# Sedra and Smith Example 3.2 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.

• If you assume that D1 is ON, and ideal than you can replace it by a short - thus point B is connected to ground. – Mike Oct 15 '14 at 11:28
• Can you show a bit better your thought process? – clabacchio Oct 15 '14 at 11:34
• 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. – Tom Windell Oct 15 '14 at 11:36
• At first, one should clarify "what means ideal"? Voltage drop of 0 volts or 0.7 volts? – LvW Oct 15 '14 at 11:50