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I'm hoping that you friendly and knowledgeable folk can help me finally figure out how I should be wiring up my LEDs in my panels.

Just a short history...

Many years ago I took over a panel building department which was in a dire state. I have managed to turn it around now having correct wiring diagrams, CE conformity, etc, etc.

I build 3 phase panels for use in the chain hoist industry and each panel has a 'handset' which controls the hoists. Each 'channel' on the handset has a toggle switch and bi-colour LED to show the direction chosen, green for up and red for down. Now ever since the dawn of time, we have fitted an AC transformer in the panel which sends 24v AC and a 0v down to the handset which goes to the toggle switches and also the LEDs.

On each toggle switch, we solder a resistor which then feeds both legs of the LED.

Now my question....

As far as I am aware it's unusual to power LEDs with an AC supply. Primarily they prefer DC. I am assuming the use of the AC power supply was because of the very long distances these handsets have been used in (some over 100 meters) and thus the AC works better.

I wish to ditch the AC transformer and fit a DIN mounted 5amp DC unit.

My question is though, will I still have to fit the resistors? I know they should be needed to smooth out the current going to each LED but just wanted to check.

Below are links to the LEDs and resistors I use in case anyone wants to check the specs.

LEDs

Resistors

I hope you can help me gain more of an understanding of LEDs.

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  • \$\begingroup\$ You can use a 5V supply to power your LED's. You will need to use smaller resistor values. I would think the transformer might be more reliable in the long run. I assume sending people out to do maintenance on cranes is expensive and irritating. But it could still make sense to use the DC supply. Just weighing pros and cons. Oh, sorry. Chain hoist, not crane. \$\endgroup\$ – mkeith Aug 29 '19 at 16:07
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    \$\begingroup\$ Are these toggle switches also feeding contactors or whatever that actuate the hoist? Or are the LEDs on a separate circuit using the 24VAC supply through second contact sets on the toggles? \$\endgroup\$ – Phil G Aug 29 '19 at 16:15
  • \$\begingroup\$ The 24v also feeds the toggle switches which in turn switch 24v relays back at the panel, this then determines the direction of the hoist using its own onboard 42v transformer, just to add we have never had any true failures of the LED's using the 24v AC transformer and resistors. \$\endgroup\$ – Alan Omega Aug 30 '19 at 8:19
  • \$\begingroup\$ Please post the schematics, not pictures of resistors. \$\endgroup\$ – Lundin Aug 30 '19 at 11:04
  • \$\begingroup\$ If you've never had a failure of the LEDs, then you need to ask yourself if a change is necessary. You are adding parts (5V power supply) and changing a circuit (resistors for LEDs) to fix a "problem" that has never caused difficulties. Additionally, the 5V power supplies are extremely over sized for a couple of LEDs. Each LED needs around 25 mA. The 3A powersupply could power over 100 LEDs. \$\endgroup\$ – JRE Aug 30 '19 at 11:06
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Now ever since the dawn of time, we have fitted an AC transformer in the panel which sends 24v AC and a 0v down to the handset which goes to the toggle switches and also the LEDS.

This is probably because when the system was designed at the dawn of time,

  1. They might have used 24 V incandescent lamps instead of LEDs.

  2. A 24 V transformer was probably lower cost than a DC supply.

I wish to ditch the AC transformer and fit a din mounted 5amp DC unit. My question is though, will I still have to fit the resistors? I know they should be needed to smooth out the current going to each LED but just wanted to check.

You will still need to fit the resistors. These limit the current through the LED so it doesn't blow up. You will probably need a different resistor value to get the same brightness from your lamps, when you switch from 24 V AC to some other DC voltage.

You can also buy LEDs or LED lamps with the resistors built in, which could save you having to assemble the resistors yourself.

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  • \$\begingroup\$ The 24v also feeds the toggle switches which in turn switch 24v relays back at the panel, this then determines the direction of the hoist using its own onboard 42v transformer, just to add we have never had any true failures of the LED's using the 24v AC transformer and resistors. It was designed, I feel, by someone who didn't really know what they were doing to a degree...I have the option of fitting a 3amp version of the DC power supply, I guess I just need to work out the value of the resistor I need based on 24v DC \$\endgroup\$ – Alan Omega Aug 30 '19 at 8:21
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In fact you should stop using AC for this task immediately. When you look up the datasheet of the LED you may notice that the LEDs have a maximaum reverse voltage of only 5 V. This is exceeded by your current setup.

I recommend 150 Ohms resistor for 5V supply. You can't put the LEDs in series as they have a common cathode.

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  • \$\begingroup\$ Initially, I feel the resistor was added to attempt to take the voltage down from 24v AC to 5v for that purpose. Like I've said above we've never had an LED failure in almost 12 years from normal operation in this current orientation i.e 24v AC Trans and 1.2K resistor. What resistor would you recommend for a 24v DC supply on that LED? \$\endgroup\$ – Alan Omega Aug 30 '19 at 8:30
  • \$\begingroup\$ Subtract 2 V from the 24 V supply and divide it by the needed current (20 mA). 1 kOhm. But you will need a resistor capable of handling 500 mW thermal power and you should make sure it can dissipate the thermal power. \$\endgroup\$ – Ariser Aug 30 '19 at 9:01
  • \$\begingroup\$ Great thank you \$\endgroup\$ – Alan Omega Aug 30 '19 at 22:38
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As far as I am aware it's unusual to power LED's with an AC supply?

