LED wiring project questions
October 18, 2021 1:26 PM   Subscribe

What is the best approach for updating the wiring/lighting in this small craft project? Resources for getting started with LED's?

My mom gave me an old advent calendar to repair that was built by my uncle decades ago. The wiring for the lighting looks like it should be completely replaced. Some of the insulation on the wiring is completely burned off. (Pictures: https://imgur.com/a/pFDdq1t)

I'm thinking I could update this using LED's, but I don't have much experience with them and the available options are overwhelming. Would there be a kit I could purchase, or do I need to assemble a kit myself from parts? What would I need to purchase, and from where? Types of bulbs, power supply, a controller?

I'd like to keep them all wired in parallel. I'm OK with some soldering. I don't need these to change colors, but having them be dimmable would be nice if it doesn't add too much expense.

Thanks for any info and resources!
posted by pilibeen to Technology (18 answers total)
Best answer: The easy way would be to get a 12VDC power supply and 12V prewired LEDs, i.e., like power supply and LED. Then you just wire them up to the existing bus structure. Do make sure to get the polarity right.
posted by doomsey at 2:52 PM on October 18, 2021 [1 favorite]

Best answer: The main thing to understand about LEDs is that unlike incandescent light bulbs they're not very good at regulating their own power draw.

The electrical resistance of an incandescent bulb has a very marked positive temperature coefficient: when the filament is cold the resistance of the bulb is very low, but the hotter the filament gets, the higher its electrical resistance becomes. This means that the bulb will draw enough power to start glowing even if connected to a supply that's well below its nominal rated operating voltage, and will also provide roughly the same light output for a relatively wide range of operating voltages either side of nominal.

So if you have multiple incandescent bulbs and you want them to glow roughly equally brightly, the easiest way to achieve that is to wire them in parallel so that they all see the same supply voltage, and let each bulb choose its own operating current. Those currents will not be identical because of manufacturing variations, but they'll be close by design.

If you wire incandescent bulbs in series, this forces them all to operate at the same current rather than the same voltage. But this configuration is actually somewhat unstable: if any particular bulb in the series string has a slightly higher resistance than the others, then the fact that they're all operating on the same current will make the voltage across the high-resistance bulb greater than that across the others, which will mean it dissipates more power than the others, which will make it brighter, which will make it hotter, which will increase its resistance even more. In a long series string, this positive feedback can be enough to cause bulb failure.

This is the same "hot spot" effect you see in old bar radiators and toasters, where the hottest spot in the radiator element is also the one that corrodes the fastest, which makes it thinner, which increases its resistance, which increases the voltage across it compared to other stretches of the element of similar length, which makes it dissipate more power than those, which makes it hotter, which makes it corrode faster until it becomes the place where the element ultimately fails.

For multiple bulbs wired in parallel, the bulb that runs brightest will be the one with the lowest filament resistance because that's the one that will draw the most current at the common supply voltage. But the tendency to draw more current increases the filament resistance, both temporarily because of the positive temperature coefficient and permanently because of faster metal evaporation from the filament surface. Over time, then, the current draw of incandescent bulbs wired in parallel tends to equalize, which also equalizes their power draw and therefore their brightness.

LEDs, by contrast, have a very sharp "knee" in their current consumed vs. voltage applied curve. Below a particular threshold voltage, which depends more on the colour of the LED than its size, it will draw almost no current at all and produce no light; above that voltage, current drawn will increase very fast as supply voltage increases. A 5% increase in supply voltage can, under certain conditions, double the amount of power dissipated by a LED run off that supply.

This means that running LEDs in a series string, ideally from a power supply designed to supply a constant current rather than a constant voltage, is the best way to get them roughly equally bright.

In practice, constant-current supplies tend not to get used. Instead, they get approximated by feeding series strings of LEDs from constant-voltage supplies via current limiting resistors. The resistor moderates the LED string's extreme sensitivity to supply voltage to an extent that keeps the LEDs inside their safe operating area.

