LED Project - Page 1
Forum section: electronics / beginners

author: Jbouss
author_rank: Contributor
timestamp: February 24, 2022, 12:30:31 am
content: Hello,This is my first post; I am relatively new to electronics and so far, it has been quite exciting to learn. Currently, I have a couple issues that I need assistance with.1. My circuit consists of 10 electronic control boxes with 10 external switches. There is one switch per box, when the switch is closed no voltage is present, if the switch opens, 3.3V is present. My goal was to wire all 10 switches together so if one opens an indicator light turns on. I connected all switches in series, increased the output voltage with a DC-DC converter, and attached a 12V LED tower light. All the switches are wired in series, so if one switch opens, the 3.3V is present, however, it now drops below 2V and the DC-DC convertor now only gets me up to 10V. That should not be a problem because when I attached my LED to a PS at 10V it works correctly. The problem seems to be related to insufficient current; when attaching my 12V LED to the supplied 10V, the light flashes once and goes so dim it is barely visible. What is a possible solution to this problem?2. The second problem is related to the above. In addition to the light tower, I am using panel LEDs to indicate which of the 10 boxes failed if my tower light is on. I was using a 3.3V LED on a single box without issue, but because of the voltage drop, they do not light anymore. Should I just look for a differently rated LED or is there something simple I can add the circuit to boost the voltage 1V without a full DC-DC convertor?Thank you in advance for your assistance; please let me know what kind of questions may need answered.Josh

author: bob91343
author_rank: Super Contributor
timestamp: February 24, 2022, 12:49:33 am
content: post a circuit diagram

author: xrunner
author_rank: Super Contributor
timestamp: February 24, 2022, 01:41:54 am
content: Quote from: Jbouss on February 24, 2022, 01:34:54 amJbouss,The attachment system has a bug. Attach your pic as the second attachment instead of the first.

author: Jbouss
author_rank: Contributor
timestamp: February 24, 2022, 02:07:03 am
content: See attached diagram, sorry for the delay and confusion. I appreciate all the help:)3.3_Warn_LED_2.pdf(37.82 kB - downloaded 49 times.)

author: Jbouss
author_rank: Contributor
timestamp: February 24, 2022, 02:15:00 am
content: Additionally, note the single LED worked with a through-hole LED with no resistor.Now it doesn't with the DC-DC step-up and the tower LED. The tower LED is a cheap green/red tower LED, something like the below links; 15 bucks on Amazon. The Tower LED features a Buzzer as well, and the buzzer works fine.https://www.amazon.com/LUBAN-Industrial-Warning-Facility-Indicator/dp/B09BW6L833/?_encoding=UTF8&pd_rd_w=h4pMj&pf_rd_p=396c0314-c1f6-41c2-b37b-91540c07423f&pf_rd_r=H7JGQBADP7500T53J9AS&pd_rd_r=3c78b2a2-7b06-4756-a60a-b85b2ec9db35&pd_rd_wg=6CceD&ref_=pd_gw_ci_mcx_mr_hp_atf_m

author: Zero999
author_rank: Super Contributor
timestamp: February 24, 2022, 09:11:13 am
content: Assuming the power supply is constant voltage, not current output, this is a bad circuit.When all of the switches are closed, there will be a short circuit across the power supply. Hopefully it'll safely limit the current, without blowing up.When one switch opens, the full power supply voltage will be across one LED. Hopefully the current will be limited to a safe level, otherwise it will destroy the LED.The switches need to go in series with the LEDs, which each need a current limiting resistor. If you need to reverse the logic, so the LED is normally on, but goes off, when a switch is activated, use normally closed switches.

author: tooki
author_rank: Super Contributor
timestamp: February 24, 2022, 05:42:40 pm
content: Quote from: Zero999 on February 24, 2022, 09:11:13 amRe-read the original post: I think the DC-DC converter isn’t the power supply for the circuit, but rather is being powered by all those unnamed modules.Frankly, the whole thing makes no sense to me; I suspect they’re trying to use the pull-ups on some kind of output as the power source.

author: Zero999
author_rank: Super Contributor
timestamp: February 24, 2022, 06:39:58 pm
content: Quote from: tooki on February 24, 2022, 05:42:40 pmDid you see the schematic attached to post #3?It will stand a chance of working, if the power supply has a constant current output and open circuit voltage greater than the forward voltages of the LEDs, but that's not the case, going by the original post.If the switches have to be normally closed, then they could short the individual LEDs, each wired in series with a current limiting resistor, but it's only practical if the forward current is a few mA. Alternatively the LEDs could be driven using transistors, with the switches shorting out the drive signal.

