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I had this problem where the anode of a diode was connected to the negative terminal of a sine generator.

I wondered how it functions so that I can get the current that goes through the diode.

If lets say Vi(V1) is the positive terminal of the generator, would the negative one have the opposite wave, as Vo = -Vi?

I know it's a very basic question but I was surprised to find out that most of my university's students didn't knew the answer to it.

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    \$\begingroup\$ What sine wave generator? What does its manual say? \$\endgroup\$ – The Photon Feb 25 at 19:04
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    \$\begingroup\$ If you hit the CircuitLab button on the editor toolbar you can draw the complete circuit and simulate it. Important: You need to add a ground symbol and connect it somewhere in your circuit. Hit "Save and Insert" when done to add in the schematic so we're clear what we are discussing. \$\endgroup\$ – Transistor Feb 25 at 19:06
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    \$\begingroup\$ funny thing about a sinewave is that the negative terminal keeps changing... \$\endgroup\$ – user16324 Feb 25 at 19:10
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    \$\begingroup\$ Not sure what the question is. If it is about current through the diode, then you first have to know the peak-to-peak amplitude of the sine wave. \$\endgroup\$ – AnalogKid Feb 25 at 19:39
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    \$\begingroup\$ You didn't add the GND symbol so, I guess, you didn't simulate either? What is V2 for? \$\endgroup\$ – Transistor Feb 25 at 20:02

You have (in the two voltage sources) an output that is Vout = \$5\sin(\omega t)-2\$ relative to ground.

The negative terminal of V1 is 0V by definition since you connected it to ground.

The positive terminal of V1 is \$5\sin(\omega t)\$, which is the same as the positive terminal of V2 since you connected them together. \$\omega\$=2\$\pi \cdot1000\$, of course.

And the negative terminal of V2 is the output so as in the first sentence. It's 2V less than the positive terminals of V1 and V2.

  • \$\begingroup\$ Thanks a lot, so if we connect the ground to the negative terminal of V2, the output of the negative terminal of V1 would be Vout = -5sin(ωt)+2, thus it graph would look like shown in the linked picture right ? drive.google.com/file/d/14g2HP1MGMKQ3BOqATbX5BHJbFpsLm6K9/view \$\endgroup\$ – Ayman Feb 26 at 11:30
  • \$\begingroup\$ That looks correct. \$\endgroup\$ – Spehro Pefhany Feb 26 at 12:22

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