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  1. #1
    Join Date
    Sep 2004
    Posts
    6

    photoelectric switch circuit

    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?

  2. #2
    Join Date
    Dec 2003
    Posts
    24221
    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.
    CNC, Mechatronics Integration and Custom Machine Design

    “Logic will get you from A to B. Imagination will take you everywhere.”
    Albert E.

  3. #3
    Join Date
    Sep 2004
    Posts
    6
    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.

  4. #4
    Join Date
    Dec 2003
    Posts
    24221
    It would appear that the rated power consumption on DC is 1 watt so on 24vdc the current would ~42ma.
    Al.
    CNC, Mechatronics Integration and Custom Machine Design

    “Logic will get you from A to B. Imagination will take you everywhere.”
    Albert E.

  5. #5
    Join Date
    Sep 2004
    Posts
    6
    Thanks for the help!

  6. #6
    Join Date
    Jan 2006
    Posts
    738
    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

  7. #7
    Join Date
    Dec 2003
    Posts
    24221
    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.
    CNC, Mechatronics Integration and Custom Machine Design

    “Logic will get you from A to B. Imagination will take you everywhere.”
    Albert E.

  8. #8
    Join Date
    Sep 2004
    Posts
    6
    Quote Originally Posted by vger View Post
    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.

  9. #9
    Join Date
    Dec 2003
    Posts
    24221
    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.
    CNC, Mechatronics Integration and Custom Machine Design

    “Logic will get you from A to B. Imagination will take you everywhere.”
    Albert E.

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