author: mariush
author_rank: Super Contributor
timestamp: February 24, 2022, 06:55:33 pm
content: A simple solution would probably be attaching an optocoupler (maybe with a resistor to limit current going through the led inside the optocoupler) in parallel with each led. When a switch is open, that circuit sends power to turn on the led, and some current will also go through the optocoupler.Connect output of each optocoupler together and to a mosfet or transistor or something, when an optocoupler turns on, some current will go in the base of a the transistor and that would enable to voltage regulator, or turn on the bigger led etc ...

author: Zero999
author_rank: Super Contributor
timestamp: February 24, 2022, 08:11:57 pm
content: Attached is the PDF converted to PNG with the contrast enhanced.It doesn't make any sense: a DC:DC converter with one LED on one side and a load of LEDs connected in series, each with a switch in parallel to the other side. There's no power source shown. If you put that into a simulator, it would give you nothing, zero voltages and currents.

author: alexanderbrevig
author_rank: Frequent Contributor
timestamp: February 24, 2022, 09:46:47 pm
content: I have a theory of what's going on here.Each block marked 3v3 actually generates power (probably has its own power supply), but because there's an LED he only gets ~2V residual out (at about 20mA perhps?).So,the chain of ten switches becomes 20V of potential. They then convert that 20V back down to 10V but the DC/DC is probably not at all ideal, and it may even get too little current to function properly.I suspect the initial flash is because of capacitance in the converter.Am I on the right track here Jbouss?

author: alexanderbrevig
author_rank: Frequent Contributor
timestamp: February 24, 2022, 10:03:27 pm
content: This actually confused me as I drew it.If all switches are open, there's about ~20V in to the DC/DC, if only one opens there's about 2V at DC/DC in.In hindsight, it's more likely that you're taking VO directly, but it sags due to current draw/cable/switch/other loss in the system perhaps? I just tried to find a way to explain some 1.3volts ish loss.Is this what you are doing Jbouss?So confused. Also, curious!(Please ignore the random part numbers, tried to find symbols that matched my intent)

author: Jbouss
author_rank: Contributor
timestamp: February 25, 2022, 12:13:47 am
content: Wow, I am so grateful for all the feedback. I created a more legit schematic and eliminated the single LEDs. Please see additional notes below.The DC to DC converter is 2V-28V 2A DC-DC MT3608.I am able to take the 1.8V up to 12V with no problem with this DC to DC booster, but when attaching the LED light tower the voltage drops in half, the light goes bright for a split second, then is barely visible. There are (4) rows of (3) LEDs in parallel that measure ~2.0V with a DMM. They are using (4) 40Ω resistors.Interestingly when I hook up the LED tower to a PS, the LED tower doesn't work either, that is, unless I eliminate the DC-DC convertor and give it enough voltage, maybe I need a better or different converter. It is hard to find something that boosts 1.8V though.The light tower also has a buzzer, and that works just fine if that is helpful.See the attached schematic for clarity, hopefully, it is more legible. Sorry, I am a newb.3.3V_Warn_LED_3.pdf(32.11 kB - downloaded 48 times.)Thanks again, everyone!

author: Jbouss
author_rank: Contributor
timestamp: February 25, 2022, 12:32:50 am
content: alexanderbrevigYou are on the right track to some degree, sorry for the confusion. I removed the single LEDs and posted a better schematic.You are correct each switch has its own PS. I am losing over 1.0V due to resistance in the wiring I believe? My main issue as of today is supplying power to an LED light tower. For my prototype I just used an LED I had lying around the lab, since it lit when one of the switches opened as expected, I thought boosting the voltage from 2.0V to 12V would make my LED tower plug and play.Thank you for your feedback.

author: alexanderbrevig
author_rank: Frequent Contributor
timestamp: February 25, 2022, 12:36:21 am
content: Remember that the pins of a microcontroller is not really meant for driving loads.The schematic is still a bit hard to understand. Are you really tying the pic32 signal out to the GND of a 3.3 V power supply? Or do you mean that there's a supply feeding 3v3 there?R1 and R2 forms a voltage divider.Your RLC filter makes me wonder if this is being driven by PWM? (Or, maybe you're just trying to protect the input, that's fineBut yeah, the schematic still leaves a bit too much for our imagination. Maybe do just one more pass?

