capacitor question

Discussion in 'The Projects Forum' started by dbenney, Jul 18, 2014.

  1. dbenney

    Thread Starter New Member

    Jul 17, 2014
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    I am in the process of building an ac to dc variable voltage power supply. I am running the 120V 60Hz ac through a stepdown transformer which will lower the voltage to 24V. I am then going through my rectifier circuit, 4 diodes arranged for full wave rectification. here is my question:

    I know that ac voltage displayed on a voltmeter is RMS voltage and components need to be sized for FULL voltage which would be peak to peak voltage...this would be 1.414*24*2. This would give me just under 70V. Now this is on the AC part of the circuit. After the rectifier circuit I will be adding a couple smoothing capacitors. The capacitors will be on the DC side of the rectifiers so can I use capacitors rated at anything over 24V because DC voltage has no peak to peak and is a constant voltage?
     
  2. paulktreg

    Distinguished Member

    Jun 2, 2008
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    If you are full wave rectifying 24VAC you will get approx. 1.414 x 24 = 34VDC.

    Your capacitors need a rating, factoring in a little safety, of at least 50VDC.
     
  3. dbenney

    Thread Starter New Member

    Jul 17, 2014
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    Wouldn't that be the peak voltage? Peak = RMS*1.414....I thought components needed to b sized for peak to peak which is 2*peak
     
  4. paulktreg

    Distinguished Member

    Jun 2, 2008
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    Yes you are partly right. If Vrms=24V then Vpk=34V and Vpk2pk=68V. The full wave bridge rectifier "flips" the negative side of the AC waveform positive so you get a series of peaks that are 34VDC but at 120Hz not 60Hz. You now need a capacitor to smooth out the peaks so 50V will do and the bigger the capacitor the less ripple you will have on your DC. Hope that makes sense?
     
  5. MaxHeadRoom

    Expert

    Jul 18, 2013
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    With a single phase bridge the output will be a 120Hz 100% ripple, Pulsating DC, the peak of which will be 1.414 x AC in.
    As soon as you add capacitor the DC value will be the peak, the x 1.414 value.
    With zero load the DC will then posses zero % ripple, the % ripple will depend on the degree of load and the capacitor values.
    How are you going to vary the voltage?
    Max.
     
  6. dbenney

    Thread Starter New Member

    Jul 17, 2014
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    AH...that makes perfect sense...I forgot about this. THANK YOU.

    I am planning to use a 10Kohm Pot.
     
  7. MaxHeadRoom

    Expert

    Jul 18, 2013
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    Not with the output from the slider surely?? :confused:
    Max.
     
  8. dbenney

    Thread Starter New Member

    Jul 17, 2014
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    slider? sorry for the ignorance. I have just begun an electrical apprenticeship and im in a digital electronics class and I figured the more circuits I can build and understand how they work the better it will make me.
     
  9. RichardO

    Well-Known Member

    May 4, 2013
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    You may be confusing the filter capacitor voltage rating with the rectifier diode voltage rating. The rectifier diodes do see 2 times the peak voltage. This is because the filter charges up to the positive peak voltage of the AC input. When the AC input goes negative the diodes see the negative peak of the AC plus the positive peak stored on the capacitor. I hope I explained this in a way that helps.
     
  10. crutschow

    Expert

    Mar 14, 2008
    13,018
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    A 10kΩ pot has an equivalent output impedance at the wiper that can vary from 5kΩ at mid-point to 0Ω at the ends as you adjust the voltage. Thus you cannot draw much current at the mid-voltage output values without significant voltage drop due to this resistance. That's why an active series voltage regulator, such as an LM317 is often used to buffer the output of the pot and allow output currents of an amp or more with little voltage drop.
     
  11. dbenney

    Thread Starter New Member

    Jul 17, 2014
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    sorry, i forgot to mention the LM317... I already have that and am connecting the pot to the middle leg of the 317.
     
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