Have you considered LED strips?
It would cost quite a bit more, but would make wiring and construction easier.
You should be able to get something like these (UK store):
http://www.rapidonline.com/Electronic-Components/330mm-Rigid-LED-Strips-12V-8x-LEDs-524998
You would 7 this size so these ones would work out at about $50, but you can probably find cheaper.

 

Markd77,
Thanks for the novel idea but I've had all the parts including the nearly 200 individual LEDs on hand now so too late to make changes on this.

I'm not real familiar with those strip LEDs but seems you would have to
lay them out in segments and attach to some backing... so would be a bit of planning and layout there too and if they are more costly... the LEDs I bought were like $14. Trying to keep cost at minimum so this will do fine.

And...elec_mech has gone way beyond expectations to help me on all
aspects of this project so I want to continue with his support to justify his extensive, invaluable and consistant guidance...Best regards...

 

Well I've got all the holes drilled into the board now and may be needing some advice on laying out or lining up the LEDs in a practical fashion . One thing I was trying to work out is which LED to assign D2 to (as a starting point). BY THE WAY why aren't the LEDs starting with a "D1" rather than D2 ? Also with each segment, you have double rows of 4 and then a "point" as the beginning and end. How would you do it? Make the first point number D2 and the last one in the segment D13 ? I see the one picture of one you built has some screw terminal barrier strips for wire junctions. Anyway it seems like this could get a bit on confusing without a clear plan..thanks for the help.

 

why aren't the LEDs starting with a "D1" rather than D2 ?

D1 denotes the 1N4001 used for reverse voltage protection to the circuit. Because the number of LEDs was not set initially and changing all the LED designators then trying to remember to change any diodes in the main circuit after the fact seemed like a recipe for error, I opted to number the LEDs last (after any diodes in the main circuit).

Also with each segment, you have double rows of 4 and then a "point" as the beginning and end. How would you do it?

If you look at the schematic, there are two sets of six LEDs. It is easier to show than explain, so please see the attached drawings. The drawings are shown from the back of the display where you'll be connecting everything to make things easier. I assume you know how to correctly wire LEDs in series and know which lead is the cathode and anode. If not, please let me know before proceeding.

You can opt to put on terminal strips to the back of the display, but unless you have some spare ones handy, you're talking $6-10 extra. I'd solder the resistor to one end on the LED terminal, then solder wires to the ends and connect them to your circuit board.

 

Perfect !....

Duh... I see what you meant about the diode numbering. I forgot about D1 as the reverse polarity saftey diode so your explanation makes perfect sense and I should have thought of that reason.

I was trying to pencil in on a copy of the earlier template, how to arrange and connect things and since it was late and getting tired, I just thought I'd be more sure to get it right if you had some good suggestions. So... the drawings with the LEDs labled and the resistor connection points are a tremendous help. Good reminder that the view is from backside so the left/right relationship is reversed when considering placement of LED positions.

Regarding those multiple 200 ohm resistors.... I think you mentioned earlier that depending on the brightness result from other variables, I could use the 200 or the 130 ohm.... so if I use the 200 ohm I would expect a bit reduced brightness?
Wondering if it would make any sense to opt for the brighter output selection of resistors and then could a resistor be put in the common
supply voltage side to reduce current (brightness) overall if a person wanted to reduce the brightness somewhat?

Or ... for that matter (excuse my stupidity here if it applies) could the resistors at the cathode end going to the ICs be eliminated and use one
resistor on the supply side (Vcc end) of the LEDs? Maybe that would
require a high wattage version producing too much heat? Anyway, just a thought.... figured that way it would minimize the parts count and number of connections too.

Yes, just to be sure, I checked the data sheet to confirm the cathode and anode leads of the LEDs ... but thanks for the reminder... I may have lots of years of experience but never hurts to check as it is easy to overlook something and make dumb mistakes, ha... (and I've made my share).

 

... afterthought...

Again, just to be clear on my understanding, on your schematic you show the drawing of the two "digits" on the bottom right of the page....

When you refer to "1st Digit", is that the "ones" position as viewed by
the user and so "2nd Digit" is the "tens" position?
In other words numbers 0 through 9 will show up on the "1st Digit" and
if the number is 10 or greater, it will include lighting of the LEDs on the
"2nd Digit" .

Yet another way to say it would be that U6 drives the "ones" position display and U7 drives the "tens" position display. Correct?

