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See the problem in the picture below. The book tells me the answer is 0.66 mA and solved using a Thevenin equivalent circuit. I don't really understand how Thevenin's theorem is used here.

To be clear, ideal diode model in this case means zero voltage drop across the diode.

enter image description here

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  • \$\begingroup\$ There are two separate branches. \$\endgroup\$
    – RussellH
    Oct 24, 2023 at 4:27

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The solution using the Thevenin theorem goes like this:

  1. Replace the 2 V source and two resistors on the left with their Thevenin equivalent.

  2. Replace the 4 V source and two resistors on the right with their Thevenin equivalent.

  3. Replace the diode with its equivalent circuit model (either an open circuit or a voltage source depending on which operating mode, forward or reverse biased, you think the diode is in)

It may help at this point to re-draw the circuit with the given substitutions.

This circuit should now be easily solvable using a single KVL equation.

If you have guessed wrong about the operating mode of the diode you will get an inconsistent result (forward bias when you used the reverse bias model or vice versa), and you will have to try again with a new guess for the operating mode of the diode.

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  • \$\begingroup\$ Thank you, I think this helped me figure it out. Can I essentially treat the ground as one large node here, and solve each side for its Thevenin equivalent? That is what I did and I got the correct solution. \$\endgroup\$
    – GBR
    Oct 24, 2023 at 5:31
  • \$\begingroup\$ @GBR, the ground symbol always indicates a common node connected to all instances of the symbol. \$\endgroup\$
    – The Photon
    Oct 24, 2023 at 5:57
  • \$\begingroup\$ @GBR The ground is always one large node, unless there are multiple grounds using different symbols. You won't see multiple grounds in basic educational circuits; that's only used in specialty stuff where you need to worry about low-noise measurements. \$\endgroup\$
    – Hearth
    Oct 24, 2023 at 11:55

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