DC Switching Circuit w/ Inductive Load

Discussion in 'General Electronics Chat' started by jcrollman, Mar 12, 2010.

  1. jcrollman

    Thread Starter Member

    May 21, 2008
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    Hello,

    I am looking for a second opinion on this circuit to turn on/off a 24VDC solendoid valve. After days of pouring through application notes and other ideas, this is what I have come up with. The OPTO_DRAIN input is a 3.3V control line from a micro.

    I am most concerned about the CeraDiode (normally for ESD protection) used as a flyback diode on the inductive load. It seems appropriate to me, but I'd be more confident if other people gave it an ok.

    I'd be very grateful for any other thoughts on the circuit as well.

    Thanks,

    Jeff
     
  2. SgtWookie

    Expert

    Jul 17, 2007
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    The emitter diode should get about 3mA current.
    Since the Vf of the diode will be somewhere around 1.2v to 1.3v, I suggest you use 680 Ohms between pin 1 of the optorelay and 3.3v.
    Rlimit >= (3.3v-LED_Vf)/3mA

    In order to saturate the 3904 transistor, you will need a base resistor (Rb) of the correct value to supply 1/10th of the desired collector current; this minimizes Vce, thus power dissipation in the transistor. (ETA: completed sentence 3/15)
    Assuming that your input signal OPTO_DRAIN switches between 0v and 3.3v:
    Rb = (Vhigh_OPTO_DRAIN - Vbe) / (Ic / 10)
    Assuming a Vbe of 0.7v, the formula becomes:
    Rb = (3.3v - 0.7v) / (3mA/10)
    Rb = 2.6 / 0.0003
    Rb = 8,667 Ohms
    8.667k is not a standard value, but 8.2k and 9.1k are. You could use either in this case; they would be close enough.

    Note the power dissipation for the optorelay will be 500mA squared times .55 Ohms = 137.5mW. You will need to dissipate this heat, as if the optorelay gets too warm, its current rating drops. At 80°C, it is only rated for 420mA current.

    I would switch the ground side of the coil instead of the high side, but I suppose switching the high side would work just as well.

    I don't know anything about the ceradiodes. If you want feedback on them, post a link to the datasheet.

    You could simply use a 1N4004 diode, or use a 24v Zener and a 1N4004 back to back. The latter would snub the relay current more quickly.
     
    Last edited: Mar 15, 2010
  3. jcrollman

    Thread Starter Member

    May 21, 2008
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    Thank you for your input. I was configuring the input transistor to have a current of 0.95mA, as the data sheet seemed to imply that the input current should be between 0.55 - 2mA. So my calculation looked like this

    Rlim = (3.3-1.2(Vf)-.2(Vce))/0.00095(Ic) = 2K

    and

    Rb = (3.3(Vhigh_OPTO_DRAIN) - .7(Vbe) / (0.00095(Ic) / 10) ~= 27K.

    However, it states for "applications requiring high temperature operation, an LED drive current of 3mA is recommended", does my 500mA load push me into that realm? I do not have an intuitive sense as to how much heat is generated with 140mW. From you input I assume it's a lot. Maybe an optotransistor isn't the way to go here.

    Attached are the datasheets for the ceradiode and the optotransistor. If you have any more comments they would be certainly appreciated.
     
  4. SgtWookie

    Expert

    Jul 17, 2007
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    I don't know in what environmental conditions you will be utilizing this circuit. I was erring on the side of caution, assuming a wide range of temperatures. If you go with the 1mA LED current, performance may suffer at elevated temperatures.

    When using a transistor as a saturated switch, you force the beta (hFE, gain) of the transistor to be 10 to ensure that the saturation voltage (Vce) will be low. This minimizes power dissipation in the transistor. If you supply significantly less than 1/10 Ic to the base of the transistor, Vce will increase, thus increasing power dissipation.

    I don't see a maximum Vdd or Vce specification for the output side of the CPC1002N. You need to contact the manufacturer about that.

    The ceradiodes are interesting, but you have a trade-off between the max DC voltage and the breakdown voltage; you'd have to use one of those rated for 30vdc, but it doesn't trip until around 50v or so with a max of 120v. The lack of info about the max Vdd/Vce spec on the CPC1002N is a concern.

    Have you looked at TVS diodes for the reverse-EMF suppression?
     
    Last edited: Mar 15, 2010
  5. jcrollman

    Thread Starter Member

    May 21, 2008
    19
    0
    I found this in the ceradiode datasheet after reading it a bit more:

    EPCOS CeraDiodes are not suitable for switching applications or voltage stabilization where
    static power dissipation is required.​
    CeraDiodes are designed for ESD protection only.

    Which I would assume makes this application unwise. I do not have a good grasp of the differences between inductive spikes and ESD, but I'll take them at their word.

    I have been looking at this beefier TVS (SMCJ36A-E3/57T). It is rated for 36V standoff and 40V breakdown, along with a power rating of 1500W. I have no idea how to determine if this power rating is fantastic overkill or not.

    I also realized that my uController can source 20mA, which probably makes the entire transistor input unecessary. I should be able to just drive the diode from the 3.3V source uC, in line with a 500-1K resistor.
     
  6. SgtWookie

    Expert

    Jul 17, 2007
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    I don't know how often you are planning on turning this solenoid on and off. If you are planning on cycling it more than perhaps 5 times a second, power dissipation in the optorelay may become a problem.

    Skip the ceradiodes.

    ESD is electrostatic discharge, typically very high voltage and low current, like getting "zapped" when touching a doorknob after walking across a carpet.

    Inductive spikes can be quite high in voltage, the current depends upon the current flowing in the inductor when it's current path is blocked (turned off). Initially, the current flow will be the same as when there was a complete current path. Your solenoid current will be around 500mA, so in a diode with a Vf of 1, that is 1/2W instantaneous power (1V x 0.5A = 1/2W). However, that is peak power, and will be of short duration.

    A part rated 1500W for this application would be like using a battleship to kill a fly.
     
  7. jcrollman

    Thread Starter Member

    May 21, 2008
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    That all makes sense. There is a 400W version of the same Vishay part, some versions go down to 100W, but not at the voltages I'm operating at (at least not those readily available at digikey). But if it works, good enough. I don't want to spend too much time searching for a flyswatter. Thanks for your help.
     
  8. SgtWookie

    Expert

    Jul 17, 2007
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    As I mentioned before, if you wish to have the solenoid disengage more quickly by stopping the recirculating current (reverse-EMF) then you can use a Zener diode back-to-back with the 1N400x-type diode.

    With just a single diode and a 0.7v-1v drop, it will take a bit of time to dissipate the power.

    If you use a Zener with the diode, the current flow will stop much more quickly.
     
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