Peak Voltage Detector

Discussion in 'The Projects Forum' started by mossman77, Feb 2, 2015.

  1. mossman77

    Thread Starter New Member

    Feb 2, 2015
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    I made a simple peak voltage detector last night to measure the ignition coil voltage on my dirt bike and I tested it on a house outlet (120Vac, 60Hz). The voltage shot up to around 120V, but then slowly decreased down to 100, 90, 80, etc. (meter is on DC voltage setting). I don't understand how the capacitor is discharging since the diode should be reverse biased on the negative cycle. The detector seems to work as expected when measuring my coil voltage (I didn't have much time to test it out fully), which I believe is pulsed DC (no negative component), but shouldn't it work with AC as well? Here is the schematic:

    [​IMG]
     
    Last edited: Feb 2, 2015
  2. wayneh

    Expert

    Sep 9, 2010
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    C1 discharges through R2 and your meter, which is ~1M also.
     
  3. mossman77

    Thread Starter New Member

    Feb 2, 2015
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    That's correct, it does discharge through the 1M resistor, but it shouldn't be discharging while still connected to the source.

    I'm pretty sure the meter impedance is 10M.
     
    Last edited: Feb 2, 2015
  4. ronv

    AAC Fanatic!

    Nov 12, 2008
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    Check your work. It should pop up to 170 volts and stay there.
     
  5. mossman

    Member

    Aug 26, 2010
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    I'll try again. Maybe the probes weren't making contact with the prongs inside the outlet.
     
  6. ScottWang

    Moderator

    Aug 23, 2012
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    Disconnect the R2 and measure again, if the voltage can stay longer then change the R2 to 10M, and try again.
     
  7. wayneh

    Expert

    Sep 9, 2010
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    The TS is trying to make a continuous reading - not just a touch to the AC mains, as I first assumed based on the fading reading - so there must be another problem.
     
  8. ScottWang

    Moderator

    Aug 23, 2012
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    I just want to make sure how the R2 affecting the discharging, if it doesn't then we can discuss some other problems.
     
  9. mossman

    Member

    Aug 26, 2010
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    The input impedance of my multimeter is 10M, so if I increase the parallel resistor much higher, my voltage reading won't be accurate. If anything, I may just remove the 1M resistor and instead install a push button switch with low value series resistor to discharge the capacitor when I'm finished with it.

    Regarding resistor R1, I think I need to move it in series with only the capacitor. Otherwise I'm creating a voltage divider between R1 and R2, which will be pretty significant at 200V+.

    [​IMG]
     
    Last edited: Feb 2, 2015
  10. Reloadron

    Active Member

    Jan 15, 2015
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    I agree it should go to about 170 volts peak and just sit there as long as power is applied. That assumes a 120 Volt 60 Hz input. There isn't a need for R1 which will cause C1 to charge slower but not enough to matter. If R2 is really 1 Meg it also won't matter, even if the measuring meter has a 1 Meg input. Now if R2 were not 1 Meg and something like 1K then problems begin. R2 could be removed as all it does is give C1 a discharge path when power is removed, with a 60 Hz line frequency and if R2 is 1 Meg it will take C1 a long time to discharge through R2.

    Ron
     
  11. mossman

    Member

    Aug 26, 2010
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    R1 is there to limit the initial current through D1. At t=0, the current will be huge won't it? I realize it's only for a short amount of time, but according to the 1N4004 datasheet, the maximum non-repetitive forward current is 30A for 8.3ms. And if the input Z of my meter was 1M, then my reading would be half of the actual voltage because I would have 1M in parallel with 1M. An R2 of 1k would give me an even more accurate reading than the 1M. However, 1k wouldn't work because of the amount of current that would flow through it (200 mA), which is 40w. R2 needs to be high enough to limit the maximum current through it while providing a discharge path for C1 when the source is disconnected. Am I completely wrong here???

    How about (meter across cap only)...

    [​IMG]
     
    Last edited: Feb 2, 2015
  12. mossman

    Member

    Aug 26, 2010
    131
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    Or better yet...

    [​IMG]
     
  13. Reloadron

    Active Member

    Jan 15, 2015
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    Your original drawing was fine. Here is what it would look like:

    Peak Detect 1.png

    As to R2 it really only comes into play when power is removed being 1 Meg. When you place a 1 Meg input meter across the circuit R2 effectively becomes 500K which won't change anything. Your original circuit should work just fine with the parts you show in the configuration you have.

    Ron
     
  14. Reloadron

    Active Member

    Jan 15, 2015
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    If you are going to do that it will also work and C1 will have a nice fast discharge time. Replace the switch with a FET, we can pulse it and design a peak responding voltmeter around it. :)

    Ron
     
  15. Alec_t

    AAC Fanatic!

    Sep 17, 2013
    5,805
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    Cringe.... Are you aiming for a Darwin Award? Keep your fingers well away. Mains voltage can KILL you! The 170V on the cap is an even greater death threat. I'm surprised the mods haven't interpreted this project as a 'transformerless supply' (for the meter), contrary to the Terms of Service of this site. ;)
     
  16. mossman

    Member

    Aug 26, 2010
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    It's safe. All the electronics are inside an enclosure and I'm probing a single phase 120V outlet with multimeter probes and using them as they're intended to be used. 170V on such a miniscule capacitor is less of a hazard than a 9V alkaline battery.

    I tried the detector again this evening and it actually does work. The probes weren't making good contact. It charges to 165V and stays there until I remove the probes, just like I intended. I still may modify it per my latest diagram.
     
  17. mossman

    Member

    Aug 26, 2010
    131
    3
    You're right. I mistakenly divided by 1,100,000 instead of 1,000,100 when calculating the voltage across the 1M resistor.

    Safer circuit:

    [​IMG]
     
    Last edited: Feb 4, 2015
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