I have circuit as below,
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Because the input source for D1 is ground, I for D1 is zero. Both diode have forward bias of 0.7V. The equation to find I for D2 is 10V - 5(I for D2) - 0.7V - 10 ( I for D1 + I for D2) = -10V . Am I right? Thanks.
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\$\begingroup\$ Please correct me on this question. \$\endgroup\$– RyanFeb 14, 2017 at 6:00
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\$\begingroup\$ Input voltage must above 0.7V in order to turn on D1, is that correct? thanks. \$\endgroup\$– RyanFeb 14, 2017 at 12:40
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2 Answers
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Nope. Sorry.
Start by assuming that D1 is simply not there. Then the current in the two resistors will be identical, and the voltage at A will be (about) 3 volts. You can calculate the exact number, but it's accurate to within about 0.2 volt. Certainly it's greater than 2 volts.
When you connect D1, it must be reverse-biased, so no current will flow, and the voltage at A will be unchanged.
answered Feb 14, 2017 at 6:40
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\$\begingroup\$ Any current from D1 flow through 10kΩ resistor? \$\endgroup\$– RyanFeb 14, 2017 at 7:52
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\$\begingroup\$ Please read the answer and pay attention. What part of " it must be reverse-biased, so no current will flow" confused you? \$\endgroup\$ Feb 14, 2017 at 13:27
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You can start by assuming either \$D_1\$ or \$D_2\$ is ON and that the other diode is OFF. Either way, you then work out whether or not it holds when you examine the situation to include the other diode. Since WhatRoughBeast took one approach, I'll take the other.
Suppose \$D_1\$ is ON and \$D_2\$ is OFF. Tentatively, that could work. The \$-10\:\textrm{V}\$ rail is lower than ground, so \$D_1\$ would provide one diode drop from ground and the rest of the voltage would be across the \$10\:\textrm{k}\Omega\$ resistor. So point B would be one diode drop below ground. Since the other rail is \$+10\:\textrm{V}\$, this would imply that \$D_2\$ would also be ON and that point A would be at ground potential. But this suggests that the current through the \$5\:\textrm{k}\Omega\$ resistor would be about twice the current in the \$10\:\textrm{k}\Omega\$ resistor, as each of the two resistors would have close to \$10\:\textrm{V}\$ across them. And that can't work. The additional current from the \$5\:\textrm{k}\Omega\$ resistor would have to sink though the \$10\:\textrm{k}\Omega\$ resistor and that would force the drop to be so large that point B would be positive relative to ground. And in that circumstance, it must be the case that \$D_1\$ is OFF.
So the assumption fails and you can be sure that \$D_1\$ is OFF.
Given that, the rest is easy as WhatRoughBeast illustrated by making the right assumption in the first place.
But you can work either way. Start wrong, prove the contradiction, and arrive at the right answer. Or start with the right answer and get lucky. Either way, you get to the same place.