Geesh, I feel like I'm really belaboring the point when I ask....
If I have my "board" with the 168 diode holes drilled into it and
I am on the "back" side inserting the LEDs..... the "ones" position
would be on my left and receive diodes numbered D86 through D169.
... and the top would include diodes D110 through D115.

Also with your numbering on the template pages you just sent, the upper segment of 1st Digit would be the LEDs D110 through D115.
And numbering on second Digit would be with the top segment using
LEDs D27 - D31... is this the correct position and oreintation of the respective "displays" ?

 

Regarding those multiple 200 ohm resistors.... I think you mentioned earlier that depending on the brightness result from other variables, I could use the 200 or the 130 ohm.... so if I use the 200 ohm I would expect a bit reduced brightness?

Relative to the 130Ω resistors, yes, the LEDs could appear dimmer. It really depends on the forward voltage of the LEDs, the voltage drop from the driver IC, and the output of the power supply. If you have both resistor sizes, your best bet would be to use one size on one digit (or just one segment) and use the other size on the other digit (or another segment), then look at them and see what you think. Also double-check the current going across each row of six for each resistor value to be sure the LEDs aren't getting too much current - extra safety measure.

Or ... for that matter (excuse my stupidity here if it applies) could the resistors at the cathode end going to the ICs be eliminated and use one resistor on the supply side (Vcc end) of the LEDs?

Not a stupid question at all, but a big NO. The reason is the LEDs will fight each other for current so you'd see some segments really bright and some really dim. Imagine a high school cafeteria with no line and no limit on portions. Some students would get too much food, some not near enough. The individual resistors on each line help ensure each segment gets the same amount of current and thus each segment will have the same brightness as the next one.

I may have lots of years of experience but never hurts to check as it is easy to overlook something and make dumb mistakes, ha... (and I've made my share).

Preaching to the choir . . .

When you refer to "1st Digit", is that the "ones" position as viewed by
the user and so "2nd Digit" is the "tens" position?
In other words numbers 0 through 9 will show up on the "1st Digit" and
if the number is 10 or greater, it will include lighting of the LEDs on the
"2nd Digit" .

Yet another way to say it would be that U6 drives the "ones" position display and U7 drives the "tens" position display. Correct?

Yes, yes, and yes.

Geesh, I feel like I'm really belaboring the point when I ask....
If I have my "board" with the 168 diode holes drilled into it and
I am on the "back" side inserting the LEDs..... the "ones" position
would be on my left and receive diodes numbered D86 through D169.
... and the top would include diodes D110 through D115.

Correct.

Also with your numbering on the template pages you just sent, the upper segment of 1st Digit would be the LEDs D110 through D115.
And numbering on second Digit would be with the top segment using
LEDs D27 - D31... is this the correct position and oreintation of the respective "displays" ?

You got it.

 

Thanks for the great explanations....a side question....would you mind explaining to a dunce how you select and indicate a paragraph / or lines you want to place as a "quote" for reference into a reply ?

Not being familiar, I have to show a copied section from your last post shown below to ask further questions on....

"Relative to the 130Ω resistors, yes, the LEDs could appear dimmer. It really depends on the forward voltage of the LEDs, the voltage drop from the driver IC, and the output of the power supply. If you have both resistor sizes, your best bet would be to use one size on one digit (or just one segment) and use the other size on the other digit (or another segment), then look at them and see what you think. Also double-check the current going across each row of six for each resistor value to be sure the LEDs aren't getting too much current - extra safety measure."

Are you able to start the use of say, just one segment, by connecting just that group of LEDs to the driver IC (including the resistors, of course) and have no other LEDs yet connected to the remaining driver output lines....? ..without damaging anything...ie.....no load condition on those lines? Otherwise it wouldn't be practical to do trials of resistor options and see the results as opposed to connecting up all 168 to see what they look like.
To repeat a dumb question, the 200 ohm resistors would give the dimmer light output but would be safer in limiting the max current through the LEDs..? By the way, being too lazy to check, do you know what the max current should be and should I be able to check just one leg of a segments diodes for that value....what might be a safe level without running them at max...perhaps 80 % of max ?

 

how you select and indicate a paragraph / or lines you want to place as a "quote" for reference into a reply ?

1. Click Post Reply.
2. Highlight and copy the desired text you want quoted.
3. Click on the body of the message so the cursor is blinking.
4. Click on the icon just above the message that looks like a caption from a comic strip.
5. You'll see QUOTE and /QUOTE surrounded by brackets with the cursor blinking between the two.
6. Paste text you copied.

