AC Voltage Rectification Formula

Discussion in 'General Electronics Chat' started by leetch02, May 17, 2005.

  1. leetch02

    Thread Starter New Member

    Mar 11, 2005
    3
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    Hello, I am an electrotech student stuck on an equation which I am unsure of. We are presently on the Power Supply Filtering, rectification topic, converting AC to a DC using 4 diodes. We also learn't that by adding a capacitor, we are able to hold the peak AC value with minimal lost, this is called the ripple effect. However, in our textbook we are given this formula Vrpp=(T*Vdc)/(Rl*C). Our teacher told us to use the peak value Vm so the equation becomes Vrpp=(T*Vm)/(Rl*C). What is the correct formula for Voltage ripple peak to peak, Vrpp?
     
  2. pebe

    AAC Fanatic!

    Oct 11, 2004
    628
    3
    I don’t know that there is an accurate formula for ripple. It depends on for how long the supply charges the capacitor and for how long it discharges. The rectifier diodes conduct when the supply AC voltage exceeds the voltage across the capacitor (plus the diode voltage drop) after the discharge period. The capacitor voltage at the end of the discharge period depends on the current drawn by the load. The diodes cease to conduct when the AC voltage is no longer high enough to charge the capacitor. So the diode ‘on’ time is a variable. I usually assume the duty cycle to be 10% ‘on’ and 90% ‘off’.

    Knowing the mains frequency and knowing whether the power supply uses full or half wave rectification, determines the period of the ripple waveform. Applying the duty cycle gives the capacitor discharge time.

    I tend not to remember all the various formulae I have encountered in the past, but try to work from first principles instead. I remember that the voltage across a 1Farad capacitor will drop by 1 volt when a current of 1ampere is drawn from it for a period of 1second. Or voltage drop = Discharge time x Current / Capacity.

    As an example, 50Hz mains full wave with 10% on/off gives a discharge time of 9mS. Assume a 1000uF capacitor with a load current of 200mA. The resulting waveform is a saw tooth.

    The Peak-peak ripple = 0.009secs x 0.2A / 1mF = 1.8volts.

    I hope that helps.
     
  3. mozikluv

    AAC Fanatic!

    Jan 22, 2004
    1,437
    1
    hi

    C = Vp / (2 * mains freq. 50hz or 60hz * load resistance * desired ripple voltage)

    Vr = Vm / (2 * 50hz or 60hz * Rl * C)

    hope that helps
     
  4. zhi_yi

    New Member

    May 15, 2005
    9
    0
    please correct me if there is any mistakes

    according to ohm law, V is always = I.R, Vripple is small disturbance at the peak when the capacitor try to filtering the rectified signal. and it's because of charging and discharging of the capacitor, so the Vripple = Idc.Xc whereas Xc is the resistance of a capacitor and it was 1/wC, so the Vripple = Idc/wC, but we use Idc/fC without 2.phi. and then, if we bring Vripple into calculation, the Vdc will be Vdc = Vm - (Vripple / 2). Idc = Vm/RL
     
  5. Tekker

    Active Member

    Apr 22, 2005
    33
    0
    Here is the formula I was taught in electronics class....

    Vrpp = I / (f * C)

    I is the load (resistor) current
    f is the ripple frequency
    C is the capacitance

    -tkr
     
  6. pebe

    AAC Fanatic!

    Oct 11, 2004
    628
    3
    That's for half wave rectification.
     
  7. Tekker

    Active Member

    Apr 22, 2005
    33
    0
    Ah, yep sorry about that... I had to skim my electronics book since it's been a while, and I found a section on bridge rectifiers and didn't realize they went back to half wave when they were showing that formula! lol

    Vrpp = I / (2f * C)

    That should be better. :D

    -tkr
     
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