Need some help building a circuit involving switchs and relays - Page 1
Forum section: electronics / projects

author: Shaydzmi
author_rank: Regular Contributor
timestamp: May 25, 2020, 06:09:27 pm
content: I need some help and suggestions to build this circuit:It has 4 sections, and a board with relays, which lacks components needed to operate the relays.How it works :Section1 : When one of the 84 switches closes, Relay1 should operate.Section2 : When one of the 84 switches closes, Relay2 should operate.Section3 : When one of the 3 switches closes, Relay3 should operate.Section1 : When one of the 3 switches opens, Relay4 should operate.Power supply : 6Amps maxLamps: 24v 1.2whttps://i.ibb.co/QD7zR60/cats-vert2.jpg

author: Benta
author_rank: Super Contributor
timestamp: May 25, 2020, 07:12:50 pm
content: "Section 4" seems to be a short circuit???

author: themadhippy
author_rank: Super Contributor
timestamp: May 25, 2020, 07:58:37 pm
content: Hope you dont intend all the lamps being on at once coz 6A aint gonna be enough,as to a solution any one make a 84 input OR gate?

author: Ian.M
author_rank: Super Contributor
timestamp: May 25, 2020, 09:14:19 pm
content: Full wave rectify the supply to all the lamps and switches so they are running on unsmoothed DC so you can use diodes to implement the two 84 input, and single three input OR gates required. Use 24V DC coil relays, and if they chatter, add just enough capacitance across each coil to hold up the coil voltage through the zero crossings. Don't use too much capacitance, as in the limit, it will increase the average coil voltage by a factor of up to 1.4 and cook the relays!As Benta has pointed out, section 4 wont work as drawn. Either swap the thee NC switches in series to three NO switches in parallel, with the whole switch group in series with the relay 4 coil, or put the coil in series with the NC switches, and wire the contacts based on its normal state being energised and releasing on any button press, or use an extra relay to energise relay 4 when it releases, or (assumiong DC supply) use a transistor, a couple of resistors and a diode to implement an inverting driver for the relay 4 coil controlled by the three NC switches.N.B this all assumes you've either got access to *BOTH* ends of the transformer secondary supplying the 24V AC, or can rewire the ground side of all the lamps to the negative terminal of the bridge rectifier. If not, so all you've got is ground-referenced AC, it gets a *LOT* harder . . .

author: Shaydzmi
author_rank: Regular Contributor
timestamp: May 25, 2020, 10:09:50 pm
content: In some cases, they will bi lit all of them, maybe I need some thing that can cut the power when it reaches the maximum current, but in the same time keeps the relay energized.

author: Shaydzmi
author_rank: Regular Contributor
timestamp: May 25, 2020, 10:17:19 pm
content: Quote from: Ian.M on May 25, 2020, 09:14:19 pmThank you for your valuable answer, I didn't quite understand it though, can you please demonstrate it in a circuit diagram?Section4 is easy: I can put the relay in series with the switches. the big trouble is in other sections.

author: Ian.M
author_rank: Super Contributor
timestamp: May 26, 2020, 07:14:31 am
content: Here's the circuit for three switches/lamps. Extend it with more diodes for extra switches/lamps. C1 is optional - to be fitted if the relay chatters when on and increased in value till it stops chattering. All the sections can share the same bridge rectifier.N.B. it still doesn't solve the problem with overloading the AC supply if too many lamps are on.

author: Shaydzmi
author_rank: Regular Contributor
timestamp: May 26, 2020, 10:09:22 am
content: Thank you for your efforts, very appreciated..First of all, the lamps and switches are a standalone device, And I don't have access to the ground, because it is conned to the machine's chassis, so I can't put any thing between the device and the ground/ chassis, or inside the device.

