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Rectifier circuit: What am I doing wrong?

  1. May 12, 2012 #1
    Hi, Everyone.

    I'm doing an AC to DC converter circuit using rectifier, filter capacitor and zener shunt regulator, to provide an output DC voltage of 4.3V. I designed the circuit in several steps, which I have attached in a PDF file. Here's the final circuit:

    2j5hxzl.jpg

    When I do a Multisim simulation, the output voltage across the 100Ω load is 4.257V, which is a good thing anyway because the error% is small. However, when I assemble the circuit in reality on a breadboard, the output is 3.95-3.98V, which is really far less than the required output. The question is, where is the problem in the design?

    I hope you can check the design calculations in the attached PDF file, and I would be pleased to take your notes.

    Thanks.
     

    Attached Files:

    Last edited: May 12, 2012
  2. jcsd
  3. May 12, 2012 #2
    I think you should check your zener diode.
     
  4. May 12, 2012 #3
    I replaced it with another one, and it's the same. Nothing changed!
     
  5. May 12, 2012 #4

    jim hardy

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    Fairchild datasheet for the diode says it'll be 4.0 to 4.6 volts at IZ of 50 milliamps.

    You are pushing ~ 50 ma through your 200 ohm resistor
    and 4 volts will push 40 of those through your 100 ohm resistor
    leaving only 10 through the zener.
    So the zener is low on current.
    Sanity check: Zener is 13 ohms at Iz(50 ma) and 40 ma X 13 ohms leaves you 0.56 volts short. You measured half that. Not bad for a nonlinear device.

    I'd say your zener current is low.

    Datasheet: http://www.google.com/url?sa=t&rct=...lPv8RcTiFArglgwxw&sig2=AKLxGVRSy67DvFsh1hdAbg
     
  6. May 12, 2012 #5
    From your post, I understand that I need to assume a bigger value for IZ(Min) so that I can end up with a lower resistance value and a bigger capacitance value, hence more current will flow through the resistor, R. Assuming a bigger value for the minimum current through the zener diode, would guarantee that the zener will have enough current to breakdown somewhere between 4.0-4.6V. Is that correct?
     
  7. May 12, 2012 #6

    jim hardy

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    something like that.

    i didnt follow step 6 where you calculated 194 ohms then ~'d it to 300

    but i think the discrepancy is in using 13 ohms for Rz at 10 ma.
    Spec sheet says it's 13 ohms at 50 ma and 500 ohms at 1ma ,
    so at 10 ma it'll be somewhere in between.

    Remember that's a dynamic impedance, the slope of the I/V curve at any one point.

    1 watt at 4.3 volts would be 232 milliamps
    and they specified 13 ohms ar 50 ma which is 21% rated. Probably middle of its good regulation regime.
    Your 10 ma is only 4% rated, not far from the knee actually. It'll regulate better in 10-70% range.
    Semtech suggests IZmin in 5 to 10% range .
    http://www.google.com/url?sa=t&rct=...xr87q9wPS_qo-tfsw&sig2=T14YmtGaQfDf2jDXbEnoLQ

    Since you have it breadboarded why not just just briefly lift the 100 ohm resistor , forcing all 50 ma through the zener ? What's its voltage then?

    Read up on Zener Dynamic Impedance. Google gave me several references but none of them struck me as Belles Lettres .

    Anyhow bravo for actually building it. You learn so much more that way.

    Good Work !

    old jim
     
  8. May 14, 2012 #7

    sophiecentaur

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    Did you consider a three pin voltage regulator? The designers will have taken care of all those grisly details.
    Shunt zener regulators are a bit prone to getting hot and they 'waste' current if you need substantial power out of them.
     
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