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Zener diode question

Discussion in 'General Electronics Chat' started by potoole64, Sep 24, 2008.

  1. potoole64

    potoole64 New Member

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    I'm using a voltage divider to drop the voltage that I sample from the ignition of my motorcycle. The divider is set to drop the ignition volts from 12v to 4v (1/3 the original). The divider consists of a 220k-ohm R in series with a 110k-ohm R, from 12v ignition source to ground. Output is taken across the 110k R, which should result in 4v. There is also a 5.1v zener diode across the output, which is supposed to limit any extraneous voltage spikes. However, my output volts is always around 3.15v, even at a steady 12v dc applied across the divider. I've measured the resistors and they all check out, even the applied volts checks out. I've varied the sizes of the resistances and it doesn't change much. When I took the zener out, the output went to the proper voltage.

    I read, recently, that a zener won't operate properly if its current is restricted to less than 3 or 5 ma. Do you think that's what is happening in this case? I'm concerned that if I take the Zener out, extraneous voltage spikes might damage my circuitry down the line, one circuit being a Basic Stamp microcomp. that needs to be restricted to inputs of 5v or less


    Thanks
    PO'T
     

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  2. Pommie

    Pommie Well-Known Member Most Helpful Member

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    It is probably the leakage current of the zener that is causing the problem. A 200k resistor is rather large for a voltage divider, I would reduce the value of the resistors by a factor of 10 or more depending on the voltage source.

    Mike.
     
  3. MrAl

    MrAl Well-Known Member Most Helpful Member

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    Hi,

    I have to agree...try a 20k and 10k resistor instead, and if that doesnt work
    try a 2k and 1k resistor. Just in case that doesnt work either, try a 1k and
    a 470 ohm resistor.

    You could also test your zener by applying a voltage through a resistor
    and see what the voltage across the zener is. You could also post
    the part number so someone here could look it up.
     
    Last edited: Sep 24, 2008
  4. dave

    Dave New Member

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  5. rjvh

    rjvh New Member

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    A zenerdiode needs to have a certain curent to work properly as written earlyer the choice of the resistor values are to high so it prohibit the current to reach a range that the zener can do its job

    lower the values to the order of 1k and 470ohm as Mr Al wrote and you will have a active zener diode

    Robert-Jan
     
  6. CarlosFandango

    CarlosFandango New Member

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    Did this suggestion work?

    Hi, this is my first post to this forum. I signed up because I have a zener diode problem almost exactly the same as described here. Basically, I am measuring the voltage of a 12V (motorcycle) system and using a micropower i2c A2D converter to do so, but I want to ensure that the input to the converter never exceeds 3.3V. In the attached circuit (the usual one for a zener) I originally used a 10K res for R1 and a 2.7K res for R2 - the intention being that I could measure up to about 17V before the zener clamped the incoming voltage to 3.3V.

    However this did not work as expected, and in fact I can increase the input voltage to 30V (in an isolated replication of this circuit, NOT attached to the A2D!) and still not reach 3.3V at the zener anode... it only ever gets to about 2.4V. Graphing the output data shows a distinctly non-linear response.

    So, in an attempt to get more current flowing in the circuit which I perceived to be the problem, I exchanged R1 for a 1K and R2 for a 220R. I'm getting different but neverless broadly similar results, in so far as the voltage still does not reach 3.3V at the anode!

    It's quite a while since I used zeners, but I don't recall having this problem before, so I am a little confused by these results. By the way I have tried different zeners in case there was a problem with that, no change was observed.

    Any ideas anyone? It's driving me nuts!

    - CF
     

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  7. audioguru

    audioguru Well-Known Member Most Helpful Member

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    Nobody says the part number of their zener diode!
    They are very different.
    A 1N4689 is 5.1V at 50uA. Its current is 10uA max at 3V.
    A 1N4684 is 3.3V at 50uA. Its current is 10uA max at 1.5V.

    A 1N4733 is 5.1V at 49mA. Its current is 10uA at only 1V.
    A 1N4728 is 3.3V at 76ma. Its current is 10uA at only 1V.

    Other zener diodes are spec'd at 5mA and 20mA.
     
  8. colin mac

    colin mac New Member

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    The maximum current at the output of the Zener is the current through R1.
     
