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Rectifier Circuit

  1. Feb 17, 2010 #1
    1. The problem statement, all variables and given/known data

    Design a full-wave rectifier that converts 110 [V], 100[Hz] ac electricity from a regular wall socket to a low DC voltage.

    a) (Fill in the blanks) To lower the voltage from 110 [V] to 11[V], we use a _________, with a turn ration of ___:___. The units of the horizontal axis (time) should be in ______ .

    2. Relevant equations

    I assume V1/V2 = N1/N2

    3. The attempt at a solution

    To lower the voltage from 110 [V] to 11[V], we use a transformer with a turn ratio of 10:1. The units of the horizontal axis (time) should be in ______ .

    1. Is the turn ratio calculated from the transformer equation V1/V2 = N1/N2?
    2. How does one determine the unit of time?
     
  2. jcsd
  3. Feb 17, 2010 #2

    vk6kro

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    You just about have it.
    N1 / N2 is the turns ratio, so it is just the same as the ratio of the two voltages.

    Now, 110 volts is the RMS value of the supply voltage, so what is the peak value?
    Peak voltage = 1.414 * RMS voltage.


    If you wanted to produce 12 volts DC after filtering, what RMS voltage from the transformer would produce this? Peak voltage = 1.414 * RMS voltage.

    (The 12 volts has to be the peak value of the transformer output minus two diode drops from the bridge rectifier. This is the value the filter capacitor will charge to. So the peak value of the output from the transformer is 12 + 0.6 + 0.6.)

    If you have a 100 Hz sinewave, how long does one cycle take to complete? This should tell you the answer to the time scale question.
     
  4. Feb 18, 2010 #3
    Thanks for the help.

    I would just need to use the period T, right? Which would be 1/100. So the time scale would be in centiseconds?
     
  5. Feb 18, 2010 #4

    vk6kro

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    You could, but milliseconds would look more elegant. You would get 5 steps per half cycle.

    Try it and see if you agree.
     
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