simulate this circuit – Schematic created using CircuitLab

How does C1 hold Vr1 when V2 is off(0-3v). I'm confused since C1 is referenced to 4V from V1. Shouldn't there be a negative charge on the "R1 Side" of C1?

  • \$\begingroup\$ The phrasing of your question is a bit confusing. Additionally, we need to know a few more things - is V1 initially charged? Has V2 been off while C1 was charging? You need to describe the situation a bit more thoroughly. \$\endgroup\$ – Billy Kalfus Nov 8 '17 at 4:55
  • \$\begingroup\$ Sorry, I'm annoyed I'm so confused by this. Lets say this is basically steady state. V1 is DC, so its been on for a while (seconds). V2 is high for a while (seconds). then V2 turns off. How does C1 hold Vr1 high? \$\endgroup\$ – BAO Nov 8 '17 at 5:01
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    \$\begingroup\$ You are asking a question about the behavior of a circuit. In a circuit like this which contains time-dependent components, you need to describe the conditions the circuit is currently in. \$\endgroup\$ – Billy Kalfus Nov 8 '17 at 5:04
  • \$\begingroup\$ Sorry I edited the above, accidentally hit enter \$\endgroup\$ – BAO Nov 8 '17 at 5:06

Let's suppose the forward voltage of the diode is 0.3V. While the capacitor is charging, the voltage at the point on the bottom of C1 (let's call this node Vc-) is held at 2.7V, and so the capacitor charges to (4-2.7) = 1.3V. When V2 shuts off, the voltage at Vc- is still 2.7V but V2 = 0, so the diode is in reverse bias and no current flows. Thus, the capacitor starts charging up to 4V and will eventually reach it (technically it's asymptotic but you get the idea) so long as V2 stays off.

This graph can better display the behavior of the circuit (simulated in LTSpice). You can see that when V2 shuts off, Vc- begins dropping because the capacitor is charging to 4V. However, if V2 were to turn back on, the node Vc- is asserted to 2.7V and the capacitor adjusts accordingly.

  • \$\begingroup\$ Ok I think I understand it now, when v2 turns off it causes current to flow out of C1 and across R1 producing a voltage. I think the current going into / out of the C1 helps. imgur.com/a/fGRNE \$\endgroup\$ – BAO Nov 8 '17 at 6:51

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