author: Ian.M
author_rank: Super Contributor
timestamp: May 26, 2020, 10:24:33 am
content: That makes it a *LOT* harder. If all you've got access to is the 24V AC feed to each lamp/switch pair with a common ground, your only option would be to add a sensitive current sensor in the feed to detect the lamp current. That wont work if the 24V AC is used for other functions in the device. It *MAY* be possible to use one current sensor for multiple devices, but it may be difficult to get enough sensitivity (in a potentially high EMI environment) to reliably discriminate between the 50mA of one lamp on, and no current when all switches are off, and still be able to handle the high current when lots of lamps are on. Also it will miserably fail to detect the switch position if the lamp fails.

author: Shaydzmi
author_rank: Regular Contributor
timestamp: May 26, 2020, 10:36:41 am
content: The part of the lamp failing, is not a problem, because inside the device, there is a resistor across every lamp.Cant I use a power transistor or a similar device to monitor the switches before going to relays?

author: Ian.M
author_rank: Super Contributor
timestamp: May 26, 2020, 10:52:46 am
content: Transistors don't work on AC so any simple current monitor using one is not going to work.To avoid further confusion, take a hilighter and on your schematic, hilight the sections of it that are purely internal in the device(s) and you have no access to, then re-post it.

author: Shaydzmi
author_rank: Regular Contributor
timestamp: May 26, 2020, 11:20:11 am
content: I've edited the picture above.

author: Renate
author_rank: Super Contributor
timestamp: May 26, 2020, 12:43:57 pm
content: Huh? I'm not soldering in 171 diodes in a machine with moving parts.Just make sure you have the 24 VAC feed that is only going to the bulbs.Put in this circuit (and the rest, an op amp to amplify and rectify and detect the 1.4 V P-P signal).The diodes are beefy enough for any number of amperes.The resistor is the right size that the diodes are still conducting when just a bulb bypass resistor is conducting.The cap cleans up potential noise.

author: Shaydzmi
author_rank: Regular Contributor
timestamp: May 26, 2020, 03:28:13 pm
content: Thank you, I didn't fully understand this circuit, can you explain to me how does it work, and the purpose of each component?Tank you.

author: Renate
author_rank: Super Contributor
timestamp: May 26, 2020, 03:48:12 pm
content: Well, it's basically a shunt for measuring current.You have to be able to measure 25 mA or so, but it has to be able to tolerate amperes going through it.You could just use a single resistor, but then if 22 bulbs went on it would drop too much voltage, heat up too much and burn out.So we have the two big diodes across it so that no matter what happens we only drop 0.7 V in either direction.We just want to know if any bulb is on, not how many.(If we wanted to count bulbs, we would use a very small value resistor and a much more sensitive/precise measuring circuit.)What is the value of the bypass resistors across the light bulb?Some rough numbers here:If the bypass were 1k then the normal current would be 24 mA (RMS)If the sense resistor was 30 ohms the voltage across it would be 0.7V (RMS without the diodes in place).The diodes would flatten the peak to 0.7V instead of the 1.0V peak.That's enough of a nice "to the wall" signal"Why did they put in those bypass resistors?Did they have their own measurement system for any of the bulbs going on?If you wanted to make things easier and you could tolerate a bit more drop on the voltage to the bulbs:You could use two (or 3?) diodes in series (and two (or 3?) diodes in series the other direction).With that you could get enough voltage to run an LED on an optoisolator.That simplies the whole problem of using an op amp with a separate floating supply.

author: Ian.M
author_rank: Super Contributor
timestamp: May 26, 2020, 04:10:35 pm
content: Hmmmm . . . I don't like the voltage drop for the optoisolator solution.I'd experiment with using a common mode choke as a saturating 1:1 current transformer. Once you've got an isolated and limited signal, you've got a lot more options. Core saturation lets you use a fairly large burden resistor for a decent amplitude at 24mA RMS without the high voltage drop issues when passing several amps.

author: Shaydzmi
author_rank: Regular Contributor
timestamp: May 26, 2020, 05:04:30 pm
content: QuoteThey put them to ensure continuity if the bulb burns.QuoteIt doesn't matter, as long as I have enough voltage to glow the bulbs, it's OK.

