Zener Diode help

Discussion in 'Homework Help' started by prescott2006, Apr 1, 2011.

  1. prescott2006

    Thread Starter Active Member

    Nov 8, 2008
    72
    1
    From the figure below, the zener diode should tap the voltage at level of 360V. Unfortunately, it does not work out. The zener diode is having the breakdown voltage at 360V. Is there any problem occur to the circuit? Thank in advance.
     
  2. StayatHomeElectronics

    Well-Known Member

    Sep 25, 2008
    864
    40
    How much current is going through the zener diode? Current over the rating of the part will cause the voltage to go higher?

    You might want to look at something like a TVS diode.
     
    Last edited: Apr 1, 2011
  3. prescott2006

    Thread Starter Active Member

    Nov 8, 2008
    72
    1
    May I ask why the DC voltage at probe 1 is 381V? Don't it suppose to be 0.636*384=244V?
     
  4. StayatHomeElectronics

    Well-Known Member

    Sep 25, 2008
    864
    40
    You have a very large capacitor following the diode bridge. Both the capacitance and load resistance determine the ripple voltage after the diodes. In your situation, the capacitor has enough capacitance to smooth out the ripple almost entirely. Try the simulation with either a smaller capacitance or smaller load resistor (larger load current) and watch the ripple increase.
     
  5. Adjuster

    Well-Known Member

    Dec 26, 2010
    2,147
    300
    No, that would be the value for an averaging filter. You would get that value with the rectifiers feeding a simple resistor, or possibly with a choke-input smoothing filter under certain conditions, but with a capacitor filter like this you would normally expect to get closer to the peak value.

    The waveform into a resistive load does have an average value of about 0.637 times the peak, if we ignore the voltage dropped in the diodes. Looking at this waveform you would see a series of undulations going right down to zero twice every cycle, but the effect of adding a large smoothing capacitor is to fill in the gaps, so that the output never drops very far below the peak value.
     
  6. BillB3857

    Senior Member

    Feb 28, 2009
    2,400
    348
    Where is the current limit resistor for the zener?
     
  7. Ron H

    AAC Fanatic!

    Apr 14, 2005
    7,050
    656
    Yeah, that's the elephant in the room.
     
  8. prescott2006

    Thread Starter Active Member

    Nov 8, 2008
    72
    1
    The first circuit is wrong. I have attached modified circuit in post #3. I have few more questions.
    1.Why is the 1640uF cap can charge although no resistor before it?
    2.How can I increase the charging time?
    3.How to compute the charging time in this circuit?
     
  9. BillB3857

    Senior Member

    Feb 28, 2009
    2,400
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    T=RC where T is in seconds, R is in ohms and C is in Farads.


    Exactly what is the end result you are trying to achieve? 360 volts is pretty high to be experimenting with.
     
  10. prescott2006

    Thread Starter Active Member

    Nov 8, 2008
    72
    1
    The circuit is to hold a constant voltage at 200V. I bet this is the simplest way to do it? For my case, I have no resistor in series with the capacitor, so my charging time is 0?:confused:
     
  11. Ron H

    AAC Fanatic!

    Apr 14, 2005
    7,050
    656
    The capacitor will charge very rapidly, limited only by diode resistance and source impedance.
    Do you really need to drive a 300 ohm load? If so, what is it?
     
  12. BillB3857

    Senior Member

    Feb 28, 2009
    2,400
    348
    You are looking at 133 Watts at the load. Your best bet would be to use a transformer to reduce the voltage.
     
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