Given the following voltage divider :
I understand that the voltage across each resister is the same due to their values being the same, and so the voltage is evenly weighted.
But if I then add a diode, like this :
I don't really understand why the voltage across each resister is now different?
My guess is that it is because the diode has enough forward voltage such that it's resistance is now very small, so the voltage divider now has more resistance in the top half, hence more voltage.
Is that correct?
My real confusion of all this though is how you would have computed the voltages manually, I find my self getting into a loop. I can't compute the voltage across r3 because I don't now the resistance across d2, and I can't compute the resistance across d2 because i don't know the current through d2 etc etc
I understand that the voltage across each resister is the same due to their values being the same, and so the voltage is evenly weighted.
But if I then add a diode, like this :
I don't really understand why the voltage across each resister is now different?
My guess is that it is because the diode has enough forward voltage such that it's resistance is now very small, so the voltage divider now has more resistance in the top half, hence more voltage.
Is that correct?
My real confusion of all this though is how you would have computed the voltages manually, I find my self getting into a loop. I can't compute the voltage across r3 because I don't now the resistance across d2, and I can't compute the resistance across d2 because i don't know the current through d2 etc etc
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