Another method would be to use a 30 ohm resistor in series with the coil and have an electrolytic capacitor across the resistor, the capacitor would present as a short circuit across the resistor at power up thereby putting the full 12v on the relay until charged up.
Welcome to the world of coil economisers. The usual simple solution is a high value capacitor to provide the inrush current, with a resistor in parallel to supply the hold current, the pair being in series with the relay coil. The best solution is an actual 'peak and hold' circuit. There are some fuel injector drivers that are ideal for the job as well as dedicated ICs or standalone PCBs. BTW, a very commonly overlooked problem is the release delay caused by using a freewheel diode across the coil. At higher switching voltages, this can lead to 'pulling' an arc. Use a power Zenner diode instead, which will allow the magnetic field to collapse faster.
Power diode? Are you referring to a Schottky diode? Afaik freewheel is just the configuration, but I get the concept, I hadn’t thought about arc closure speed before, thanks
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.
Nice simple solution! The bulb is only getting about 5 volts (7 volts drop across the relay) so the temperature and therefore the resistance isn’t as high as when it had the whole 12 volts across it.
1. heat the relay in the housing 2. measure the temperature immediately after removing the housing. 3. Treat temperature as the temperature difference with the surroundings. If it has now warmed up by 40 degrees, in a shed where it may be 30 degrees it will also warm up by 40 degrees, i.e. to 70 degrees or more.
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
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.
I think he should do a resistor and capacitor instead. He needs the voltage drop across the coil to be lower, maybe closer to 6v and he'll have trouble doing that when the bulb is not getting its full rated current through it. He might add two bulbs in serries and together they may rise up to 30 ohms resistance but their start resistance is 6 ohms which is starting to become a serous fraction of the coil's resistance. The nearly 0 ohm impedence of a capacitor in parallel with the resistor or in parallel with two serries light bulbs drops the initial impedence close to 0 ohms and the impedence eventually rises to the ohms from the resistance alone.
Nice easy and effective solution - I like it! I have a relay like this in one of my vehicles and was never happy that it got so hot - I think I’ll implement this solution - thanks!!
Nice way of dealing with current. I had a similar situation where I needed to reduce the power a relay was using so I PWM'd it once it had actuated. Not idea but has worked fine a decade or so.
Comment before finishing video: Multiple Diode dropper with parallel capacitor. Capacitor gives you the current spike to pull it in. Resistor could be used in place of the diodes of course.
Might need a resistor to discharge the capacitor. If you turn it off and on again the capacitor will already be charged the second time, and the relay might not get enough voltage to pull.
I have used a resistor with a parallel capacitor. The relay get full voltage and current at the start, then the capacitor charges up to whatever the voltage the resistor drops. like, ~33ohm coil with a 33ohm resistor. It start with 12v and then it goes down to 6 volt.
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.
I have connected a resistor with a 1000µF electrolytic capacitor in parallel in series with the relay. Due to the capacitor, the full voltage was present when the relay was switched on and dropped to the holding voltage of approx. 8V after < 0.5 seconds. Just like your solution with the bulb in series.
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.
just use a Fet, full power to pull the contact in and then pulses to reduce the current for the hold in. There was a device called a "cool cube" that we fitted to solenoid's to give a high power pulse then reduced the power with PWM. I did a video about the leakage current in a cap which was enough to hold a relay in once it had been energised, but not enough to pull in. it was a antenna tuner unit, poor old micro could not work out what was going on, as once it had operated the relay once it just stay on. but it did not even get warm even though it was engaged all the time.
Relay Contacts are always springloaded in some some way. If the electro magnet closes the gap to the attracted part completely the contactpressure is only a function of the force stored in the displacement of the spring. So if you can’t see or measure a gap on top of the electromagnet you can assume it to be fully closed with the correct contact pressure.
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.
