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Can someone explain the behaviour and the output(at B) for the following circuit? What I think is that, C will charge and at one point stop conducting. So at B it will be HIGH output at that instant and till it charges. Now after it is fully charged, it will discharge throught R, so the output should decrease with a slope to 0,and this should repeat. I am not sure if this is right. Can someone please confirm that? I tried building the circuit, but the rise and fall happens only once. Can someone explain why would it not repeat itself?

schematic

simulate this circuit – Schematic created using CircuitLab

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No, as the voltage across the capacitor increases the voltage across the resistor will fall from 5V to 0V. At that point current stops flowing and the voltages remain constant...forever.

EDIT: For the capacitor to discharge you need a resistive path from one terminal of the capacitor to the other (without the 5V source in series). There's just no path to discharge the capacitor. I don't know how to explain it more simply.

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  • \$\begingroup\$ Is there a way to get this to repeat? (apart from using a Multivibrator). \$\endgroup\$ – Sherby Jan 24 '14 at 1:50
  • \$\begingroup\$ There are of course many ways to make it repeat. Some would say that all of them are multivibrators. \$\endgroup\$ – Joe Hass Jan 24 '14 at 1:52
  • \$\begingroup\$ It will never quite get to zero, but after about 10*R*C = 50msec it will be pretty darn close (+227uV) but still quite measurable. \$\endgroup\$ – Spehro Pefhany Jan 24 '14 at 3:12
  • \$\begingroup\$ @SpehroPefhany I didn't feel that throwing in an exponential transient response would help the OP understand the situation. Since electrons are countable, it could be debated that at some point the voltage would be so small that it could not move another electron and current flow would in fact stop. \$\endgroup\$ – Joe Hass Jan 24 '14 at 3:19
  • \$\begingroup\$ I think we could use the Johnson noise to predict when it would no longer be measurable (as a function of temperature). ;-) Something about a mathematician and an engineer joke in there. \$\endgroup\$ – Spehro Pefhany Jan 24 '14 at 3:23
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You have a fixed supply, a fixed capacitor, a fixed resistor, and no feedback. The circuit will find a steady state (in this case the capacitor will be fully charged) and change no more.

If you want it to repeat then you need to have both feedback and a control mechanism. Your circuit has neither.

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What you have is a high pass filter, basically allow high frequency signal to pass and block low frequency ones

Power supplies are DC source, ie. 0 frequency, so they would be blocked

The brief 'high' you get is the transient of the circuit and depends on the resistance and capacitance the components

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