Zener Voltage Regulator (1N53)

Discussion in 'General Electronics Chat' started by MadJ, Feb 18, 2011.

  1. MadJ

    Thread Starter New Member

    Feb 18, 2011
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    I have built a simple voltage regulator using the 1N53 zener diode ( 3.9v )
    The input source is 5v at 500mA, at the output i need 3.9v.
    But when i measure the voltage across the diode, it shows same as input voltage (5v).
    I tried changing the zener diode, but to no avail.
    What could be the issue ?
     
  2. JDT

    Well-Known Member

    Feb 12, 2009
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    Have you got a series resistor?
     
  3. MadJ

    Thread Starter New Member

    Feb 18, 2011
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    yup, a 2.2Ohms resistor in series
     
  4. JDT

    Well-Known Member

    Feb 12, 2009
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    How big is this zener diode?
    At no load, you are dissipating 1.95W with 500mA through the diode.
     
  5. MadJ

    Thread Starter New Member

    Feb 18, 2011
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    its a 5W zener
     
  6. JDT

    Well-Known Member

    Feb 12, 2009
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    Should be OK then. If you are getting 5V out, there is no current in the diode.

    Have to say, I looked up 1N53 and it wasn't a zener. Are you sure? have you got a data sheet?
     
  7. MadJ

    Thread Starter New Member

    Feb 18, 2011
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    Please find the datasheet attached
    the one i have is 1N5335B
     
  8. JDT

    Well-Known Member

    Feb 12, 2009
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    Well, according to the datasheet it has a nominal Vz of 3.9V at 320mA so it should work. The voltage you will get will be higher than 3.9V due to the effective zener resistance of 2Ω. But, yes, it should be doing something. Certainly should be conducting current and getting quite warm.

    You can't get a circuit much simpler than this so I can't explain why it's not working. Sorry.

    Unless the diode is open circuit. Does it conduct in the forward direction? (with about 0.6V drop - just like a normal diode?)
     
  9. MadJ

    Thread Starter New Member

    Feb 18, 2011
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    Yes, it does conduct in forward direction.

    Also, i tried replacing the diode with a 1W zener ( glass casing it seems ), this diode would warm up and at the output i get 3.9v
     
  10. Wendy

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  11. MadJ

    Thread Starter New Member

    Feb 18, 2011
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    I specifically asked for a zener diode.
    The text on the body of the diode says 1N53 35B
    Did you check the datasheet i have attached in my previous posts ?
     
  12. SgtWookie

    Expert

    Jul 17, 2007
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    Are you certain that you have the Zener connected correctly?
    Zeners are shunt regulators.
    The +5v output should connect to one side of the current limiting resistor, the other side of the resistor connects to the cathode of the Zener, and the anode of the Zener connects to the supply return (ground); the output voltage is taken across the Zener.

    Using a power Zener in this manner will result in terrible efficiency; most of the power in your circuit will be dissipated in the current limiting resistor and the Zener itself. Output regulation will probably not be very good.

    Why don't you tell us more about your load requirements? What are you trying to power with the 3.9v output; what current does it (your load) require?
     
  13. MadJ

    Thread Starter New Member

    Feb 18, 2011
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    Yup, i have connected the Zener just in the manner you described above (reverse biased).

    The load would be 4 IR Lasers, each running at 3.9v @ 60mA
     
  14. SgtWookie

    Expert

    Jul 17, 2007
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    You really need to control the IR lasers by current rather than by voltage, and they each need their own current limiters.

    Is your 500mA 5v supply well regulated?
    If so, you might use individual resistors for each IR laser.
    Rlimit >= (Vsupply - VfLaser) / DesiredCurrent
    Rlimit >= (5v-3.9v) / 60mA
    Rlimit >= 1.1v / 0.06A
    Rlimit >= 18.333... Ohms.
    That is not a standard value of resistance. 20 Ohms is the closest standard value.
    I=E/R, so 1.1v/20 Ohms = 55mA.

    Power dissipation in the resistors will be 1.1v*55mA = 60.5mW, double for reliability = 121mW; you can use 1/8W or higher rated resistors.

    Four IR lasers in parallel would then use 55mA x 4 = 220mA


    If the supply is not regulated, you'll need to use a different approach.
     
  15. MadJ

    Thread Starter New Member

    Feb 18, 2011
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    Ohhk.
    Yes, the 5v supply is through 7805, so it is regulated.
    And ill even go with just a 20Ω resistor in series.

    But i am just curious as to why the zener regulator is not producing a desired output !
     
  16. SgtWookie

    Expert

    Jul 17, 2007
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    OK. Do you have the required 0.33uF cap on the input to ground and 0.1uF on the output to ground, as close as possible to the regulator? These are not optional.

    What is the input voltage to the 7805?
    OK.
    Keep in mind that you need to verify the Vf of the lasers by measuring the voltage drop across the current limiting resistors. If you are getting more than 1.1v across the 20 Ohm resistors, then you risk early failure of the IR laser diodes.
    E=IR, so if the 7805 is giving exactly 5v output, then the most you should read across the resistors is 0.06A x 20 Ohms = 1.2v.

    It might be mis-marked, it could be a counterfeit part if you didn't buy it from an authorized distributor, you may have an error in your wiring - only three of several possibilities.
     
  17. MadJ

    Thread Starter New Member

    Feb 18, 2011
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    No, i don't have it. I did go through the Sticky Post on Bypass Caps. I will try adding it and let you know if that makes any change.

    6V - 10V ( its from a wallwart, which contains a multi tapped transformer, a bridge rectifier and a huge capacitor inside )


    oh Yes.


    Quite possible
     
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