Discussion in 'LEDs and Optoelectronics' started by lxlramlxl, Jul 6, 2016.

1. ### lxlramlxl

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Jul 5, 2016
Essentially I'm trying to make a dummy load that will pick a current operated relay.

The relay requires 300mA to pick and the feed is 110v.

Would I be correct in thinking that 33W would need to be drawn?
In which case what value resistor could I use to be able to use a normal LED

Edit: Each LED has a separate feed but for all intents and purposes links are good enough for this representation

Last edited: Jul 6, 2016
2. ### BobK

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Jan 5, 2010
I cannot make any sense of your post. Please try to explain what you are trying to do.

Bob

3. ### lxlramlxl

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Jul 5, 2016

Something like this. For the relay to operate then a minimum of 300mA must be drawn.

4. ### CDRIVEHauling 10' pipe on a Trek Shift3

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May 8, 2012
lxlramlxl, welcome to Electronics Point.

Many times a schematic like you posted answers a lot of questions. Often though, it doesn't. This may be one of those times. Your circuit appears to be powered directly from the AC Mains but I may be wrong.

The best way to get accurate replies is to describe in detail what your trying to accomplish. Links to components your using is also helpful. Your 300mA relay would be a good candidate for such a link.

Chris

5. ### lxlramlxl

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Jul 5, 2016
It's hard to explain, or rather I'm not good at explaining.
Basically in a real life situation you have a Traffic light or signal that is fed from a power box. For safety reasons you need to prove that a light is lit. This is where the current proving relay comes in, and so is cut into every lamp.

It is possible to check the functionality of the power box without a signal being there to prove that the current proving relay actually works. To do this you need a dummy load to draw said current.

The circuit diagram that I posted is misleading as the relay and the feed are external and you would croc-clip on the circuit to some links.
In reality or you would have is a resistor and an LED in series with clips on either end to put on the positive and negative feed to represent a load.

As for the relay, if the information is still relevant then: http://www.urbanengg.com/relay_lamp.asp

6. ### AnalogKid

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Jun 10, 2015
1. 300 mA at 110 Vac is a *lot* of relay coil power. 33 W. Unless the relay is the size of a baseball, this probably is not correct. One option is that the relay coil takes 300 mA at something like 12 V.
2. The LEDs in your drawing will act as half-wave rectifiers, altering the relay coil voltage and current.
3. The LEDs will have 160 Vpeak across them during the negative half-cycle of the input 110 Vac. Can they handle that?

It sounds like the idea here is for the LED current to go through a relay coil, so the contacts are closed when the LED is on and open if the LED fails (open circuit). If so, then we need the true nature of the LEDs. Operating voltage, current, do they have current limiting built in, are they actually LED arrays or series/parallel groups, etc.

ak

7. ### CDRIVEHauling 10' pipe on a Trek Shift3

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May 8, 2012
If I'm reading those relay specs correctly this relay coil will drop about 3.24VAC/50Hz @ 108mA.

Chris

8. ### CDRIVEHauling 10' pipe on a Trek Shift3

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May 8, 2012
I should have caught this before making my last post but your description seems to indicate that you want to build a portable test instrument to take to the field with you. It also suggests that you don't want to be illuminating the traffic lights to make the test. Hence the need for the dummy load.

Did I get this right?

Chris

9. ### lxlramlxl

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Jul 5, 2016
This is absolutely correct.

In the absence of a real signal to draw the current It'd be extremely useful to have a portable dummy to replace it.

10. ### lxlramlxl

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Jul 5, 2016

Is a good example.

11. ### hevans1944Hop - AC8NS

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Jun 21, 2012
This thread appears to be related to Indian rail systems, specifically the LED signals used to safely direct the train engineer along the tracks. The nomenclature used for the relays may be unfamiliar. For example, the "back contacts" refer to normally-closed contacts when the relay coil is de-energized; "front contacts" refer to normally open contacts that close when the relay is energized. More information can be found in this PDF article here.

