Simple Diode Problem

Discussion in 'Homework Help' started by toofatt4u, Dec 7, 2010.

1. toofatt4u Thread Starter New Member

Nov 9, 2010
2
0

The circuit looks like the link at the bottom except there is source of 120 V rms between the pulse signal and the ground signal.

Vrms= 120V
f=60 Hz
R= 10 KΩ
C=600x10^-6

Find Vdc, Idc, and Vripple. Explain how to solve this please. Im not sure of the formulas. The answers should be: 169.47, 16.947, .471 respectively.

Thank you!

The circuit looks like the link at the bottom except there is source of 120 V rms between the pulse signal and the ground signal.

Vrms= 120V
f=60 Hz
R= 10 KΩ
C=600x10^-6

Find Vdc, Idc, and Vripple. Explain how to solve this please. Im not sure of the formulas. The answers should be: 169.47, 16.947, .471 respectively.

Thank you!

2. blah2222 Well-Known Member

May 3, 2010
552
33
Hi, okay so before solving this, or memorizing some formulas, you should probably know what is going on in this circuit first.

The pulse voltage is only actively seen at $V_{dc}$ during it's positive half-cycle, due to the diode. The capacitor gets charged-up to the peak voltage of the pulse voltage. This capacitor voltage is in parallel with the resistor and in this case, it is your $V_{dc}$. As soon as the diode "kicks in" and disallows the negative-cycle of the pulse to move through the circuit, current $I_{dc}$ flows from $V_{dc}$ through the resistor to ground. Also, since the capacitor is charged-up and nothing is charging it, it discharges exponentially through the resistor as well. $V_{ripple}$ is the difference in voltage between when the capacitor is fully charged, and the capacitor voltage at the end of the cycle.

Some terminology:

$f = 60 Hz$

$T = \frac{f}{60} = \frac{1}{60} s$

$V_{rms} = 120 V$

$V_{peak} = 120\sqrt{2} V =$ 169.705 V

$tau = RC = (10000)(600)(10^{-6}) =$ 6 s (time constant for cap discharge)

$V_{ripple} = V_{peak}exp(\frac{0}{tau}) - V_{peak}exp(\frac{T}{tau}) = 169.705 - 169.235 =$ 0.471 V

If you look at the transposed waveform of $V_{dc}$, you see that it looks like a rounded sawtooth wave that goes between $V_{peak}$ and $[V_{peak} - V_{ripple}]$. $V_{dc}$ is the average value between those two voltage levels.

$V_{dc} = V_{peak} - \frac{V_{ripple}}{2} = 169.705 - \frac{0.471}{2} =$ 169.47 V

$I_{dc} = \frac{V_{dc}}{R} = \frac{169.47}{10000} =$ 16.947 mA

Draw the waveform of the voltage around the resistor and ground to get a better visual understanding of this.

Last edited: Dec 7, 2010
3. toofatt4u Thread Starter New Member

Nov 9, 2010
2
0
You are my hero. Thank you!