LEDs, also unlike incandescent bulbs, are electrically asymmetric: they only conduct current flowing in one direction. If you wire them in backwards, they won't conduct any current until you jack the applied voltage up high enough to destroy them instantly. So LEDs must be supplied with direct current and wired in with the correct polarity; you can't wire them straight to an AC supply the way you can with incandescent bulbs.

Referring to this table of forward voltages for various colours of LED and taking blue LEDs as an example (white LEDs are just blue ones with an extra phosphor coating to make some of the blue light yellower) we see a range from 2.5 to 3.7V. The "knee" in the current vs voltage curve means that at 2.5V the LED will be only just glowing, if at all, while at 3.7V it will be running hard enough to risk burning out. A typical operating point for a white LED would be something like 3.35V at 30mA, but check the data sheet for whatever LEDs you end up with.

So if you want to run white LEDs off a 12 volt DC supply, the standard way to do it is to put three of them in series, which will drop 3.35V * 3 = 10.05V at their operating voltage, and then add a series resistor sized to drop the 2V difference between that and the supply voltage at 30mA. Ohm's Law for resistors is resistance in ohms = voltage in volts / current in amps, so you'd want a resistor of 2V / 0.03A = 67 ohms. Closest standard part would be 68 ohms. Power dissipation in that resistor would then be 2V * 0.03A = 0.06W, so an ordinary quarter-watt type would be completely suitable.

When running reasonably close to their designed operating point, to a pretty good first approximation the LEDs will maintain the same forward voltage. So if the supply voltage is low by 5% then we'd see 12V * 95% - 10V = 1.4V across the series resistor, which would make the current through it drop to 1.4V / 68 ohms = 20mA. In fact the LEDs would reduce their forward voltage a smidge at this reduced current, which would increase the drop across the resistor, which would tend to bump the current back up again; the string would settle on an operating current somewhere between 20 and 30mA and continue to operate quite safely.

Likewise, at 5% over nominal we'd expect 12V * 105% - 10V = 2.6V across the series resistor, yielding 2.6V / 68 ohms = 38mA string current. In practice the LEDs would increase their forward voltage a bit and the current wouldn't get quite that high. Still pretty safe.

To run more than 3 LEDs off that 12VDC supply you'd run them in groups of three, each with its own 68 ohm current limiting resistor, and wire the groups in parallel across the supply.

We could improve the consistency of the operating point in the face of supply voltage variations by putting fewer LEDs in the series strings and using correspondingly higher limiting resistances, but this would waste power. In the limiting case, with the series strings each consisting of a single LED and its own dedicated limiting resistor, the resistor would need to drop 12V - 3.35V = 8.65V at 30mA, and would therefore dissipate 8.65V * 30mA = slightly over a quarter watt. So we'd need beefier resistors, they'd each run warm, and we'd need three times as many of them. This is essentially what you get if you use the pre-wired 12V LEDs that doomsey linked above.

A much better idea is to use a nicely regulated 12VDC supply to keep the supply voltage stable regardless of what the mains voltage does. Fortunately, suitable 12V wall warts with efficient little switch-mode DC supplies built in are readily available; the one that doomsey linked would be perfect for this job.

Now let's talk about dimming.

Incandescent bulbs, unlike LEDs, are quite happy to run off an alternating current supply. Standard AC lamp dimmers work by interrupting the AC supply, connecting the bulbs to the supply for only part of each AC cycle. This will continue to work even if, as is the case for your existing low-voltage appliance, the bulbs are supplied via a 120V to 12V AC step-down transformer.

It won't work for LEDs. For starters, that nicely regulated 12VDC switching wall wart is going to be doing its damndest to maintain a stable 12VDC on its output regardless of what violence you're doing to the mains supply that feeds it. So we're going to need to dim the LEDs downstream of that little power supply.

The most efficient way to do that is to use a similar supply-chopping trick directly on the DC power supplied to the LED strings. This also has the advantage that instead of switching the LEDs on and off at 120Hz, as a mains dimmer would do, we can easily switch them much faster and avoid all visible flicker.

