This motorcycle starter relay is only intended for short duration use. Here's a technique to allow longer duration operation.
Hello! Um, a little while ago I showed this motorcycle starter relay. Um, it's got 100 amp main contacts and a 12vt DC coil. Uh, in fact, let's open it up and have a look inside. So in here, it looks like this.
So there's the coil connected to these two uh Spade terminals and there's the thick braid connected to one of these terminal posts via this u-shaped copper thing I Think the original photograph had two braids, hence the U shape. Uh, this only has one and there's the main contact. Now, a couple of people in the comments section pointed out that this is a starter relay, and therefore it's only intended to be energized for short periods of time and that it's possible that if I energize this for an hour or two, which is what I intend, that the coil will well, I presume overheat would be the uh the main problem. So what I want to do today is well, I'm going to put 12 volts on this and see how hot this gets, but also look at a possible a possible mitigation strategy for uh, backing the current off in this relay.
uh, not initially. So initially it'll have enough current to pull the relay in, but then it'll back the current off so that, um, the holding Force which will be smaller of course, because the metal is nearer the electromagnet um will be at a lower current and therefore this thing won't get as hot. but let's stick 12 volts on it uh, or perhaps a bit less and just see how warm it gets. So I found suitable cables? Uh, let's plug in couple of spay terminals.
Those go to these banana plugs, so let's fit couple of adapters and then that goes to the output of my power supply. as does that. Okay, let's get this powered up and stick. Well, let's say 10 volts initially? Uh, see if it'll pull in right? That's got 10 volts on the output.
Let's make sure this pulls in. Yeah, it does. Actually, it' be interesting to see what voltage this doesn't pull in, so let's try 8 vol. Vols Okay, 8 Vols Yeah, it still pulls in at 8 Vols Quite, Uh, it feels quite solid.
Okay, let's try 6 volts and no, that doesn't pull in, but will it hold? Yes, it will so it won't pull in, but if I press it in with my finger, then it does hold at 6. Vols So what you really want is, uh, let's look at voltage: a high voltage initially to pull the relay in and then back it off to something lower where it'll just hold like that and that holding force is quite strong. CU If I tried to bend this away, you can see that the metal slab is not lifting off the electromagnet, so that's got a very firm holding Force there. I've done it now, even if it hasn't got enough uh, electromagnetic force to initially pull it in right? How warm is it getting? So thermal imaging camera? Uh oh yes, there is some heat there.
Ah That's interesting. Uh 43 is the hottest point? Yeah, so if I have 12 volts in this, um, that's going to get a lot hotter. Let's do some experiments. Okay, Uh, 7.99 or 8 volts and it's at 51 and possibly getting warmer.
Uh, 10 volts and it's now at 60 and getting warmer. 62 63 Okay, let's 64. Let's shove it up to 12 volts and now, uh, 12 volts. Oh yeah, now that's getting, um, very warm And I would probably agree with those commenters who said, uh, it's probably not intended for continuous use at 12 Vols Uh, 84 that is going to go up up to 90 isn't it? Okay, that's been a minute or two and yes, that has effectively hit 100. There we are 100 so that's quite hot isn't it? And if you put the cover on this thing, then Heat's going to build up inside here. So I do think a mitigation strategy um, to pres present an initial High current, but then back it off off probably is necessary for this. I Mean it hasn't melted yet, but the you can see the former is made of plastic. Let's just check that temperature again.
Yeah, it's now up to 105. See quite warm. So here's my idea. Put a light bulb in series with the coil because a light bulb has an interesting property in that it's it's a positive temperature coefficient resistor.
Um, its cold resistance is much lower than its warm resistance. So this is my old friend, the 21 wat uh, 12v vehicle brake light bulb. So really, what we want to do is find out what's the resistance of this coil and what's the resistance of this light bulb cuz I'd Quite like since we saw that the relay will hold in perfectly well at 6 Vols I'd quite like to have about the same resistance. Now that's of course going to be the hot resistance.
but let's just measure the resistance of these two things. To start with, So I'm measuring the resistance of the coil the relay at 33.6 ohms. but I'm just put a fan on it to cool it down because it had got quite warm. so I'll let that cool down.
