Why is a Van de Graaff generator a current source? Why does it deliver a constant current regardless of the load?

I've seen people say that this is due to the fact the belt delivers the same amount of charges per a period of time, but why would the current flowing through the circuit necessarily be equal to the rate of charge deposition what prevents some of those charges accumulate on the metal globe?

  • 1
    \$\begingroup\$ Your title says constant voltage. The body of the question says constant current. \$\endgroup\$ Commented Apr 28 at 19:40
  • \$\begingroup\$ @MathKeepsMeBusy Sorry ,edited it. \$\endgroup\$
    – John greg
    Commented Apr 28 at 20:04
  • \$\begingroup\$ Think about steady-state not transient conditions. \$\endgroup\$ Commented Apr 28 at 20:07

1 Answer 1


If you keep building up more and more charge on the metal globe, then eventually the voltage will get so high, it will start arcing to anything nearby. Essentially, the globe is a capacitor, and you can't keep charging it forever.

Assuming that isn't happening, then in the long run, all the charges generated by the Van de Graaff generator must be going somewhere. And, by definition, a constant rate of charges flowing is a constant current.


simulate this circuit – Schematic created using CircuitLab

Consider this circuit diagram. The belt is delivering a constant current to the globe (the capacitor here).

With SW1 off, the charge on C1 will keep building up until something goes bang. Much of the time, this is what people want when demonstrating a Van der Graaf generator. Big sparks. The current in the sparks is definitely not constant. The globe keeps charging up until the voltage is high enough to arc to something.

But turn SW1 on, and the capacitor never gets to charge to any significant voltage. Instead, all the current goes through R1. And that current through R1 is near enough constant.

Keep turning SW1 on and off, and the current through R1 won't be constant again, as pulses of current will go through it.

But note that in each case, I1 is constant. It's the discharging of the capacitor that gives the pulses of current.

  • \$\begingroup\$ Okay thanks. But why doesn't it just build up charge until it arcs?Is that somewhat similar to why charges in normal electric circuits don't build up in a wire? \$\endgroup\$
    – John greg
    Commented Apr 29 at 0:04
  • \$\begingroup\$ @Johngreg See edit. You can't build up much charge in a random bit of wire because it's capacitance is too low. Electrons repel each other. Try to shove too many into a small space, and they start shoving back. In more technical terms, the voltage on a capacitor is given by V = Q/C, where C is the capacitance, and Q the amount of charge it's been charged with. If C is tiny, then only a small Q creates a large voltage, which opposes the voltage you're using to try to charge up the capacitor. \$\endgroup\$
    – Simon B
    Commented Apr 29 at 20:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.