This little circuit, designed around the cheap and readily available 555 timer chip, will do that and should be easy enough to build up on a bit of perforated strip board. As you can see from the circuit diagram, it uses the same three-LED plus 68 ohm strings explained above, so you could easily add it on after getting the essential LED wiring and power supply done and tested first.

This is an excellent little project on which to sharpen up your soldering skills. Good luck!
posted by flabdablet at 5:37 PM on October 18, 2021 [7 favorites]

That would be the not easy way :)
posted by doomsey at 5:41 PM on October 18, 2021 [1 favorite]

To be fair, pilbeen did ask for the best way, not the easiest :)
posted by flabdablet at 5:46 PM on October 18, 2021 [2 favorites]

If you want the absolutely easiest dimmable LED replacement for what you already have, use doomsey's pre-wired 12V LEDs as drop-in replacements for your incandescent bulbs and run them off your existing 12VAC transformer via a bridge rectifier to convert the transformer output to pulsed DC (if you want to build your own bridge out of four diodes, the 1N5404 is cheap and more than adequate).

That should let you use a standard light dimmer upstream of the transformer, assuming you could work out a way to make the mains wiring workable. But this arrangement really is kind of janky, and you'd definitely see flicker.
posted by flabdablet at 6:10 PM on October 18, 2021 [1 favorite]

Slightly less janky would be a standard lightstrip dimmer. Would work fine in the context. Example
posted by doomsey at 7:49 PM on October 18, 2021

Response by poster: Thanks for the links doomsey, and thanks for the solid primer flabdablet - much appreciated.

I'm going to experiment with Tinkercad Circuits to test out some of the options, and trying to figure out the resistance in ohms for the pre-wired 12V LED's that doomsey linked. It looks like the resistors are rated at 0.25watts. I found a formula P=V²/R, which I used to come up with 36 ohms on the resistor? Does that make sense?

Correct me if I'm misunderstanding anything:

doomsey's suggestion of using the 12V LED's in parallel off the existing bus would be the simplest solution, but it's a less efficient circuit and the resistors will run warm.

flabdablet's idea of doing the series/parellel circuit with groups of 3 LED's wired in series with a resistor, and those groups wired in parallel, would be a more efficient design, only downside is more soldering.
posted by pilibeen at 6:19 AM on October 22, 2021

Response by poster: I just checked my math on the resistor in the pre-wired 12V LED's, and the formula P=V²/R actually gives me 576 ohms on the resistor??

Regardless, I'm going to build flabdablet's circuit suggestion (less the dimmer), since it seems like the ideal solution and I need the soldering practice.

Thanks again for the help!
posted by pilibeen at 8:22 AM on October 22, 2021

Best answer: only downside is more soldering

Not a huge amount more.

With paralleled integrated LEDs, you need two solder joints per LED, between the ends of the pre-supplied wire leads and the supply busbars. So every three LEDs will need six solder joints.

With bare LEDs and resistors and supplying your own linkage wires, you still need two solder joints per LED. Five of those in every group of three will be from a LED pigtail to a bit of wire, and one from a LED pigtail to a resistor pigtail. Then you need one more to join the resistor pigtail to its own bit of wire, then two more to connect the wires at the ends of the string back to the supply busbars: nine joints for every three LEDs.

And at least some of those LEDs are going to be close enough to the busbars that you might not even need wire between the busbar and the LED, which saves you one joint, and/or between the second resistor pigtail and the other busbar, which saves you another. So some of the strings are going to need only eight or maybe even seven joints per three LEDs. You might even be able to connect some of the LEDs in the string directly pigtail to pigtail, without an intervening bit of wire; if you manage that for all three and the busbar connections at the ends, that's only five joints for the string - one less than the parallel configuration.

Your main issue is going to be protecting the joints and routing the wires neatly so you don't get shorts. Heat shrink tubing and Kapton tape will be your friends there. Don't forget to slip the unshrunk heatshrink over your wires before soldering them onto device pigtails :-)

the formula P=V²/R actually gives me 576 ohms on the resistor??