  9. MrAl

    MrAl Well-Known Member Most Helpful Member

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    Hi Carlos,


    You need to increase the value of the 220 ohm resistor, say to 470
    or maybe even 680 ohms or something like that. The problem is
    that with a 220 resistor on bottom and 1k on top the resistive
    divider only puts out a little over 2v, which is not enough to
    reach the zener voltage of 3.3 volts. When you want a circuit
    like this to work you have to make sure the divider voltage is
    higher than the zener voltage, at least.
     
  10. audioguru

    audioguru Well-Known Member Most Helpful Member

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    With a 30V input, the 10k resistor is providing a current of 2.76mA to the 2.7k resitor and the zener diode. The 2.7k resistor has 2.4V across it so its current is only 0.89mA. Therefore the zener diode has a current of only 1.97ma. A 3.3v zener diode measured at 5ma might work but a 1N4728 that is 3.3V at 76mA might be conducting 1.97mA at 2.4V.
     
  11. Willbe

    Willbe New Member

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    Looking to the right of the zener, the Thevenin equivalent is a 4v source in series with a 73kΩ resistor. To drop (4-3.2) = 0.8v in a resistor this large takes a current of 0.8/73k = 11µA.
    I guess the Zener V-I curve from the manu. will show this current through the diode comes out to about 3v and the slope of the V-I curve @ 3v & 11µA should show a resistance of 290kΩ, if they spec the diode at this low of a current.
     
    Last edited: Sep 26, 2008
  12. CarlosFandango

    CarlosFandango New Member

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    I thought Vout = Vin x (R2 / (R1 + R2))? That would mean an input voltage of approximately 18.5V would produce a voltage of 3.3V at the middle of the divider. Unless I'm wrong about that as well.
     
  13. CarlosFandango

    CarlosFandango New Member

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    You're absolutely right, and this information should of course be provided. I have tried several types from those that I have here, I think they are ZVNs. Honestly, I only identified them by reading the voltage from the side of the case... The actual one in the circuit I'm trying to get to work is a surface mount type, I tried to cross reference the catalogue no. to retrieve the actual part number as a decent reply to your post, but the RS site is down for maintenance...??!
     
  14. CarlosFandango

    CarlosFandango New Member

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    I've obviously got a deep misunderstanding of the characteristics of these devices so I'm struggling to get my head around it. The device in question (I managed to find the name after all contrary to an earlier post) is a MMSZ5226BT1G. Characteristics are nominal voltage Vz = 3.3, Izt = 20mA, Zzt @Izt = 28Ω, leakage current Ir at 1V = 25µA.

    What I was hoping for, and indeed expecting, was a strictly linear response from the resistor divider from 0V up to the zener voltage, and then a limiting of any further voltage rise at the circuit output. This is the ideal anyway. Is there any way of achieving such an ideal?

    [This is why I tend to stick to digital systems - there are analogue considerations even then, but generally they are not this bothersome!]

    -CF
     
  15. MrAl

    MrAl Well-Known Member Most Helpful Member

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    HI Carlos,

    Well then you better fix your schematic because it shows a 12v input :)

    Also, you can not apply 18.5v either to that circuit and expect it to work.
    The reason again being that right around 18.5v or just below it the current
    to the zener is too low. Thus, you need to either raise the 220 ohm or
    decrease the 1k.

    If you need something that tracks from 0 to 18.5v and then clamps when
    the input equals 18.5v (3.3v output) then i think you would be better off
    with an ACTIVE clamp not a PASSIVE clamp like the zener is.
    You might look into the LM431 ic and use it as a shunt regulator,
    or another ic that works as a shunt regulator. The idea is to make
    the shunt regulator kick in at 3.3v (and do it sharply) and thus limit the
    voltage.
    Another idea is to use an op amp and clamp the output using a zener,
    which will work much better.

    If you dont need that much accuracy, perhaps you can get away with
    lowering the impedance of the whole network. This will make the
    zener appear like a sharper device.

    Another idea is to use a higher voltage zener (5v or so) and use a
    divider AFTER the zener with a single resistor to the zener.
    Many of the higher voltage zeners have a sharper knee so this might
    work good enough. I would bet a 12v zener works much better, with
    a divider AFTER the zener. You could make the network impedance
    lower that way too.
     
