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Full Wave Rectifier - Capacitor Filter and Output Ripple

  1. Oct 3, 2015 #1
    1. The problem statement, all variables and given/known data
    MjhXiti.png
    I'm having trouble understanding part C of this question.
    2. Relevant equations
    ΔVo = Vout/2fRC

    3. The attempt at a solution
    So to get ~12V at the output, the turns ratio should be 14 because 120√2/12 = 14.14.

    Using a turns ratio of 14, the output voltage will be 12.12 V.

    To find the the value of the resistance, I can use P = V2rms / R, where P = 24 W and Vrms = 12.12/√2.

    So R = 3 Ω.

    But then how do I find ΔVo so I can find the capacitance?
     
  2. jcsd
  3. Oct 4, 2015 #2

    rude man

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    OK
    Not OK. The output voltage is 12VDC, the power is 24W, what is R? Not 3Ω.
    You are allowed a 3% droop from the 12VDC every time the ac input drops off its + or - peak.
     
  4. Oct 4, 2015 #3
    Oh okay so then with my turns ratio, ΔVo = 12.12*0.03 = 0.3636 V
    Hm I'm a bit confused, how else would I calculate the power absorbed by the resistor here? Everything else we've done so far has it as either P = V2rms/R or P = I2rmsR. Or because it's DC, should it just be 12.122/R without converting to rms?
     
  5. Oct 4, 2015 #4

    rude man

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    right
    Yes, and you'd better understand why fully. If you had a 12V car battery running a load dissipating 24W, what would the load resistance have to be?
     
  6. Oct 4, 2015 #5
    Using the equation it'd be 6 Ω.

    But is it different for half-wave rectifiers then? I'm looking at this example in my book and it's using RMS to calculate the power (part b):
    4FhsKtF.png
    But the output of a HWR is DC too isn't it?
     
  7. Oct 4, 2015 #6

    rude man

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    right.
    The hwr is a half-sine wave WITHOUT A CAPACITIVE FILTER. Your full-wave rectifier circuit INCLUDES a CAPACITIVE FILTER. The latter is practically like the output of a car battery; the former is a half-sine wave. If your fwr circuit didn't include the capacitor, the rms voltage would be double the half-wave unfiltered output.

    You badly need to understand what rms means fundamentally.
    In the case of the car battery, 12V is the rms voltage.
    What is rms voltage? It's the voltage, when squared and divided by resistance, gives power.
    For a 115V home voltage source the rms is 115V. For a 12V car battery it's 12V.
    In general, the rms voltage is √∫V2(t)dt/T over any integer number of periods T. In the case of a DC voltage there is no period so you can integrate over any time interval T you like. A trivial case.
    The idea is you get the same amount of power dissipation for any periodic or DC voltage if you use the rms value of that voltage.
     
  8. Oct 4, 2015 #7
    Right, I understand that.
    Okay
    I'm sorry but I'm still having trouble understanding. I know what rms means but it's confusing me in this instance.

    In a full-wave rectifier without cap filter, the output voltage would be the fully rectified input sinusoid, and the rms would be double the hwr, like you said.
    How does that change with a capacitor? It's not a fully rectified sinusoid anymore, but the fwr is still periodic with the period being from when one diode pair is on to when both diode pairs are off (and the capacitor starts discharging), right? Why wouldn't I take the rms of this for power calculations? You mentioned above that 12V IS the rms voltage, but how/why? Do I just assume it's rms when they tell me what the output voltage should be like in this question?
     
  9. Oct 4, 2015 #8

    rude man

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    OK, good question. And in fact you're right. The output of your filtered (capacitor added to satisfy the 3% error criterion) is not a pure 12VDC voltage. It's a periodic voltage with sudden jumps of 0.36V (when the diodes conduct) and exponential decays back down 0.36V until the next diode conduction. So it's a voltage with just under 12V average with peak-to-peak "ripple" of 0.36V. If you perform my rms formula you will get close to 12V as the rms voltage - you should try it. Notice that the capacitor size matters a lot, which is why the problem asks you to come up with a value big enough to meet the 3% droop requirement. As you lower the capacitance from this value you will get more and more decay before the next diode conduction event, until when C=0 there is just the fwr rectified sine-wave.
     
  10. Oct 4, 2015 #9
    Ohh okay I think I'm starting to get it now. Ideally, a FWR with cap filter would output a pure DC voltage. That's why (depending on the capacitor value) I can use the DC power formulas (P = IV, P = V²/R), even though there's a small voltage ripple. But with a FWR without a cap filter, I need to use AC power formulas (P = V²rms / R, P = I²rms*R) because the output is sinusoidal. Is that correct?
     
  11. Oct 4, 2015 #10

    rude man

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    Yes, but keep in mind what i said: rms voltage always gives power = Vrms2/R, no matter what the shape of the voltage waveform. In fact, that's the definition of rms voltage: to give the same power (averaged over 1 or more cycles) irrespective of the waveform shape.
    The general formula for Vrms for any periodic waveform is the integral I stated previously. You should try to understand why that integral always gives the correct power averaged over 1 or more cycles.
     
  12. Oct 4, 2015 #11

    gneill

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    Pardon my jumping in here, but has anyone considered that a practical design should take into account the voltage drops that occur on the diodes? There are two basic rectification arrangements, one using a bridge rectifier (two diode drops) and the other the classic full wave rectifier using a center-tapped transformer (one diode drop).
     
  13. Oct 4, 2015 #12

    rude man

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    You could say that, and you could also point out that transformer resistances are not zero, core is nonlinear, etc. but for purposes of this exercise I think assuming ideal components is the way to go. It's a very introductory course obviously.
     
  14. Oct 4, 2015 #13

    gneill

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    That's fine, so long as the idealization is acknowledged in the submitted answer.
     
  15. Oct 4, 2015 #14

    rude man

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    Yes, but a circuit was not described in the post so I'd assume the default is ideal diodes. Otherwise the answer would depend on the circuit configuration.
     
  16. Oct 4, 2015 #15
    Yeah we're assuming ideal components. The non-idealities come into play for the lab portion of this class.
     
  17. Oct 5, 2015 #16

    CWatters

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    The output must remain within +/-3% of 12V and that includes the ripple.

    Q = CV
    I = Cdv/dt
    dv/dt = I/C

    You know the allowed dv, estimate dt and you know I.

    3% is actually quite a tight spec.
     
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