Not the right formula, I think. 1/4 watt will be the resistor's rating for maximum continuous power dissipation - that's a limit, not a guide to an operating point.

R = V/I (Ohm's Law) is the formula you need for this, and in order to use it you need to make some assumptions about the LED's operating point. If you're running your LEDs at 3.35V and 30mA, which is typical for a bright white, then you need the resistor to drop the difference between the 3.35V across the LED and the 12V supply while passing 30mA (0.03A). So that gives you a resistance of (12V - 3.35V) / 0.03A = 288 ohms.

Next highest standard value is 330 ohms, which (assuming to begin with that the LED still runs close to 3.35V for reasonable operating currents) will give you an operating current of (12V - 3.3V) / 330 ohms = 26mA. Power dissipation in the resistor is then (12V - 3.3V) * 26mA = 226mW which just squeaks under the quarter watt dissipation limit.

Next standard value up is 390 ohms, for which the same calculations yield 22mA operating current and 194mW power dissipation. Still plenty bright and a bit less close to letting the magic smoke out. That's the value I'd pick if I were designing those things. Plus I'd use half-watt resistors instead of quarter-watt ones and charge one cent more for my LEDs.

Even though we're designing our 3-LED strings around 68 ohm current limiting resistors, which should get you roughly 30mA LED current with a 12V supply, it wouldn't hurt to check out how bright the LEDs are if you use an 82 ohm or 100 ohm or 120 ohm resistor instead. You might find that white LEDs run at 30mA are jarringly brighter than the little incandescent bulbs originally fitted, and higher value series resistors will calm that down. Could be important if you're not going to implement an adjustable dimmer.
posted by flabdablet at 12:48 PM on October 22, 2021

Also, the tidiest way to solder stranded wire to a solid device pigtail is to start with a bit of nice lightweight hookup wire, slip a bit of heatshrink 6mm longer than the pigtail down to the middle of the wire, strip maybe 15mm of insulation off the end, then wrap the bared strands tightly around the pigtail starting very close to the end and working in toward the device. When the wrap is complete, bend and gently tug the wire where it first enters the wrap so that it's pointing straight off the end of the pigtail. That way you get a joint that's mechanically stable enough not to need you to hold it together by hand while you're soldering it and wants to wick solder into itself when you make it hot with the iron.

Apply iron and solder to the wrapped joint from opposite sides, letting only just enough solder flow to wick all the way through the wrapped strands; you want the end result to be close to the thickness of the pre-soldered wrap, not big and blobby. Let it cool on its own for a few seconds without moving it or blowing on it, so that the solder solidifies bright and shiny. Slip the heatshrink all the way down to cover the entire device pigtail and the joint and about 6mm of the wire insulation.

When soldering pigtail to pigtail, tin about 6mm of both ends before you start, then press the tinned parts together and apply more heat to reflow the solder. Given the right diameter of heatshrink, you can then slip it all the way over the body of a resistor and still have it shrink down nicely to insulate the entirety of a resistor to LED pigtail pair. LED to LED pigtail joints are going to be pretty unavoidably bare, so avoid putting them near anything else; use Kapton tape to fix them in place if they seem inclined to wander.
posted by flabdablet at 1:15 PM on October 22, 2021

Also also, a third hand is going to be a tremendous help for the pigtail-to-pigtail joints and won't hurt for the pigtail-to-wire ones either.
posted by flabdablet at 1:23 PM on October 22, 2021

Response by poster: Thanks for clearing some things up for me.

The LED's I was looking at are these LED's, which have a 3.0V forward voltage and are rated at 20mA max current. When I plug the values into this LED circuit calculator it suggests using groups of 4 LED's in series with a 1 ohm resistor, 6x groups in parallel.
12V - 3.0V*4 = 0V. I assume the 1 ohm resistor is just there to account for a swing on the 12V supply? Are the LED forward voltage ratings generally trusted? What larger resistor values would you experiment with to control brightness in this circuit?

