Suppose we have two identical ideal diodes (without any resistance or voltage drop). We connect these two diodes in series, in the same direction, and reverse-bias them with voltage V. Now we have two series reverse-biased ideal diodes. There isn't any other element in the circuit.

What is the voltage value at the node between the diodes? Maybe one would say V/2 because of the symmetry, but any answer between 0 and V cannot be refuted. A reverse-biased diode can (ideally) have any voltage.

  • \$\begingroup\$ Hi! Welcome to EE.SE! I think your question might be a duplicate of this one. Please check it out and clarify if you're asking something else. \$\endgroup\$
    – K H
    Commented Jan 12, 2019 at 5:42
  • \$\begingroup\$ The Voltage depends more on the capacitance of the diode at some bias and then it becomes a capacitance divider until the leakage takes over to become a resistance divider. But then the diode with the higher capacitance tends to have the lower leakage resistance, so it may not shift in the opposite direction. Now what is the time constant of the drift in the centre? \$\endgroup\$ Commented Jan 12, 2019 at 6:17
  • \$\begingroup\$ Imagine two 1N4148 with 20nA @ Vr=20V yet 5uA at -75V and C=4pF @ 0V yet C= 0.1pF at -20V that varies on log scale. What is the centre drift T=RC with 2 same in series reversed at 40V? \$\endgroup\$ Commented Jan 12, 2019 at 6:24
  • \$\begingroup\$ Yet actually the one with the lower Rs in the forward direction like the 1N4448 that has 10x the current @ 1V over the 1N4148 yet only 1/2 the capacitance at 0V. So if you put a 1N4148 in series with a 1N4448, with 40V reverse, What is the node voltage and how long does it take for T? But they have identical reverse current leakage. \$\endgroup\$ Commented Jan 12, 2019 at 6:34
  • \$\begingroup\$ @Sparky256 this is not a duplicate of that question. That is an engineering real world question and diodes are not ideal there. This is an abstract mathematical situation solution of which was interesting to me. I wanted an answer like those of Dennis Ernst. \$\endgroup\$
    – Alex L
    Commented Jan 12, 2019 at 6:38

2 Answers 2


In case of ideal diodes you can see them as open switches if reverse biased, since there is no leaking current and so on. So there is no voltage at this point. Its floating.

  • \$\begingroup\$ I edited your answer because we need to keep them in the realm of science. Adding emotional content should be avoided. Stern but explained warnings to those working with deadly voltages (surge generators, large high-voltage capacitors, etc) are ok. \$\endgroup\$
    – user105652
    Commented Jan 12, 2019 at 6:17
  • \$\begingroup\$ So how do you satisfy KVL? \$\endgroup\$
    – Alex L
    Commented Jan 12, 2019 at 6:41
  • \$\begingroup\$ Because reverse biased ideal diode is an open switch, there is no complete circuit and KVL doesn't apply. \$\endgroup\$
    – TemeV
    Commented Jan 12, 2019 at 7:39
  • 2
    \$\begingroup\$ @TemeV KVL does apply. The junction of the diodes is floating, say, \$v\$ volts. Therefore KVL gives \$V= (V-v)+(v-0)\$ \$\endgroup\$
    – Chu
    Commented Jan 12, 2019 at 8:02
  • \$\begingroup\$ @Chu KVL states literally "The directed sum of the potential differences (voltages) around any closed loop is zero." (From Wikipedia) When there is an open switch, there is no closed loop. If you were to measure that voltage between two open switches, what result will you get? What if you first measure from V to v and then v to 0, will you get the same result? \$\endgroup\$
    – TemeV
    Commented Jan 12, 2019 at 9:39

Ideal diodes will be two open circuits, hence the junction of the diodes is floating at, say, a random \$\small v\$ volts. Therefore KVL gives: \$\small V=(V−v)+(v−0)\$, where \$\small V\$ is the voltage across the series combination.

  • \$\begingroup\$ It is not possible to define voltage between the two open switches. Voltage between two points is defined as the work needed per unit of charge to move a test charge between the two points. 1 volt = 1 joule (of work) per 1 coulomb (of charge). en.wikipedia.org/wiki/Voltage As there is no path to move the charge (circuit is open) the amount of work needed to move it is, I guess, infinite. So voltage would also be infinite. \$\endgroup\$
    – TemeV
    Commented Jan 12, 2019 at 10:10
  • \$\begingroup\$ @TemeV Interesting point. Can an isolated conductor have an instantaneous voltage relative to an arbitrary ground reference? I guess the conductor can acquire a charge, and that charge would need work to move it to the arbitrary ground. Hence it has a voltage. \$\endgroup\$
    – Chu
    Commented Jan 12, 2019 at 10:25
  • \$\begingroup\$ If there is a path for the charge, then yes there is a voltage. This is of course always the case in reality, because perfect insulators don't exist. There is always some parasitic impedance where the charge can travel given enough work. But this topic is about ideal components. \$\endgroup\$
    – TemeV
    Commented Jan 12, 2019 at 10:31
  • \$\begingroup\$ @TemeV Yes, ideal components but not necessarily ideal environment. A random ideal conductor will acquire noise signals. \$\endgroup\$
    – Chu
    Commented Jan 12, 2019 at 10:46
  • 1
    \$\begingroup\$ There is no point on assuming ideal components if the environment is not also assumed ideal. \$\endgroup\$
    – TemeV
    Commented Jan 12, 2019 at 12:35

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