author: Jbouss
author_rank: Contributor
timestamp: February 25, 2022, 12:50:59 am
content: Sorry, I am so close to the project, I assume everyone sees what I see. I will review and modify more when I can, and appreciate the feedback.The 3.3V is coming from a control box that uses a 12V 5A PS. The J connectors are part of the circuit I am trying to tap into. J1-J10 are all the same as the J circuit provided and I cannot change that portion of the schematic. I should note the switches are pressure switches. Basically if the processor senses the switch is closed it does nothing, if the system loses pressure, the switch opens, and 3.3V is present.

author: Zero999
author_rank: Super Contributor
timestamp: February 25, 2022, 10:58:50 am
content: Quote from: alexanderbrevig on February 24, 2022, 10:03:27 pmThat doesn't make much sense either. When any of those switches are closed, the current limiting resistor to the LED will be bypassed, which will allow the regulator to push as much current through the LED, as it can.Quote from: Jbouss on February 25, 2022, 12:13:47 amSorry, that still doesn't make any sense. What's with that 3.3V supply, with the negative terminal going to R1 and the positive not connected to anything? Just connecting a capacitor directly to the output pin of a micro-controller is normally a bad idea.Please start from the beginning, rather than solutionising. What are you trying to do?

author: alexanderbrevig
author_rank: Frequent Contributor
timestamp: February 25, 2022, 11:58:36 am
content: To be fair, I did not say it made sense. Actually the opposite. I was just trying to figure out what was going on here.Quote from: Zero999 on February 25, 2022, 10:58:50 amThe other side of the R will be GND, so the R takes all the current, yes? The LED will surely turn off...?Quote from: Zero999 on February 25, 2022, 10:58:50 amThis.

author: Zero999
author_rank: Super Contributor
timestamp: February 25, 2022, 02:11:28 pm
content: Quote from: Jbouss on February 24, 2022, 12:30:31 amYour mistake appears to be trying to power the LED directly from the 3.3V signals.How much current does the 12V LED need?You could use a relay to turn on the LED. A transistor and diode OR gate could be used to activate the relay from the 3.3V signals.Here's an example. The relay will turn on, if one or more of the inputs goes high. Duplicate the diodes for more inputs.

author: tooki
author_rank: Super Contributor
timestamp: February 25, 2022, 02:34:35 pm
content: Quote from: Zero999 on February 24, 2022, 06:39:58 pmYes, I did, which is why I said you needed to go back.And going by your subsequent reply…Quote from: Zero999 on February 24, 2022, 08:11:57 pm… it sounds like you figured out what I was getting at.

author: Cerebus
author_rank: Super Contributor
timestamp: February 25, 2022, 03:29:20 pm
content: This looks like a classicXY problem.For those not familiar with the expressionXY Problemhere is a synopsis:Quote from:https://en.wikipedia.org/wiki/XY_problemSo @Jbouss, what is the original problem that you're trying to solve?To me it seems that you have 10 switches (you haven't said what function those switches are supposed to perform), and you want a light to light up if one or more of those switches are in one position ('open') you want the light to be on. If all the switches are in the other position ('closed') you want the light to be off. Also these switches seem to also be somehow each associated with a signal from a PIC32.It would seem that these 10 switches are in 10 completely different devices and you're just trying to produce some overall warning/monitoring solution to what state these 10 devices have been set into. Perhaps the switches represent an override or an interlock or position sensing switch. I suspect that there's no guarantee about the relation between ground and supply voltages of these 10 different devices. I further suspect that they represent a method of disabling a relay, for which L1 is the relay coil, by lifting the relay's ground. I also suspect that there's a possible logical flaw here, in that the signal you're sensing may not truly sensejustthe state of the switch but is also dependent on whether the PIC32 output is turned on. (Of course that PIC32 signal may be an input, but then I don't see the sense of the coil and diode, but the pullup to 3V3 doesn't make much sense in terms of it being an output so I'm a little nonplussed by which it really is.)I'm guessing that you have little or no control over the content of the "10 electronic control boxes" and you have either added the switches as some sort of override, or the switches are already part of the "10 electronic control boxes" and you're trying to pick off the signal generated by the existing switches for your own monitoring.The logical function being sought here appears to be a simple logical OR of the state of the 10 switches, which is then used to turn on a light. There are a number of ways to do this, all simpler and more reliable than the solution that you're trying to use. The simplest solution is probably 10 pairs of wires from the "10 electronic control boxes" driving 10 optoisolators with the optoisolator outputs wire ORed, with the resultant signal used to indirectly drive the indicator LED.@Jbouss - Please tell us exactlywhat problemyou are trying to solve,nothow you've tried to solve it. It's probably obvious by now that the attempted solution has problems, and if all is as I suspect, any attempts to make that solution work are going to be flawed.Please tell us:What are the "10 electronic control boxes"?If you know, are the "10 electronic control boxes" electrically isolated from one another, or do they share a common DC supply or a common DC ground?What is the function of the switch on each of them that you're interested in the state of?What is your final output LED/light intended to indicate?Have you added the switches yourself, and if so, what are you intending to achieve with them?