Yes.

Primarily they prefer DC?

It's not so much that they prefer DC, as that LED's don't like having current shoved through them in reverse. And, in fact, if these are wired the way I suspect they are (just the resistor in series with the LED) then every half-cycle of AC they're seeing just about as much reverse current as they're seeing forward current during the other half cycle. I'm surprised you don't have aging problems with them.

I am assuming the use of the AC power supply was because of the very long distances these handsets have been used in, some over 100mtrs! and thus the AC works better.

A simple transformer would have been less expensive than a converter. I'm not sure at what point the cost would have reached parity for industrial stuff, but it was about ten years ago for wall-warts.

I wish to ditch the AC transformer and fit a din mounted 5amp DC unit.

5A is way more than you need for a few itty bitty LEDs. 24V / 1.2k\$\Omega\$ = 20mA. If all you're doing is powering the lights, you need 20mA per light.

My question is though, will I still have to fit the resistors? I know they should be needed to smooth out the current going to each LED but just wanted to check.

It's not to smooth out the current, it's to limit the current. A bare LED (which is what you have) drops a fairly constant voltage. If you try to supply a constant voltage, then there will always be some mismatch, and you'll either get too little current or too much. So you need resistors; those resistors are about right for a 24V supply, but you'll need to scale them for different supplies.

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  • \$\begingroup\$ Why would the LEDs conduct in reverse on the "wrong" half cycle? Are you thinking of the reverse recovery problem? \$\endgroup\$ – The Photon Aug 29 '19 at 16:16
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    \$\begingroup\$ I'm thinking that LEDs generally have a reverse breakdown voltage between 5 and 10 volts, and that 24 volts is bigger than 10 volts. \$\endgroup\$ – TimWescott Aug 29 '19 at 16:19
  • \$\begingroup\$ Re, "...just about as much reverse current..." The D in LED stands for "diode." A typical indicator LED is meant to light up with a few tens of milliamperes in the forward direction. If you push that much current through an LED in the reverse direction, then you are operating it in reverse breakdown. I'm not an expert on the subject, but I've heard that that mode of operation does not promote long life in PN diodes. \$\endgroup\$ – Solomon Slow Aug 29 '19 at 17:22
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    \$\begingroup\$ One obvious solution is to put two LEDs in anti-parallel with each other. There's two benefits: (a) One or the other is illuminated on every half-cycle of the AC waveform, which can reduce the visible flicker if they are close enough to each other, and (b), Neither LED can see a higher reverse voltage than the forward voltage of the other. \$\endgroup\$ – Solomon Slow Aug 29 '19 at 17:25
  • \$\begingroup\$ as you re-design your product, keep in mind the need to keep the LEDs COOL so they last. Incandescent bulbs are much more tolerance than LEDs. Ditto for power supplies versus transformers. Keep the DC supplies COOL , so you don't have warrantee problems. Use sheets of metal to absorb heat and spread heat, and venting slots for heat flows. \$\endgroup\$ – analogsystemsrf Aug 29 '19 at 17:48
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The LEDs from the link use indeed DC (afaik all LEDs do, but I don't have that much experience with different kind of leds).

Since those LEDs have 3 legs, these are two colored LEDs.

You will need resistors. It is not a normal lamp where the resistance is part of the lamp itself.

The resistors you use are way too much, if you use them, most of the power will go into heating up the resistors, which is not necessary.

As a basic rule, use:

V = I * R 

Mostly those LEDs can have a continuous current of 20 mA, so if you use a 5 V DC power adapter, you need a resistor of V - Vfd = I * R <=> (5 - 2.2 = 0.02 * R) <=> R = 110 ohm. To be safe, use 120 ohm. 2.2 V is the so called 'forward voltage' and needs to be reduced (similar like a diode, a LED 'is' a specific type of diode).

The power through those resistors will be P = V * I = 5 * 0.02 = 0.1 W, which most through hole resistors can handle.

If you use a 5V adapter, you also can put two LEDs in series (with one resistor), than you will get V - Vfd = I * R <=> 5 - 2 * 2.2 = 0.02 * R <=> R = 30 R. However, since not every diode is equal, to use a slightly larger resistor. This gives you less power loss.

Also note, that LEDs have two different sides, so make sure you put the anode and cathode side correctly.

Update

First, I'm not experienced with LED chains myself.

For a 24 V supply, using 1200 R resistors, you will have:

V = I * R <=> 24 = I * 1200 <=> I = 20 mA

This is exactly the max current a LED can get so you are ok.

The power is:

P = V * I = 24 * 0.02 = 480 mW, which is just below 0.5W

So yes, one 1200 R resistors fit with one LED (or actually, one per color), but if you want to have less resistors you can put multiple LEDs in series with one resistor, and also you lose less power in resistors (heating).