author: Shaydzmi
author_rank: Regular Contributor
timestamp: May 26, 2020, 05:09:09 pm
content: Quote from: Ian.M on May 26, 2020, 04:10:35 pmA little schematic diagram would be better for me to understand as a beginner.

author: Benta
author_rank: Super Contributor
timestamp: May 26, 2020, 06:33:33 pm
content: Much simpler: let each 24 VAC line run through a current transformer. Detecting current/no current on the secondary side is a simple exercise, and you even get isolation for free. I'd suggest a 1:50...1:200 turns ratio, but it's uncritical.https://www.quora.com/How-is-a-current-transformer-connected

author: Renate
author_rank: Super Contributor
timestamp: May 26, 2020, 06:57:28 pm
content: Quote from: Shaydzmi on May 26, 2020, 05:04:30 pmYes, but that presumes that they have interest in continuity.I can't believe that this loom? has 168 "thread out" indicators and no common indicator.Was that an option that wasn't installed or got broken?

author: Shaydzmi
author_rank: Regular Contributor
timestamp: May 26, 2020, 07:05:26 pm
content: What do you mean by "common indicator"?

author: Renate
author_rank: Super Contributor
timestamp: May 26, 2020, 07:07:25 pm
content: Quote from: Shaydzmi on May 26, 2020, 07:05:26 pmThe same thing as what you are trying to build. A light bulb/relay/siren that goes on when any bulb goes on.

author: Shaydzmi
author_rank: Regular Contributor
timestamp: May 26, 2020, 07:14:26 pm
content: Oh, OK, there is one not working as it should, but I'm trying to build my own for the purpose of learning, and to replace the old one if I succeed in this building.Sorry for the bad English, I learned it by myself no teachers, no schools involved.

author: Benta
author_rank: Super Contributor
timestamp: May 26, 2020, 07:27:09 pm
content: Quote from: Shaydzmi on May 26, 2020, 07:14:26 pmIn that case, you're doing extremely well. There's absolutely nothing "bad" about your English

author: Shaydzmi
author_rank: Regular Contributor
timestamp: May 26, 2020, 09:16:37 pm
content: Quote from: Renate onTodayat 06:57:28 pmThis is a circular knitting machine:https://www.xdknitmachinery.com/circular-knitting-machines-an-ultimate-guide/

author: Renate
author_rank: Super Contributor
timestamp: May 26, 2020, 10:42:52 pm
content: Well, could you take a photo of the original detector circuit board?I'm sure we'd all be interested in how they did it.The current transformer idea would work fine.I just like my own idea.

author: Shaydzmi
author_rank: Regular Contributor
timestamp: May 27, 2020, 08:57:19 am
content: Ok

author: Renate
author_rank: Super Contributor
timestamp: May 27, 2020, 03:30:24 pm
content: That's interesting.First, the stacked power resistors are bypassed with a big solder blob.I'm still confused.Looking at those three bridge rectifiers/capacitor/top resistor/bottom resistor/optocouplers:I can see that they are designed to detect a 24VAC signal.Those half watt 3.3k (top) resistors make sense at 24 VAC = 7mAStill, that 470 ohm on the bottom seems to steal too much of that 7 mA.(I can't really tell because I don't know what the LED forward voltage is.)It looks to me that there must be three external current transformers?Something's going on here that I don't understand.

author: Shaydzmi
author_rank: Regular Contributor
timestamp: May 27, 2020, 03:49:46 pm
content: You can look at only one section: Bridge, cap, 3.3k, 470, optocouplerI think that 470 ohm is to reduce the voltage going to optocoupler.The optocoupler gets 1volt, when the switch is closed.

author: Benta
author_rank: Super Contributor
timestamp: May 27, 2020, 03:53:20 pm
content: The stacked resistors are 330 ohms, not 3.3 k.Current transformers may very well be external and mounted directly on the power leads, that would be the natural place.

