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Full wave Bridge rectifier and smoothing circuit

  1. Feb 21, 2016 #1
    L1. The problem statement, all variables and given/known data
    A Transformer with a stepdown ratio of 10:1 is supplied with a 230 RMS voltage at 50hz connected to a four diode full wave bridge rectifier and smoothing circuit. If the smoothing capacitor value is 1000uF and the load resistance RL is 25 ohms, find:

    the secondary winding rms and peak voltage,
    the peak voltage across the capacitor
    the peak to peak and rms ripple voltages
    the load dc voltage and current
    2. Relevant equations


    3. The attempt at a solution
    Ok the peak secondary winding voltage is 23V +1.4V(assuming 2 diodes at 0.7v each)=24.4V

    The secondary rms is 24.4/sqrt2= 17.253V

    Peak capacitor voltage VCP is secondary rms - 1.4V = 17.253-1.4= 15.853V

    Peak to Peak ripple voltage is VRPP= (1/(2fCRL))(VCP-(VRPP/2). Solving for VRPP, VRPP=0.4(15.853-VRPP/2),
    VRPP=5.28V

    RMS ripple voltage VR(RMS) is 5.28/(2*3sqrt)= 1.52V

    DC load voltage is VCP - (VRPP/2) = 15.853 - (5.28/2)= 13.213V

    Load Current IL is VL/RL= 13.213/25= 0.528A

    Can someone verify the methods I have used.
     
  2. jcsd
  3. Feb 21, 2016 #2

    Merlin3189

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    The following work looks ok, though I'm not familiar with these ripple approximations. (Perhaps you should quote the relevant equations to help people see what you are doing? I think you can find them in Hyperphys here.)
     
  4. Feb 21, 2016 #3
    The peak secondary voltage is the desired output VO + the diode voltage drops. i.e 23V + 1.4V. The question has not given me the values of the forward voltage drops across the diodes, Ive assumed them. I know that when 2 are conducting, the other pair are reversed biased or "off."

    The secondary RMS is the secondary peak voltage didvided by the square root of 2 I think.

    The secondary voltage has to be higher than one diode drop (or in this case, 2). Unless its the peak VC is secondary peak voltage - 2 diode drops.
     
  5. Feb 21, 2016 #4

    collinsmark

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    I have a couple of comments.

    1) Shouldn't you be considering the voltage drop of the diodes when considering the peak voltage of the rectifier's output, rather than its input? I'm unclear as to your reasoning why the rectifier will affect the voltage on the transformer's windings directly (assuming an ideal transformer).

    2) The diodes produce a voltage drop. "Drop" being the key word here. Shouldn't you be subtracting rather than adding?

    [Edit: I'm assuming the circuit topology is 230 RMS voltage source --> transformer --> rectifier --> smoothing capacitor & load resistor (the load resistor and smoothing capacitor are connected in parallel). Correct me if I am mistaken on this assumption.]
     
    Last edited: Feb 21, 2016
  6. Feb 21, 2016 #5

    collinsmark

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    Oh, and welcome to PF! :smile: :welcome:
     
  7. Feb 21, 2016 #6

    Merlin3189

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    I see where you're coming from now. This would be true, if you were designing a power supply to provide a desired output. But
    here you are given a power supply and asked to work out the output.
    You have to work forwards through the given components.
    You can't say the transformer must provide 1.4V more than the peak capacitor voltage, because you don't yet know what that is.
    You do know what the input of the transformer is, and you carry on from there.

    Edit: changed output to input in last sentence.
     
  8. Feb 22, 2016 #7
    Thanks for your responses guys, this is essentially the circuit if we assume the source input is actually a transformer with a 10:1 ratio at 230 rms at 50 Hz;
    diode23.gif

    I can understand where your coming from about adding the diodes to the secondary rms voltage; the equation I used is a design equation indeed.

    I did have original figures before I changed the equations as follows,

    The secondary rms voltage is 23V as the primary is 230V rms at a ratio of 10:1. The peak secondary voltage is of course 23*√2= 32.53V


    The peak capacitance voltage is always less than the applied peak voltage because of the voltage drops acrossed the diodes therefore Vc(peak) is peak secondary voltage - 1.4V(2VD) (the equation from my textbook) which gives me 31.13V. I'm pretty confident about this seeing as though 0.7V drop per diode is the theoretical norm.

    The peak to peak ripple voltage or VPPR equation is 1/(2fCRL)(Vc(peak)-(VPPR/2))=VPPR.
    If I plug in the values and solve for VPPR, I get 10.38V. And VR(rms), because of a sawtooth wave form, is VPPR/(2√3)= 2.99V

    DC load voltage is Vc(peak) - half of the peak to peak ripple voltage or 31.13-(10.38/2)= 25.94V and so the load current is simply ohms law, 25.94/25=1.03A
     
  9. Feb 24, 2016 #8

    NascentOxygen

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    Staff: Mentor

    Those values look good to me.
     
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