Okay, so when this relay is cold, it's coil resist resistance is 30 ohms and the resistance of this 21 wat. uh, brake light bulb is measuring at half an OHM. So 30 ohms for the Relay coil half an OHM for this brake light bulb. Now of course that's the cold resistance for the hot resistance.
We can actually calculate it. so I've got uh I equal W over V. That should probably be P for pal. So IAL power over volts.
That's because V * IAL Watts Uh, we know that Ral v I from Oh's law. So I've come up with Ral v^2 over power. Um, so a 21 watt bulb v^2 is 12 squ so that's 144 is that over 21 which is 6.85 ohms and a 5 wat bulb uh is 144 over 5 which is 28.8 Ohms. Well, that's much nearer the 30 ohms of this relay.
So what I really want is a 5w bulb, right? One of these bulb kits from I think this was B&M or Home Bargains or something like that. Uh, here's a 5 wat bulb. So from the calculation, the Uh warm resistance. the hot resistance is 30 ohms.
Let's measure the cold resistance right. The cold resistance is let's put that on there and that on there and that is 3 ohms. Okay, so 3 ohms as a proportion of 30 ohms is only 10% so it shouldn't affect the pull in power of this relay too much. But when this gets hot, it'll go to 30 ohms. And that mean means that we've got a roughly equal Uh resistance split 30 ohms for the Relay 30 ohms for the bulb. so they should each get Uh 6 volts from the 12 volts. Now of course I'm putting this on lithium ion phosphate, which could start at 14 volts, but that would still only be 7 Volts for each Uh device. Now I'm going to solder some wires to this bulb, um, put it in series with this relay and then check that it pulls in reliably at Uh 12 volts and that that voltage then Uh reduces to 6 volts or thereabout when The Bulb Has uh warmed up.
Okay, so I have soldered, um, a cable to the bulb to go to the relay. Got a banana plug there? Uh, so that's going to be negative, isn't it? Let's bring that around to there positive of my 12 volts goes to the relay. Now the question is, does the relay with bulb, uh, pull in at 12 volts and then settle back to a uh, lower voltage? Well, let's just see if it pulls in first. Uh, so 12 volt switch it on and yeah, it pulls in and you can see It takes quite a long time for the bulb to come on, so it should have almost the full 12 volts.
Uh, I think it was 10% goes into the bulb when it's cold, wasn't it? So it'll get um, 10.8 or something like that. And then when the bulb warms up, it should be a a 50/50 split of voltage. But certainly that pulls in perfectly adequately. And now is it holding in? Yeah, it seems to be holding in pretty well.
I can. Oh, it seems to be getting a bit more than the 6 volts because it pulls back in with the bulb next to to it. The question now is uh well. let's look at the voltage across the relay and it is uh oh, why am I not getting volts cuz I'm on AC Yeah, it's annoying.
You have to press the function button to get to DC volts. Okay, it's a little bit more than half, but that's okay. Uh, 7.1 Vols on the re lay. Now let's check with the thermal imaging camera and that's a much more reasonable Uh 50 C that's going to settle at I think than the 100 C Um, when it was getting 12 Vols I think it could probably handle 50 C I'll just leave it on for a bit and see how hot it does get and it's now at 55 degrees C But it's going up now.
Oh why is that suddenly jump to 57 I think I put my hand in front of it. No, no, it is 57 where the hottest part of the image is, but it doesn't seem to be getting any hotter than that. That's probably okay. and uh yeah.
so I think that's probably it. That's probably the solution to uh, one of these relays which is only intended to be turned on for a few seconds. Uh, when left on friendly length of time gets up to 100. Now it's only getting up to sort of 60.
and uh yeah, all it needs in series with it is a five watt Uh Car 12vt bulb that allows it to pull in hard and close the contact and then it slightly releases the power inside the relay. Well, we could calculate power inside the relay. It probably releases it quite a bit. Uh, the question then is is there enough contact UM between there I think I bent that up slightly so I have just bent it back down with a pair pliers. um to hold this 100 amps that I'm going to be putting through the Uh relay switch contacts. but I think uh, that's the principle and that's it for this video. So Cheerio.
The reason you don't quite get 50/50 is because the bulb doesn't heat up to 5W so its resistance is lower than calculated. A 4W bulb (if one exists) would match better.
