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The current flowing through a capacitor equals C*dV/dt, I'm aware of that. What I don't understand is the physics of the process. Why does a capacitor pass pulsed DC (0-10V for example) when charge carriers don't change their direction?

Even if I use the "water analogy" it doesn't make sense. The flow moves in one direction, so the "diaphragm" will not be able to move back and forth.

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    \$\begingroup\$ Ah, but they do change direction. \$\endgroup\$
    – Majenko
    Apr 8, 2015 at 22:24
  • \$\begingroup\$ Capacitor 10V, leads 0V, what happens? \$\endgroup\$
    – PlasmaHH
    Apr 8, 2015 at 22:28
  • \$\begingroup\$ When the voltage drops from 10V doesn't the velocity of charge carriers decrease until there is no current flow at all (0V)? \$\endgroup\$
    – v.m.
    Apr 8, 2015 at 22:31
  • \$\begingroup\$ No, the direction of the charge carriers reverses until the two potentials equal out. \$\endgroup\$
    – Majenko
    Apr 8, 2015 at 22:37
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    \$\begingroup\$ As Majenko said, the charges do change direction. On the rising edge of the pulse dV/dt is positive and on the falling negative. \$\endgroup\$ Apr 8, 2015 at 23:31

6 Answers 6

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Zero volts doesn't mean zero current. Assume your circuit looks like this:

schematic

simulate this circuit – Schematic created using CircuitLab

When the switch turns on (connects to 10V), current flows to the right and charges the capacitor up to 10V. Once that happens, the current stops*. When the switch turns off (connects to ground/0V), current flows to the left and discharges the capacitor. (The capacitor acts like a voltage supply.) The current stops when the capacitor reaches 0V.

Short version: Pulsed DC is actually AC.

*The charge and discharge are actually exponential decays, so mathematically, the current never really stops. It approaches zero asymptotically at a rate determined by the resistance and capacitance.

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  • \$\begingroup\$ That makes sense if one pole of the switch is connected to ground, but what if it's 'floating'? \$\endgroup\$
    – v.m.
    Apr 8, 2015 at 23:05
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    \$\begingroup\$ @v.m. Then, in a perfect capacitor, absolutely nothing happens - it will hold the charge and will remain in the exact same state. In reality though there is no such thing as a perfect capacitor, and leakage current means that the plates will slowly discharge and equalize in potential. So when you next connect the power there will be a potential difference and current will flow again until that potential is equalized. \$\endgroup\$
    – Majenko
    Apr 8, 2015 at 23:13
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    \$\begingroup\$ All right, my mistake was thinking that charge carriers will reverse their direction only when the voltage changes polarity. \$\endgroup\$
    – v.m.
    Apr 8, 2015 at 23:22
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    \$\begingroup\$ Think of two buckets, linked together with a hose connected to the bottom of each bucket. Fill one with water and raise it up higher than the other - the water flows into the lower one. Lift the lower one up higher and it flows back into the first. Both are above ground, so at positive potential - it's the difference in the two potentials that is key though. \$\endgroup\$
    – Majenko
    Apr 8, 2015 at 23:25
  • \$\begingroup\$ In this case, the voltage is reversing polarity. Look at the voltage across the resistor. With the switch connected to +10V, the voltage across the resistor is 10V - Vc, a positive number. With the switch connected to ground, the voltage is 0V - Vc, a negative number. \$\endgroup\$
    – Adam Haun
    Apr 9, 2015 at 0:37
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The voltage across a capacitor cannot change instantly - it takes some time, determined by the capacitance, and resistances in the circuit.

If the pulses in your pulsed DC are sufficiently short relative to the circuit's time constant, the voltage across the capacitor will not have time to change significantly during the pulse (the capacitor will charge or discharge very little), so the voltage changes on the output side of the capacitor will closely follow the voltage changes on the input side. Therefore, it will appear that the DC pulses passed through the capacitor.

This effect is used in "coupling capacitors" in analog circuits, among other places.

