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I need help to understand this circuit and calculate voltage \$V_o\$

enter image description here

First what will \$V_o\$ be if \$V_2=1v\$ and \$v_1=0.5v\$

I do not understand how to think here. The voltage at both terminals of the amp should be the same and if we see the voltage at positive terminal as from the voltage from potential divider then it should be 0.25 V.

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  • \$\begingroup\$ You should apply the superposition theorem: determine \$V_{01}\$ when input \$V_1\$ is grounded, then determine \$V_{02}\$ when input \$V_2\$ is grounded. The output voltage is the sum of these two results. Superposition theorem works with linear circuits and should be part of your analysis toolbox with Thévenin and Norton - plus a few more : ) \$\endgroup\$
    – Verbal Kint
    Nov 29, 2017 at 15:16
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    \$\begingroup\$ Before "doing Maths" let's first just look at the circuit to see what happens. The + input of the opamp is easy, there's no feedback so we can indeed say the + input is at 0.25 V. The - input is more complex, there is feedback. Let's assume that the opamp somehow manages (by using the output) to force the - input to 0.25 V as well. Then we know the voltage across the 5 k resistor, we know the current, that current must also flow through the 10 kohm resistor to the output. That gives the voltage across that 10k resistor. We know that the - input is at 0.25 V so the output voltage is... \$\endgroup\$
    – Bimpelrekkie
    Nov 29, 2017 at 15:21

2 Answers 2

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enter image description here

There are many ways to solve. One way of solving is:

Voltage at node b = 0.25 V (Voltage Divider) It should be equal to the voltage at node a too, in ideal op-amp. Hence $$V_a = V_b = 0.25 V$$ Now current through 5k resistor would be $$(V_2 - V_a)/5k = 150 \mu A$$ The same current flows through the 10k feed back resistor. Because, from node a, no current flows into the -ve terminal of the ideal op-amp. Therefore: $$(V_a - V _o)/10k = 150 \mu A$$ On solving: $$V_o = - 1.25 V$$

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Consider the following diagram (very similar to yours)

enter image description here

The transfer function for this amplifier would be:

enter image description here

The voltages (V1, V2) are swapped in the above diagram, so make sure to take this into consideration in your calculations.

Please follow this link, or let me know if you need further assistance.

Hope this helps.

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  • \$\begingroup\$ I recognize the font and picture from Electronic Tutorials. Can you link that webpage so that if the OP has any questions, they can refer to this website? \$\endgroup\$
    – user103380
    Nov 29, 2017 at 16:14
  • \$\begingroup\$ Absolutely, edited. Good point! \$\endgroup\$
    – rrz0
    Nov 29, 2017 at 16:14

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