author: Jbouss
author_rank: Contributor
timestamp: February 26, 2022, 02:20:52 am
content: Sorry for all the confusion, as I said I am a newbie. I really appreciate all the direction and questions. For the most part, Cerebus's post is right on; see the below answers to Cerebus’s questions, as I think they will clear up most of the confusion.So @Jbouss, what is the original problem that you're trying to solve?The original problem is, I want to use a warning light to tell me if one of my 10 switches opens. If any switch opens, the light comes on.What are the "10 electronic control boxes"?The 10 boxes are electrically isolated from one another and do not share a common power supply. Each box has a 12V 4A power PS and utilizes a PIC32 to perform many features. The switch portion is just a small section of the box’s circuitry.What is the function of the switch on each of them that you're interested in the state of?Please note these are pressure switches and they are mounted outside the control box ~18in away. The switches are attached to the control box with a two-wires, two-pin quick connection.The function of the switch is if the switch is closed the control box has a closed circuit and everything works as intended; J1_p1 and J1_p2 have no voltage. However, if the switch opens, the circuit is now open and a 3.3V signal is present across J1_p1 and J1_p2. That signal is sent to the microprocessor to enable the firmware to throw a warning code to an LCD. The J portion of my schematic represents the connection on the control box's PCBA. I only included the J portion of my schematic because I thought that it would provide further insight as to where the voltage that I was trying to boost was coming from. Unfortunately, I cannot change the J portion of the circuit, I just want to use the signal that occurs if a switch opens to light a warning LED tower.I am interested in visually monitoring if one of the 10 switches that I’ve wired in series opens. If a switch opens, 3.3V is present (~1.8V with my current wiring), and I want to use that 3.3V(~1.8V) to light an LEDTo summarize, I am trying to create a visual warning to monitor if a switch opens. The reason 10 switches are in series is that I didn't think I needed to tap into all 10 switches separately. I thought I could grab the voltage from any closed switch, boost the voltage to light a warning light. I am assuming this is not possible because there is not enough current to drive the LED? I also wanted to maximize the parts used to make wiring easier and have fewer components to troubleshoot if something fails. During my initial testing, if I opened any of the 10 switches I got voltage, albeit it is was dragged down, I think from resistance?Zero999 proposal to treat the 3.3V(~1.8V out) as a signal rather than voltage, seems like a viable approach. This way, if a switch opens, the signal flips a transistor that closes a relay and supplies 12V to my LED tower. I am just not sure the best way to do this but feel I am getting closer with all the insight gained from this thread. I am grateful I reached out and joined this forum:) You guys are a great resource!*The 3.3V is coming from a 3.3V regulator(not the PIC32) that is sourced from the 12V 4A power supply of whatever control box has an open switch. If multiple boxes have open switches the voltage doesn't appear to change much.

author: Damianos
author_rank: Frequent Contributor
timestamp: February 26, 2022, 05:30:54 am
content: To light up a LED, except of voltage, it is also needed a current. This current is provided via the R1 and of course there is a major voltage drop; even if it's value is only 1kΩ there will be a drop of 1V per mA. Drawing any current from there will also disturb the signal going to the microcontroller.Does this "magic box" provide any signal that is relative to what is needed?Can these boxes be connected with a common ground?In this case, two hex buffers may give drive to dedicated indicators and be OR-ed to drive also a common indicator. Alternatively an OR-ing circuit with high impedance inputs can drive a common indicator. This needs power from somewhere and probably some filtering.Just out of curiosity: are these boxes "absolutely isolated"? It is not there any leakage? There is not any disturbance while they are connected in series from the inputs (each "ground" to the input of the next)? What happened to the operation of each box, while they were connected as indicated, when one or more switches opened drawing a current? Are there any critical operations of the system, especially from these signals, that need special treatment/protection? If it's a toy then play and learn freely, but if it can create any danger be careful![Assuming that I understand what is discussed here!]

