Leakage current

[PLC Pie Guy]

Hello all.
I have a scenario where I am trying to operate a 120Volt coil relay from a transistor output photoeye. I am getting a voltage of 36 to 38 Volts on the coil through the switch even when not activated. I am wondering if somebody might know what size resistor I would need to dissipate this voltage.

Thanks

 

[jvdcande]

120V on a transistor output photo-eye? Never heard of that situation over here in Europe. Normally when working with 120V it is AC. Transistors are ment to be used on DC and don't do well on AC. Can you give more information to clear this situation:

• what type photo-eye
• what exact voltage (including AC or DC)
• what type of relay (reference, type, brand, coil resistance, ...)
• ...

Maybe with all that known we can get a glimpse of the exact problem.

Kind regards,

 

[PLC Pie Guy]

You are right. My mistake, just looking at to many photoeyes in one day. The particular one I am dealling with is a Banner diffuse proximity switch. Output type is a solid state AC switch. I have tried a couple relays here but all are chattering when the switch is not made. Also I detect almost 40 volts on the output of the sensor.

Thanks

 

[jvdcande]

Yeah, that sounds like the typical effects for leakage current from a solid state AC output. The specs should mention the maximum leakage current. Combined with the maximum off-state voltage for the relay and the current it consumes you can start to calculate.

What you can also do is to measure the current the relay is consuming in the on-state (or use the specs). Take half the maximum load current for the photo-eye and subtract the current consumed by the relay in the on-state. Calculate the resistor (to be placed in parallel with the relay) from 120V divided by the just calculated current and it should do. In the on-state the current will be half of what's allowed for the photo-eye. In the off-state the remaining voltage should be low enough to stop the relay from cluttering. Don't forget to calculate also the power dissipated in the resistor and make sure the resistor can take it.

Beware, this is only a rule of thumb. It might work with a third or a quarter of the rated current too. If it doesn't work, step down the resistor value, but make sure to remain within the current rating for the photo-eye.

Kind regards,

 

[PLC Pie Guy]

I ahve the data sheet here for the sensor. It says that my off state leakage current:less than 1.7mA.
on state vltage drop is :<5volts at 300mA, <10volts at 15mA load.
Output rating is:Minimum load current 5mA; maximum steady-state load capability 300mA to 50deg.C.

Input voltage is 120Volts AC

 

[PLC Pie Guy]

So I used the resistor calculator on the AB web site. It indicates that I should use a 71 OHM, 2 Watt resistor. Or the next higher available resistive value. I do have an 84 Ohm resistor here but it has a 1/4 Watt energy value. What would be the implications of using one rated 1/4 Watt rather than the recomended 2 Watt?

 

[keithkyll]

It will burn. 71 Ohms is way off. I expect 10K Ohms at 2 Watts.

 

[PLC Pie Guy]

Okay, I do have a 10K resistor but it is still at a 1/4 Watt. Will this work?

 

[jvdcande]

Okay, I do have a 10K resistor but it is still at a 1/4 Watt. Will this work?

NO, if you try to dissipate 1.21W into a 1/4W resistor it will definitely burn.

The spec states that if you use 15mA the remaining voltage will be less than 10V. To get to this situation you need a resistor of 110V/15mA=7.33k. 10k is bigger, so the current will be lower and, most probably, the remaining voltage somewhat higher. Supposing that with 10k you however you still end up with 10V remaining voltage, then the power dissipated by that resistor is 110*110/10000 or 1.21W.

Now suppose that the combination of the relay in parallel with the 10k resistor gets the remaining voltage as low as 5 V, then the power dissipated in the resistor will become 115*115/10000 = 1.32W.

As you can see, your resistor of 1/4W will never withstand the heat dissipated in it and WILL burn.

Regards,

 

[PLC Pie Guy]

So to get this all straight, I need to use a 10K resistor with a 2 Watt power rating to bleed off a maximum of 1.7mA.(40volts) to do it without burning.

 

[sbaum]

Why not just use a relay with built-in leakage suppression?

For example... AB 700-HLT1L1

 

[keithkyll]

So to get this all straight, I need to use a 10K resistor with a 2 Watt power rating to bleed off a maximum of 1.7mA.(40volts) to do it without burning.

The bleed off isn't the part we're using to rate the resistor. When the relay is on, it will have 120 VAC across it. So will the resistor. We need to rate it for that.
Ohms law for DC can be used. 120V /10K = 12 mA.
.012 Amps * 120Volts = 1.44 Watts

 

[webbie]

To remove the leakage current I would be inclined to use a Diac in series with the relay coil. A Diac is an AC switch with a break-over voltage > 30VAC. When the relay is on there is no power being dissipated due to the resistor. Check it out

 

[ComfortCoveCottage]

Remember, the maximum current is when the voltage is at it's peak... 120 V + its upper tolerance (+/- 10 %) so that's the maximum power you're resistor will have to dissipate.

If you don't pick a resistor that doesn't have a high enough power rating, it will eventually degrade.

We build things that last so make sure to spec it for the worst-case scenario.

 

[shooter]

so your relay is on with only 1.7 mA????

or use two relays in parallel the voltage should be down to 30 volts if you are correct.