JohnInTx,

here is, hopefully, my final wiring diagram. The only thing I'm unsure about is R4. I emailed the company that makes the 24vdc LED lights what the forward voltage or voltage drop is across the LED. All that was on the data sheet was 80 mA operating current and 24vdc. I attached the two data sheets incase you see something I missed. But let me know what you think about the final wiring diagram. Thanks again!!

P.S. I hope you enjoyed your holiday weekend!

 

The only thing I'm unsure about is R4. I emailed the company that makes the 24vdc LED lights what the forward voltage or voltage drop is across the LED. All that was on the data sheet was 80 mA operating current and 24vdc.

Based on the datasheets, I'd say that any necessary current limit is built into the indicator unit. The wiring diagram on pp15 shows none needed. Industrial stuff like this is usually pretty self contained. If it needed a resistor, it would not have a 24 volt spec, only an operating current and minimum Vf.

If you have doubts, make R4 300 ohms at 2W and turn it on. Measure the voltage across R4. If its about 12V you don't need R4 (sharing voltage drop with the internal resistor). Alternately, If you have a variable power supply connect the lamp and run the voltage up slowly. If a resistor is already present, you'll be able to run the voltage to 24V with no issues. If the light gets very bright at very low voltages, you need R4. BUT - I'm 99% sure its built in.

The schematic looks good - ship it!
Let us know how it finally worked out.

P.S. I hope you enjoyed your holiday weekend!

Thanks! 19hour slow-smoked brisket was melt-in-your-mouth great!

 

Based on the datasheets, I'd say that any necessary current limit is built into the indicator unit. The wiring diagram on pp15 shows none needed. Industrial stuff like this is usually pretty self contained. If it needed a resistor, it would not have a 24 volt spec, only an operating current and minimum Vf.

If you have doubts, make R4 300 ohms at 2W and turn it on. Measure the voltage across R4. If its about 12V you don't need R4 (sharing voltage drop with the internal resistor). Alternately, If you have a variable power supply connect the lamp and run the voltage up slowly. If a resistor is already present, you'll be able to run the voltage to 24V with no issues. If the light gets very bright at very low voltages, you need R4. BUT - I'm 99% sure its built in.

The schematic looks good - ship it!
Let us know how it finally worked out.

Thanks! 19hour slow-smoked brisket was melt-in-your-mouth great!

Thanks JohnInTx, I will let you know! and if I have any further questions I'll let you know! Thanks again!!

 

If you have doubts, make R4 300 ohms at 2W and turn it on. Measure the voltage across R4. If its about 12V you don't need R4

Thinking on that.. its not a real good test since there are probably several LEDs inside in series which will add up to close to 24V plus some internal current limiting. I bet you are good without R4.

 

Thinking on that.. its not a real good test since there are probably several LEDs inside in series which will add up to close to 24V plus some internal current limiting. I bet you are good without R4.

I think so too, given there is no additional information on the data sheet. But if I hear back from the manufacturer I will let you know.

 

JohnInTx,

I think I have an issue. Looking at the light that I am using, there is only one ground for both the lights. So if I close the ground then both lights will go on. I believe the only way to turn on each individual light is to send 24volts to it. Since there is only one ground per two lights, then both lights will go on if I complete the ground.

I have one of the mounts with me now. It has one wire for ground and 5 wires for up to 5 different lights. Since I'm going to have two lights on one base, how can I control the lights by controlling the 24v and not the ground? Or is there a way to still use the photo-couplers controlling the ground? Please let me know your input. Thank you

Attached is more schematics I have drawn up. Also, I know I told you already, but the part number for the light base is Litestak model: LSB-024-240 Federal Signal

View attachment 86658 View attachment 86659


View attachment 86658

 

Is there a way to do this with only 2 relays? Can I just parallel the signals some how? attached is what I have drawn up with 4 phoenix opto's

 

The sketch attached should fix things.
The two beacons are run in parallel from 2 DEK relays (1 each red and green).
The outputs of the DEKs are reconfigured to source 24V rather than sink - so your common ground in the beacon will work. I should have caught that, the other beacons I've worked with work the same way..
I haven't shown R4 et al - I don't think they are necessary.
Good luck.

