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Full wave rectification

  1. Mar 10, 2012 #1

    D44

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    Hi guys

    I'm working on a few questions regarding the full wave rectifier circuit similar to the attached image.

    The AC voltage is 12v at 110Hz, the ripple voltage is 1v peak to peak and the load current is 24mA. I am to assume each diode has a forward voltage drop of 1v when conducting.

    I have to find:

    1) The length of time the capacitor is supplying current to the load for if the diode conduction angle is 20 degrees.

    2) Value of capacitor required to produce the 1v peak to peak ripple voltage.

    3) Average power dissipated in the load.


    For question 1, I found the period = 1/(2*110) = 1/220 = 4.55ms, then dt = 4.55-(4.55/6) = 3.79ms. Where (4.55/6) was the period of the capacitor charging.

    For question 2, I then used I=c(dv/dt) and transposed for c = I(dt/dv). Therefore c = 24x10-3*(3.79x10-3/1) = 90.96 micro Farads.

    For question 3, I used P = I*(Vac - Vdrop) = 24x10-3*(12-2) = 0.24W. This doesn't seem right to me.

    I'd appreciate your help as I'm not confident that any of them are correct.

    Thanks
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     

    Attached Files:

  2. jcsd
  3. Mar 10, 2012 #2

    rude man

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    Is the 12V rms , amplitude, pk-pk or what?
    Is the 24mA the average load current?
    Define "conduction angle". Is it 20 deg of 180 deg or 360 deg? (NOTE: 360 deg would be 1/110 Hz which is the period of the excitation voltage. Be real careful with this.
     
  4. Mar 10, 2012 #3

    NascentOxygen

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    Method looks right, though I haven't checked your arithmetic.
    Method looks right, though I haven't checked your arithmetic.
    Obviously not right. You will get a much clearer idea of what you are dealing with if you sketch the graph of the time-varying voltage across the load resistor. Hint: the voltage across the load is identical with the voltage across the capacitor.
     
  5. Mar 11, 2012 #4

    D44

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    Hi

    Thanks for the replies.

    I think I'm to take the 12V as RMS and the current is the current through the load (average I guess??). I also think it's 30 degrees of 180.

    So for the power, its the dt of the I = c(dv/dt) I'm trying to find? In which case this would be dv = dt*(I/c)...

    v = 3.79x10^-3*(24x10^-3/90.96x10^-6) = 1v.......hmmm...This is just the ripple voltage then?

    I'm not exactly sure how to draw the graph and find what I need from this.
     
  6. Mar 11, 2012 #5

    D44

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    Something like this?
     

    Attached Files:

  7. Mar 11, 2012 #6

    NascentOxygen

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    That's probably the waveform. Once you know the voltage waveform, you can estimate the power.

    Remind me, how many volts are lost across your rectifier?

    Wasn't it 20° at the start of this question?
     
  8. Mar 11, 2012 #7

    D44

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    Sorry that must have been a typo. It's 30 degrees.

    There is a voltage drop of 1v across each diode, so 2v.
     
  9. Mar 11, 2012 #8

    D44

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    Any help on the power please?
     
  10. Mar 11, 2012 #9

    rude man

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    The input voltage is 12√2sin(wt) V, ω = 2π*110 Hz.

    So the voltage while 2 of the diodes are conducting is (12√2 - 2 V)sin(ωt) across the load, centered around the peak voltage = 12√2 - 2 V. You know that voltage appears 20/180 = 11.11% of the time. The rest of the time all diodes are off and the load is discharging the capacitor. So you can figure out where the sine voltage applies its voltage across the load, and when it doesn't. When it doesn't, the capacitor discharges per V(t) = V0e-t/RC and V0 is the voltage across the load at the point where the diodes start and stop conducting.

    So you have a waveform that looks like (12√2 - 2)sin(ωt) for 11.11% of the time, centerd around the peak voltage of 12√2 - 2 V, and
    V0e-t/RC the rest of the time. This is your V(t) over the entire half-cycle (1/220 s).