A properly placed capacitor would allow one to use more current limiting higher resistance light bulbs while still allowing high initial relay closing current. If one puts a large capacitor in parallel with the light bulb, the capacitor starts off discharged through the bulb. But when the relay is switched on, the capacitor represents a very very very very low impedence path that counteracts not only the inductive impedence in the coil and the inductive impedence in the bulb's coiled filiment, but initially represents 0 impedence. Thus the capacitor helps one get high initial current which helps quickly close the relay. As the light bulb warms up and its resistance increases to say 30 ohms, then the capacitor is charged up to the voltage drop across the light bulb, resulting in the capacitor's impedence approach infinity (not really but close enough for pratical purposes). As little to no more current flows through the capacitor, the capacitor and light bulb network impedence then becomes effectivly close to the 30 ohms resistance of the bulb. But as you observed, because the bulb isnt getting the fully current its rated for, the bulb did not heat up fully and did not reach its full 30 ohm resistance. 12v supply -7.12v across the relay coil means just 4.88v is being dropped across the bulb. Mater of fact, if the voltage drop across the 30ohm coil is 7.12v and the voltage drop across the bulb is 4.88v. Then the bulb's low current hot resistance must be as little as (12v×30ohm)/7.12v - 30 ohms = 20.56 ohms. If you us a higher battery charging voltage i suspect that you could or rather should should use two light bulbs in serries with one large capacitor across the two light bulbs. Briefly, as the capacitor charges the initial impedence will be close to 0 ohms. but as it reaches its full charge with two light bulbs in serries, the hot resistance of each bulb will probably be close to 15-17 ohms a piece or 30-34 ohms in serries with each other. And lastly, this way, you'll be closer to the operating conditions you were looking for, and producing much less heat and only dropping 6v across the coil rather than 7.12v. That in turn gives you room to apply a charging voltage without overheating the coil inside a close relay box.
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
Forgive me if you have already, but have you considered a Latching Relay, yes it does complicate the application requiring an input to disconnect etc...
What about placing a capacitor and resistor in parallel, then adding these in series with your relay? The inrush current to charge the capacitor should be enough to latch the relay, then the resistor will hold it there?
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?
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.
Great Video Julian! Thanks very much! It would be interesting, for example, to switch on xxx ms with 12V and then, for example, 1 KHz (or similar) with 12V PWM 50:50 (or similar).
Prefer to have the diode as azener across the transistor, just a lower voltage than absolute max for the transistor. Faster turn off, and less arcing on the relay contacts.
The relay will last longer, the lamp, running at reduced volts will last till the heat death of the universe and you have an indicator lamp during actuation - brilliant.
An issue not discussed is that of the relay's contact voltage rating. Specifically it's DC contact voltage rating. Switching higher voltage solar panels may not be reliable long term.
You are forgetting that 30 ohm resistance of the relay. This means the actual wattage of the bulb in circuit will be much lower and it wont get as hot and therefore it's resistance will be a bit lower. Hence the 7+ volts across the relay.
Neat idea, but I think I'd still prefer to use a 100amp contactor that's designed to run for long periods of time, or at least some logic based solution. Seems a bit wasteful to be burning the difference!
The normal way to do this is to use a capacitor and resistor. The reduced contact pressure will also certainly increase the contact resistance and result in damage early failure and overheating. This is really a great example of what not to do with a relay. Just get something correctly rated for the job. Given your trying to switch huge current this really is a very silly idea.
@@JrgenDurkeHansen no you are wrong and it’s very easy to prove having done the same thing in the past and witnessed an actual failure due to low holding current.
I'd guess the temperature will rise quite a bit when you put the cover on. Perhaps you should make a new one so that you get some circulation (3d printer or similar)?
The downside of a "real" PTC is the thermal mass. If you turn the relay off and then back on the resistance will be too high and the relay won't close. The bulb takes perhaps half a second to cool down so this is less likely to happen.
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 wonder if a 10w bulb would work better ? You should still have enough power to keep the relay latched but it would also reduce the cureent through the coil.
