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In an series RC circuit why does a capacitor charge faster when there is a diode connected in parallel with the resistor?

enter image description here

I tested on an oscilloscope, using a square signal and this is what I got.

With diode: enter image description here

Without diode: enter image description here

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  • \$\begingroup\$ Is the diode forward or reverse biased? Could you provide a schematic or a picture of one? \$\endgroup\$ – vini_i Mar 12 '19 at 15:34
  • \$\begingroup\$ @vini_i i updated the question \$\endgroup\$ – Pedro Mar 12 '19 at 15:37
  • \$\begingroup\$ What happens when the voltage source tries to push more than 0.7V across the resistor? \$\endgroup\$ – TimWescott Mar 12 '19 at 15:41
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A diode is effectively a one-way electrical check valve. It's more complicated than that but it's the essence.

In one direction the valve opens and give a low resistance path for the current to travel. This effectively shunts current around the resistor and give rapid charging. In the other direction, the valve is closed and almost nothing flows. This forces all the current through the resistor.

Diodes are not perfect. To open a diode (forward bias) requires some amount of voltage, usually around 0.7v for a silicon diode. The diode also has some resistance other than zero. When closed (reverse bias) the diode does not perfectly stop the current, there is some slight leakage.

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When the diode is on, it provides a really low impedance path for the current to flow (along with a drop of about 0.7V for a silicon diode). This essentially bypasses any resistance you have in parallel with it.

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Additional to previous answers that in forward bias, the diode has about 0.7 voltage drop and very low impedance which ignores the parallel resistor, then according to capacitor charge, time=R*C, it will charge quickly because of low R (impedance of diode).

If you remove the diode the charge time will be long because here R is the resistor which is high.

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Your question does not apply if there is a DC source. With a DC source the charge (electron) flow is always in the same direction so a diode, if connected in the correct direction, will have no consequential effect on the time it takes the capacitor to reach maximum charge. Charge flow will always be into the capacitor until the capacitor reaches maximum charge. If the diode is in the wrong direction, not considering inconsequential leakage through the diode, the capacitor will not charge because the diode will then be blocking flow all the time.

If there is an AC source, and no diode, the capacitor will charge during the first half cycle then that charge will be pushed back to the source during the second half cycle (where the charge flows in the opposite direction). The capacitor will alternate from the maximum charge achieved during the first half cycle to practically zero charge at the end of the second half cycle. Adding a diode will allow the capacitor to reach maximum charge because the diode only allows charge to flow into the diode (for half the cycle).

Also, since the DC source is presumably at a constant voltage without reversing direction, the capacitor will charge faster because 1) the voltage source is always at a constant maximum value and 2) is not interrupted every half cycle as would happen with a diode in the circuit.

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  • \$\begingroup\$ Your answer is assuming the diode is in series. The question specifically asks about a diode in parallel with the resistor of an RC circuit. \$\endgroup\$ – Hearth Mar 12 '19 at 16:03
  • \$\begingroup\$ The resistor does not invalidate my answer. When flow is in the forward direction the diode is essentially a short circuit across the resistor and when in reverse direction the resistor, depending on its value, just increases the leakage flow through the diode allowing more charge to be pushed out of the capacitor, reducing the maximum charge the capacitor will get to. \$\endgroup\$ – Barryng Mar 12 '19 at 16:16

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