simulate this circuit – Schematic created using CircuitLab

So I have this circuit here. My assumption was that both of these are ON and then I go ahead and solve the circuit. The only thing is though, I'm not sure if my work is right. I did nodal analysis but I'm not really sure if I applied it correctly.

My Work:

Does that look okay or did I mess up somewhere?

  • \$\begingroup\$ Well, if you assume that the diodes are ideal (0 forward bias Voltage) I guess your analysis is correct. If you use a realistic V/I curve for the diodes, it will be difficult to solve using closed form equations. Easier to simply iterate with approximate solutions until the answer is close enough. \$\endgroup\$
    – user57037
    Sep 6, 2015 at 20:26
  • 1
    \$\begingroup\$ @mkeith is right - this is correct for the ideal diode you have described, and since you explicitly state your assumption it's hard to argue with! However, if I were analysing this myself, I would include a 0.7V fixed voltage drop across each forward-biased diode. This is about correct for a silicon PN junction diode - different exact models and operating conditions will change it, but most engineers use this value (or 0.6V, or 0.65V depending on their preference) for back-of-a-cigarette-packet analysis. \$\endgroup\$
    – stefandz
    Sep 6, 2015 at 21:15
  • \$\begingroup\$ Ahh, we're not that far into i-v curves yet haha. We JUST started classes on Monday and we're only on the ideal diode so far. That's next week! Thank you so much though! \$\endgroup\$ Sep 6, 2015 at 21:41

2 Answers 2


No, you can't just assume both diodes will be on. You have to decide that first as it will impact the rest of the analisys. Start by ignoring D1 and then see if its anode voltage would be high enough to turn it on. Then you proceed with one of two branches in analyzing the circuit, depending on whether D1 is on or not.

When D1 is on, then its anode will be one diode voltage above ground. You then assume D2 drops the same voltage, so that its cathode is also at ground. Now you have all the voltages in the circuit, so you can find the currents by Ohm's Law.

When D1 is off, then just cross it off and pretend it's not there. Now you just have two resistors and a diode in series. Figure the diode looks like a voltage source of one diode drop, and the rest is just a resistor divider.

By the way, you should be able to tell from inspection whether D1 is conducting or not, but do the math if you're not sure.


The way to double check it is to solve it using some other mehod, let's try:

  1. D2 should be on, so assume it is, so, the voltage across it is zero. Ignore D1 for a while.
  2. The current trough D2 is 10V/15k=0.66mA
  3. The voltage on D1 is 5V-0.66mA*5k=5V-3.3V=1.7V
  4. So, D1 is also on. Reboot.
  5. Voltage on D1 is zero.
  6. Voltage across R1 is 5V (one end to power supply, the other grounded)
  7. Current trough R1 is 5V/5k-1mA.
  8. Voltage across R2 is 5V
  9. Current trough R2 is 5V/10k=0.5mA
  10. Current trough D1 is 1mA (IR1) - 0.5mA (IR2) = 0.5mA

Now, let's simulate:


NOTE: I could not set the diode voltage drop (vf) to zero, but 0.6microvolts is good enough.

  • \$\begingroup\$ That's very helpful! Does the fact there is a -5V on the bottom right change the sign or anything? \$\endgroup\$ Sep 6, 2015 at 21:44
  • \$\begingroup\$ Also, in step 3, why is it 5V-0.66*5K. Should it not be 5V+0.66*5K? \$\endgroup\$ Sep 6, 2015 at 21:51
  • \$\begingroup\$ @SubhashisChakraborty 5V is the power supply, the voltage on the diode is the voltage of the power supply minus the voltage drop on the resistor, which is 0.66ma*5k. The polarity of the power supplies show where the current flows, but it would be very similar if the +5 was +10, D1 was connected to +5 instead of 0 and the -5 was 0. \$\endgroup\$
    – Pentium100
    Sep 7, 2015 at 2:57

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