10w bulb over a 5 watt bulb? No it would not work better in this case. Given 12v rated bulbs 10w is twice the power over 5 watts, thus twice the current flows through half the resistance. Then in a 30ohm coil and resistor (light bulb) voltage divider, the higher watt lower resistance light bulb drops a smaller faction of the 12v leaving a larger fraction to be dropped across the coil causing the coil to heat up more.
thank you very much Julian from a person smart enough to realize how stupid he is lol - all jokes aside, I can understand this! so invaluable to me, especially reading (well trying to read and comprehend) the alternate methods/terminology in the comments - humbly grateful for your content!
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.
Ah yes, you'll have a bad time if you try switching a much higher voltage or something particularly inductive. I made this mistake before, and the relay would switch-- and weld the contacts together.
You should never self pull in blokes undervoltage, that's how you end up with children! But you may last longer if they look after you is how I read this video ROFL. Seems like simple ohms logic to me.
Another method would be to use a 30 ohm resistor in series with the coil and have an electrolytic capacitor across the resistor, the capacitor would present as a short circuit across the resistor at power up thereby putting the full 12v on the relay until charged up.
Welcome to the world of coil economisers. The usual simple solution is a high value capacitor to provide the inrush current, with a resistor in parallel to supply the hold current, the pair being in series with the relay coil. The best solution is an actual 'peak and hold' circuit. There are some fuel injector drivers that are ideal for the job as well as dedicated ICs or standalone PCBs.
BTW, a very commonly overlooked problem is the release delay caused by using a freewheel diode across the coil. At higher switching voltages, this can lead to 'pulling' an arc. Use a power Zenner diode instead, which will allow the magnetic field to collapse faster.
Power diode? Are you referring to a Schottky diode? Afaik freewheel is just the configuration, but I get the concept, I hadn’t thought about arc closure speed before, thanks
@@____________________________.x Sorry, the word Zenner got missed out and it's opening speed that's the problem.
@@EVguru Thanks, I'll look into it
This is what I did as well, works great. You do need a beefy resistor though, which I stuck to the metal case with some thermal paste
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.
Nice simple solution! The bulb is only getting about 5 volts (7 volts drop across the relay) so the temperature and therefore the resistance isn’t as high as when it had the whole 12 volts across it.
Yes, he did the math (more or less) for a 6V 5W bulb 😃
1. heat the relay in the housing 2. measure the temperature immediately after removing the housing. 3. Treat temperature as the temperature difference with the surroundings. If it has now warmed up by 40 degrees, in a shed where it may be 30 degrees it will also warm up by 40 degrees, i.e. to 70 degrees or more.
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
Brilliant solution, simple and effective.
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.
Clever! I would have done a resistor and capacitor, but I like this option better as you get a pilot light too!
I think he should do a resistor and capacitor instead. He needs the voltage drop across the coil to be lower, maybe closer to 6v and he'll have trouble doing that when the bulb is not getting its full rated current through it. He might add two bulbs in serries and together they may rise up to 30 ohms resistance but their start resistance is 6 ohms which is starting to become a serous fraction of the coil's resistance. The nearly 0 ohm impedence of a capacitor in parallel with the resistor or in parallel with two serries light bulbs drops the initial impedence close to 0 ohms and the impedence eventually rises to the ohms from the resistance alone.
Nice easy and effective solution - I like it! I have a relay like this in one of my vehicles and was never happy that it got so hot - I think I’ll implement this solution - thanks!!
Nice way of dealing with current. I had a similar situation where I needed to reduce the power a relay was using so I PWM'd it once it had actuated. Not idea but has worked fine a decade or so.
Comment before finishing video: Multiple Diode dropper with parallel capacitor. Capacitor gives you the current spike to pull it in. Resistor could be used in place of the diodes of course.
Might need a resistor to discharge the capacitor. If you turn it off and on again the capacitor will already be charged the second time, and the relay might not get enough voltage to pull.
A VERY interesting experiment. Thanks, good info. 👍
I have used a resistor with a parallel capacitor. The relay get full voltage and current at the start, then the capacitor charges up to whatever the voltage the resistor drops.
like, ~33ohm coil with a 33ohm resistor. It start with 12v and then it goes down to 6 volt.
