Request for an explanation - zener voltage clamp - my debug process...

Discussion in 'General Electronics Chat' started by kingdano, Apr 26, 2010.

  1. kingdano

    Thread Starter Member

    Apr 14, 2010
    377
    19
    To any who viewed my royer oscillator thread - this thread is a continuation of sorts.

    My plan for making the circuit as safe as possible was to clamp the input voltage such that the output could never rise above the required 500 VRMS.

    I did some circuit characterization and found that at around 4.5-4.75V with a loaded or un-loaded output the royer oscialltor measured around 525-550 VRMS on a DMM.

    Lucky me, a 4.7V zener is a common component - so i went about designing a simple input stage of components.

    I came up with a circuit, and went straight to breadboarding it prior to simulation...breadboarding did not work - the clamp worked independent of the royer, and the royer worked independently of the clamp..but once i put them together neither part worked - and a lot of good components gave their life as i tried again and again. In the schematics i will post refer to the C1/R2 node, this point never went above 0.6VDC...i had no idea what was going on.

    So i went back to simulation to try and see what was going on...the story begins here.

    (note:
    green trace = V1/R1 node
    blue trace = D1/R1 node
    red trace = C1/R2 node)

    Schematic 1 was a close representation of my system - my clamp circuit comprised V1 (wall wart), R1, D1 and C1 - in the schematic R2 is the resistor which is on the royer circuit i was given, and I1 is intended to be the draw on the components (this max' out at 0.5ADC at the right output voltage)

    [​IMG]

    I made the load intentionally tiny (10 mA), because i just wanted to get SOMETHING to work. My first instinct was that the load on the output was too much.

    Looking at the transient analysis, there is some odd behavior at start-up, but then normal "zenering" action as one would expect.

    [​IMG]

    So I increased the simulated load to 100 mA - and boom, it all went to <snip>. Interesting to me, but at least it made me feel less bad about my lab results.

    [​IMG]

    [​IMG]

    One thing that i had played with in the lab, was changing R1 values - dropping R1 to a low value had made the circuit somewhat functional - but <snip> did that thing get HOT! I eventually used a 10Ω 25W resistor in the real world.

    So in schematic 3 i did the same, dropped R1, and left the load high. The simulation showed the zener working, but the output voltage (red) going to <snip> still. <snip>

    [​IMG]

    [​IMG]

    (continued on next post)
     
    Last edited by a moderator: Apr 27, 2010
  2. kingdano

    Thread Starter Member

    Apr 14, 2010
    377
    19
    I started to look at the waveforms and realized that the current needed to be steered towards the load, and away from everything else.

    So in schematic 4 i added a schottky diode to block current flow, leaving the load at 100 mA.

    [​IMG]

    The simulation was a nice bit of progress - success!

    [​IMG]

    So now my curiosity had the best of me - i swapped R1 back to 330Ω thinking everything would be fine now...NOPE! everything went to <snip> again!

    [​IMG]

    [​IMG]

    So i am thoroughly confused - can anyone explain this to me? Or explain the general circuit behavior that i am clearly not seeing which is causing it all to go to <snip>?

    At any rate - i think i have settled on a final circuit (next post)
     
    Last edited by a moderator: Apr 27, 2010
  3. kingdano

    Thread Starter Member

    Apr 14, 2010
    377
    19
    In this schematic i have the low R1 value (no idea why this is necessary) - the 4.7V zener clamp - the current steering schottky, and a high frequency filter cap all feeding the 330Ω resistor and current load which represent the royer.

    [​IMG]

    I did a DC sweep on the load and got the results i expected, knowing that this zener is rated for 500 mW - much less than i need. The response drops off as you pull more current past the zener.

    [​IMG]

    The transient also looks good - knowing that the load exceeds the zener model I backed off to 200 mA. At 500 mA it "zeners" at around 3.8VDC - but at 200 mA it looks fine. (200 mA below)

    [​IMG]

    So can anyone explain to me why this is behaving this way?

    Specifically why does R1 need to be so small to make the circuit function...i would love to know, rather than just blindly believing.

    Also, as a side note - i originally had an inductor in line with R1, but ended up with an RLC oscillator in the simulation... :D slightly embarrassing.

    (edit - in the circuit i will use i am using a 5W 4.7V zener, not the 200 mA model in the spice circuit)

    (does anyone know how to model a royer oscillator in ltspice without having a transformer model....i kid, i kid)
     
    Last edited: Apr 26, 2010
  4. kkazem

    Active Member

    Jul 23, 2009
    160
    26
    Hi,
    Although I didn't get to thoroughly look at your complete ckt w/ oscillator, I can tell you that in your simulation, your "load" at the right end of the schematics seems to be a current source, but without the actual SPICE netlist I can't tell for sure. Assuming it is, and noting that the way the arrow is pointing down, the polarity of the current source is positive where the arrow is coming out. Therefore, the zener isn't operating in the zener or avalanch mode, but in a normal diode mode, which will put about a 0.65V across the zener, with the polarity of the 0.65V being positive at the ANODE end (bottom-side). If you're trying to use the current source as a load, what you really want is a current sink, not a current source. This can easily be made with a single supply op-amp, small value current sense resistor (1 Ohm), a cheap voltage reference (like a TL-431) and a small N-Channel MOSFET (like a BSS138) with the source going to the op-amp ground, the gate going to the op-amp output and the drain being the current-sink input. This should simulate just fine, as well as work good in your breadboard. Please see the attached PDF schematic.