 

I'm pretty sure this would work. I'm just thinking to hard about this whole circuit. If they both run on the same 24v supply, should that be ok?

 

Ok, I just saw your schematic. That makes sense. Instead of putting the LEDs on the source side, put them on the bottom of the "switch". Also, so that all the grounds are the same with all LEDS, There will only be on ground between the 24v source and the 2 light stacks? I see you added a pill up resistor between the inverters. Is that a 33k pull up resistor?

 

Yep.
One common ground for all the beacons returning to the ground of the 24V supply.
The pullup is the same one we added when we went to the stronger open-collector inverters.
EDIT: And to be clear on terminology, the LEDs are now on the source side. Previously, they were on the sink side. Source or sink refers to what the driver is doing - outputting current (sourcing) or providing a switched path to ground (sinking).

 

OK Great. Thanks!!! I'll finish my schematic and post it and see what you think!! Thanks again!!

 

Here is hopefully the final circuit. Let me know what you think and if you approve?

 

I think its good. You've noted R4 as dependent on power supply.. On 5V you probably don't need it? But not a bad idea to provide the space if you have doubts. When you wire it, you'll know for sure.
Other than that, its the way I would do something like this so I'm on board (obviously!)

 

I'm glad you said that. I meant to put R4 in line with the 24v light outputs. Again, I don't think I'll need it, but when it gets wired I'll find out Thanks again for all your assistance and patience!

 

Given the light had 80 mA operating current, depending of the current output of the power supply, not sure what R4 has to be

View attachment 86672

 

As I said, I don't think you need the resistor. The install drawings for the signal don't show any plus its spec'd at 24V/80ma. That implies that the necessary current limiting is built into the lamp. If you want to be sure, prototype it with R4 300 ohms at 2Watts. I think you'll find that its not needed i.e. the lamp will be quite dim due to 2X the current limit circuitry (yours and the built-in stuff).

I haven't used Federal but all the PatLites I've used work this way i.e. built in current limiting - just apply rated voltage with enough current capacity and away it goes. The lamp module will self-limit to what it needs.

 

I rewired my circuit to the way we just drew it out and now the inverter wont invert the signal. I am still using the 74ls04's (the 74ls06's come in the mail today). Would this new way of wiring make it not work with the 04's? I thought it was maybe the chip so I pulled it out and it was very hot. The voltage going to the inverter is correct. Once it goes through the first inverter, it reads about 3.5v. Then the next inverter output it reads about 3.5 v as well. Which turns both of the LED's on. Hopefully it will work better with the 74ls06's. Let me know what you think. Thank you JohnInTx!!


also when one 700 ohm resistor is connect to the opto through 5 volts. The first inverter signal turns off the one led. But the 2nd led is always on with an inverted signal of 3.5v (with or without any of the 6 first optos energized)

 

Review the last sketch and make sure the DEK inputs are wired like that. Your drawing shows different. You only need one A2 lead to the output of the inverter. The way it looks like you have it connects the inverter output directly to the +5 supply. The hot chip is the giveaway.

Recheck the other wiring as well.

While you are at it, redraw the schematic so that there are only TWO terminals on each power supply + and ground. Its hard to tell what's what.

 

Circuit1 has one 700 ohm resistor connected to leaving only one LED on. The next two pictures have it disconnected and both LEDs are one. With it disconnected, Inverters input is 5v. The inverted signal is 3.5v. That is jumped to another inverter input which outputs 3.5v.

I took out all the other wires to pull the other inverters lows so you can see the photo.

I am still using the PS2501-1's, but that doesn't explain why the inverter isn't inverting the signal.

I just pulled the inverter chip out just not while it was on and it was warm again. It had the exact wiring in the picture. The only difference in wiring is the last two optos. The chip never got hot before. What do you think it could be?