Are you able to start the use of say, just one segment, by connecting just that group of LEDs to the driver IC (including the resistors, of course) and have no other LEDs yet connected to the remaining driver output lines....? ..without damaging anything...ie.....no load condition on those lines? Otherwise it wouldn't be practical to do trials of resistor options and see the results as opposed to connecting up all 168 to see what they look like.

If I understand your question, yes. No damage will befall the circuit if only one segment or even just one row of LEDs is connected and the others are not. You can think of the driver IC like a switch. If nothing is connected to the output, nothing happens, but no harm is done either.

If you've tested your circuit, you can simply connect, say both "a" segments for each digit, one connected through 130Ω, the other 200Ω, power the circuit and trigger the reset briefly so 00 will be shown which will turn on both a segments.

To repeat a dumb question, the 200 ohm resistors would give the dimmer light output but would be safer in limiting the max current through the LEDs..?

200Ω would limit the current more than 130Ω, yes. Very roughly speaking, the less current passed through the LEDs, the longer they will last. But as long as the current draw is at or below the manufacturer's recommendations, you'll get many years of service.

By the way, being too lazy to check, do you know what the max current should be and should I be able to check just one leg of a segments diodes for that value....what might be a safe level without running them at max...perhaps 80 % of max ?

According to the datasheet, absolute max is 25mA, so 20mA would be ideal.

You're only concerned about the current through one row of six. You can check between the driver IC output and the resistor - easy to do since you have the terminal blocks.

 

Well.... I got the top segment of both digits put together. I wired one with the 130 ohm and the other 200 ohm resistors. Was thrilled to see that the circuit board and LEDs worked right away ! Kinda fun to see all that effort actually produce a working product (although far from complete). Most of my life has been in troubleshooting problems in manufactured finished products.
Anyway I found that the current draw with the 130 ohms was 48 milliamps,
the 200 ohm version drew 35 mills. I experimented and found that by wiring the two in series to use 330 ohms yielded about 24 mills. However,
this was using a variable bench power supply and had it at about 17.5 volts. Thought I'd better connect up the actual "wall wart" that will be used and unfortunately with 330 ohm resistance, it shows a draw now of
28 milliamps !
So may have to use a 390 or 430 resistance... dang... done a lot of rewiring just in swapping the other values in and out .. soldered them all to insure good connections in this trial. Main downer is that I will have to probably order 30-35 of the final value since I have no nearby place to buy them and so will delay me nearly a week on getting it completed. Oh well,
there is no way to be sure without some trial and error and I need to build the "box" and mess with the R/C situation and build a housing for it's up/down/reset buttons and was thinking of attaching a little telescoping whip antenna in stead of just the loose dangling transmitter antenna. Yes, I will have to see if it is long enough fully extended or to what extension to produce best results knowing it should be some fractional multiple of the wavelength for 27 Mhz.

Just thought I'd update you on status but am very happy with the appearance and function thus far. That scrap plastic (I think it is ABS) from the TV back cover was ideal... drills nicely and not fragile and using the template you provide with centerpunched holes, I fortunately didn't mis-drill and ruin anything.

 

Anyway I found that the current draw with the 130 ohms was 48 milliamps, the 200 ohm version drew 35 mills. I experimented and found that by wiring the two in series to use 330 ohms yielded about 24 mills.

Hmm, something sounds way off here. Perhaps I royally messed up the math.

Let's step through it.

According to the datasheet, the typical forward voltage is 2.2VDC with a maximum of 2.5VDC. Six LEDs in series will thus require 6 x 2.2 = 13.2V.

Now, there is roughly a 1V drop through the driver IC, which means we'll need at least 13.2V + 1V = 14.2V to drive the LEDs.

If you go through D1 for reverse voltage protection, 0.7V is dropped across the diode and we'll need 14.2 + 0.7 = 14.9V.

We've selected a 18VDC power supply, so 18 - 14.2 = 3.8V needs to be dropped across the resistor. We want to limit the current through the LEDs to 20mA, thus 3.8V/0.02 = 190Ω.

This is assuming we bypass D1. If not, (18-14.9)/0.02 = 155Ω. I don't think I took this into account originally.

Now, the datasheet also says the max forward voltage could go as high as 2.5V. I figured 2.4V was more realistic. Thus, 6 x 2.4 = 14.4V then with the drop across the driver IC, 14.4 + 1 = 15.4V.

18-15.4V = 2V, 2.6/0.02 = 130Ω. Again, didn't factor in D1.

