MHB Op-Amps Circuits: Explaining Prof's Method

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The discussion revolves around understanding a professor's method for solving op-amp circuits without extensive nodal analysis. Participants highlight the use of Thevenin's theorem, which simplifies the circuit to a voltage source and resistor. One contributor notes that when the output node is open, the current through two resistors must be equal, leading to a relationship between the voltages across them. This approach allows for a more elegant solution compared to traditional nodal analysis. The conversation emphasizes the effectiveness of Thevenin's equivalent in simplifying circuit analysis.
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Can anyone explain to me what my professor did? I am able to get the same answer, but not without a page of nodal analysis. It seems quite elegant and none of my friends know what method she used. Thanks!
 

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Rido12 said:
Can anyone explain to me what my professor did? I am able to get the same answer, but not without a page of nodal analysis. It seems quite elegant and none of my friends know what method she used. Thanks!

The Thevenin equivalent of a linear electrical network is composed by a voltage generator $V_{T}$ in series with a resistor $R_{T}$...$V_{T}$ is the output voltage when the output node is open [output current is 0...], $I_{T}$ il the otput current when the output node il shorted to ground [output voltage is 0...] and finally $R_{T} = \frac{V_{T}}{I_{T}}$...

Kind regards

$\chi$ $\sigma$
 
Hi $\chi \sigma$,

I have done that and applied nodal analysis at node $V_T$ and at ground, but I am unable to follow the logic presented by my professor:

$$V_A=\frac{10}{40}V_s=\frac{V_s}{4}$$
How did she get this?
 
Rido12 said:
Hi $\chi \sigma$,

I have done that and applied nodal analysis at node $V_T$ and at ground, but I am unable to follow the logic presented by my professor:

$$V_A=\frac{10}{40}V_s=\frac{V_s}{4}$$
How did she get this?

If the output node is left open, then the current in the 10 k and 30 k resistors must be the same, and that means that the voltage on the 30 k resistor is three times the voltage on the 10 k resistor...

Kind regards

$\chi$ $\sigma$
 
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