Need help picking proper Zener diode

Discussion in 'The Projects Forum' started by justtrying, Aug 9, 2011.

  1. justtrying

    Thread Starter Active Member

    Mar 9, 2011
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    Please see circuit attached...

    This tricks me because there is an overall voltage drop of 24V across two diodes. The ground is in the middle. The way this will work with the rest of the circuit though (my on-going PID project), is that either of the diodes can be removed to make the input unipolar (either positive or negative). So I am trying to figure out whether I should be looking at the 24V drop and both diodes should have a value of 27 or 30, or I should be looking at individual half-circuits and then the diodes would have a much lower value of 16 or 18? But would that not affect circuit when both both diodes are in or because the ground is in the middle, each diode should never "see" the 24V?

    Any input is, as always taken in with gratitude.
     
  2. SgtWookie

    Expert

    Jul 17, 2007
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    You can't put two in series with no current limiting resistor. It's very likely that either one or both will be destroyed due to high power dissipation, or the regulation will be terrible because the total voltage is lower than the Zener voltage.

    You don't say how much current the "virtual ground" might need to source or sink, or if you need to swing the "virtual ground" all the way to positive or all the way to negative.

    If you used two resistors of equal resistance across the 24v, you will wind up with the total voltage/2 at the junction of the resistors. Then if you shorted across one of the resistors, you would have the entire voltage across the opposite resistor.
     
  3. justtrying

    Thread Starter Active Member

    Mar 9, 2011
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    Yes, you are right. It is wonderful how everything works without resistors in multisim. Here is the correction. I guess there are the resistors, value unknown to me still... This is a very grey area for me.

    What I think I know... current is coming from a supply and will probably be under 0.5 A.

    Actually, I don't think I can even get the current flow on this thing...

    Also, do you mean that based on diode values, the virtual ground would shift?
     
  4. SgtWookie

    Expert

    Jul 17, 2007
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    Ahhh - you still have two Zener diodes in series without a current limiting resistor. :rolleyes: That's just not going to work.

    Instead of pasting your image into a Word document, paste it into MS Paint, and save it as a .png file. That eliminates the need to download it, and then start Word to load up an unnecessarily large file.
     
  5. justtrying

    Thread Starter Active Member

    Mar 9, 2011
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    will use MS paint from now on, thanks for the tip.

    And I of course forgot to include the current limiting diodes... give it one more shot. So R4 and R5 are current limiting. There is another potentiometer in parallel with the one in the circuit. I am playing around with this on breadboard with some random diodes that have 200V rating (since they are in my arsenal) and using the half circuit. It seems to do what it was designed to do , but I need to figure out diode values for the PCB and so far am stuck... A glorified voltage divider.

    Still the question is which voltage do I look at to try and figure out diode values.
     
  6. SgtWookie

    Expert

    Jul 17, 2007
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    Here, have a look at the attached.

    I've used a couple of optoisolators to short out one resistor, then the other. You can see by the plots below what the voltages are at the corresponding points on the schematic.

    Note that this won't handle any current to speak of; if you were going to take measurements, they would have to be via high impedance sensing.
     
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  7. Wendy

    Moderator

    Mar 24, 2008
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    justtrying likes this.
  8. justtrying

    Thread Starter Active Member

    Mar 9, 2011
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    OK. Let my supply provide .5A of current. Then, if I use 2k limiting resistors, 1V is dropped across each and I can use zeners with 11V rating. I have to figure out power rating, but do not now where to start with that ground to try and get current through the diodes. Is my thinking completely off?

    Another question, when I breadboarded this, I expected the voltage at the potentiometer to vary between the supply values (minus limiting resistor, I used 3.3), instead it was limited to the good old forward bias zener +/-0.7V. I think it is because, with respect to the power supply zeners are forward biased. Or I may have messed up the ground, I don't now.
     
  9. Wendy

    Moderator

    Mar 24, 2008
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    Zener diodes act like regular diode if they are forward biased. You may have yours backwards. You can attach digital pictures to your posts, you may want to consider this option for your project.
     
  10. justtrying

    Thread Starter Active Member

    Mar 9, 2011
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    They were in backwards. I also borrowed a 6.6V diodes to play around with. I think I am understanding the behavior now. The value of current limiting resistors will determine the voltage drop across the diodes and thus will define the range of values that can be set with the potentiometer. I guess I have to decide how close zener voltage should be to the supply minus voltage drop across current limiting resistors.

    On to the logarithmic amplifier...
     
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