If you see an error in the math, please let me know.

48mA and 35mA? Very odd indeed. Could you check the voltage drop across the following points for both resistor values?

1. Across the six LEDs (cathode and anode at ends of row - just the LEDs, not the resistor or driver IC).
2. Across a single LED within the row. Please do this for each of the six within a row to verify voltage drop is about the same for each.
3. Across the resistor.
4. Between GND and the driver output.
5. Between the driver IC output pin and Vcc (anode of LED row).
6. Across D1.
7. Across the input - verifying the input voltage is indeed 17.5V or 18V.

Is the anode of each LED row connected directly to your 17.5/18V input or after D1?

When you say you put two rows in series, I'm picturing a total of twelve LEDs in series with one 330Ω resistor. Is that correct? And you then saw 24mA with a 17.5VDC supply? Or do you mean you simply used a 330Ω resistor in series with each row of six LEDs (in place of the 130/200Ω resistor)?

Lastly, where are you taking the current measurement from? I don't doubt your abilities in the least, I just want to cover all bases.

How far away do you need the remote to work from the display?

I've got several spare extension antennas, happy to mail one if you'd like once you get a length in mind. Offer for the green cellophane still stands as well.

You may want to consider gluing the LEDs into the plastic so they don't come out in the event the display is dropped. Super glue or any CA should be okay, but I'd strongly suggest testing the glue with a spare LED and scrap piece of your ABS and waiting a day or two to verify the LED doesn't come out.

 

follow-up...

I discovered that a 470 ohm resistor shows a current draw of 20.9 milliamps so guess I will have to go with that as a final value.

BTW... maybe mentioned before but regarding the functions... when you hit reset, does that result in two zeros being lighted / displayed or does it have all LEDs off?
Also when using only digits up to 9, will the left (tens) side be unlighted
(not a zero... just blank)? It would have been nice to have the tens side
dead when not in use so it would add to the validity that the displayed number is NOT greater than 9 from a distant glance...

If there is a simple re-wire to have the left side blank when not showing
a value, that would be helpful otherwise just something not thought about soon enough but may also have added a lot more to the circuitry.

Will be awaiting your thoughts... thanks much !

 

Seems that I don't always get the email notification of a new posting so somehow the messages might be out of order on the last couple. I didn't see your post before I sent this previous one starting with ... follow-up.

Sooo I will review all your comments later and try to provide all the information you would like. Very possibly I'm doing something wrong.

With it being Sunday morning, we are heading for church services and have a busy afternoon... just became grandparents a couple days ago !

If I have a chance later today I will try to check things out and reply. Your math seems right after a quick read through. Regarding a little antenna, I found an old trashed cordless phone and took off the telescoping antennas that were on both the base and handset plus discovered another in my junk drawer so think I'm in good shape there but thanks much for the offer.

TTYL

 

BTW... maybe mentioned before but regarding the functions... when you hit reset, does that result in two zeros being lighted / displayed or does it have all LEDs off?
Also when using only digits up to 9, will the left (tens) side be unlighted
(not a zero... just blank)? It would have been nice to have the tens side
dead when not in use so it would add to the validity that the displayed number is NOT greater than 9 from a distant glance...

If there is a simple re-wire to have the left side blank when not showing
a value, that would be helpful otherwise just something not thought about soon enough but may also have added a lot more to the circuitry.

Reset will force both digits to show zero. The CD40110 does not have a blank enable pin, so there is no way through the IC to force one digit off.

You could add a transistor to the anode on the tens digit, but then you'd need to add some logic (using an IC or several diodes). Not sure if there would be enough room on the board or not.

While I see your logic in desiring only one digit lit when the value is under ten, I've never seen this done in a commercial counter. I suspect the reason is if a digit failed outright , you'd know immediately that there was an issue with the counter since both are always lit. Also, I suspect most people have seen these in action and seeing only one digit lit might make them question if the counter was working correctly.

I would suggest against turning off one digit, but if it is something you really want to do, I can put some ideas together.

 

Hi, I have "egg on my face".... but first regarding this group of phrases from your message....

[/When you say you put two rows in series, I'm picturing a total of twelve LEDs in series with one 330Ω resistor. Is that correct? And you then saw 24mA with a 17.5VDC supply? Or do you mean you simply used a 330Ω resistor in series with each row of six LEDs (in place of the 130/200Ω resistor)? QUOTE]

What I said was....

I experimented and found that by wiring the two in series to use 330 ohms yielded about 24 mills.

