zener diode conducting before Vbr

Discussion in 'General Electronics Chat' started by jiyluv, Feb 3, 2011.

  1. jiyluv

    Thread Starter New Member

    Aug 22, 2010
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    Hey guys,

    My portion of the circuit has been trying to use a simple idea of using a full bridge rectifier and an AC voltage source to charge a capacitor, and to have a zener diode connected to it so that when the charged voltage of the capacitor reaches a certain value that I want, the zener will let the current through and therefore supply the output resistor. Refer to the screenshot I put as reference(values I used are different though).

    Now, in order to implement the zener diode set up, I connected two 1n5337B zener diode in series (Vbr of 4.7V). The Izt indicated in the datasheet showed that it carries a current of 260mA each, so I figured that I will need:

    (4.7*2 = 9.4V)/ (260mA*2)=18 ohm resistor as my resistor value (R5 in the diagram). I didn't have exactly 18 ohms available, so I substituted with two resistors in series that established 14.7 ohms and I changed the calculation accordingly to see the new voltage:

    14.7*260mA*2=7.644V.

    I supposed that after I hooked this circuit up, as I ramp up the amplitude of the voltage source, I shouldn't see voltage on the output Resistor(R5) until the AC source reaches 7.644V.

    however, when I hooked the circuit up, two issues came up:
    1. The zener diode was already conducting at 7.6V- it was giving about 200mV through the resistor. As I dialed down the voltage amplitude to see when it started conducting, it was at about 5.6V when it started to conduct.
    2. Also-as you can see in the oscilloscope readout, the capacitor and diode readouts have a big spike, in the same phase as the positive ramp of the voltage source. I don't understand- isn't the whole point of the bridge rectifier to use both the plus and minus side of the voltage source? Why would this be happening? I found out that substituting the charging capacitor with some huge value (like 1000 uF) made the spikes disappear, but the max voltage, capacitor voltage, and the zener diode voltage all went down in magnitude by about 1V or so. Besides, having a 1000uF cap might be too big(in terms of size)- i was hoping for nothing bigger than 100uF.

    I looked also at this post
    http://forum.allaboutcircuits.com/showthread.php?t=7209
    to look for answers, but I couldn't find the answer to my issue at the moment.

    Right now, one thing that comes to mind is that changing the resistor value might have altered the Vbr- however, should it change it that drastically?(from 9.4V to 5.6V)
    another thing that I noticed is that the diodes itself hold a Rz value(as indicated in the linked post), and it said in the datasheet that it was about 2 ohms each. However, would that have also changed the results that drastically? I actually substituted the resistor to a 20 ohm to reflect this (18 ohms + 2*2 ohms = 22 ohms- matched it as much as possible), and it raised the conducting voltage by about 1V (so 6.6V), but still not at the rated voltage value of 9.4V.

    Any suggestions or help will be appreciated, as always.

    Thanks for your help!
     
  2. Ron H

    AAC Fanatic!

    Apr 14, 2005
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    Why are you using 5 watt (i.e., high current) zeners? And what are you really trying to do?
    Is this homework?
     
  3. jiyluv

    Thread Starter New Member

    Aug 22, 2010
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    The part that I used, I used with no significant feature about current usage in mind. Are there zener diodes available that maintain Vbr with low current?

    My goal for using the zener is to set up a system that will trigger when a certain voltage value has been charged on the capacitor, and to have it do something when it reaches that voltage (turn on an LED when it does, for example). I am trying to make the circuit as low in power as possible.

    thanks!
     
  4. Ron H

    AAC Fanatic!

    Apr 14, 2005
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    What is the maximum peak voltage you expect at the input? Is it from a transformer?
     
  5. Adjuster

    Well-Known Member

    Dec 26, 2010
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    Low power zeners like BZX84 (among others) would attain their working voltage at currents of around a few mA.
    http://www.nxp.com/documents/data_sheet/BZX84_SERIES.pdf

    Some voltage reference devices called band-gap references are available in versions which can work at still lower currents. They can be very much more accurate, but generally are only available in a more restricted range of voltages. Here is an example from Linear Technology - again there are many other types.
    http://cds.linear.com/docs/Datasheet/1634fe.pdf
     
  6. jiyluv

    Thread Starter New Member

    Aug 22, 2010
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    Thanks for all the input. At the moment my target voltage is 10v but was using the 7.6 v was to learn more about how the diode works. I will also look into the different parts that u mentioned thanks.

    Meanwhile- can anybody help me to point me atthe reason of the behavior of my voltage curve mentioned in my beginning post?

    Btw my input is coming from a full bridge hookup from the ac voltage source as shown in the schematic.

    Thanks!
     
  7. Ron H

    AAC Fanatic!

    Apr 14, 2005
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    In order to duplicate your simulation-
    In your simulator, when you place a 15V sine wave, is that 15V RMS, or 15V peak?
    The 7.6V you are referring to makes no sense, because as far as I can tell, your assumptions that led to that number are totally wrong.:(
     
  8. thatoneguy

    AAC Fanatic!

    Feb 19, 2009
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    Some of the super cheap "lighter jack" voltage meters for cars are made with 3 sets of a zener diode, an LED, and a resistor. The idea is the more LEDs that light up, the higher the charge. When "Mid" is on, sometimes "high" will also glow dimly without being fully on though.

