Simple Diode Problem

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

  1. toofatt4u

    Thread Starter New Member

    Nov 9, 2010
    2
    0
    Please help for my finals tomorrow

    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!

    Please help for my finals tomorrow

    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!

    http://www.google.com/imgres?imgurl...&ndsp=17&ved=1t:429,r:5,s:18&biw=1072&bih=530
     
  2. blah2222

    Well-Known Member

    May 3, 2010
    554
    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!
     
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