author: Shaydzmi
author_rank: Regular Contributor
timestamp: May 27, 2020, 04:12:31 pm
content: There is nothing else other than this board. This is how I see it :

author: Renate
author_rank: Super Contributor
timestamp: May 27, 2020, 04:27:33 pm
content: I agree with @Shaydzmi on the interpretation.(I had mentioned the "sandbox" power resistors only in passing, the ones I care about were the 1/2W 3.3k)So, we went looking for a current sensor and we found something that is a 24 VAC voltage sensor.So, being more exact, 24VAC * sqrt(2) = 34 VDC - 1V (led drop) = 33 VDC / 3.3k = 10 mA - (1V / 470 ohm) = 8 mAThis seems peculiarly designed for a voltage source, not a current transformer.It would have to be a very small ratio transformer to get 10 mA out of a 50 mA main current, or?

author: Renate
author_rank: Super Contributor
timestamp: May 27, 2020, 04:54:23 pm
content: Are you quite sure that the thread-out switches are not actually double pole, DPST?I say that they are double pole.The cable that runs along the feeder bar is two conductor.It has 24 VAC supply and the common alarm.The feeder connects in with insulation piercing to the two conductors.The chassis provides return.Am I wrong?

author: Shaydzmi
author_rank: Regular Contributor
timestamp: May 27, 2020, 07:03:10 pm
content: Quote from: Renate on May 27, 2020, 04:54:23 pmYou are absolutely right, and because of that I put two separated circuits 1 and 2.This does not change any thing in our problem.

author: Renate
author_rank: Super Contributor
timestamp: May 27, 2020, 08:29:56 pm
content: Quote from: Shaydzmi on May 27, 2020, 07:03:10 pmWell, some of this information should have been in the first post.Your schematic does not show any of this.Was the goal to switch to current sensing?How are you even sure that there are bulb bypass resistors? What is their value?Why would you think that current monitoring would be more reliable than the existing voltage bus?Is there a problem with the existing boards? With the second poles of the thread-out switches?

author: Shaydzmi
author_rank: Regular Contributor
timestamp: May 27, 2020, 09:26:29 pm
content: OK:The machine consists of 4 circuits that control:1-Yarn in.2-Yarn out.3- Broken needles.4- Safety door open.but in reality they combined the (1-Yarn in) and (2-Yarn out) in one circuit, which is not good, because (2) needs a specific behavior is different than the others.because of that, I did what I did.Now you have an answer to this:QuoteQuoteThe goal is to monitor these circuits regardless of the method.QuoteI saw it, I don't remember its value, it's just above the bulb resistance.QuoteI don't know, the members who answered my question, suggested that.QuoteYes the part (3- Broken needles) and (4- Safety door open ) doesn't work.QuoteThey combined the (1-Yarn in) and (2-Yarn out) in one circuit, I want to separate them.

author: Shaydzmi
author_rank: Regular Contributor
timestamp: May 27, 2020, 09:48:40 pm
content: Correction:Quote from: Renate on May 27, 2020, 04:54:23 pmEvery device has 2 separated switches, so we can call them: DPDT.

author: Renate
author_rank: Super Contributor
timestamp: May 28, 2020, 12:06:10 am
content: Alright, I think I'm beginning to get a picture.You've got 84 feeders total, each with an entrance and an exit yarn switch.The feeder bar cable is a two conductor.So what you're planning on doing is modifying 84 feeders to split the two switches?Then what? You've only got two conductors and ground going out.You could add a second bulb to the feeder (or switch to DC and use diodes).Then you would have two buses on which you would have to use current detection.The broken needle detection is simple.It's just a question of the sensor itself and a latch.The same goes for cabinet interlock.Do I have the right picture?Graphic updated to show two bulbs. And yet more modified

author: Shaydzmi
author_rank: Regular Contributor
timestamp: May 28, 2020, 03:41:01 am
content: QuoteExactlyQuoteThe feeders already have two switches and two bulbs each, all they did was they merged the two conducting wires to go to the board as one wire, so simply I'm going to separate them.By the way, how can you achieve that with this? (Just for learning purposes)QuoteQuoteYep.QuoteYes.QuoteYes.Now, I think it is your turn to give us your ingenious solution!

