Problem Statement: A simple rectifier circuit consists of a diode and a capacito

Question

Problem Statement: A simple rectifier circuit consists of a diode and a capacitor as shown above: The circuit's operation can be simulated by the circuit on the right.The switch in this circuit is closed momentarily over a very short period of time every 1/60 second. The short interval is long enough to charge the capacitor voltage V0(t) to V(s). The purpose of the capacitor is to "hold" the voltage V0(t) over the time period while the switch is open. A 10ohm resistor simulates the circuit  that draws power from the capacitor.

-Find the minimum value of the capacitor in Farads,such that V0(t) does not drop below 90% of V(s) over the time period of 1/60 sec. IF the capacitor is too small the 60Hz "ripple" on the V0(t) can be high enough to cause a problem.</strong></p>

charging time constant is very short). The purpose of the capacitor is to time period 1/60 second while the switch is open. A 1o ohm resistor simultaneously amplifier circuit) that draws power from the capacitor. Find the minimum such that Vo(t) does not drop below 90% of Vs over the time period of 1/60 small, the 60Hz ^'ripple^' on the Vo(t) can be high enough to cause problem.

Explanation / Answer

Equation for voltage across a capacitor in an RC arrangement, as a function of time, is V(t)= Vs*exp(-t/RC). We need: exp(-t/RC) = V(t)/Vs > 0.9 for the cap to hold 90% of the voltage at t secs. With t=1/60, R=10, we have exp(-1/600C) > 0.9, 1/600C < -log(0.9) to base e. 1/600C < 0.1054, or C > 0.0158 Farad.

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