author: tooki
author_rank: Super Contributor
timestamp: February 26, 2022, 09:57:49 am
content: OP, given your difficulty in even describing the situation, plus the way your proposed solution does things that just don’t make sense, I don’t think your electronics skills are yet at a point where you understand the potential ramifications of what you’re trying to do. (This is in no way intended as an insult; we all started as novices!)To expand on what I think Damianos is getting at: With a high degree of certainty, the 3.3V you’re measuring across the connector of the control boxes are simply the pull-up on a data line. It isn’t a power supply, and isn’t intended to be used as one. A pull-up is created with a pull-up resistor to the internal power supply, often in the range of 1kΩ. You’re really going to struggle to boost that up, since the current draw is higher on the input side of a boost converter. Worst-case, it’s aninternalpull-up in a microcontroller, in which case excessive current draw might fry the microcontroller.You say the controls are fully isolated, but honestly (and again no insult intended) I don’t trust your skill level to be high enough to accurately judge if this is actually the case, nor what the ramifications are if you’re mistaken. So the prudent way forward is to assume they’re not isolated and go from there.What we also don’t know is what voltages the pressure switches are designed to handle. We only know they’re capable of switching 3.3V logic.Your goal is a beacon that indicates that any of the switches has gone open.So what I would do is to assume the worst case and design a system that ensures isolation. A 12V power supply is used for the LED beacon. I’d use a 3.3V power supply to run 3.3V DPDT relays, one for each pressure switch: the pressure switch holds the relay energized. One normally-open relay contact (i.e. closed when energized) would then be used to bridge the sensor’s control box input. The other relay contact would be used normally-closed (so open when energized) in parallel with the same contacts of all the other relays, to turn on the LED beacon. If any pressure sensor goes open, its relay de-energizes, opening the contact to its control box, and closing the contacts to energize the beacon.Note that the 12V and 3.3V power supplies here must be separate from whatever operates the control boxes. You could use a dual-output power supply to get the two voltages.

author: Jbouss
author_rank: Contributor
timestamp: February 26, 2022, 05:04:02 pm
content: Thanks again for all the feedback, lots to consider.Tooki, I agree with you, my electronic skills are low level at best. You are correct, R1 is a 1KΩ pull-up resistor. Also, the signal from Jp2 is not from the PIC32, it is sent to the PIC32. I have attached a revised schematic showing this more clearly; I also left a switch open indicating ~1.8V is present.I feel certain the controls are fully isolatedoutside of the switches that I have wired in series. Each box is powered by a separate 110VAC power cord, each has its own 12V PS, each has its own LCD, and each runs its own respective system. The only way they are connected is through the switches wired in series. The boxes are used to control a water pump, the switch verifies the water is flowing by detecting water flow. If the water flow stops, the switch opens, voltage is present.It is becoming clear that I will need to power the LCD from its own 12V PS. What I am still trying to resolve is how I tell the 12V to turn on the light. I am confident, I could use a Raspberry Pi or something similar to achieve this, but for some reason do not like that idea and want to do it through hardware rather than software, if that makes sense? I was hoping for something more like Zero999 suggestion.Everyone has been so helpful, and I appreciate your not making me feel bad for my lack of knowledge.[ Specified attachment is not available ]

author: tooki
author_rank: Super Contributor
timestamp: February 26, 2022, 05:47:38 pm
content: I already gave you a complete hardware-based solution.

author: Jbouss
author_rank: Contributor
timestamp: February 26, 2022, 06:45:09 pm
content: Thank you Tooki, I see that, sorry, it took me a couple times to understand what you were suggesting. Your suggestion appears to be a great solution the more I think it over. My problem: I needed to throw away the idea of having the switches in series and tapping into a weak ~1.8V signal and go in a totally different direction. I guess scrapping my original idea is where I got hung up:(To clarify, you suggest using a DPDT relay per switch that is held closed by 3.3V(something I know the switch can handle based on the circuit) and if a switch opens the relay flips to a 12V supply that is connected to my warning light?I will draw it up as best I can, and post it when I am done. Again appreciate all the patients and help with this.

