Some context for why I am asking the question:

I have application in which I am to charge a supercapacitor from a voltage source which is equal to the the cap's rated voltage. As such I considered using a diode in series with the voltage source in order to realise a reduction in capacitor voltage. This had me thinking about how such a real circuit would work, and how its ideal counterpart would work.

The ideal problem

In the following schematic please assume all circuit components to exhibit ideal behaviour:

Charging an ideal capacitor through a diode

Assuming the voltage source starts at 0V, and is then set to 3.3V, the capacitor will begin charging. The capacitor voltage will be equal to the source voltage less the diode forward voltage drop, Vfwd. Vfwd decreases with current, a behaviour that is described by the Shockley diode equation: enter image description here


  • I is the diode current,
  • I_S is the reverse bias saturation current (or scale current)
  • VD is the voltage across the diode
  • VT is the thermal voltage kT/q (Boltzmann constant times temperature divided by electron charge)
  • n is the ideality factor, also known as the quality factor or sometimes emission coefficient.

This equation suggests that given any voltage across the diode, some current should flow. If left forever, the voltage on the ideal capacitor should therefore approach equilibrium with the voltage source.

So before I move on to the real problem - in the ideal circuit above, what would the voltage on the capacitor, Vc, tend to if left forever?

  • 1
    \$\begingroup\$ Point of order: people seem to have different definitions of "ideal diode", not all of which imply Shockley behaviour. \$\endgroup\$
    – pjc50
    Sep 12, 2017 at 9:02
  • \$\begingroup\$ Very true; there are a few ideal diode models; that's why OP explicitly defined what ideal diode model he is using: Sheckley's diode model. \$\endgroup\$
    – Curd
    Sep 12, 2017 at 9:53

2 Answers 2


The Shockley diode equation yields non-zero current for non-zero voltage, therefore the capacitor will charge up as long as there is a difference between capacitor and source voltage (\$V_C < V_1\$), i.e. \$V_C\$ will get arbitrarily close to \$V_1\$ if you just wait long enough.

  • \$\begingroup\$ Thank you. Extending it to the real case, leakage current through the capacitor would induce a voltage drop on the diode. Practically, would this set the the limit on the capacitor voltage? \$\endgroup\$
    – Gerry
    Sep 12, 2017 at 8:53
  • \$\begingroup\$ In practice a diode will often leak much more than a good capacitor. Only when you would use a (electrolytic) capacitor there is a chance that it will leak more than a diode. \$\endgroup\$ Sep 12, 2017 at 8:59
  • \$\begingroup\$ @Gerry: Yes; leakage current of the capacitor means that some current going through the diode (causing voltage drop) is not used to charge the capacitor. Therefore it can not charge completely to \$V_1\$. \$\endgroup\$
    – Curd
    Sep 12, 2017 at 9:04
  • \$\begingroup\$ @Bimpelrekkie: as I understand the problem the diode in this case will always be biased in forward direction; i.e. there is no diode leakage. \$\endgroup\$
    – Curd
    Sep 12, 2017 at 9:06
  • \$\begingroup\$ As fas as I know a diode is not forward biased unless some current is flowing. Also there is always leakage in any mode (foward, reverse, zero-voltage). Only in forward mode the forward current is so much larger than the leakage current that the leakage current is no longer visible. \$\endgroup\$ Sep 12, 2017 at 9:12

The situation in which the circuit ends up after a long time, almost no current will flow.

The Shockley diode equation is valid for a situation where a current does flow.

The Shockley diode equation assumes a constant value of \$I_s\$ which is "good enough" for calculations when some current is flowing, but not when the current is smaller than the value of \$I_s\$.

So other effects come into play when the capacitor has been charged to a voltage close to the value of V1. Then leakage and drift in the diode will make it behave more or less like a high-value resistor so the voltage will drift towards V1.

  • 1
    \$\begingroup\$ When using Shockley equation it doesn't make sense to distinguish the cases "when current is flowing" and not. As said there is always current flowing (if there is voltage) even if current is smaller than \$I_s\$ (not only according to the Shockley model, also in a reality). It is still a much better diode model than an on/off-diode-model. Why do you say Shockley equation cannot be used when \$I<I_s\$?. That's not true in general. \$\endgroup\$
    – Curd
    Sep 12, 2017 at 9:50

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