  16. CarlosFandango

    CarlosFandango New Member

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    Thanks for the reply and the very helpful ideas proposed. I'm going to think about all of this for a while, as well as my own line of thought detailed below.

    And yes you're right :) I should have been a bit clearer on input voltage, the input ranges between 0 and some maximum value (if it were 18.5V then the vehicle's electrical system will fry anyway, this is just for theoretical purposes!).

    OK, I'm starting to understand this whole thing a lot better now. What I'm observing, I think, is the zener starting to conduct - break down - as the current builds in the circuit due to higher input voltages. I've been doing some math, and some practical experimentation. If I were to design (time will tell if I am capable of this or not!) a circuit that reaches 3.3V for some maximum permissable input voltage - whatever it's decided that should be - then using this simple circuit across the expected voltage range would require fairly high power resistors. Active circuits such as those you suggest are far better, but increase component count, overall cost, and require more board space.

    However, my 8 bit A2D converter for a range of 0-18V (say) would have a resolution of 0.07V per bit. This is stupid in my application, as 0.5V would probably be perfectly adequate. So, by this logic, I only actually NEED a 36 bit range to measure 0-18V. 36 bits is 14% of the total range, so the output voltage of my measuring circuit only needs to be 14% of 3.3V or just under a half a volt. In practice a slightly better resolution to reduce the effect of single bit errors is desirable, but, you know, in a 'perfect' world...!

    All I am really worried about you see is that in a vehicle, which is a dirty environment electrically, the input of my A2D will get popped. That's really the only reason for the zener of course. What I believe I need to do is make sure that in the event of spikes reaching 50-60-100V (I don't know what they might be, by the way) there will be enough current for the zener to START conducting and prevent such catastrophies. By the time that range is reached, the zener would be interfering with 'correct' measurements, i.e., by introducing a non-linearity to the measurement system, but that wouldn't matter.

    I'm not sure if this logic is really correct, or whether I'm just setting myself up for another fall :-( I guess I'm just looking for a bit of confirmation that this line of thought makes any sense, at all!

    -CF
     
  17. audioguru

    audioguru Well-Known Member Most Helpful Member

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    A 3.3V zener diode is a very poor voltage clamp. It starts conducting when its voltage is less than 1V. At 2V then it conducts a lot. It is a poor voltage regulator. Its voltage and current change when the temperature changes.
    Yours is 3.3V when it has a current of 20mA. Maybe it is 2V when its current is a couple of mA.
     
  18. Willbe

    Willbe New Member

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    I think you can more closely approximate an ideal Zener with an LMXXX voltage reference and a generic diode connected to it so that the composite starts to conduct at 0.5v above the ref. V.
    But, the LM reference needs to be able to sink current, just like a real Zener.
     
  19. CarlosFandango

    CarlosFandango New Member

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    Yep. I get it now I think - your earlier reply really helped me with this problem. I think I have it sorted, I changed R1 for a 3.9K and R2 for a 330R - this means effectively that the output from the resistor divider is almost totally as expected (and linear) up to the point where there is enough current in the circuit to cause my darned zener to start conducting. This is well outside the range that I intend to measure, but still protects the input from high voltage spikes. Well, I believe it should - I suppose this can only be confirmed when this unit is installed on the vehicle, and DOESN'T explode... ;-) There's nothing as real as the real world.

    So thanks again to everyone who helped me (and of course the originator of the thread in the first place!) with this, I hope this thread is of use to anyone having problems with their zeners in the future. Brilliant.
     
  20. Willbe

    Willbe New Member

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    No diode responds instantly. While you're at it, I guess you should try to get a spec. on the response time of Zeners while clipping transients.
    Automotive spikes can't have very fast risetimes, but who knows?

    And, any transient suppressor should have an energy absorbing spec., the amount of joules it can take and still work.
     
  21. Diver300

    Diver300 Well-Known Member

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    I have 3 possible suggestions.

    1) Leave out the zener. The A2D converter probably has clamping diode so that the current that you can get through a 220k resistor won't damage anything.

    2) Use 3 LEDs in series instead of the zener. They actually cut off sharper than zeners.

    3) Use a 15 V zener, with a 10 resistor in series, then a voltage divider after that to get down to 5 V.
     

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