photoelectric switch circuit

[Malph]

I want to be able to activate the coil of a relay with a photoelectric switch.
The switch is 24VDC, 100ma load. The relay is also 24VDC, 1.5VA (1W) consumption.
Am I correct in calculating that the relay has a draw of .04 amps and that the sensor will be able to drive this relay? Anything else I should be aware of?

[Al_The_Man]

DC power is IxV, VA is usually used to indicate AC power rating due to inductive effects.
If the P.E. switch is 100ma max, then the coil on 24vdc requires a resistance of 240ohms or less, most manufacturers state coil resistance in the specs.
Al.

[Malph]

I don't see a ohm rating in the specs.

Input circuit
Supply voltage:
24V DC terminals A1(+)-A2
voltage selector engaged
24V AC terminals A1-A2
voltage selector engaged
110 to 240V AC terminals A1-A2
voltage selector not engaged
Tolerance:
24V DC ±10%
24V AC -15% to +10%
110 to 240V AC -15% to +10%
Rated frequency: 48 to 63Hz
Rated consumption:
24V AC/DC 1.5VA (1W)
110V AC 2VA (1W)
230V AC 8VA (1.3W)
Duration of operation: 100%
Reset time: 100ms
Residual ripple for DC: 10%
Drop-out voltage: >30% of the supply voltage

http://www.maxicont.hu/doc/termekek/...TA/en_d6dq.pdf

Thanks for your help.

[Al_The_Man]

It would appear that the rated power consumption on DC is 1 watt so on 24vdc the current would ~42ma.
Al.

[Malph]

Thanks for the help!

[vger]

Put a heafty diode across the coil (A1 A2) in reverse polarity or you may quickly kill the output of the photoswitch. The inductive kickback from the coil can be a very high voltage spike of opposite polariity to the supplied voltage. This is common practice.

http://www.allaboutcircuits.com/vol_3/chpt_3/9.html

Hope you read this in time.

Steve

[Al_The_Man]

I would suspect that you may only need the back EMF diode if fitting an external relay as in the dotted line. As I imagine that if using the internal relay that would be taken care of?
Al.

[Malph]

Originally Posted by vger Put a heafty diode across the coil (A1 A2) in reverse polarity or you may quickly kill the output of the photoswitch. The inductive kickback from the coil can be a very high voltage spike of opposite polariity to the supplied voltage. This is common practice. Steve

So would a IN4001A diode do the job?

I would suspect that you may only need the back EMF diode if fitting an external relay as in the dotted line. As I imagine that if using the internal relay that would be taken care of? Al.

I'm not sure what dotted line you're referring to.

[Al_The_Man]

The diagram appears to show the internal relay and contacts, there is a dotted line with S switch and a external relay to the R.H. side.
Presumably to switch a heavier duty relay? (1VA A2-B1)
If you use the external type a IN4001 should be OK.
Al.