    Is this similar to what you are trying to acheive?
     
  9. Adjuster

    Well-Known Member

    Dec 26, 2010
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    I note that your "AC input" measurement comes from a single voltage probe. This is effectively ground referenced. Since the bridge negative output is returned to ground, you will not get a good representation of the input voltage in this way. As can be clearly seen, the result is not a sine wave.
     
  10. SgtWookie

    Expert

    Jul 17, 2007
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    You should always post an accurate schematic of your circuit(s); otherwise you may receive many comments that are misleading. Most people will ignore the majority of the text, and head straight for the schematic which tells the tale in a universal language.

    You went wrong there.
    Current in series is the same; it's equal through all components.
    Your voltage source is not higher than the voltage rating of your Zeners in series, so you have another problem there.
    You calculated R5 for Zeners in parallel, which is another problem.

    What exactly is it that you're trying to do?
    A Zener diode is a shunt regulator; it does pretty well if the current through it is pretty constant. However, your capacitor is only 47uF, which is quite small, and dissipating lots of power across Zeners fell out of vogue a long time ago.
     
  11. jiyluv

    Thread Starter New Member

    Aug 22, 2010
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    Hey guys thanks for your opinions.
    I just noticed Sgtwookie's response as I was putting out my reply so i am editing this post once more.

    @Ron h- the value I listed in my simulation diagram is not the actual value i used- the value i used is a 10v source amplitude sine wave(generating a 20v peak to peak wave.)
    If my calculations seem very inaccurate- can you point me in the right direction- maybe a way to get my calculation correct?

    @Thatoneguy- yes it is- just think of my situation as where the LEDs are already showing high when they should be only showing mid- and also where I see the LEDs go high to off in a flashing pattern, when all i want is the led to shine when it reaches the vbr voltage.

    @adjuster- yup that was what i thought. In the simulation program- putting a second probe in the negative terminal shows the negative side of the wave being rectified as well, in the simulation program (I don't know if I can do that in real life by putting two probes pointing opposite to each other because I would imagine that will short the output voltage out).

    But one way or the other- why would i see the same wavelike behavior in the Zener output shown on the left readout image? Disconnecting the Zener diode from the circuit and measuring gives the normal rectified behavior like the readout image shown on the right on the attached image.

    @sgtwookie- Here is what I am trying to do:

    1. I am trying to use an AC input voltage source, and charge up a capacitor of 100uF. In order to achieve this, I am implementing a full wave bridge rectifier.
    2. I am hoping to use the zener diode as a "gate" (sorry for lack of a better way to describe it) that will let current through and conduct when the voltage from the capacitor reaches my target voltage of 10V. Maybe hooking up an LED in series with the resistor to light up when the target voltage is reached sounds like a good plan as well.

    Here are some notes that I put down as I read your other feedback:

    - As with the schematic, I understand. I thought my explanation was enough, but I will definitely be clearer and consistent with my diagrams as well next time.

    - As for your comment with the zener diodes- I see what my problem was. So instead of adding up the current for the zener diode in series, I should have used just the IzT value for one? I combined the two for my calculations because I wanted the zener diode voltage to be as close to 10V as possible, but the zener diodes I have available to test it out are 4.7V and 3.3V.

    - If I set my voltage source at 10V amplitude, and my target zener diode voltage that I am trying to achieve is at 9.4V (since all I have available are 4.7V zener diodes), how would I set up my circuit? Can you guys just help me on how to get started? It might be better than you guys trying to decipher my approach.

    - Is there a more power efficient way to achieve what I am trying to do (to sense when a voltage reaches a certain value then let current through when it reaches it)?

    Hopefully my explanation helped to clear things up more- I have been struggling with this as well, and therefore I am trying to explain to the best that I can, but it might seem like gibberish at times. thanks for hanging with me here.

    thanks for your continued interest.
     
    Last edited: Feb 4, 2011
  12. Ron H

    AAC Fanatic!

    Apr 14, 2005
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    Are you aware that, if you are really trying to sense the output of a FW rectifier, driven by a low impedance source such as a power transformer, then you are wasting your time? The capacitor will be fully charged within a half cycle of the input.
     
  13. thatoneguy

    AAC Fanatic!

    Feb 19, 2009
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    What you want is called a "Window Comparator"

    If I understand the question, you only want the LED on between two voltage levels. At line frequencies, it will appear to be continually lit.
     
  14. Ron H

    AAC Fanatic!

    Apr 14, 2005
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    I may be wrong, but I think you have misunderstood.

    I think this is all he wants.
     
  15. Audioguru

    New Member

    Dec 20, 2007
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    A low voltage zener diode does not switch on and off. It still conducts a small current when the voltage across it is much lower than its rating.

    I use a 5.6V/25mA zener diode in series with an LED and a current-limiting resistor as a 9V indicator/battery-voltage LED. The LED is nice and bright when the battery is 9V and dims as the battery voltage runs down. When the battery is 6V then the 5.6V zener diode still conducts 1mA at about 3.5V which is enough to see the LED dimmed.
     
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