What Zener Diode to use to reduce release delay?

Discussion in 'The Projects Forum' started by jellytot, Jan 7, 2015.

  1. jellytot

    Thread Starter Member

    May 20, 2014
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    I have a Solenoid controlled by a TIP120 transistor (which is controlled by a microcontroller). I added a diode to prevent flyback voltage but it seems the diode has increased the release time of the solenoid.

    I researched the subject, and it seems the best solution is to add a Zener diode (currently using a resistor which helps but I need to reduce the delay as much as possible). What I can't figure out is the Voltage and Power Dissipation rating of the Zener I need.

    For voltage, I read that it should be between the Transistor rating and the supply voltage. But for a 60V rated Tip 120 and 9 volt power supply, then what Zener Voltage should I choose, and what's the difference?

    And can anybody shed light on what Power Dissipation I should look for? Thanks.
     
  2. ericgibbs

    Senior Member

    Jan 29, 2010
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    hi,
    I would allow a 5v to 10V below the transistors Vce rating, so that would ~50Vz for you application if you want to minimise the relay release delay.
    Don't forget to have a regular diode in series with the zener.

    E
     
  3. Dr.killjoy

    Well-Known Member

    Apr 28, 2013
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    Are you using a pull down resistor on the transistor ??
     
  4. JohnInTX

    Moderator

    Jun 26, 2012
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    This should answer your questions.
     
  5. jellytot

    Thread Starter Member

    May 20, 2014
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    Yes. I'm using also another resistor (the one I mentioned) with the diode to reduce release delay until I can get a proper Zener.

    OK, so I read somewhere that I should choose a Power Dissipation equal to load current. Using my multimeter, the solenoid draws 1.2 amps. With a 9v supply, I need to find a 11Watt Zener? That can't be right. The solenoid isn't fired very often (like only a few times a minute), so can I go with a lower Power Dissipation Wattage?
     
  6. RichardO

    Well-Known Member

    May 4, 2013
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    Or you could do this:

    Solenoid driver.png

    You can use just about any low power Zener diode of about 39 volt rating in this circuit. A 200 mw or 1/2 watt diode will work, for instance. Keep in mind that the TIP120 is now acting as a high power Zener diode so it may need increased heat sinking.
     
    ronv likes this.
  7. GopherT

    AAC Fanatic!

    Nov 23, 2012
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    Use an H-bridge to drive your solenoid. Slam it in the reverse current to prevent any delay.
     
  8. crutschow

    Expert

    Mar 14, 2008
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    You read wrong. :rolleyes:
    The power dissipated is proportional to the firing rate of the solenoid and the current and inductance of the solenoid. The power to be dissipated is equal to 1/2 LI^2 times the firing rate per second.
     
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  9. jellytot

    Thread Starter Member

    May 20, 2014
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    I'm going to need help with this please. My knowledge of electronics is minimal.
    Assumptions:
    1) Formula output unit is Watts
    2) L = inductance of solenoid coil in Henries. Roughly calculated as 1 Henry (4cm long, diameter 1cm, 800 turns).
    3) I = current drawn by solenoid.
    Please correct my assumptions if they are wrong.
    "times firing rate per second" is very low (0.1 times a second). The solenoid draws 1.2A, then we have Joules = (1/2*1*1.2^2)*(0.1fps) = 0.72joules *0.1fps.

    Watts = 0.72, so I need a Zener with a Power Dissipation rating higher than 720 milliwatts. Does that... look right?
    And with a voltage rating of 50 volts? Note that the Zener is placed backwards in series with a flyback diode (the Zener diode is not the flyback diode).
     
    Last edited: Jan 8, 2015
  10. crutschow

    Expert

    Mar 14, 2008
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    Not quite. You missed a decimal point. It's 0.72 joules times 0.1 = 72mW.
    A 50V zener rating should work
    Just make sure the zener can handle a surge current of 1.2A.
     
  11. ErnieM

    AAC Fanatic!

    Apr 24, 2011
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    I happen to have experience here. Several firms I've worked for actually make suppressor networks for various relay manufacturers. One I actually did the layout for. In all cases these were those small metal can relays, 28V control with 10A contacts. And all used a 1000V PIV diode with a 36V zener, same as Figure 3 of the above app note. (Aside: TE was one of the companies I did this work for.)

    Energy in the coil (inductor) is 1/2 L I^2 so 1H and 1.2A is .72 joules, and that needs to be dissipated each and every time the coil turns off. So I am unsure what the "times firing rate per second" mean.

    If this means the relay is only on for 0.1 second very occasionally the power is very low and you should rate the diodes by the single impulse ratings.

    If this means it fires 10 times a second you indeed multiply by 10 to get .72 J * 10/sec for 7.2J/S or 7.2 watts.
     
  12. crutschow

    Expert

    Mar 14, 2008
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    By "firing rate per second" I mean the operation frequency of the coil.
    In this case the OP stated it was 0.1 times per second or once every 10 seconds, thus the average power dissipated would be 0.72J * 0.1Hz = 72mW.
     
  13. ErnieM

    AAC Fanatic!

    Apr 24, 2011
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    If that is what the OP means then the power is very low, I'd still check out the single event energy (who's nomenclature escapes me at the moment).
     
  14. ronv

    AAC Fanatic!

    Nov 12, 2008
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    Non repetitive surge current. Probably wants to be about a 5 watt zener. I still like Richards, except to work with the Darlington it needs a reverse standard diode.
     
  15. RichardO

    Well-Known Member

    May 4, 2013
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    Are you sure? Rohm does not think so. ;)
    They don't show the diode you mention... but they may not be showing the whole equivalent circuit in their spec sheet. You do have me wondering if the Zener will keep the transistor from saturating hard. I will have to think about it.

    I first found the Rohm part used as a pin driver in the print head of an old pin-printer. I wish I had thought of it myself.
     
  16. alfacliff

    Well-Known Member

    Dec 13, 2013
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    another way is to place a thin paper or tape between the pole piece of the relay magnet and the armature. the added space allows for less residual magnetism effects.
     
  17. ronv

    AAC Fanatic!

    Nov 12, 2008
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    You know, I think your right. :rolleyes: The fact that it is an inductor in the collector keeps much current from flowing before the transistor turns on.
     
  18. crutschow

    Expert

    Mar 14, 2008
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    But not enough to prevent a significant inductive spike when the coil is de-energized.
     
  19. alfacliff

    Well-Known Member

    Dec 13, 2013
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    it dosn't reduce spikes it reduces residual magnetism.
     
  20. jellytot

    Thread Starter Member

    May 20, 2014
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    So I wanted to say thanks to everyone that posted! Thanks!
    Just a quick update: I replaced my 100 ohm resistor with a backwards-installed 47V Zener (with 5Watt power Dissipation). I also replaced my 1N4004 flyback diode with a 2Amp diode, since the solenoid draws 1.2 amps. It... seems to work better. It's hard to tell exactly what the release delay is now because I don't have proper testing equipment. But I doubt I can do better than what I have in there now so I'm moving on.
     
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