I also picked up some 22ga stranded hook-up wire. I was having trouble finding resistors in small quantities in the correct sizes. Is there a good source for small quantities of different size resistors?

Lastly, I was thinking about picking up a new soldering iron to replace my ancient $10 Radioshack iron. Is there much benefit in moving to a unit with a temp controller vs. fixed temp? I'll pick-up a 3rd hand tool as suggested.

Thanks again!
posted by pilibeen at 7:23 AM on October 25, 2021

Best answer: I assume the 1 ohm resistor is just there to account for a swing on the 12V supply?

The 1 ohm resistor is there because that circuit calculator needs to put something there. In fact 1 ohm is so low a resistance as to be virtually useless for current limiting. Let's see what that same circuit calculator does if we vary the 12V supply 5% either side of nominal, which would be the typical tolerance that 12V switching power supplies are designed to.

Plug 12V * 95% = 11.4V into the calculator and you'll see that it suddenly switches to three LEDs per string, with a 120 ohm series resistor. And with 12V * 105% = 12.6V, it sticks with four LEDs but the series resistor goes up to 30 ohms.

The fact that a 5% increase in power supply voltage has increased the recommended series resistor for a four-LED string by a factor of 30 tells you that we're sailing way too close to the magic smoke with that design. And when the same calculator tells you it can't run a four-LED string at all when you drop the supply voltage by 5% tells you that the assumptions baked into the calculator's design are no substitute for an actual understanding of what's going on.

The figure to concentrate on here is the maximum LED current; you don't want your design to exceed that under any conditions. According to the data sheet for those LEDs, the absolute maximum continuous current they're rated for is 20mA, their recommended operating current is 10 to 20mA, and their forward voltage at 20mA is typically 3.0V but could be as high as 3.4V.

A bit further down that same datasheet there's a forward voltage vs forward current graph for various kinds of LED, none of which appear to be the one the datasheet is for so that's shitty and probably best ignored.

The thing to keep in mind about LEDs is that their forward voltage vs operating current curve is approximately exponential rather than linear as a resistor's would be. Once you get into the operating region where a LED is actually emitting light, it only takes a tiny increase in forward voltage to cause a massive increase in LED current. Conversely, if you engineer a modest variation in LED current, you'll only see a very small change in LED voltage.

If you've got a LED that drops 3.0V at 20mA, and you push 10mA though it instead, it will drop a bit less than 3.0V but not much. Might go down to 2.9V, say.

So even though the actual voltage vs current graph on this particular LED's datasheet is kind of useless, we can still design around the specified maximum operating current (20mA), the typical recommended operating current (10 to 20mA), and the forward voltage when run at 20mA (typically 3.0V but could be as high as 3.4V).

As we've seen above, four LEDs in a series string makes the required resistor value way too variable. Let's try three LEDs, and do the calculations for various resistor values, and see what we can achieve.

Let's start by assuming the maximum allowable power supply voltage that 5% tolerance on a nominally 12V supply allows, which is 12.6V, and try to make our LEDs consume as close to 20mA as we can get without going over.

If we get all the way to 20mA, the LED forward voltage is going to be at least 3.0V and this will be the highest-current case for any series resistor (because if the LEDs drop more voltage, the resistor will see less, and the current through it will be proportionately lower). So a string of three LEDs, each dropping 3.0V, gives us 9.0V across the LEDs. The power supply voltage is 12.6V, so that's 3.6V across the resistor. We want 20mA to flow through the whole series string including the resistor, so that means the resistor needs to be 3.6V / 20mA = 180 ohms, which is (conveniently) a standard E12 resistor value and therefore cheap and easily available.

Now let's see what happens to that current if it turns out that our LEDs are actually dropping their specified maximum of 3.4V each and the power supply is on the low side, at 12V * 95% = 11.4V. Three LEDs at 3.4V is 10.2V across the LEDs, which leaves 11.4V - 10.2V = 1.2V across the resistor, which means that 1.2V / 180 ohms = 6.7mA will flow through it. That's not the 10mA minimum we're working towards, but it's about the best we can do given the tolerances we're working with.

