Full wave rectifier - diode voltage and current

Discussion in 'General Electronics Chat' started by Dritech, Nov 8, 2015.

  1. Dritech

    Thread Starter Well-Known Member

    Sep 21, 2011
    756
    5
    Hi all,
    For a circuit as the one shown below, how can I calculate the parameters of the diodes and the load?
    Does anyone know of any tutorial which include calculations for determining there ratings?
    Thanks in advance.

    [​IMG]
     
  2. Papabravo

    Expert

    Feb 24, 2006
    10,144
    1,791
    Are you looking to calculate Absolute Maximum Ratings, or are you more interested in typical ratings?
    I'm not sure which ratings you think are calculated as opposed to measured.
    In any case you want to have an adjustable AC source and an adjustable load. Then you want to have two more meters connected as ammeters.
    Now you can construct a set of characteristic curves describing the voltage and current as you vary either the AC source, or the load. It is tedious, but instructive work.
    For example:
    1. Set the RMS AC source voltage at 30VAC
    2. Set the load at 60Ω
    3. Measure the RMS AC current
    4. Measure the Peak AC voltage across the load
    5. Measure the Peak AC current across the load
    6. Change the the load
    7. Go back to step 3, n times
    Plot the data.

    From this data you should be able to tell the forward voltage drop of the diodes at various current levels.
    With a scope you might be able to see the reverse recovery time.
    One time only you might be able to find the maximum reverse blocking voltage
    You get the idea.
     
  3. Dritech

    Thread Starter Well-Known Member

    Sep 21, 2011
    756
    5
    Thanks for the reply. If the Load is 10ohms, what will be the current on the resistor? When using an ammeter in simulation, I am getting 338mA.
     
  4. Dodgydave

    Distinguished Member

    Jun 22, 2012
    4,986
    745
    10 ohms with 338mA is 3.38 V, + 1.4V for the bridge rectifier drop, gives 4.78 vAC.
     
  5. wayneh

    Expert

    Sep 9, 2010
    12,126
    3,048
    Dodgydave has probably given you all you need. I'll just add that a diode is not ideal - its properties change with current level and temperature, which are in turn related to each other. So if you need precision, things get more complicated. Also, the 4.78V AC noted by Dave is the RMS voltage. The peak-to-peak voltage is higher. You would see this if you were to add a filter capacitor to the DC side of the rectifier.
     
  6. Papabravo

    Expert

    Feb 24, 2006
    10,144
    1,791
    Adding the capacitor will make any DC measurements easier to make, but may impede your ability to characterize the diodes in the bridge rectifier. I was thinking that you wanted a complete set of IV curves for the diodes, rather than just the solution to a particular setup.
     
  7. Dritech

    Thread Starter Well-Known Member

    Sep 21, 2011
    756
    5
    Thanks for the replies. I had a mistake in the simulation, that's why I was getting 338mA when the load was 10ohms and the input supply was 240VAC RMS. Now I am getting 33A. 33A*10ohms = 330V + 2.2V of the diode. This is far higher than the 240VAC of the main supply. What can be wrong?
    Also, for a diode that can handle 33A, what parameter should I look for?
    (Note: this is only for simulation and it will be not implemented)
     
  8. Papabravo

    Expert

    Feb 24, 2006
    10,144
    1,791
    240 VAC(rms) * \sqrt{2} = 339.36 VAC(pk)
     
    Dritech likes this.
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