Connect with us

Current Operated Relay

Discussion in 'General Electronics Discussion' started by lxlramlxl, Sep 19, 2017.

Scroll to continue with content
  1. lxlramlxl

    lxlramlxl

    28
    1
    Jul 5, 2016
    basically I need to be able to operate a relay which is picked my around 200-400mA.

    I'm trying to do this by passing a load through a resistor and an LED from a 110v AC supply. For example some
    LEDS that I've seen have a 50mA operating current with a 2.5v drop.

    I'm not sure how to work out the required resistor To add the required current draw.
     
  2. Externet

    Externet

    775
    168
    Aug 24, 2009
    110VAC-------------/\/\/\/----------LED---------relaycoil----------load-----------neutral

    Is this what you meant ?
    For a LED capable of conducting 200-400mA ?
     
  3. lxlramlxl

    lxlramlxl

    28
    1
    Jul 5, 2016
    What I'm wanting is essentially a dummy load.

    In the real world the actual load would be a 110v Lamp. But to simulate this I'd like to have a small box with some LEDS in and I assume I'd have to have a resistor to limit the voltage going through the LED to provide a 200-400mA load.

    If there was a compact LED that could draw 200-400mA then it would be idea but I'm aware that this probably isn't possible.

    I can upload a circuit diagram of what I think I need tomorrow morning.
     
  4. BobK

    BobK

    7,682
    1,688
    Jan 5, 2010
    Why an LED? If you want a load in the range of 300-400 mA, use a 40W bulb.

    Please explain more about what you are truing to do. Where does rhis relay coil enter into it?

    Bob
     
    Last edited: Sep 19, 2017
  5. Externet

    Externet

    775
    168
    Aug 24, 2009
    The Sunbeam 9W led bulbs I have in the house are labeled 135mA 125VAC, available for $1 at the 'Dollar tree" store. And yes, can be paralleled. Will see your schematic..."
     
  6. lxlramlxl

    lxlramlxl

    28
    1
    Jul 5, 2016
    [​IMG]

    Sorry, the load only needs to be around 180mA for circuit 1.

    The second circuit requires a minimum of 3 LEDS lit for the current operated relay to pick. I believe the schematic should work fine, again its just working out the resistor values. This circuit requires 780mA and needs to lose roughly 260mA from the last two LEDS being disconnected.
     
  7. kellys_eye

    kellys_eye

    4,275
    1,147
    Jun 25, 2010
    5mm LEDs require around 20mA to operate.

    Circuit 1 will draw 60mA assuming you have the correct dropper resistors fitted.
    Circuit 2 will draw 60mA with all LEDs lit.

    To run them from AC you will require a diode (reverse biased) across each LED too.

    Circuit 2 won't work (well L7 and L8 won't light) - see if you can spot why.
     
  8. BobK

    BobK

    7,682
    1,688
    Jan 5, 2010
    This thread is labeled "current operated relay".

    I don't see any relays in your schematic.

    I don't see any load requiring 180mA in you schematic.

    Why are you are wasting our time by keeping information from us?

    Bob
     
  9. lxlramlxl

    lxlramlxl

    28
    1
    Jul 5, 2016
    This is the issue that I'm trying to overcome. 20mA isn't enough so I need a way to draw more. The LED is simply a visual indication to show that the circuit is working.

    I'm not hiding anything nor trying to waste your time. The relay isn't in the schematic as all it would show is the feed going through the relay coils and straight out towards the LEDS. I assumed that people would know how it works, which is my fault.

    The purpose of the dummy load is to test the functionality of the current proving relay.
    Think proving that a lamp is light within traffic lights, you would be able to test the functionality without the traffic light being installed.
     
  10. BobK

    BobK

    7,682
    1,688
    Jan 5, 2010
    It is not at all obvious. What you are doing is very odd.

    One does not normally put a load, dummy or otherwise, in series with a relay coil. To prove the operation of a relay, the indicator (not dimmy load) would be in series with the relay contacts, not the coil.

    And the resistor for an LED, run from 110V at 180mA is going to dissipate 20W, which is why I suggested an incadescant bulb instead.

    And please do pay attention to Kellys_eye, or you will very quickly have deceased LEDs.
    Bob
     
  11. lxlramlxl

    lxlramlxl

    28
    1
    Jul 5, 2016
    You're probably right that it is odd. However in the rail industry it would be something that would be extremely useful due to installation constantly being behind due to network constraints and other reasons.

    Having a signal (traffic lights) in series with the lamp proving relay is the only way for the relay to operate (pick) and is standard by Network Rail.
     
  12. BobK

    BobK

    7,682
    1,688
    Jan 5, 2010
    Ah, okay, now it makes more sense.

    Bob
     
  13. kellys_eye

    kellys_eye

    4,275
    1,147
    Jun 25, 2010
    The LEDs can be connected across the 110V AC (with diode and suitable resistor of course) and the whole arrangement would have another resistor across it to develop the load you require. Each leg of the LED arrangement will consume 20mA so for circuit 1 your load will already be 60mA.

    For circuit 1 the LED series resistor will be 2750Ω (nearest standard value is 2700Ω) with a wattage of 1W (use a 2W resistor to be safe). A diode (1N4007) should be reverse-connected across each LED.

    To get a total load of 200mA a further 140mA would be required to be drawn and this can be achieved with a resistor of 785Ω. This would need to be connected across the whole of circuit 1.

    The resistor would need to be rated at 20W to be safe (aluminium-clad resistor bolted to the metal case of the instrument would be best). Of course you can't GET a 785Ω resistor (easily) so make one up from a 680Ω and 100Ω in series (total 780Ω but near enough).

    The value of the resistor needed is calculated simply R(Ω's) = V(volts)/I(amps) - 110/0.14 = 785 ohms.

    and the wattage of the resistor by V(volts) x I(amps) - 110 x 0.14 = 15.8 (safely rounded up to 20W)

    I leave it to the student to calculate the resistance and wattage for other currents!
     
    lxlramlxl likes this.
  14. lxlramlxl

    lxlramlxl

    28
    1
    Jul 5, 2016
    Thanks for the extremely detailed reply.
    Circuit 1 will only have a single LED lit at any given time so I'd require a 611 resistor (110/0.18) I believe. This would be fine to place just before the neutral return I'd imagine since it is a common path so should provide a load for each LED?

    Thanks again and I will inform you of any progress I make!
     
  15. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    25,499
    2,839
    Jan 21, 2010
    I imagine the progress is going to be that you report a blown up LED.
     
  16. lxlramlxl

    lxlramlxl

    28
    1
    Jul 5, 2016
    Whys that?
     
  17. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    25,499
    2,839
    Jan 21, 2010
    Because you are applying AC to a LED. When forward biased, the forward current may be enough to damage the LED, but let's ignore that.

    When reverse voltage is applied, 170V or so will appear across the LED with a breakdown voltage of around 7V. **POOF**
     
  18. kellys_eye

    kellys_eye

    4,275
    1,147
    Jun 25, 2010
    I also believe I reiterated that in a later post reply - I 'think' the OP noted the fact but it's just as well it's been brought up again in case it was forgotten.
     
  19. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    25,499
    2,839
    Jan 21, 2010
    Fair enough. I didn't see them in his schematics.
     
  20. BobK

    BobK

    7,682
    1,688
    Jan 5, 2010
    And I reminded him as well, but have seen no confirmation that he will fix it.

    Bob
     
Ask a Question
Want to reply to this thread or ask your own question?
You'll need to choose a username for the site, which only take a couple of moments (here). After that, you can post your question and our members will help you out.
Electronics Point Logo
Continue to site
Quote of the day

-