In fact a white LED running at 6.7mA would be very unlikely to drop the maximum 3.4V it's specced at for 20mA.

In order to get the minimum 10mA we're after through a 180 ohm resistor, we'd need 10mA * 180 ohms = 1.8V across it. With our 11.4V supply, that would leave 9.6V across our three LEDs. Assuming they're all matched, that would put 3.2V across each one. That looks pretty realistic to me, for a LED that would drop 3.4V if you ran 20mA through it.

What would the LEDs need to drop, each, to get 15mA through this string? 15mA * 180 ohms = 2.7V. So with our 11.4V minimum power supply that would be 8.7V across the LED string, or 2.9V per LED, which is not outrageous for something specced for 3V at 20mA. With a 12.6V power supply it would work out to 3.3V per LED, again not outrageous.

So this configuration probably would, in practice, light the LEDs up satisfactorily; and because it was designed in the first place not to blow them out with worst case applied voltage and assumed LED voltage drop, it's going to light them safely as well.

Go for strings of three of those LEDs, with a 180 ohm resistor per string. Oh, and the resistor power dissipation will be at worst 20mA * 3.6V = 72mW, so anything rated for at least 1/8W will be fine.

A $10 Radio Shack iron will be just fine for this project if you keep the tip clean by giving it a wipe on a damp cloth or sponge right before making each joint, keep your device pigtails reasonably long, and don't dilly-dally with the iron on the joint; leave it there only just long enough to make the solder flow and wick nicely. This advice does not apply if it's a 100+ watt brute of a thing designed for plumbing.
posted by flabdablet at 12:39 AM on October 26, 2021

One tip for using that LED circuit calculator: don't give it as much scope to fuck things up as it did for you. Limit it to numbers of LEDs that force it to use just one series string and it does pretty good work of automating the kinds of manual calculation I've done above.

Just make sure to test each of its suggested designs at all extremes of power supply and LED voltage tolerance. If it starts suggesting wildly different resistor values under those conditions, reduce the LED count so it's forced to start with higher-valued resistors.
posted by flabdablet at 12:43 AM on October 26, 2021

Response by poster: Flabdablet - thanks again for all of your help.

I got some different LED and resistor sizes to try and settled on these 3mm diffuse white LED's, using groups of 3 LED's in series with a 220 ohm resistor. I built this circuit and measuring with a meter I'm seeing 16.5mA in the series with 2.78V drop across each LED (3.58V drop across the resistor).

The LED's are plenty bright in this circuit for my needs. Is there any reason to worry about not hitting the forward voltage rating of 3.0V-3.4V on the LED's?

The other question I had is about the practical wiring of this. Right now I'm thinking about wiring the series groups off the existing bus structure (copper tape). This seems easiest, but not very clean. Is there a better way to approach this you'd suggest, where I can manage the wires in a tidier fashion? Terminal strip maybe?
posted by pilibeen at 7:28 AM on November 1, 2021

Is there any reason to worry about not hitting the forward voltage rating of 3.0V-3.4V on the LED's?


Although LEDs have an entirely nonlinear relationship between the forward voltage across them and the currents that pass through them, for any given device it remains a fixed relationship. So there is nothing you can do to control a LED's forward voltage independently of its operating current.

Your design gets you an operating current of 16.5mA for each LED string, and your LEDs are bright enough, and 16.5mA is well below the suggested operating current for the LEDs you're using which means it's also going to be below their maximum allowable current (which you don't have a spec for because no data sheet is provided, booooo! Adafruit) so you're golden.

With 16.5mA running through them they're just gonna drop whatever they drop. If that's 2.78V each then so be it. It's their operating current you need to care about; their forward voltage really only matters to the extent that you need to assume something about it to make a start on choosing the series resistance that sets the operating current you want to achieve.

