# Diode circuit analysis using ideal diodes exam problem

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

• I think your assumption that there is no current flowing through the lower right 5k resistor is incorrect. – Joe Hass Apr 30 '14 at 15:39
• Yes, the diode connects it to a 10 Kohm resistor. Only if the diode connected the top of the 5 Kohm to ground would the assumption that I=0 through the 5 Kohm be valid. – horta Apr 30 '14 at 15:49

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

$$V=\frac{15V\cdot(5k||10k)}{10k+(5k||10k)}=3.75V$$

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

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

• L Thanks for the in depth answer! For case 3 however, how does 5V at the anode of D2 correspond to D2 being off? Did you already know that the voltage at the cathode of D2 is 10V? – goli12 May 1 '14 at 0:01
• @goli12: Yes, with this assumption V=10 volts (see case 3 in my answer), so there's a negative voltage across the diode (10 volts at the cathode and 5 volts at the anode, which gives -5 volts from anode to cathode). – Matt L. May 1 '14 at 6:48

Remove the middle diode and solve for the voltage at the nodes where it was connected. If the cathode voltage is higher, then adding the diode will have no effect, and the current through the diode will be zero.

• So using nodal analysis, the cathode voltage is 10V (ie the voltage across the bottom left 10KΩ to ground) while the anode voltage is 5V (the voltage across the 5KΩ to ground). So the diode is actually in reverse bias, and thus, I=0A yes? – goli12 Apr 30 '14 at 23:33
• @goli12 Yes, exactly. – John D May 1 '14 at 3:56