author: Renate
author_rank: Super Contributor
timestamp: May 28, 2020, 01:34:00 pm
content: Note: I updated the drawing above.Quote from: Shaydzmi on May 28, 2020, 03:41:01 amSo you want to remove a feeder from a knitting machine, sit down at a bench, open it up, modify the wiring, put it back together and reinstall it84 times?It's worth that much to you to tell the difference between the yarn being out at the entrance or the exit of the feeder?You'll end up with a non-standard feeder that could cause a lot of mischief if it were mixed with others.And then you have to make a new current sensing board.Is this a project you're doing or is it a machine that is knitting in production right now?It's going to take you a while to do this.

author: Shaydzmi
author_rank: Regular Contributor
timestamp: May 28, 2020, 01:59:40 pm
content: Oh no no no... Please see the picture:

author: Renate
author_rank: Super Contributor
timestamp: May 28, 2020, 05:17:02 pm
content: There's some confusion here.I updated the graphic above to show a second bulb.If this is not 100% correct, please correct it.The cable on the feeder ring is two conductors.Both conductors are already being used.Both conductors already go to the board.We're talking in circles here and the problem is not language.

author: Shaydzmi
author_rank: Regular Contributor
timestamp: May 28, 2020, 05:56:06 pm
content: There you go :There is no thing such an alarm wire.

author: Shaydzmi
author_rank: Regular Contributor
timestamp: May 28, 2020, 06:05:55 pm
content: And there are some photos from inside the feeder, showing the swiches and the resistors across the lamps: (You can tell me how ohms they have)

author: Renate
author_rank: Super Contributor
timestamp: May 28, 2020, 07:53:53 pm
content: Those are 1.2 k ohm resistors.If the full 24 VAC was across them it would be 1/2 Watt dissipation.Since those are 1/4 Watt, that's not what is happening here.Tell me, from your experience, or actually testing now:Stick a finger on a switch and make a bulb go on.Stick a finger in another feeder switch and make another bulb go on.Did the bulbs get dimmer when you put another switch on?See, if there are no more resistors anywhere then all the bulb current goes through that 3.3 k ohm resistor on the board.That would mean that the bulb is running on much less than 24 VAC.Do you have a replacement bulb number?Do you have the service manual for this beast?Note: I've updated again the drawing above.

author: Shaydzmi
author_rank: Regular Contributor
timestamp: May 28, 2020, 10:49:36 pm
content: QuoteI didn't understand, can you explain this to me?QuoteYes, they get dimmer and dimmer until they bacome off when 5 or 6 are triggered.QuoteLook, those 168 bulbs/resistors aren't going to the 3.3k ohm, they are going to the 3 big ceramic resistors (you can see them on the board).The 3.3k ohm is handling just 3 big leds (I assume... tomorrow I'll take a picture of one of them).QuoteWhen they burn out, usually I replace them with this:https://www.narva.com.au/products/47508QuoteNo.QuoteCorrect.Thank you.

author: Renate
author_rank: Super Contributor
timestamp: May 29, 2020, 12:02:57 am
content: 24VAC ^ 2 / 1200 ohm = 0.48WI guess that they always presumed that there would never be full 24 VAC there.17 VAC ^2 / 1200 ohm = 0.24WOk, that control board has been all screwed up.As I said, the stacked power resistors are bypassed.There is only the 470 ohm resistor actually doing anything.But it looks like one end is ground as it goes to the minus side of the big cap.The AC comes in on a 2 pin connector and each side wraps around the board with screw holes!The 3 sensor circuits seem to be designed for 24 VAC vs 0V.It doesn't seem that they would work very well for working with a voltage over a drop resistor or a load.Can you measure continuity between the feeder cable and the plug that goes into the control board?Can you see where somebody shorted the two conductors of the feeder cable?

