solid state relay off-state leakage with transformer

Discussion in 'The Projects Forum' started by relay, Jan 15, 2010.

  1. relay

    Thread Starter New Member

    Jan 15, 2010
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    I have a solid state relay (SSR) that controls a transformer that powers a resistive heater. Schematic attached below.

    The problem is that in the SSR off-state I get 200 mA leakage across the heater. I measure only 9 mA in series to the SSR output, which is within rating for the SSR (data sheet below). However, I'm not sure how the 9 mA leakage on the SSR translates to a (large) 200 mA on the heater and how to correct this. I do see this is explained by the (large) 15.6VAC on each input coil, but I'm not sure how that voltage comes about. I suspect the inductance on the transformer is the culprit, but I'm not sure how that works in terms of equations.

    The relay is an Opto 22 Model Z240D10 [ http://www.opto22.com/documents/0859_Solid_State_Relays_data_sheet.pdf ]. It's rated for 24-280VAC output, with 12mA max off-state leakage.

    The transformer is a A41-25-16 [ http://www.signaltransformer.com/Data/Datasheets/A41.pdf ]. It has two independent coil pairs. Each pair normally is intended for 120VAC -> 8VAC.
     
  2. Duane P Wetick

    Active Member

    Apr 23, 2009
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    You have to consider the algebraic sum of the resistance and reactance of the transformer primary winding, 1677 ohms indicated. So there is current flowing thru the primary winding and that produces a magnetizing current that results in a voltage and current (6 ohm load) in the secondary winding.

    Cheers, DPW [ Everyone's knowledge is in-complete...Albert Einstein]
     
    Last edited: Jan 16, 2010
  3. relay

    Thread Starter New Member

    Jan 15, 2010
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    Thank you, that makes more sense now with reactance taken into account.

    The next question is how to most easily reduce the 200 mA leakage current on the heater by 50% to 75%. I think putting two relays in series would reduce it by 50%. I also think that putting a 1733 ohm resistor in parallel to the primary winding would reduce it by 50%, but that may impose an excessive load (33 W) during the ON state. Adding a second transformer with coils in parallel to the first maybe would too, but I don't want two transformers.

    Does that sound correct? Are there better solutions?
     
  4. retched

    AAC Fanatic!

    Dec 5, 2009
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    A diode would drop your voltage after the inducted voltage rise, in turn dropping the multiplier.
     
  5. relay

    Thread Starter New Member

    Jan 15, 2010
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    Would the diode be placed parallel to the inductive coil? Since this is AC, would diodes be needed in both directions?
     
  6. eblc1388

    Senior Member

    Nov 28, 2008
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    You cannot reduce the leakage current from outside the SSR. The current path is internal due to a RC snubber(protection circuit).

    There exists type of solid state relay that does not has this snubber circuit built-in so will only leaks less than 1mA in the off state.

    Or use a mechanical relay instead.
     
  7. retched

    AAC Fanatic!

    Dec 5, 2009
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    I was under the assumption, it wasn't actually the leakage current inside the SSR but the resulting gain after the relay to the heater he wanted to drop.

    RELAY: You want to reduce current going to your heater, correct?
    If so, the best resistor combination will reduce the current to desired levels whilst increasing the wattage load.

    Were you thinking of using another transformer or another relay NOT to use as a relay but just for current drop?
     
  8. relay

    Thread Starter New Member

    Jan 15, 2010
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    Correct.

    Still, replacing the relay could be an acceptable solution. This relay turns on/off once a second for long periods of time, so I thought an SSR would be preferred.

    As far as I see, to halve the off-state heater current requires an approximately 1700 ohm resistor in parallel to the primary winding. During the ON state, that causes a 34W load on the resistor.

    Yes. If I place two relays in series (connected to the same control line), I suspect that during the OFF state this would nearly double the impedance in the primary loop, halving the current, halving the voltage drop across the transformer primary windings, and halving the leakage through the heater. During the ON state, the system should still operate nearly as before.
     
    Last edited: Jan 16, 2010
  9. retched

    AAC Fanatic!

    Dec 5, 2009
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    Here is a page on coils:
    http://info.ee.surrey.ac.uk/Workshop/advice/coils/air_coils.html

    It gives you the formulas and how-tos to create a coil for your own transformer or simple inductance loads. It can help. It is a good page to bookmark for great knowledge.

    however:
    You would (probably) be better off using a standard mechanical relay for ease of design. They are rated for millions of cycles now-a-days and you have less problems sneaking up on you.

    The power has to go somewhere, be it heat or light.
     
  10. eblc1388

    Senior Member

    Nov 28, 2008
    1,542
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    There is no point turning on the heater via the transformer.

    The best way is to let the transformer supply permanently ON, then uses TRIAC to turn ON the heater. Much easier to achieve.
     
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