If you were to bump that resistor down to 180 ohms and your LEDs still dropped 2.78V each, then the 3.58V drop across the resistor would yield 3.58V / 180 ohms = a whisker under 20mA. In fact running those same LEDs at 20mA would make each of them drop a little bit more than 2.78V, which would mean there'd be a bit less than 3.58V left for the resistor to drop, which means you'd need less than 180 ohms to get your actual 20mA, which means that if you did use 180 ohms you'd still be under 20mA and your LEDs would run a fraction brighter than they do right now.

But 220 ohms gives your design more scope to deal safely with over-spec power supply voltages, and if the LEDs are bright enough as-is you've got no good reason to change it.

The fact that the voltage drop across the resistor is in the same ballpark as that across each LED is going to result in pretty good stability with respect to power supply voltage variation. If you were to go with series strings of four LEDs rather than three, then to get the same 16.5mA at 2.78V across each LED, you'd have only 3.58V - 2.78V = 0.8V across the resistor. I'm not going to bother working out what resistor value you'd need to get you 16.5mA at 0.8V, because ±5% of your nominally 12V supply is ±0.6V which is large compared to 0.8V. The design would therefore end up with an operating current that was super sensitive to supply voltage variation. Not as sensitive as the original 1 ohm idiocy you got out of the LED calculator page, but definitely heading that way.

Right now I'm thinking about wiring the series groups off the existing bus structure (copper tape). This seems easiest, but not very clean.

That's what I'd do. You should be able to keep things reasonably tidy by placing your 3-strings judiciously, covering as much exposed pigtail with heat shrink tubing as you can, keeping such bits of hookup wire as you use as short as possible, and using a few spots of hot glue or kapton tape to stop any longish wires from flapping about in the breeze once they're all soldered in. If you don't end up with wiring that's nicer looking than the old incandescent setup I'd be surprised. Post pictures when you're done!

Afterthought: solder wick is probably the right thing to use to get the existing solder joints mostly off that copper tape. Those pttthhOKK handheld solder suckers are frustratingly non-thorough for that kind of job.
posted by flabdablet at 10:10 AM on November 1, 2021

Response by poster: I tackled the soldering this weekend. I removed as much old solder as I could, and tried to locate my new connections on fresh copper whenever possible. I also put Kapton tape over the horizontal extensions of the copper bus, and anywhere that the wires had to cross over the bus. It's probably overkill but I figure it couldn't hurt.

Here are some picture of how it turned out: https://imgur.com/a/RvuC6kX

I still need to secure the wires in place to the back...was thinking of just using electrical tape in spots for that? The Kapton tape is great but doesn't seem like it has enough tack to hold wires in place against their will permanently.

When I measure the total current of the circuit I get around 130mA. I was thinking I might install a fuse to protect against overcurrent. Do you have any recommendations for type of fuse to use, and how to size? If the circuit draws 130mA, should I go with a .25A fuse (if fuse ratings that small even exist)? I was thinking about maybe getting an in-line switch like this and modifying it so it also has an in-line fuse? Does that seem reasonable?

Thanks again for all of your help on this project!
posted by pilibeen at 10:44 AM on November 22, 2021 [1 favorite]

Nice tidy job. Well done!

doesn't seem like it has enough tack to hold wires in place against their will permanently.

A few small blobs of hot melt glue at strategic spots should be plenty. Electrical tape will be disappointing here; unlike the adhesive they use on the Kapton tape, electrical tape adhesive absorbs moisture from the air and goes gummy and gooey after a year or so.

Alternatively, you might want to make a cover out of cardboard or plastic that's the same shape as the back board and sandwich all the wiring under that.

I honestly wouldn't bother with fusing this thing; any switching 12V wall wart will already have overcurrent protection inbuilt. Your best protection against overcurrent is already there, with all the attention you've paid to insulating everything so nicely.

That said, 250mA fuses are readily available so you'd have no trouble finding the parts if you did decide to go that way.
posted by flabdablet at 11:13 PM on November 22, 2021

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