Electronic – Diode circuit analysis using ideal diodes exam problem


I've been reviewing some past exam and stumbled across this question
diode problem

So I need to find the current I through the middle diode and the voltage across the bottom left 10KΩ resistor.

To try solve this problem i used the technique, "assumed states for analysis of ideal switch model diode circuits" (not sure if thats the actual name of the technique).

So for my first state, i assumed that both D1 and D2 are on

Since the current are not negative for both diodes, the assumption for both diodes being on must be correct. Thus the required current across the diode is:

(15V – 11.25V)/10KΩ = 0.000375A

However upon looking the answers, it says that the current across the diode is 0A. This means that the diode is actually "off" (while the other diode is on), and thus my analysis is wrong.

Is there a problem with my reasoning or am i missing something crucial?

Thank you for your time!

Best Answer

I think the easiest method to solve such problems is to assume that the diodes are off (both, and then one of the two), compute the voltages across the diodes and see if there's a contradiction with your assumption. Let's call the top left diode \$D_1\$ and the diode in the middle \$D_2\$.

Case 1: \$D_1\$ off, \$D_2\$ off: Since \$D_1\$ is off there is no current through the top 5k resistor, and since \$D_2\$ is off, there is also no current through the bottom left 10k resistor. So \$V=0\$ and the voltage at the anode of \$D_1\$ is 15 Volts. Contradiction! (\$D_1\$ should be on).

Case 2: \$D_1\$ off, \$D_2\$ on: again no current through top 5k resistor. Voltage \$V\$ is


Contradiction! (Because the voltage across \$D_1\$ would be \$15V-3.75V=11.25V\$ and it should be on.)

Case 3: \$D_1\$ on, \$D_2\$ off: Voltage \$V\$ is

$$V=\frac{15V\cdot 10k}{5k+10k}=10V$$

The voltage at the anode of \$D_2\$ is \$15V\cdot 5k/15k=5V\$. This agrees with our assumption, because with these voltages \$D_2\$ must be off. So your solution is

$$I=0A,\quad V=10V$$