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Suppose the following inverting op-amp.

How does one calculate the voltage output when we connect a DC voltage source of for example 750mV to it.

With an ideal amplifier we can just multiply the gain with the input voltage. But that is not the case with inverting op-amps. But i don't know how to then calculate the output voltage with a inverting op-amp when we give a dc voltage input. I know that in this case the output voltage should be 2.05V. But i do not know how this is calculated.

schematic

simulate this circuit – Schematic created using CircuitLab

Since there is a negative feedback there is no current going into the input terminals of the op-amp. Thus the currents of R1 and R2 are the same. U = IR, so I = U/R.

For R1, I = 750mV/100 and for R2 I = \$V_{out}\$/300. These are equal to each other

So we get \$ 750mV/100 = V_{out}/300\$

which leads to \$ V_{out} = 2250mV\$, which is still not the correct answer

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  • \$\begingroup\$ There are two principles that are used in analyzing such a circuit: 1) When there is a negative feedback then the voltage on the two input terminals is equal. 2) There is no current flow into/from the input terminals. Now, using these principles write down the voltages and currents on this schematic. \$\endgroup\$ – Eugene Sh. Oct 30 '19 at 19:39
  • \$\begingroup\$ I have edited the post, but with your information i get to 2250mV instead of 2050mV \$\endgroup\$ – elcie Oct 30 '19 at 19:50
  • \$\begingroup\$ You need to take account which direction current flows, or which way you look voltage levels. 2.25V is from negative input to output pin - so -2.25V to ground level. \$\endgroup\$ – jalaffo Oct 30 '19 at 19:54
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    \$\begingroup\$ Apart from the sign, you calculated it correctly -2.25V is the correct answer. Note that \$I_{R1} = -I_{R2}\$. I think your reference "I know that in this case the output voltage should be 2.05V" is incorrect. That answer is lacking a sign too, BTW \$\endgroup\$ – Huisman Oct 30 '19 at 19:58
  • \$\begingroup\$ You don't really show it in the schematic, but is the op-amp + terminal connected to ground? \$\endgroup\$ – user4574 Oct 29 '20 at 0:11
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In this case Opamp's high gain will drive both positive and negative inputs to same potential. it means that you have 0.75V over R1 or 7.5 mA. Same current will also flow thru R2 so 0.0075A*300R = 2.25V (ohms law)..

Output voltage is -Vin * R2/R1.

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  • \$\begingroup\$ But the voltage should be 2.05V, not 2.25 \$\endgroup\$ – elcie Oct 30 '19 at 19:52
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    \$\begingroup\$ The inverting and non-inverting inputs will only be at the same voltage if the circuit has negative feedback and has not saturated...high gain alone does not guarantee this. Also, the output voltage is not merely the ratio of resistor values. \$\endgroup\$ – Elliot Alderson Oct 30 '19 at 20:06
  • \$\begingroup\$ -3 * 0.75 V = -2.25 V \$\endgroup\$ – jalaffo Oct 30 '19 at 20:10
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    \$\begingroup\$ -R2/R1 does not equal -2.25 \$\endgroup\$ – Elliot Alderson Oct 30 '19 at 20:11
  • \$\begingroup\$ @Elliot I corrected the formula (oops), and the statement to mean this case, which I meant. \$\endgroup\$ – jalaffo Oct 30 '19 at 20:12

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