But don't let the heat from the resistor or relay winding bake the capacitor.
@@iblesbosuok The resistor and capacitor can be at the switch instead of the relay.
Capacitors are also rated atleast 85C so they can take some heat
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.
I have connected a resistor with a 1000µF electrolytic capacitor in parallel in series with the relay. Due to the capacitor, the full voltage was present when the relay was switched on and dropped to the holding voltage of approx. 8V after < 0.5 seconds.
Just like your solution with the bulb in series.
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.
just use a Fet, full power to pull the contact in and then pulses to reduce the current for the hold in.
There was a device called a "cool cube" that we fitted to solenoid's to give a high power pulse then reduced the power with PWM.
I did a video about the leakage current in a cap which was enough to hold a relay in once it had been energised, but not enough to pull in. it was a antenna tuner unit, poor old micro could not work out what was going on, as once it had operated the relay once it just stay on. but it did not even get warm even though it was engaged all the time.
Relay Contacts are always springloaded in some some way. If the electro magnet closes the gap to the attracted part completely the contactpressure is only a function of the force stored in the displacement of the spring.
So if you can’t see or measure a gap on top of the electromagnet you can assume it to be fully closed with the correct contact pressure.
09:00 You calculation is slightly off. 5W at 12V is 28 ohm, but the same light bulb at 6V is not. That's PTC in a nutshell (light bulb). 🙂
Is the bulb technically acting as a PTC and at 100A are the contacts going to produce added heat to the overall temperature of the relay?
Yes
About 2 or 3 yrs ago, I had one of these exact relays melt running a 8" fan..
That's very cleaver. Nice idea!
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.
A properly placed capacitor would allow one to use more current limiting higher resistance light bulbs while still allowing high initial relay closing current. If one puts a large capacitor in parallel with the light bulb, the capacitor starts off discharged through the bulb. But when the relay is switched on, the capacitor represents a very very very very low impedence path that counteracts not only the inductive impedence in the coil and the inductive impedence in the bulb's coiled filiment, but initially represents 0 impedence. Thus the capacitor helps one get high initial current which helps quickly close the relay.
As the light bulb warms up and its resistance increases to say 30 ohms, then the capacitor is charged up to the voltage drop across the light bulb, resulting in the capacitor's impedence approach infinity (not really but close enough for pratical purposes). As little to no more current flows through the capacitor, the capacitor and light bulb network impedence then becomes effectivly close to the 30 ohms resistance of the bulb.
But as you observed, because the bulb isnt getting the fully current its rated for, the bulb did not heat up fully and did not reach its full 30 ohm resistance. 12v supply -7.12v across the relay coil means just 4.88v is being dropped across the bulb. Mater of fact, if the voltage drop across the 30ohm coil is 7.12v and the voltage drop across the bulb is 4.88v. Then the bulb's low current hot resistance must be as little as (12v×30ohm)/7.12v - 30 ohms = 20.56 ohms.
If you us a higher battery charging voltage i suspect that you could or rather should should use two light bulbs in serries with one large capacitor across the two light bulbs. Briefly, as the capacitor charges the initial impedence will be close to 0 ohms. but as it reaches its full charge with two light bulbs in serries, the hot resistance of each bulb will probably be close to 15-17 ohms a piece or 30-34 ohms in serries with each other. And lastly, this way, you'll be closer to the operating conditions you were looking for, and producing much less heat and only dropping 6v across the coil rather than 7.12v. That in turn gives you room to apply a charging voltage without overheating the coil inside a close relay box.
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
How about sneaking a bead thermistor into relay case with wires accessible for the occasional check?
Nice use of resistance to drop voltage with the use of the bulb, Hmm do you have a use for the relay????
Hey Julian, what is that module you are playing with in the upper left of the screen?
Would the bulb fit inside the case? Tidy but would presumably increase the temperature by more than you would like.
You can get an "Audio" relay type to use for high amp continuous handling...mine goes to 500A and has been running for 17 months so far.