E.g. Assume you want to use max 1/8 W resistors (default through hole), assuming 6 LED colors per chain:

0.125 = V * 6 * 0.02 = 1 V

This is the voltage through the resistor, this means the resistor should be:

V - Vfd = I * R <=> 24 - 6 * 2.2 = 6 * 0.02 * R <=> 10.8 = 0.12 * R <=> R = 90 R

Remember your bi color LEDs use 0.04 A per full led, and you might take some safety margin, e.g. the power should be 0.11 W instead of 0.125 what I used above.

But this way you save a lot of heat and number of resistors. Make sure the resistance is not almost 0, as the forward voltage might change slightly because of temperature, same as each resistor has a tolerance.

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  • \$\begingroup\$ We've never had any issues with the LED's themselves failing with the 24v AC supply and the resistors we use? Ideally, I need to use a 24v DC supply as this feed is shared by the toggle switches to switch 24v relays back at the panel. Would you be able to recommend a resistor that would work with the 24v DC? \$\endgroup\$ – Alan Omega Aug 30 '19 at 8:24
  • \$\begingroup\$ I updated my answer, 1200 R is 'perfect' for a single LED (color), but you can improve some \$\endgroup\$ – Michel Keijzers Aug 30 '19 at 9:13
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It has to be DC since this gives a stable current without flicker. Also, since a diode is a semiconductor, it will only allow current in one direction, unlike a bulb. If fed too much reverse voltage, it will break.

All LEDs have a rated current used for writing down the specification - 20mA is the industry standard for that. Absolute maximum rating is somewhere >25mA. Higher currents than that and the LED will blow up. Thus you need a current limiting resistor in series.

Forget about using 20mA though, you might need that much for crap-off-the-shelf LEDs like the one you linked. But you should go for a high intensity one instead, they cost the same but need far less current to produce the same light. Buying brighter LEDs ~200-300mcd for the same price means that you can cut the current consumption at least 20 times. Kingbright and others have these.

The next thing you need to note is that LEDs of different colors often have different forward voltage. Using Ohm's law is an over-simplification, the correct formula is (Vin - Vforward) / Iforward. So if you have 24VDC, a forward voltage of 2.5V for a certain color, and aim to get around 5mA, you should pick

(24-2.5)/(5*10^-3) = 4k3 ohm

4k7 might be a more sensible value, standard E series value.


And finally, you mention CE marking. To CE mark a control panel for hoists must conform to mandatory harmonized standards for electrical equipment. Namely EN 60204-32 regarding electronics for hoisting machines, which is harmonized under the Machinery Directive. This standard states which colors that should be used for certain situations (10.3.2). Red is reserved for emergencies and hazardous situations.

In order to dodge this text in the standard, you would need to raise a strong argument why red light is suitable for your product - for example if your customer has addressed this in their risk assessment. Which they probably haven't. So this is something that machinery inspection from any government is likely to leave remarks about.

In addition, it is very common that men have trouble distinguishing between the colors red and green in particular (women don't have this problem, but operators in traditional industry are mostly men still).

I would therefore recommend to use a different color than red. Green + yellow or green + white, something like that. You might have to use RGB diodes instead of bi-color.

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  • \$\begingroup\$ The LED's I use, and posted, are Kingbright not crap at all? Regarding the CE marking it's funny you should mention that, it was raised at the time but we were fine to use Red/Green. I remember seeing the paperwork stating why but I can't quite remember the reason why. Thanks for all the help \$\endgroup\$ – Alan Omega Aug 30 '19 at 22:49
  • \$\begingroup\$ @AlanOmega It's not about the brand but luminous intensity. These have a guaranteed 20 mcd, typical 60 mcd for both colors. As mentioned you can get parts with 200-300 mcd for the same price, the only difference being less current consumption. \$\endgroup\$ – Lundin Sep 2 '19 at 6:26
  • \$\begingroup\$ @AlanOmega And regarding red light and hoisting equipment, I'd say you are basically at the mercy on the individual inspector. If they are pedantic, they can decide to cause problems for you. \$\endgroup\$ – Lundin Sep 2 '19 at 6:28
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Since you say you've never had a failure in the LEDs driven by 24VAC, I would think the old adage "never touch a running system" applies.

From the description, though, you are operating the LEDs in a way that could cause failure. The LEDs you linked to have a maximum reverse voltage of 5V. You are subjecting them to way lots more than that.

Your description sounds like you have this:

schematic

simulate this circuit – Schematic created using CircuitLab

If you are concerned about the LEDs being exposed to the high reverse voltage, then you could do this:

schematic

simulate this circuit

D3 conducts during the half cycle of the AC when the LEDs would normally be exposed to the high reverse voltage. It prevents the reverse voltage from getting above about 0.7V. When the LEDs light up, they are only exposed to their normal forward voltage.

If you do that, you will want to change R1 from a 1 watt part to may a 2 or 3 watt part because it will be conducting current for a larger part of the time. You'll also want to use something like a 1N4002 - it is rated for higher voltage and current than required so you won't have reliability problems.


In the end, though, I suggest you consult an engineer locally. He could look at your circuits and tell you if you need to worry about the LEDs - or, more importantly, worry about and fix something that you may not have thought of.

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