# How do I analyze this circuit with diodes?

If diodes in this circuit are ideal diodes:

I'm making assumptions to solve this circuit for I and Vo but I have some questions: 1. What is the voltage at the cathode of the upper diode if the other two diodes are off (i.e. open circuit) ? 2. If all diodes are on (i.e. short circuit), then will the circuit look like a resistor with voltages 16 and 12 at its ends?

• The problem I have with this question and the subsequent answers is the definition of 'ideal diode' given that the diodes are clearly and specifically marked as silicon devices. This would introduce a forward voltage drop of about 0.6V across each diode in series rather than the 'short circuit' as assumed, giving Vo as 16 - 1.2V (=14.8V) and I = (14.8 - 12)/4700 = approx 600 uA Oct 17, 2014 at 16:48

1. As shown in the original post, the lower diodes can never be turned OFF because they'll never be reverse biased.

2. "Ideal" diodes can be likened to switches with zero resistance between the contacts when ON, and infinite resistance between the contacts when OFF.

3. Redrawing the circuit from that point of view, with perfect switches and with the original post's context allowing us to turn OFF the two bottom diodes, we have:

• Some further details explaining how you got to these answers would be useful for future viewers who want to understand the correct techniques. Oct 17, 2014 at 15:20

If the diodes are ideal (note, it also applies for the threshold model) there is enough voltage difference to keep them on, therefore they will behave like short circuits (or, in the threshold model, as voltage drops).

In the case you describe, yes, you can just tie Vo to 16 V.

What is the voltage at the cathode of the upper diode if the other two diodes are off (i.e. open circuit) ?

Here's how to reason to the correct answer.

(1) If the other two diodes are off, the current through the upper diode is zero

(2) If the current through the upper diode is zero, the voltage across the upper diode is zero or negative.

(3) The voltage across the diode is the difference of the anode voltage and cathode voltage:

$$V_{diode} = V_{anode} - V_{cathode}$$

From (1), (2), and (3), it follows that the voltage at the cathode of the upper diode is 16V (or more).

But this is inconsistent with the other two diodes being off; if the cathode of the upper diode is 16V (or more), the voltage across the other two diodes is positive since their cathodes are at 12V (there is no current through the resistor).

Since we've reached a contradiction, we know that the other two diodes must in fact be on rather than off.