author: Renate
author_rank: Super Contributor
timestamp: May 29, 2020, 12:42:05 am
content: Ok, I figured it out.The 3 rectifier bridges: Those are for regular 24 VAC signals like broken needle detector, door open, etcThe control board is only designed for a single feeder bus.The 470 ohm is the 24 VAC supply.The detection is done by the ST BTA16 triac.It triggers on the voltage drop across the 470 ohm.You know the good thing about these forums? It gives you a lot of chances to be wrong!So, the big triac, the 47 ohm, the two zeners and the big yellow thermistor are there to trigger and provide the full 24 VAC to all the light bulbs.It is an independent, self-contained sub-circuit that bypasses the 470 ohm.Since it triggers late in the AC cycle there will be some spikes across the 470 ohm resistor if a light bulb is on.It's the TO-92 transistors that are further along that do the actual detection.There's no reason that the bulbs should dim if 4 or 5 of them are on.The triac/thermistor bypass could be broken.

author: Shaydzmi
author_rank: Regular Contributor
timestamp: May 29, 2020, 03:55:34 am
content: Yes, the 3 bridge rectifiers are for: 1- Needle detector. 2- Open gate. 3- Revolution counter.FYI: Only the revolution counter is working, the rest are broken.QuoteThat is you, me I'm always wrong.The ceramic resistors ARE NOT bypassed, you can look at the board again.QuoteI didn't understand any of this.QuoteThere must be a reason though.

author: Shaydzmi
author_rank: Regular Contributor
timestamp: May 29, 2020, 03:59:59 am
content: QuoteThe two conductors are twisted right at their end, you can see where by looking at one of the previous photos.

author: Shaydzmi
author_rank: Regular Contributor
timestamp: May 29, 2020, 08:14:53 am
content: This is the big led that I promised to give you a photo of it:

author: Renate
author_rank: Super Contributor
timestamp: May 29, 2020, 12:20:39 pm
content: This is a cute little circuit that feeds the 24 VAC to the light bulbs.It's "hard" (low resistance) in the sense that the triac is on most of the time.It's "soft" because the rest of the time it's just a 470 ohm resistor.A 50 mA bulb will cause a 12 V drop when the triac is off.(Ok, the triac is off only in the low voltage part of the AC cycle so it doesn't generate that much, maybe a volt or two.)The thermistor will reduce the current if too many bulbs are on.There's no reason that a dozen bulbs can't be on at the same time.The triac also can be a dimmer for all the bulbs, it's just a question of the voltage of the two zener diodes.Edit: Graphic corrected, added sensing, added 24V supply

author: Shaydzmi
author_rank: Regular Contributor
timestamp: May 29, 2020, 12:53:13 pm
content: Where are the in and the out? And where are the 3 big resistors?

author: Renate
author_rank: Super Contributor
timestamp: May 29, 2020, 10:43:26 pm
content: I corrected and updated the drawing above.As I said, the two stacked power resistors are bypassed with a solder blob.The 24VDC supply is riding on top of the 24VAC (which is a bit unusual to see).They have to do that so that they can measure the sense.The "floating ground" is the minus side of the 24VDC supply.

author: Shaydzmi
author_rank: Regular Contributor
timestamp: May 31, 2020, 09:44:14 pm
content: Hi,I tried to de- bypass the two resistors, the bulbs almost would not even glow.The zeners D2 and D3 have this: 1N5225 (if they have the same value).The thermistor gets hot even if one bulb is on.Does this information help?