I'm just here for the Ohm's law gymnastics.
Do you think it would be a problem to activate the relay while dunked in a bucket of oil for cooling? Do you think there will be an issue with the contacts not working correctly being operated in oil?
You should replace the cover and run it for a few hours and see how much heat builds up with no air flow around it.
After reading lots of comments I did not see mine. I usually drop the voltage needed to operate a relay by adding a small magnet on top of it. After some positioning, you can get the relay to work properly (without the relay sticking top or bottom). I think this super cheap and only requires some research per relay as it is a bit fiddling. When done, glue it tight and that's it.
A better way would be a either self powering (if the feed to the relais is separate from what it relays, you can use the relay to keep itself switched on, as long as you can switch it of otherwise). Another solution, not this relay though, is a latching relay. One that has two positions and with a negative pulse you can disengage it, and with a positive pulse it engages.
Just some random thoughts.
Surely the resistance of the coil decreases as it heats up? And more so with the cover on? Thermal runaway here we come 😉
Seems to me that what you need is a latching relay.
You never state the purpose of the relay. Are you switching an inductive, or resistive load? Is it an AC or DC load? Why waste all this time and effort (and math) to use something that is clearly NOT designed for your purpose. It's not just the intermittent duty coil you have to worry about, it's also the intermittent duty contacts. You could go to your local HVAC contractor and pick up a definite purpose contactor for a 3 or 4 ton condenser for about $15US. They are designed for high current AND continuous duty. One thing you did not address, is the heat that will be generated by your switched load which could negate your efforts to reduce heat.
It is comum to use a capacitor in series and a resistor of the same resistance as the coil in parallel with the cap. This is a bit more elegant and stable solution.
There was a nice circuit published in wireless world in the 90s? It used a transistor to drive the relay at 12v fed from a capacitor, and then it dropped to ~8v to supply the holding current. I’ve always thought it odd that holding current is almost never implemented in relay driver designs. Also there was an industrial circuit that fed the driver from a pulse train, if the pulse stuck high or low (ie = to a uP failure, the relay would fail safe. It was used in Power Station control systems
I use a second relay to activate the larger relay but this smaller relay has a transistor with an R-C circuit making it slow to come on it has a resistor across its normally closed contacts so when it operates and its contacts open there is a resistor in series with the larger relay coil reducing the current through it. This means the large relay operates quickly but does not overheat it can also be done with transistors but my larger relay operates at 48v and has 10mm diameter contacts
J another brill solution. plenty of those auto contactor cheaply on EB. very simple cct but uses 5W. Maybe a pump cct primer just to start by doubling momentarily 7v to say 14v, charges a capacitor enough to pull in the contactor? . Now we need to know it the relay will break 100A at 24V without arc flash continuing. 48V probably not, the gap would need to be much bigger, 20mm not 6mm. But very interesting Pls keep going
Interesting video, don't forget to fit a diode across the coil to stop back EMF destroying what ever device you use to energise the relay.
What about contact heating?
Similar thing is done with things like mains electronic timers. The power supply for the electronics is a dropper capacitor, inrush resistor and zener diode and filter capacitor, providing power to a linear regulator. But the relay coil and its switching mosfet is in parallel with the zener diode and capacitor. When the relay is energized, the capacitor provides the current to close it, but the sizing of the dropper cap is such that it can only provide enough current to keep the relay pulled in. This is fine as long as there is enough voltage to keep that linear regulator working. They often use 36v zeners and relay coils, and a 12v linear regulator. When energized, there's often only 15 volts across that relay, but that's enough to keep it pulled in.
In the RC hobby this is called a smoke stopper. Used when you power up a model for the first time. The bulb keeps the receiver and servos going, but stops any dead shorts setting the model and battery on fire.
As someone has already suggested a resistor in parallel with a capacitor then in series with the coil is the best solution and one I've seen used on solenoids. The problem with the lamp is it will eventually fail, pretty much obsolete and a short loss of supply will leave a lot of resistance from the lamp in series with the supply to the coil. With the RC solution you can adjust the pull in current and the holding current independently.
Reason these kinds of devices don't need much holding current is when a relay or contactor is closed the magnetic circuit is completed increasing the strength of the magnetic field.