    You first need to come up with V0.

    You now have to integrate V2(t)/R over the entire half-cycle to obtain the energy dissipated in the load during that half-cycle, then divide by the half-period which as I said is 1/220 s. That gives you the power per half-cycle, and since all half-cycles put the same waveform across the load, that is your answer.

    There is another, more indirect way to go: since you know the average load current and the peak-to-peak voltage, there is one and only one combination of the sine peaks and exponential decay waveforms that satisfy those two criteria. You could obtain the complete V(t) that way also. You'd still need to integrate V2/R overt a half-cycle.
     
  11. Mar 11, 2012 #10

    D44

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    Vo sounds like the ripple voltage still though. I guess that's wrong though?
     
  12. Mar 11, 2012 #11

    rude man

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    It just occurred to me that you should make a simplifying assumption from what I said above, otherwise this problem really is a bear: assume the capacitor does not discharge substantially while the diodes aren't conducting. Another way to put this is to assume
    RC >> 1/220. In fact, most power supplies would be designed that way, to minimize ripple noise and also capacitor dissipation (real capacitors do have a dissipative component, especially big ones used for power supply filtering as in this case.)

    So V0 is a constant load voltage level, the level at which the diodes just begin to conduct and, under this assumption, also the level at which they just stopped conducting. So your ripple is just the sinusoidal peaks around 12√2 - 2 V to give an 11.11% duty cycle within each half-period.

    Your pk-pk ripple is then (12√2 - 2 - V0) volts, and it consists exclusively of the sinusoidal peaks.
     
  13. Mar 12, 2012 #12

    NascentOxygen

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    Vo is a large DC level with your ripple sitting on top. Output voltage and current have the same waveform since load is resistive. In a properly-designed power supply the ripple is usually so small in comparison with the average (the DC level) that in power calculations the ripple can be ignored. (Where it can't be ignored, the ripple component can be approximated by a sawtooth or triangular waveform.)

    In your circuit, you are looking at an output voltage that seesaws between two levels that are known to be 1 volt apart. What are those levels?
     
  14. Mar 12, 2012 #13

    rude man

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    I misread the questions you were asked. Sorry. You'll have to ignore my previous post altogether (#11).

    Question 1 is easy to answer, seeing as they gave you the conduction angle. I would double-check that the 20 degrees is 20/180, not 20/360, where 360 represents the period 1/110 s.

    At this point you know: T1, the time the diodes conduct;
    T2, the time they don't conduct;
    Vpeak, the peak voltage across the load;
    Vtrough, the lowest voltage across the load;
    the shape of the load voltage during conduction;
    the shape of the load voltage during non-conduction;
    and the average (dc) current.

    Then you can solve for C, the load capacitance that satisfies the above conditions.
    Then also you have the final complete waveform and can perform the necessary integration to find the average load power.

    I'll try to find time to solve this but will withhold the answer until I see what you or anyone else has come up with. This is not a trivial problem and I'm not sure it's been completely defined. I'm referring to the conduction angle, the input voltage and the 24 mA load current specifications in particular.
     
    Last edited: Mar 12, 2012
  15. Mar 12, 2012 #14

    D44

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    The two levels would be 11.5 and 12.5v, right?

    I haven't integrated this stuff before but I'll try and follow the advice and see what happens!
     
  16. Mar 12, 2012 #15

    NascentOxygen

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    No. Remember, the AC supply is given as 12v RMS.
    I doubt that any integration is intended in this exercise.
     
  17. Mar 12, 2012 #16

    D44

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    Ah ok. So then I need to multiply by the square root of 2. I also need to take away 2v for the diodes and then 0.5v for the average of the ripple?

    I think you're right about the integration by the way.
     
  18. Mar 12, 2012 #17

    NascentOxygen

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    That should give a very close result.
     
  19. Mar 12, 2012 #18

    rude man

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    this is close but the decay does not start at the peaks.
     
    Last edited: Mar 12, 2012
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