So sorry for poorly stating that. I tried to arrive at a resistor of a total of 330 ohms for experimenting by just putting the 130 and 200 ohm resistor in series.

Now..... I did all the testing you asked for and I could give you all the details if you still like. However, what dawned on me... pardon the stupidity ( I'm really not as dumb as I appear sometimes, LOL) was that when I measured the current flow supposedly through the LEDs, I put the ammeter in series with the 18 volt supply and the anode junction of both legs of LEDs !!!!

So I was reading the total current draw through both legs and therefore my actual value through EACH of the two paths should have been stated as HALF of the value given.... ie.. when I read 48ma, there was about 24 ma through each leg. And... half of the 35 mills with the 200 ohm would mean about 17.5 through each.

Well with the little experimenting and notes I made from your request, I measured 21ma using the 200 ohm resistor... my earlier measurement may not have been as accurate when I said the combined was about 35.
So.... if I get one leg (6 LEDs and 1 resistor) to draw 21ma that should be a safe value based on the max should not go over 25ma. Therefore with both legs connected and measuring the draw between supply voltage and the anode junction of both rows of LEDs, it is fine to measure a draw of about 42mills, correct?

Sorry about all the confusion... should have just put the meter in between the resistor and LED cathode to begin with but got off on the tangent of connecting up both portions of that segment, awaiting the excitement (sorry... I don't get out much LOL ) of seeing two rows of LEDs lighting up, not thinking about the current draw reading is both legs and not just one... duhhhhh.

For some reason that fires "Quote" attempt didn't show up right on my preview but to lazy to try and fix it, ha.

Soooo I should probably go back to using the 200 ohm resistors in each leg. I will check it once more before leaving this issue.

By the way with your explanaion of the "math" and "walking" back from Vdd... first you mention going through the LEDs and their voltage drops,
(skipping the resistor since that was the "unknown" value we needed to solve for) then the IC voltage drop. Then you went on to discuss the D1 being included or not. Here again is my weak theory... doesn't the "walk-thru" explanation you gave end up with passing through the IC and then to ground?... Isn't the 12volts just there for supplying the other ICs whose logic out "triggers" the driver ICs? Again.. I was thinking the 18 volt and 12 volt supplies are working independantly other than that they come from a "common" source? Maybe I just need to think it through better, ha.

Regarding the other questions about what LEDs are on when reset, etc... I understand and we will leave things as they are... no problem. Yet another question. In the interest of saving "juice", should I consider putting this system on a "timer" or at leas plugged into a power strip to turn it off as during non-work hours? Actually more concerned if leaving this run 24/7 will cut much from the life of the LEDs... your thoughts?
Thanks!

 

Another way to measure current when you have known resistors is to measure the voltage across them and then use Ohms law
eg. if voltage across 200 ohm resistor is 1.5V, current = 1.5/200 = 0.0075A = 7.5mA
Most resistors are 1% or better accuracy and this is often a better method than measuring current directly, because the meter has a small resistance in current mode, which reduces the voltage to the circuit that you are measuring.

 

I tried to arrive at a resistor of a total of 330 ohms for experimenting by just putting the 130 and 200 ohm resistor in series.

Ah, the two resistors. I thought you were referring to the two rows . . .

So.... if I get one leg (6 LEDs and 1 resistor) to draw 21ma that should be a safe value based on the max should not go over 25ma. Therefore with both legs connected and measuring the draw between supply voltage and the anode junction of both rows of LEDs, it is fine to measure a draw of about 42mills, correct?

Much, much better. Because I've done plenty of silly things myself, I've learned the hard way to verify the things we take for granted like measuring current or making sure the on switch is on, etc.

Then you went on to discuss the D1 being included or not.

I had to look at the schematic. As drawn, the anode of the LED rows goes directly to +18VDC. This is fine and works, but it does not afford the LEDs any protection against reverse voltage. I *think* they'll be fine as is, but you could connect the anode from the LED rows to the cathode of D1 giving them reverse voltage protection but losing 0.7V to the LEDs. If you opt to do this, I'd suggest using a second 1N4001 just to limit the current going through D1. This is optional and up to you though. The only real fear is if someone loses the power supply and tries to connect one with the polarity reversed.

Yet another question. In the interest of saving "juice", should I consider putting this system on a "timer" or at leas plugged into a power strip to turn it off as during non-work hours? Actually more concerned if leaving this run 24/7 will cut much from the life of the LEDs... your thoughts?

