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I was trying to simulate an instrumentaion amplifier in ltspice. enter image description here

the problem i am having is the output when measured across C1 is wrong. since i have an amplification factor of 11, i was expecting to get (V2-V3)*11. which should be 11V. but i get 5.5V instead. if i change the power supply for the opamps from single supply (15V,GND) to dual supply +/-15V, then it works perfectly. any help?

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2 Answers 2

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A few quick calculations might make it a little more clear what is happening.

When not considering the supply voltages of the op-amps the voltage across R1 should be 1 V. The resulting voltages at the output nodes would be 8 V and -3 V as illustrated in this picture:

Calculation without considering supply voltages

The negative voltage at the output of U2 can not be achieved without supplying a negative supply voltage.

As a result, the output of U2 is forced to its negative supply rail which is 0 V. This essentially means that the voltage drop across R1 and R4 is 3 V and an output voltage of 5.5 V is the result of that as can be seen in this picture:

Calculation with zero volt negative supply in mind

Notice that when the difference between the input voltages is smaller compared to the absolute voltages, the circuit is operating with a gain of 11 even without differential supply voltages. So depending on the input voltage ranges differential supply might be required or not.

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  • \$\begingroup\$ Thank You very much for Taking time to answer my question Lars. I cant thank you enough for it. That solved my problem with all the remaining questions i had in my head as a follow up. \$\endgroup\$
    – lulusha
    Commented Oct 22, 2020 at 14:22
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Without the negative supply, the lower op-amp (U2) circuit can't work correctly because it is driven with a signal of +2 volts at its non-inverting input but, there is an overriding +3 volts forcing its way from the upper op-amp (U1) circuit via R1 to its inverting terminal. The net effect is that U2 naturally will want to produce a negative output voltage.

Using a negative supply instead of 0 volts rectifies this.

You can check this for yourself by looking at U2's output voltage relative to ground rather than trying to figure out things from the ambiguous (and non-informative) differential output voltage.

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  • \$\begingroup\$ Thank You very much for Taking time to answer my question Andy. I know understand why. thank you again. \$\endgroup\$
    – lulusha
    Commented Oct 22, 2020 at 14:23

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