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
Forgive me if you have already, but have you considered a Latching Relay, yes it does complicate the application requiring an input to disconnect etc...
What about placing a capacitor and resistor in parallel, then adding these in series with your relay?
The inrush current to charge the capacitor should be enough to latch the relay, then the resistor will hold it there?
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?
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.
Great Video Julian! Thanks very much! It would be interesting, for example, to switch on xxx ms with 12V and then, for example, 1 KHz (or similar) with 12V PWM 50:50 (or similar).
This would be a more efficient design. Dumping power into the light vs. coil is a solution but certainly not an elegant one.
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.
Prefer to have the diode as azener across the transistor, just a lower voltage than absolute max for the transistor. Faster turn off, and less arcing on the relay contacts.
You could also run a fan off it to cool itself down for extra insurance.
The relay will last longer, the lamp, running at reduced volts will last till the heat death of the universe and you have an indicator lamp during actuation - brilliant.
An issue not discussed is that of the relay's contact voltage rating. Specifically it's DC contact voltage rating. Switching higher voltage solar panels may not be reliable long term.
You are forgetting that 30 ohm resistance of the relay. This means the actual wattage of the bulb in circuit will be much lower and it wont get as hot and therefore it's resistance will be a bit lower. Hence the 7+ volts across the relay.
The heat build-up will be higher with the cover reinstalled over the relay. You might want to leave it exposed to the air?
Neat idea, but I think I'd still prefer to use a 100amp contactor that's designed to run for long periods of time, or at least some logic based solution. Seems a bit wasteful to be burning the difference!
The normal way to do this is to use a capacitor and resistor. The reduced contact pressure will also certainly increase the contact resistance and result in damage early failure and overheating. This is really a great example of what not to do with a relay. Just get something correctly rated for the job. Given your trying to switch huge current this really is a very silly idea.
Wrong. If the relay is fully pulled then adding extra force to the coil doesn't add extra force to the contacts as they are spring loaded.
@@JrgenDurkeHansen no you are wrong and it’s very easy to prove having done the same thing in the past and witnessed an actual failure due to low holding current.
I'd guess the temperature will rise quite a bit when you put the cover on. Perhaps you should make a new one so that you get some circulation (3d printer or similar)?
A PTC thermistor like used on tube TV sets for the degauss coil would do the same and you could tailor the resistances.
The downside of a "real" PTC is the thermal mass. If you turn the relay off and then back on the resistance will be too high and the relay won't close. The bulb takes perhaps half a second to cool down so this is less likely to happen.
I have the 200A and I leave it on 6-8 hours daily in summer as its connecting 2 battery packs when theres enough solar
I'm just here for the Ohm's law gymnastics.
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.
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.
What about contact heating?
very clever
I wonder if a 10w bulb would work better ? You should still have enough power to keep the relay latched but it would also reduce the cureent through the coil.
10w bulb over a 5 watt bulb? No it would not work better in this case.
Given 12v rated bulbs 10w is twice the power over 5 watts, thus twice the current flows through half the resistance.
Then in a 30ohm coil and resistor (light bulb) voltage divider, the higher watt lower resistance light bulb drops a smaller faction of the 12v leaving a larger fraction to be dropped across the coil causing the coil to heat up more.
@@kreynolds1123 My maths is bloody abysmal. You're right. 👍
The problem with the bulb is if you turn off then on quickly, the resistance of the bulb remain high. So the relay might not close in that case.
thank you very much Julian from a person smart enough to realize how stupid he is lol - all jokes aside, I can understand this! so invaluable to me, especially reading (well trying to read and comprehend) the alternate methods/terminology in the comments - humbly grateful for your content!
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.
Ah yes, you'll have a bad time if you try switching a much higher voltage or something particularly inductive. I made this mistake before, and the relay would switch-- and weld the contacts together.
You should never self pull in blokes undervoltage, that's how you end up with children! But you may last longer if they look after you is how I read this video ROFL. Seems like simple ohms logic to me.