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I think this becomes much clearer if you look at it in the frequency domain.

The impedance of a capacitor is $$\frac{1}{j\omega C}$$

So far so good: with a frequency of 0 Hz, the impedance goes to infinity (or even eleven)

But what kind of signal are you applying? A rectangular pulse:

http://en.wikipedia.org/wiki/File:Rectangular_function.svg

Image from: http://en.wikipedia.org/wiki/File:Rectangular_function.svg

The fourier transform of the pulse is this, with x being the frequency \$\omega\$: http://en.wikipedia.org/wiki/File:Sinc_function_%28normalized%29.svg

Image from: http://en.wikipedia.org/wiki/File:Sinc_function_%28normalized%29.svg

I think this makes it easy to see that there are components in the signal that have frequencies other than 0 Hz and that means that the impedance is not infinite, hence a current flows.

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Assume that you are sending repetitive rectangular pulses at frequency \$f\$ to the capacitor as seen below.

schematic

simulate this circuit – Schematic created using CircuitLab

You are not sending a single sinusoidal at frequency \$f\$. But sinusoidals at frequencies \$f\$, \$3f\$, \$5f\$, \$7f\$, \$9f\$, ...

If you guarantee that the sinusoidal with the frequency \$f\$ can pass the capacitor without distorting (i.e.; \$\frac{1}{2\pi f C}\ \!\! << \!\! R\$), the others will pass even easier. That's how DC pulses pass the capacitor in a correct circuit design.

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  • \$\begingroup\$ What happens if I add a diode in series with the capacitor? Does that mean that the falling edge of the square wave will no longer be present (due to charge carriers not being able to reverse direction in this case)? \$\endgroup\$
    – v.m.
    Apr 9, 2015 at 22:21
  • \$\begingroup\$ @v.m. If the initial voltage of the capacitor is \$V_0\$, amplitude of the pulse is \$V_s\$ and the pulse duration is \$T_p\$, then the new capacitor voltage will be \$V_1 = V_s - (V_0-V_s)e^{-\frac{T_p}{RC}}\$. The voltage you see on the resistor will be \$V_s-V_1\$ during the pulse duration. When the second pulse arrives, the capacitor voltage will be \$V_2 = V_s - (V_1-V_s)e^{-\frac{T_p}{RC}}\$, and the resistor voltage will be once again \$V_s-V_2\$. The capacitor voltage will stay constant during the off time of the pulses, the capacitor voltage will converge to \$V_s\$. \$\endgroup\$ Apr 10, 2015 at 5:58
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Your statement that "charge carriers don't change direction is not true. For you to see this more clearly, use a battery of 5v and connect a DC supply that provides + and - 5v. When the supply provides +5v, the total will be 10v and when the supply provides -5v the total will be 0v. From the point of view of the capacitor, it sees a signal changing from 10v to 0v to 10v... etc. but from the point of view of the 5v battery, it is providing 5v bias and an AC signal that goes from +5v to -5v

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The physics of this process is that the dielectric of the capacitor needs the energy of the pulse to orient the dipolar molecules that make up the dielectric, at the time that the dipoles are oriented, the dielectric is a short circuit, it is not an insulator yet because its dipoles have not been oriented, once they are oriented, the dielectric obtains its insulating character and the capacitor begins to work as such. A practical example of this property can be seen in the starting capacitors of single-phase electric motors. These have a running winding and a starting winding, a capacitor is connected in series to the starting winding to allow the power to pass through, only at the time of starting and after a while it blocks the passage of current, making the switch function.

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  • \$\begingroup\$ A capacitor is not a short circuit with zero resistance; it is an "open circuit" with infinite resistance. Both the resistor and the capacitor affect the current, but the resistor affects the current through resistance and the capacitor through voltage. The difference with a true open circuit is that this one has a constant voltage between the two terminals. So, when we change the voltage of one terminal, the other one follows it and this creates the illusion of a short circuit between the two terminals. Still nice answer. \$\endgroup\$ Oct 3, 2023 at 14:54

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