What will be voltage V if all diodes are considered ideal


simulate this circuit – Schematic created using CircuitLab

  • \$\begingroup\$ What work have you done to calculate this? What is your explanation of what is happening? \$\endgroup\$
    – gbulmer
    Sep 6, 2014 at 16:14
  • \$\begingroup\$ Consider the action of current flowing through the diodes.... \$\endgroup\$
    – Spoon
    Sep 6, 2014 at 16:22
  • \$\begingroup\$ i think all the diode become forward biased and the 3v,2v,1v will become parallel \$\endgroup\$
    – mahes
    Sep 6, 2014 at 16:38

5 Answers 5


An ideal diode can only conduct in one direction, has no forward voltage drop across it and zero resistance. Given that this is a purely theoretical circuit by defining the diodes as 'ideal' (so voltages are just voltages) I would expect this to happen:

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The lowest voltage (+1V) at the cathode end of D3 would pull the diode/resistor node down to 1V. This would reverse bias D1 and D2 so they would have no effect. The voltage drop across R1 would be 4V giving a current through it of 4mA.

  • 1
    \$\begingroup\$ This is the correct answer but I think the first sentence should be clarified - the ideal diode can have no forward (positive) voltage across but it may have a reverse (negative) voltage across. That is: \$v_D = 0\ \; \mathrm{for}\; i_D \ge 0\$ and \$i_D = 0 \; \mathrm{for}\; v_D \le 0\$. \$\endgroup\$ Sep 6, 2014 at 17:34
  • \$\begingroup\$ @AlfredCentauri yes I should have clarified that it is the forward drop that is zero (+ 1) will edit accordingly \$\endgroup\$ Sep 6, 2014 at 17:40

The correct answer has been given so I'll just add the general method for solving ideal diode problems.

With three ideal diodes, there are \$2^3 = 8\$ possible on-off combinations with only one consistent combination.

Essentially, one chooses a combination and then checks for consistency until one has found the one consistent combination.

If one chooses a diode to be off, one replaces the diode with an open circuit and then, after solving the circuit, checks that \$V_D \le 0\$ for that diode. If the voltage is positive, the result is inconsistent with the assumption that the diode is off.

If one chooses a diode to be on, one replaces the diode with a short circuit and, after solving the circuit, checks that \$I_D \ge 0\$ for that diode. If the current is negative, the result is inconsistent with the assumption that the diode is on.

So, for example, choose the combination where all three diodes are off. Replace the diodes with open circuits and find that \$V = 5V\$. But this implies that all three diodes have positive voltages across which is inconsistent with the assumption that they're all off. So, eliminate that combination.

As Jim answers, the only consistent combination is D1 off, D2 off, D3 on. (In fact, we know by inspection that \$V\$ cannot be greater than 1V and we've eliminated the possibility that all diodes are off).

To check, replace D1 and D2 with open circuits and replace D3 with a short circuit.

The solution is then

$$V = 1V$$

$$V_{D1} = -2V$$

$$V_{D2} = -1V$$

$$I_{D1} = 4\mathrm{mA}$$

which is consistent with the assumed states of the diodes.


Given the assumption that only the diodes are ideal..

If the +1,+2, +3 V sources are supplied by typical linear regulators (that cannot sink significant current) the voltage V will be ~5V.

  • 1
    \$\begingroup\$ Sorry Spehro I can't agree with your answer. This would put a forward voltage across each of the diodes (D1,2V, D2,3V and D3,4V). As they are defined as 'ideal' this is a contradiction. I think making assumptions about the specific nature of the voltage sources is an error. \$\endgroup\$ Sep 6, 2014 at 17:29
  • \$\begingroup\$ @JImDearden It could certainly lead to it being marked as such by some cretin, but I think it's a useful thought experiment anyhow. Perhaps more useful than doing someone's homework for them. \$\endgroup\$ Sep 6, 2014 at 17:32
  • 1
    \$\begingroup\$ I could see where you were coming from (that's why I didn't down vote) and it would be an interesting experiment to see what would happen with linear regulators and 'practical' diodes. I don't normally fully answer homework questions but I thought this one raised an interesting general point about diode clamping. \$\endgroup\$ Sep 6, 2014 at 17:52

Could V be greater than 1V?

If it were, then at least D3 would be forward biased and the voltage across it greater than zero, so it would be sinking infinite current. But this can't be true because that current would have to come from the resistor R, dropping infinite voltage and therefore making V infinitely negative, contradicting the assumption.

Could V be lower than 1V?

If it were, all the diodes would be reverse biased and wouldn't sink any current. But for V to be less than one it needs to have been dropped across R, which means that current must be flowing through it. But this can't be true because it has nowhere to go (since the diodes are reverse biased), contradicting the original assumption.

So we're left with V being exactly 1V. Is it possible?

D3 would be in conduction, the voltage across it would be the ideal zero volts, and would sink just enough current for R to drop the voltage from 5V to 1V. So V must be 1V.


Lets solve this in another way, consider +5V as non ideal source, so it means that voltage is increasing by 0.1V(say).
As we know that for diode to be forward bias(to turn on), potential at anode should be greater than the potential at cathode.
Now at some point of time, potential at V will be 1.1V which is greater than the potential at cathode of D3, but less than that of D1 and D2. So D3 starts to conduct i.e D3 acts as a short (as diodes are ideal) hence V is now +1V which is nothing but potential at cathode of D3.


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