The LED railroad signals operate from AC or DC, probably at 250 V but maybe less. I suspect (but have no way to be certain) that pairs of LEDs connected in inverse-parallel are used in these signals, with several of the pairs connected in series strings to obtain the overall voltage specification. Of course each string would need its own current-limiting resistor. In the USA LED traffic lights contain perhaps fifty to a hundred LEDs, depending on size, and may be retro-fits to incandescent lamp signals that are controlled (remotely) from 115 V AC control boxes. I think the "dummy load" should emulate an actual LED signal in terms of applied voltage causing a certain specified current. The LEDs should consist of at least one inverse-parallel connected pair (for AC operation) with an appropriate current-limiting resistor and LED power rating for the anticipated test current.

It is unclear how the test set will be used. How is it powered? How is the LED signal to be disconnected from its control relay to allow testing? Is the only purpose of the dummy load to ensure the test set is operational? Why use a relay coil at all to essentially measure continuity of the LED signal?

12. ### BobK

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Jan 5, 2010
So it sounds like the lights are powered from 110V mains in series with the relay coil, and the relay pulls in whenever the lights are drawing at least 300mA.

So, to test this you need a dummy load that pulls more than 300mA. A 40 Watt incandescent light bulb would seem to fill the bill.

Bob

hevans1944 likes this.
13. ### lxlramlxl

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Jul 5, 2016

This is infact a British signal head and they are powered off of 110v.

You can't measure continuity if the signal head isn't there, right? However you can test the functionality of the signal head circuit and ECR, which is the purpose of this circuit.

14. ### lxlramlxl

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Jul 5, 2016
That's true, however 4 incandescent bulbs aren't easy to carry around in a tool bag or to transport in general.

15. ### duke37

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Jan 9, 2011
If the supply is AC, you could use a capacitor to pass the current without any heating.
What is the voltage and frequency of the supply.

16. ### lxlramlxl

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Jul 5, 2016
110v 50 or 60Hz.
I'm pretty confident it is 50 though.

17. ### hevans1944Hop - AC8NS

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Jun 21, 2012
Nah, my only contact with railroads is I once traveled from California to Pennsylvania with a stop in Chicago while returning via steamship from Okinawa in the 1950s. i was a lot younger then so I didn't really appreciate the ride as much as I would today. Unlike India, the US rail system (except on the East Coast) is mostly oriented toward carrying freight instead of passengers and freight.

So, if your dummy load is to emulate a signal head, then you should wire up a few dozen pairs of LEDs "back-to-back" or inverse-parallel so they each conduct on alternate half-cycles of the AC line. Or use a 40 watt incandescent lamp as @BobK suggested in post #12. Light is light, and the signal head circuit and ECR probably don't care what the load is as long as the current is in spec.

18. ### lxlramlxl

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Jul 5, 2016
Technically it doesn't have to be light. However having a little LED showing which aspect it is showing is useful.
As i said before carrying an incandescent bulb around isn't easy safely, and they're hardly the smallest thing to put in a tool bag.

19. ### duke37

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Jan 9, 2011
110V 50Hz 0.3A
C = 0.3/(2*pi*50*110) = 8.7μF

20. ### hevans1944Hop - AC8NS

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Jun 21, 2012
Yeah, a 40 watt glass light bulb is not what I want to stuff into my go-to bag. So LED it is. You need to find an LED that will pass 300 mA through the relay to simulate the signal head load.

Most LEDs are rated at much lower currents, on the order of 10 mA or so. You could parallel a single pair of inverse-parallel connected LEDs (with an appropriate series current-limiting resistor) with another power resistor that would bring the total current up to 300 mA drawn through the relay. That would still allow you to verify the test set is working, and then you could do whatever it is you do with the clip leads. I am still unclear on how you actually perform your test or tests. Is the object to test the signal head containing the LEDs, or to test the signal head circuit and ECR (Electronic Control Relays?), or all three? Please tell us what you are trying to DO.