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Problem: Assuming that diodes are ideal, find I and V.

This is a worked example in the textbook and it says that if you assume both diodes to be conducting then V=0 and VB=0 (nodal voltage at B).

I don't understand how the voltage at B is 0.

What I'm getting confused about is how can there be a current through D1 if it's connected to ground and also does ID2 go to ground or does it flow into the node? I get that if you assume the diodes are conducting you short them, but I can't see how that allows one to assume that the potential difference between B and ground is 0? I get that the sum of the currents flowing into B must be zero, and that the diodes conducting means you short them, but does this mean that the voltage drop (increase) across the 5k resistor is 10V and the voltage drop across the 10k resistor is 10V?

Any explanation would be much appreciated.

  • 2
    \$\begingroup\$ If you assume that D1 is ON, and ideal than you can replace it by a short - thus point B is connected to ground. \$\endgroup\$ – Mike Oct 15 '14 at 11:28
  • \$\begingroup\$ Can you show a bit better your thought process? \$\endgroup\$ – clabacchio Oct 15 '14 at 11:34
  • \$\begingroup\$ But what about the voltages in the other two branches connected to the node? I feel like there's something fundamental that I'm missing. \$\endgroup\$ – Tom Windell Oct 15 '14 at 11:36
  • \$\begingroup\$ At first, one should clarify "what means ideal"? Voltage drop of 0 volts or 0.7 volts? \$\endgroup\$ – LvW Oct 15 '14 at 11:50

You stated that the diodes are ideal. Therefore, if D1 is conducting, the voltage across it is 0 by definition of a ideal diode. If the voltage across it is 0, then the voltage on both sides is the same, and point B must be 0 V since the other side of the diode is at 0 V (by definition of ground).

So, the question comes down to whether D1 will conduct. To see if it will, pretend it is open and see what the voltage across it would be. You have a voltage divider with 10 kΩ as the top resistor and 5 kΩ as the bottom resistor. The voltage on the bottom resistor will be 1/3 the total, which is 6.7 V in your case. Since the bottom is at -10 V, the output of the divider will be -3.3 V absolute (relative to ground). This would forward-bias D1, so we can conclude it is conducting and the voltage accross it 0.

  • \$\begingroup\$ Is there really a general agreement about the voltage drop of 0 volts across an "ideal diode"? \$\endgroup\$ – LvW Oct 15 '14 at 13:18
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    \$\begingroup\$ @LvW: Yes. Ideal diodes have no forward drop, can withstand infinite reverse voltage, have 0 recovery time, and lot's of other nice parameters. Unfortunately, Mouser and DigiKey don't seem to carry them. \$\endgroup\$ – Olin Lathrop Oct 15 '14 at 13:30
  • \$\begingroup\$ @LvW In academia the term ideal diode usually just means it drops 0.7V across it if it is forward biased. I believe this textbook also uses this assumption when solving its own example problems. IE assume the voltage at node B is -0.7V and you will come to the correct answer \$\endgroup\$ – ACD Oct 15 '14 at 17:09
  • \$\begingroup\$ So - who is right? O.Lathrop or ACD? Or shouldn`t the degree of ideality not be mentioned explicitely for each exercise? \$\endgroup\$ – LvW Oct 15 '14 at 17:55
  • \$\begingroup\$ @LvW: "Ideal" diodes are assumed to have 0 forward drop. After all, the common 700 mV figure is for silicon diodes only. Schottkys are different, so are germaniums, vacuum tube diodes, etc. If a ideal diode has non-zero forward drop, then that must be specified. Also in this context, since the answer was supplied and the question is about how it was obtained, 0 forward drop clearly is assumed. There is no doubt at all in this case. The mention of 700 mV forward drop only adds confusion to this question. \$\endgroup\$ – Olin Lathrop Oct 15 '14 at 18:24

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