Need Help in Designing a Full Wave Bridge Rectifier

Discussion in 'Homework Help' started by MarsG, Sep 12, 2015.

  1. MarsG

    Thread Starter New Member

    Sep 12, 2015
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    Im new to circuits, so bear with me lol.
    My homework assignment is to design a Full Wave Bridge Rectifier (Attached the circuit I built in pspice and the full description of the assignment below.) The rectifier is to have a filter Capacitor across the Load.
    Given:
    Input = 120v-rms, 60Hz
    Load = 155 Ohms
    Voltage Across individual diodes = .7V (1.4 together)


    The Avg Outpt Voltage is to be 15v with a ripple of +- 1V peak

    I have to determine the Outpt Voltage, the peak to peak ripple voltage and the Avg/Max diode currents.

    120V-rms = 170v pk (Vrms= Vpk/sqrt2)

    with the voltage drop from the diodes (1.4v)
    (120-1.4)sqrt2=Vpk=168Vpk

    The ripple frequency is double the source
    60Hz----> 120Hz

    Vripple = Iload/fC

    The problem im having is that I do not know the value of the capacitor to use nor the Load current, and im not sure how to incorporate the transformer. (was told to use 1Henry for the L1 Value not sure about L2)
     
  2. shteii01

    AAC Fanatic!

    Feb 19, 2010
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  3. WBahn

    Moderator

    Mar 31, 2012
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    By L1 and L2, I'm assuming you are talking about the input and output sides of the transformer. You can probably assume that the ratio will be the same as the square of the turns ratio. That will be the primary way of setting the average voltage. The capacitor will set the ripple voltage.

    If the average voltage across the load is to be 15 V and the load resistance is 155 Ω, what more do you need in order to at least get a good estimate of the average load current?
     
  4. MarsG

    Thread Starter New Member

    Sep 12, 2015
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    My primary concern is the transformer, we haven't really covered transformers yet so im not sure on how to determine the turns ratio or step down voltage. Dint really make my way down to the load because of the fact im stuck at the transformer but yea I could use ohms law on the load.
     
  5. WBahn

    Moderator

    Mar 31, 2012
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    When you head out on a trip somewhere, do you just start driving in some random direction and because you haven't made your way down to figuring out where you are trying to end up?

    When you design a circuit you start from what you want to achieve. What you want to achieve is a particular voltage specification at the load, so you start your design at the load. From that you figure out what the voltage is you need at the input to bridge circuit, which is also the output of the transformer. If you know what you need at the output of the transformer and you know what you have at the input of the transformer, figuring out the turns ratio should be fairly straightforward.
     
    Johann and MarsG like this.
  6. MarsG

    Thread Starter New Member

    Sep 12, 2015
    24
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    I understand now, because I have 170V(120V-rms) going into the transformer and I need about 17V(12V-rms) to compensate for the diode voltage drops, the ratio would then be 170:17 and L1 being 1 Henry would make L2 0.1 Henry.
    Thank you for the explanation and your time.
     
  7. WBahn

    Moderator

    Mar 31, 2012
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    Remember, the inductance ratio is the square of the turns ratio.
     
  8. MarsG

    Thread Starter New Member

    Sep 12, 2015
    24
    1
    Thanks, went back and corrected lol.
     
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