Hmm, good point. It would be ideal to cut power when not in use - will certainly help with LED life. You have a few options here:

1. Add a switch to the power supply - you could use one of those cord lamp switches found in hardware stores.
2. Use a wall strip with a power switch as you mentioned.
3. Plug the power supply into a timer that plugs into the wall outlet.
4. Add a light sensor to the circuit that cuts the power to the LEDs when the office lights are turned off (assuming the office cuts off the lights at COB).

Options 1, 2, & 3 are easy, but they rely on someone performing a task and can be bypassed. Option 4 is more involved, but you eliminate reliance on the user and prevent accidental bypassing. If you want to give this a shot, let me know and I'll see what I can come up with. I haven't done a light/dark circuit, but I have plenty of examples on my bookshelf.

 

So you are saying to parallel D1 with another like it to split the current load between them?

I'd suggest using a second 1N4001 just to limit the current going through D1. This is optional and up to you though. The only real fear is if someone loses the power supply and tries to connect one with the polarity reversed.

How dumb would it be to "house" the wall wart inside the cabinet with the
display unit and plug it into an extension cord. Guess I could see how long that cord should be and if I needed a longer one than "off the shelf", I could do a good splice with heat shrink insulation just inside the housing and secure the wall wart and make the cord however long I need it. At least there would unlikley be anyone reversing polarity unless they go inside and mess with things.

Well... guess I can get back to putting more LEDs into the board... golly that is tedious along with forming and soldering and clipping the lead wires.

As far as needing anything to keep the LEDs from coming out of the board.. I have the holes drilled with a #10 bit and one needs to use
a plier gripping the leads with the nose of the plier resting against the base of the LED .... then push as hard as possible and they seat in very tightly so no fear of coming out.
I found a product, new to me, that works very well gluing plastics... it is
by JB Weld and is in one of those dual syrenges. It adheres very, very well to the ABS and holds these small plastic connectors I salvaged, very strongly. So much better than any previous epoxies or plastic cement products I've used in the past. Only thing close is the JB Weld regular two tube gray product.

Well.. on to working in the shop again...

 

So you are saying to parallel D1 with another like it to split the current load between them?

If you want to add the reverse protection to the LEDs, then yes. But note you may need to use the 130Ω in place of the 200Ω due to the voltage drop across the 1N4001. I'd test the diode first before soldering in any more resistors - don't want to do all that work then find you have to replace all the 200Ω resistors.

How dumb would it be to "house" the wall wart inside the cabinet with the display unit and plug it into an extension cord. Guess I could see how long that cord should be and if I needed a longer one than "off the shelf", I could do a good splice with heat shrink insulation just inside the housing and secure the wall wart and make the cord however long I need it. At least there would unlikely be anyone reversing polarity unless they go inside and mess with things.

Not a bad idea, but the only way someone could reverse polarity is by losing the power supply and attempting to replace it with another one where polarity was reversed at the tip. No one can accidently reverse the polarity of your existing power supply - the concern is only if they lose the power supply. Now, putting the supply inside the housing would solve that issue, but another concern is heat generated by the power supply - you may need to vent the housing.

Well... guess I can get back to putting more LEDs into the board... golly that is tedious along with forming and soldering and clipping the lead wires.

Bah, you should be an expert after the first two rows. I just bend the leads so they are parallel to each other and touching, cut off any excess if needed and solder. A little tedious, yes, but it should go quickly.

I've found many LEDs have a very, very slight taper to them. This means the base will initially hold well into a tight hole, but only the widest part of the taper is holding to the material, not the entire height of the LED. A few knocks, drops, or just jostles during transport, not to mention thermal fatigue, will cause the LED to fall back if not out right pop out.

JB Weld is fine. You may want to scratch or score the surface of the ABS around the holes with an X-acto or sandpaper - this helps the adhesive get a better "hold" of the plastic.

 

Soooo would there be anything wrong with connecting a reverse polarity prevention diode in line between the 18volt supply and the anodes of all the LEDs ? My estimate would be max current with all LEDs on at about
1.4 amps so would need a 2 amp diode?

Pretty hard to get another in parallel across D1 on the board unless I did some trace cutting in another area to isolate and then put both there with
jumper wire back to where D1 was in ckt.

Then those two 1 amp diodes in parallel would allow handle the 1.4 amps.
I noticed with connecting a diode as mentioned in the first paragraph, it dropped the current draw by about 3 mills... to about 18-19 mills.

.... just thinking, ha.