author: Renate
author_rank: Super Contributor
timestamp: May 31, 2020, 11:23:32 pm
content: The 1N5225 are 3.0V Zeners.That means that the peak voltage across the triac should be 3V before it fires.It sounds like the thermistor has failed with a high resistance.Measure the resistance in-circuit (it should be so low that the other stuff doesn't matter).I would expect the resistance to be less than an ohm.The point of the thermistor is for protection.If you want to try as a test replace the thermistor with a 10 ohm 1/2 watt resistor.This is not a safe or a permanent replacement.(The resistor will burn up if you have a short in the feeder cable.)Or you could just put in a 1A fuse.You know, the triac is good for 16A, you said the supply was 6A?There is no reason anything should be complaining at 50 mA.

author: Shaydzmi
author_rank: Regular Contributor
timestamp: June 02, 2020, 08:46:17 pm
content: I'm sorry I didn't see your reply until now.QuoteI didn't understand what you mean.QuoteTo protect what? what are the things that may be damaged if no thermistor is installed?QuoteMaybe I'll do this, but I think that if the fuse blows up, the machine won't stop if a yarn is broken.QuoteYes, actually the transformer supplying the board has 2 ouputs (24V 4A) and (24V 8A), but I don't which one is which, because no labels are on the transformer anymore.QuoteCan you please explain to me what does the triac do exactly in this circuit?

author: Renate
author_rank: Super Contributor
timestamp: June 03, 2020, 02:27:49 pm
content: Quote from: Shaydzmi on June 02, 2020, 08:46:17 pmTake your meter and measure the resistance of the thermistor without un-soldering it from the circuit.It should probably be almost a dead short so that the value of the other components hardly matter.As I've said, the triac is the main piece that feeds the light bulb current.If it's not working correctly, then the only current to the lightbulb is through the 470 ohm resistor.That would make the bulbs glow dimly and not work at all when 10 bulbs were on.I never said to leave the resistor or fuse or anything in circuit permanently.I only meant to use it as a test to prove that the thermistor is bad.Just measure the thermistor, ok?

author: Shaydzmi
author_rank: Regular Contributor
timestamp: June 03, 2020, 03:13:06 pm
content: QuoteIt measures 103.9.QuoteProbably that's the case. How can I measure the triac to know if it's working or not?

author: Renate
author_rank: Super Contributor
timestamp: June 03, 2020, 03:55:03 pm
content: 100 ohms is far too high for a cold thermistor. It's bad.Consider a single bulb on that (50 mA) it would drop 5V.But it would also get warm and the resistance would even be higher.The point of the thermistor is to protect if one of the feeders or the cable got shorted to ground.You want to limit the current in such a case.The triac is not broken open because the thermistor gets warm.It might be shorted (but I don't think so). If it were shorted the lights would work, but the relay wouldn't.As I said, replace it with a low value resistor or a fuseas a test.The bulbs should be bright and the relay should trigger correctly.

author: Shaydzmi
author_rank: Regular Contributor
timestamp: June 03, 2020, 04:22:41 pm
content: I replaced with a 10resistor, burned out.

author: Renate
author_rank: Super Contributor
timestamp: June 03, 2020, 07:15:44 pm
content: Ok, then something is wrong here.A 1/4 W resistor should be able to take a load of 3 lightbulbs no problem3 * 50 mA = 150 mA(150 mA)^2 * 10 ohms = 225 mWWith the control board disconnected, measure the resistance from the feeder wires to ground.Since you know where they are spliced you can measure them individually.They should both be infinite ohms with all the yarns in place.

author: Shaydzmi
author_rank: Regular Contributor
timestamp: June 03, 2020, 07:42:15 pm
content: I don't think that there is a problem, I think the burn of the resistor is a cause of the fact that here were no yarn in the feeders, so their switches were closed all of them.What I did after is making a short between the thermistor's legs, and tested76bulbs , and after that152bulbs, they didn't dim at all, however the triac got little bit hot.I measured the current when152bulb on was around5amps.

author: Renate
author_rank: Super Contributor
timestamp: June 03, 2020, 11:33:15 pm
content: You didn't mention that the yarn was out in half the feeders.The thermistor might not have a very long life if it often sits there in overcurrent mode.Why is this machine powered up when half the yarn isn't there?Does it need to be on when you are setting it up?This makes a good design a little bit more complicated.All 168 bulbs at full brightness would be 8.4 Amps.That represents a combined load of 2.857 ohms!Throw in those 1.2k bypass resistors and you have an actual load of 2.041 ohms.