author: Jbouss
author_rank: Contributor
timestamp: February 27, 2022, 09:17:04 am
content: Thank you, everyone, you have all been so helpful! I believe the best solution is to use a SPDT type C relay NO. Parts are on order, I will let everyone know how it works out.I decided to use only one control box to monitor the 10 switches. I placed a short on the other 9 boxes, so they will still run their respective systems without issue. With one control box monitoring all the switches, I get a solid 3V out and will attempt to use that signal to activate the relay. If that doesn't work, I will try the relay NC with 3.3V running through the switches. Either way, I think this solution will work.Revised schematic attached.[ Specified attachment is not available ]

author: tooki
author_rank: Super Contributor
timestamp: February 27, 2022, 09:59:46 am
content: Quote from: Jbouss on February 26, 2022, 06:45:09 pmCorrect. Here's what it looks like (channels 1,2, and 10 shown, repeat the channel segment for the remaining channels):Note that EVERYTHING is shown isolated: 3.3V is totally isolated, 12V is totally isolated, and the relay contacts to the control boxes are totally isolated. You could easily connect the 3.3V and 12V grounds together, but I made them separate so that you can see how you could use this setup for complete isolation, for example if you wanted to use a 120V AC beacon instead of a 12V one, or if it needed to control the input of another control box of some type.[Edit: redid schematic to make the isolation even clearer.)

author: tooki
author_rank: Super Contributor
timestamp: February 27, 2022, 10:12:22 am
content: Quote from: Jbouss on February 27, 2022, 09:17:04 amThat's a completely different approach unrelated to both your original and my proposed solutions. Can't say whether it'll work since the control box's function is unknown to me.Is defeating the pressure switches on the other 9 control boxes wise? What are the safety and process ramifications of that? It's not something I'd do without FULLY understanding ALL the risks it brings.My solution is fully safe, under the assumption that a pressure switch opening is the fault condition: if any switch fails, or the 3.3V power supply fails, or any relay coil fails, the control box sees the pressure switch go open. (The only fault condition not sensed here is relay contacts welding shut, which is extraordinarily unlikely to happen with just a simple 3.3V logic signal across them.) Failure of the 12V power supply causes the LED beacon to fail, but the control box would still see the pressure switch go open.

author: Jbouss
author_rank: Contributor
timestamp: February 27, 2022, 10:52:09 am
content: I really appreciate the schematic, I was having a hard time seeing the DPDT, and now it is much clearer.Is defeating the pressure switches on the other 9 control boxes wise? What are the safety and process ramifications of that? It's not something I'd do without FULLY understanding ALL the risks it brings.I am 100% that this is OK to do, I do it all the time for other tests. Thank you for verifying.In fact, I use another set-up that uses only 1 control box to verify 10 switches for a different application, let's call that set-up B. I think I will test my solution on set-up B first.For the 10 control box set-up, let's call that set-up A, I would have a single tower light like discussed but ideally there would also be smaller indicator LEDs for each control box, making troubleshooting a failure easier. I believe your schematic allows for this addition. This was issue #2 in my original post, but that really made things confusing, so I eliminated that idea for the time being. I have come a long way since then; this forum is a great resource. I am glad I decided to reach out, and look forward to following the discussion board.I will post updates as my project progresses.

author: tooki
author_rank: Super Contributor
timestamp: February 27, 2022, 11:15:32 am
content: Yep, here's a bunch of places you could add indicator LEDs: one for each power rail, and then per channel one for the relay coil, one for the 12V relay contact, and one for the signal contact (assuming the control box can supply enough current for an LED.)(Note: to avoid having to redraw the whole thing, I've shown the LEDswithoutresistors. In reality, every LED requires in series with it a current-limiting resistor appropriate for the supply voltage.)

author: Cerebus
author_rank: Super Contributor
timestamp: February 27, 2022, 02:47:39 pm
content: It's worth pointing out that there are implied current limiting resistors in there for all those LEDs, and there's one vital component missing:Without a flyback diode on those relay coils one is very likely to exceed the reverse breakdown voltage of those LEDs in a destructive fashion. Reverse breakdown voltages on LEDs are low, typically 5V or so, the reverse voltage pulse from a relay coil can easily exceed hundreds of volts. Also there's a minor potential for turning all the long wires involved into quite effective antennas at RF as they see the reverse voltage transient pulses from the relay coils as they're turned off.