    I hope this helps. With the current sink, it will act more like a resistive load and not pump current into your circuit, but simply sink available current from your circuit.

    Regards,
    Kamran Kazem
     
  5. kingdano

    Thread Starter Member

    Apr 14, 2010
    377
    19
    Code ( (Unknown Language)):
    1. * C:\Documents and Settings\dbinnun\My Documents\RTD\Lhotse\Electrode Electronics\Simulations\VoltageClamp.asc
    2. V1 N001 0 9V
    3. C1 N003 0 1µ V=10
    4. D1 N005 N002 1N750
    5. R2 N004 N003 330 tol=10 pwr=5
    6. R1 N001 N002 10 tol=1 pwr=5
    7. D2 N002 N003 B530C
    8. I1 N004 0 .4
    9. R3 N005 0 1000
    10. .model D D
    11. .lib C:\Program Files\LTC\LTspiceIV\lib\cmp\standard.dio
    12. .tran 0 300u 0 0.1u startup
    13. .backanno
    14. .end
    15.  
    There is the SPICE netlist...i assumed that putting down a "load" component in SPICE was self explanatory....apparently not
     
  6. SgtWookie

    Expert

    Jul 17, 2007
    22,182
    1,728
    Take a look at what happens when you are sinking a constant 100mA through a 330 Ohm resistor.

    Ohm's Law says: E=IR
    So, 100mA x 330 Ohms = 33V. But, it's negative, because you're sinking the current from the resistor.

    If you're sinking more from the right side of R2 than you can source from R1, you're going to wind up with a negative voltage on C1.

    You'd probably be better off just replacing the current source I1 with a connection to ground; that way you won't confuse yourself.
     
  7. kingdano

    Thread Starter Member

    Apr 14, 2010
    377
    19
    Makes perfect sense...now that i can see it!

    So how <snip> can i simulate a load - the only thing i know about it is that it draws up to 0.5ADC...

    I can't try the suggested current sink - but the parts arent available in LTSpice. I will look in the library for comparable parts
     
    Last edited by a moderator: Apr 27, 2010
  8. Markd77

    Senior Member

    Sep 7, 2009
    2,803
    594
    Would a variable resistor be good for the load?
     
  9. kingdano

    Thread Starter Member

    Apr 14, 2010
    377
    19
    i will try to take some DMM measurements of the circuit from the input point to ground...and use components that way.

    whats that old saying?

    if you ask a question you may look stupid, but if you dont ask you will stay stupid.

    this thread will be buried in a few weeks, right?
     
  10. SgtWookie

    Expert

    Jul 17, 2007
    22,182
    1,728
    If you wanted to, you could use an LM317 regulator model with a 2.5 Ohm resistor from OUT to ADJ, and ground the ADJ terminal.

    Once the IN terminal had >3V on it, the LM317 would sink a constant 500mA.

    I posted a model for it a week or so ago, but here it is again.

    Note that Vref in the model is on the high side at around 1.29v; you'll need to use a 2.59 Ohm resistor to get approximately 500mA.
     
    Last edited: Apr 26, 2010
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  11. kingdano

    Thread Starter Member

    Apr 14, 2010
    377
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    1. thanks for the tip

    2. what did i say that was not g rated?
     
  12. SgtWookie

    Expert

    Jul 17, 2007
    22,182
    1,728
    Language. Please keep it clean.
     
  13. kingdano

    Thread Starter Member

    Apr 14, 2010
    377
    19
    the 'h' word?

    send me a PM and i will remove.
     
  14. SgtWookie

    Expert

    Jul 17, 2007
    22,182
    1,728
    You have a PM.
     
  15. ifixit

    Distinguished Member

    Nov 20, 2008
    638
    108
    Open the current source and set the 'Parasitic Properties' to: This is an active load.

    The current source will now draw less current as the voltage across it becomes less than ~1 volt.

    Regards,
    Ifixit
     
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  16. SgtWookie

    Expert

    Jul 17, 2007
    22,182
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    Nice - I learned something. Thanky, Ifixit. :) I think that might help to fix a problem that's existed with the LM2907 model I'm currently fooling with.
     
  17. kingdano

    Thread Starter Member

    Apr 14, 2010
    377
    19
    awesome tip.

    love ltspice even more now
     
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