author: Shaydzmi
author_rank: Regular Contributor
timestamp: June 04, 2020, 09:44:19 am
content: I was wrong counting bulbs above, I corrected it.QuoteIt was out only when I replaced the thermistor not before.QuoteUsually ther will be less that 10 bulbs on, but in some cases half of them or even all of them, but this case is rare,for example when a new worker doesn't know what he is doing, or when the belt broke.QuoteYes.QuoteThe brightness doesn't matter that much when most of the bulbs are on, we need it only when 20 or less are on .Btw: Other brand machines when 7 are on automatically turned off, and blink once every 5 seconds or so.

author: Renate
author_rank: Super Contributor
timestamp: June 04, 2020, 01:32:47 pm
content: Well, the original circuit is not bad, as long as you replace the thermistor when it fails.The good thing is that even if the thermistor fails the relay will operate correctly.I think that that 3 pin connector on the edge that has a big solder blob on itwas designed to go to a light bulb on a panel to indicate "too many bulbs on".Yes, you definitely want a circuit that can differentiate:No bulbs onSome bulbs on (less than 10 or 20)Too many bulbs onEven flashing the bulbs would take as much as 12 A.If the circuit was DC this would be easier to supply a peak current.Do you know if the models that flash use DC instead of AC?

author: Shaydzmi
author_rank: Regular Contributor
timestamp: June 04, 2020, 03:04:33 pm
content: Thank you for following this long thread!QuoteCan you please explain to me in details what does this triac do in this circuit? Please consider that I'm not an electrician by any mean, I'm just a learner, so don't assume that I am knowing much.QuoteDid you get this value from where?152x50mA= 7.6 or am I wrong?QuoteYes the wire to the bulbs is carrying AC, the board you know it it AC an DC, maybe, if you want, I can bring that board here also, and you can teach me how it works.

author: Renate
author_rank: Super Contributor
timestamp: June 04, 2020, 05:27:33 pm
content: Let's say that you want to build a weight scale that can weigh both a mouse and an elephant.It will have to use a very light spring to detect a mouse.But somehow you want a very strong spring that can tolerate an elephant.The triac circuit is like that.The 470 ohm resistor is the light spring.If you draw 50 mA (light bulb) or even just 25 mA (the bypass resistor) you'll be able to see the effect.(25 mA * 25 mA = 11.75 V)But what if a couple of Amps want to flow through that?That's where the triac steps in. It says "Wow, that little spring is being stretched all to heck, I better step in and start pulling."12 Amps is all the light bulbs AND all the bypass resistors.There are ways that we can cut down all the current when too many bulbs go on, but we also want to make sure that the operator can see that is what is happening.

author: Shaydzmi
author_rank: Regular Contributor
timestamp: June 04, 2020, 10:36:36 pm
content: Thank you.Ok, I see, the triac then only feeds the bulbs, it has nothing to do with operating the relay. The resistor exists to feed the transistors driving the relay?QuoteThe bypass resistors shouldn't be considered, because they won't work as long as the bulbs are working, and we won't let all of them burn .

author: Renate
author_rank: Super Contributor
timestamp: June 05, 2020, 07:12:07 pm
content: Quote from: Shaydzmi on June 04, 2020, 10:36:36 pmThey must be considered or you will hurt their feelings!The bypass resistors are in parallel with the light bulbs.Current goes through them whether the bulb is there or not.Some of my numbers were a bit off too, so let's do it again.The light bulbs are 1.2W, 24V.Dividing, we get a current of 50 mA.The bypass resistors ar 1.2k.24V / 1.2k = 20 mAThat's a total of 70 mA.76 feeders, 2 circuits each = 152 * 70 mA = 10.64 Amps.