voltage as 7V and found the output voltage was 8.5V. Then I changed my input to 14V (double of the initial input) and found out that the output voltage was still 8.5 Volts. Now I am totally lost since my initial hypothesis is wrong.

Can someone guide me on where I was wrong?

enter image description here

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    \$\begingroup\$ Is this an ideal or practical op-amp? Are you running it in simulation or actually connecting it? Also, do you realize that you have it in comparator mode, where the gain is ideally infinite? \$\endgroup\$
    – Reinderien
    Apr 14, 2020 at 16:34
  • \$\begingroup\$ Finally: is your op-amp rail-to-rail? And what are the rails? \$\endgroup\$
    – Reinderien
    Apr 14, 2020 at 16:34
  • \$\begingroup\$ Hi Reinderien, it would be a practical op-amp. i am running it in simulation. i did not realize i have it in comparator mode, is this because my inverting voltage is 0 ? I do not understand what rail to tail means. \$\endgroup\$ Apr 14, 2020 at 16:37
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    \$\begingroup\$ It is in comparator mode because you have no feedback resistance. The rails are the supplies to your op-amp. What is VCC? \$\endgroup\$
    – Reinderien
    Apr 14, 2020 at 16:39
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    \$\begingroup\$ The LM386 is not an op-amp, therefore all answers assuming it is a standard op-amp are cannot be right. \$\endgroup\$
    – Justme
    Apr 14, 2020 at 18:45

4 Answers 4

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Your output is not changing because it is saturated at the upper rail of the op-amp, which is somewhat below 9V. You will find that any voltage you give this thing above zero will show the same result. That is because it is acting as a comparator: the output will be "high" as long as the positive input exceeds the negative input. If you want different behaviour, you have to connect feedback circuitry to modify the gain, because currently the gain is ideally infinite. I do not know what simulation software you are using nor its model for an op-amp, so it's a bit of guesswork what the gain is actually limited to, but for your purposes it's infinite.

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Yeah, the gain pins at pin 1 and pin 8 show that it is an LM386 audio power amplifier, not an opamp. The LM386 already has negative feedback for a gain of 20 and biasing so that with no input then its output is at half the supply voltage.

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The differential gain of your op-amp is likely in the ~80dB range which corresponds to a gain of 10,000. Meaning if you have one input at 0.1mV, and the other at 0V, the output voltage should be around 1V. When you are doing your simulations, your op-amp is saturating at it's max output voltage. The op-amp cannot gain a 7V differential signal up by a factor of 10,000, especially if it only has a supply voltage of ~9V.

If you want to have your circuit do something useful, you have to add a form of negative feedback. Look up non-inverting / inverting op-amp configurations.

One last thing, op-amps have a input limit range (common-mode input). In this case meaning you cannot apply say 14V when your supply is less than that.

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The LM386 is not an op-amp. It is an audio amplifier chip. It cannot be analyzed as an op-amp.

And you are seriously misusing it.

The analog input voltage has a range of -0.4V to +0.4V.

You are feeding it 14V, so who knows what the simulator model will give out.

A real chip would give out smoke, but the simulator can't.

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