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Homework Help: Design of a rectifier

  1. Nov 17, 2011 #1
    1. The problem statement, all variables and given/known data

    A classical rectifier is to be designed to provide 5V +/- 1% for a logic circuit, with output power of up to 12W. The input is conventional 220V, 50Hz source.
    i) Choose a transformer ratio
    ii) Choose a diode configuration
    iii) Calculate the capacitor value to meet the requirements
    iv) Draw the load voltage, the load current and the source current (secondary side) waveforms for maximum output power (12W).

    2. Relevant equations

    3. The attempt at a solution

    I am a complete beginner in this. I have no clue how to solve. Can you please give me a hint?
  2. jcsd
  3. Nov 17, 2011 #2


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    Staff: Mentor

    It's a meaty topic. Start by a google search on: rectifier "filter capacitor"
  4. Nov 19, 2011 #3
    I need to use a full wave rectifier bridge with a Capacitor parallel to my load. Is this correct?

    What I don't understand is, do I put a normal transformer before the bridge?
  5. Nov 19, 2011 #4
    You are correct. Use a full wave bridge arrangement and you need a transformer to step down 220V to 5V
  6. Nov 19, 2011 #5
    Would it also be possible to use just a Capacitor Filter, this seems to be the easiest way to solve the problem.
    Like shown here: http://www.tpub.com/neets/book7/0249.GIF

    What is the advantage of the bridge?
  7. Nov 19, 2011 #6
    A single diode allows 0ne half of the AC current to flow. Imagine an AC wave which goes + and -. A single diode will cut off the - sections (or the +sections!!). This gives the familiar 'half wave rectification waveform'... can you picture it?
    The full wave combination of diodes needs 4 diodes connected in a 'bridge' and this allows all of the AC current to flow in the same direction....it is smoother.
  8. Nov 19, 2011 #7
    i) transformation ratio

    In the end I want a DC Voltage of 5V. Using a full-wave bridge this means
    V_{dc}=\frac{2 V_{2,max}}{π}
    V_{2,max} = \frac{π V_{dc}}{2} = 7.85V

    Since I need a transformer that is giving this peak voltage
    Transformer ratio =
    \frac{V_{1,max}}{V_{2,max}} = \frac{V_{1}*\sqrt{2}}{V_{2,max}} = \frac{220V*\sqrt{2}}{7.85V} = 39.6

    Can you tell me if this is correct?
  9. Nov 19, 2011 #8
  10. Nov 19, 2011 #9
    The output voltage is specified as 5V +/-1%
    This means that the max change in voltage allowed is 0.05V
    The capacitor stores electric charge so that current can be maintained between the peaks of the voltage, this is when the voltage is decreasing.
    The power that needs to be supplied is 12W .
    Can you calculate thew current which must be supplied?
    The time between peaks is 0.01s (half a cycle)
    Can you calculate the charge that flows in this 0.01s?
    This charge comes from the capacitor. If you know the voltage change is 0.05V and also know the charge that must be provided, can you calculate the capacitor value.
    Get as far as you can and if you get stuck come back!!!
  11. Nov 19, 2011 #10
    I put my numbers in my procedure and got C = 0.5F
    That formula you have gave 0.48F
    Looks like you we on the right track. The formula works and I hope that my steps show how the formula comes about !!!
  12. Nov 19, 2011 #11
    Thank you very much for your quick reply.

    So you mean:
    I,load = P / V,dc = 12W / 5V = 2.4 A
    Q,c = I,load / (0.5*T) = 2.4 A * 0.01s
    C = Q/U = 0.024C / 0.05V = 0.48F

    Is this the correct way?
    Is it coincidence that is the same result as with my first approach?
  13. Nov 19, 2011 #12
    What about the transformer ratio, was this thought correct?
    Thank you very much!
  14. Nov 19, 2011 #13
    Yes, you have done more than I would have!
    I would just say ratio = Vin/Vout = 220/5 = 44 to 1
    One thing to realise that these calculations, in practice, only need to be 'near enough'.
    To produce a supply of exactly 5V +/-1% would probably involve using a further electronic circuit.... but that is another matter.
    Also the capacitor value we calculated is a minimum value anything bigger than 0.5F would do a better job.
    Well done
  15. Nov 19, 2011 #14
    also...your post #11 is spot on. I think it is better to work through the physics to get the answer, once you can do this then it is OK to use the formula.
    well done
  16. Nov 19, 2011 #15


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    Staff: Mentor

    Don't forget to account for the diode voltage drops! The 5V +/-1% specification puts a pretty tight constraint on the allowed peak voltage (Vp), which will follow the peak voltage supplied by the bridge rectifier.


    Attached Files:

    • Fig1.jpg
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    Last edited: Nov 19, 2011
  17. Nov 19, 2011 #16
    To calculate C I did this:
    charge Q = I x t where t is the time for 1/2 cycle (full wave rectification) =1/2f
    Q = I/2f
    C = Q/V (V is the ripple voltage)
    C = 1/2fV

    You equation is V = 1/2fC

    Both methods are exactly the same
  18. Nov 20, 2011 #17


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    Staff: Mentor

    I'm guessing that you averaged the full wave rectified waveform to determine this DC value? That analysis only applies where there is no filter capacitor, so is not applicable to your circuit. In your circuit, the big filter capacitor charges to the peak value of the AC and holds this voltage until the next peak. So the DC output will be roughly equal to the peak of the AC (less diode drops).

    You calculated 0.5F. That's 500,000 microfarads so is a very large capacitance.
  19. Nov 20, 2011 #18
    I know 0.5F is huge.
    But can you tell me how to calculate it correctly? I thought that in this case the transformer ratio question can be solved independent from the question of the capacitor.
  20. Nov 20, 2011 #19


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    As a 'mathematical' question, it's fair to say there is no need to examine the spec of the transformer in detail. However, transformers will output different voltages according to their state of load, so you need to make sure 12W would have no influence on the level of saturation in the transformer, else I expect it would run high if the 12W load was removed.

    (Suffice to say, in the real world you'd pick a, say, 10V transformer, use a small cap that might allow, say, 3V of ripple, and stick a voltage regulator in. So your values for capacitance will look artificially high, due to the artificial nature of the question.)
    Last edited: Nov 20, 2011
  21. Nov 20, 2011 #20


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    Staff: Mentor

    Yes, transformer and filter capacitor are determined separately.

    The method you used looked right (though I didn't check the figures). Q=CV where Q is the charge drained by supplying the load current for half the AC period, and V is the allowable droop in voltage. It gives an adequate estimate.

    In a practical sense, electrolytic caps are not precision elements. Typically, they may have tolerances of +100%, -40% so you don't have to worry about 3 sig figs.
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