selecting wire gauge for pulses current

Thread Starter

Momentory

Joined Sep 4, 2022
67
How much copper mass do you have in the coil?

How much energy is dumped into the coil with each pulse?

What is the thermal capacity of copper?

How much will each pulse raise the temperature of that much copper, assuming none of it goes anywhere else?

If that temperature rise is below the limit of the wire, then you don't need to worry about the individual pulses, but rather on their cumulative effect. In that case, looking at the RMS power dissipated in the coil is what you want to focus on.
How much copper mass do you have in the coil?

How much energy is dumped into the coil with each pulse?

What is the thermal capacity of copper?

How much will each pulse raise the temperature of that much copper, assuming none of it goes anywhere else?

If that temperature rise is below the limit of the wire, then you don't need to worry about the individual pulses, but rather on their cumulative effect. In that case, looking at the RMS power dissipated in the coil is what you want to focus on.
My worries about cumulative pulsed temperature and since the wire has no obvious datasheet so i decide just want to add cooling system even it will not get much high temperature (as safety)
 

WBahn

Joined Mar 31, 2012
27,935
My worries about cumulative pulsed temperature and since the wire has no obvious datasheet so i decide just want to add cooling system even it will not get much high temperature (as safety)
But the wire is copper, right?

You know the size of the wire, so you can look up the diameter of it, right?

You can determine how much volume of copper there is in your coil, right?

You can look up the mass density of copper and the thermal capacity of copper, right?

You are certainly free to put a cooling system on it if you want without determining if you need one, but how will you know that your cooling system is adequate without determining how much cooling you need?
 

MrAl

Joined Jun 17, 2014
9,638
@Momentory
Coil in circuit #31 dissipates 30 J energy during one pulse.
Such energy can increase temperature of 1.172 kg of copper only on 0.067 °C, not melt it.
If you want to know what is fusing 2 ms rectangular pulse current for AWG 18 wire,
use this formula:
Ifuse = Area * SQRT(LOG((Tmelt - Tambient) / (234 - Tambient) + 1) / Time * 33)),
where
Tmelt -melting temp of wire in °C [1083],
Tambient -ambient temp in °C [25],
Time -melting time in seconds [0.002],
Area -wire area in circular mils [1620],
Ifuse -fusing current in amps,
then
Ifuse = 1620 * SQRT(LOG((1083 - 25) / (234 - 25)+1) / 0.002 * 33)),
Ifuse = 1898 A.
Hello,

I didnt check that formula but it's nice of you to post this info.
Now if you care too, set Ifuse=100 amps and solve for the time, which would be the time when the wire melts and breaks apart or nearly so. That should be interesting. That would be the absolute maximum time for any pulse starting with a wire that is always at ambient temperature just before the single pulse is applied.
This is still for a single pulse assuming that the off time will allow the wire to cool back down to 25C or nearly so, but still interesting.

It's also interesting that as the pulse is applied and time progresses, the wire will heat up before melting and that will cause the resistance to go up, and with a constant current of 100 amps that means the power in the wire will go up, and that means even more wire heating power, which of course causes a very fast wire heat up.
In real life though this would probably be a voltage drive so the power would actually come down as the resistance increases. Would be interesting to plot too.
 

MrAl

Joined Jun 17, 2014
9,638
My worries about cumulative pulsed temperature and since the wire has no obvious datasheet so i decide just want to add cooling system even it will not get much high temperature (as safety)
Yes the accumulated heat buildup is hard to predict without a test. This is actually true of a lot of devices we dont always think about.

In the industry transformers and inductors are often vacuum varnished where the vacuum chamber pulls out all the air spaces and thus increases the thermal conductance of the coil as a whole. This of course keeps the internal coil wires cooler although they will always be hotter than the outer layers.
Also in the industry of power supplies, an over night test is run to make sure the construction does not melt down. The various insulation layers can melt and cause a lot of problems not only with the material getting all over the place as it drips out like candle wax :)

So once you decide on some construction and you think you can use it, you have to test it at full power for several hours, probably 12 to 24 hours. Full power in this case though would be with the required pulse time and required off time, not full current for the entire time obviously.

I remember one converter back in the 1980's that melted down overnight. It was a unit of about 5000 watts, synthesized sine output (true sine output). Not only did the output transformer melt down and leak material all over the place, the 4 bridge transistors blew out and at the time they were around $100 each. The whole thing had to be repaired and a new transformer design used to fix the issue. It's not a pretty sight for sure.

The rule of thumb is sometimes you just cant be sure until you test it. This is also the case with wall warts, if a new design. They have to be run over night to be sure they can survive. There used to be a place in Northern New Jersey that did this routinely not sure if they are around anymore though. They shipped thousands of wall warts of various kinds probably within a week or less but i dont remember the time frame now.
 

Danko

Joined Nov 22, 2017
1,493
I didnt check that formula
I think this source is reliable enough (see attachment).

@Momentory:
Calculated parameters of your coil:
L=3.5 mH
Inner diameter of coil=15 mm
Outer diameter of coil=45 mm
Length of winding=60 mm
Number of turns=673
Number of winding layers=13
Magnet wire: copper, AWG 18
Length of wire=62 m (not 160 m)
Resistance of coil=1.3 Ω (not 3.2 Ω)
Weight of wire=455.43 g

So, your coil will dissipate 6.1 W power in circuit below.
Contactors coils dissipate similar power, therefore,
especially if take in account that work session will 10 min only,
you can try to test it, carefully.

Circuit (NOTE: capacitor C1 must be film, not electrolytic!):
1000V 100UF DC-link Film Capacitor
US $25.95
ADDED:
KP-200A AC 1000V 200A Stud Thyristor Silicon Control Rectifier SCR
US $15.00
1665827662358.png
 

Attachments

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Thread Starter

Momentory

Joined Sep 4, 2022
67
I think this source is reliable enough (see attachment).

@Momentory:
Calculated parameters of your coil:
L=3.5 mH
Inner diameter of coil=15 mm
Outer diameter of coil=45 mm
Length of winding=60 mm
Number of turns=673
Number of winding layers=13
Magnet wire: copper, AWG 18
Length of wire=62 m (not 160 m)
Resistance of coil=1.3 Ω (not 3.2 Ω)
Weight of wire=455.43 g

So, your coil will dissipate 6.1 W energy in circuit below.
Contactors coils dissipate similar energy, therefore,
especially take in account that work session will 10 min only,
you can try to test it, carefully.

Circuit (NOTE: capacitor C1 must be film, not electrolytic!):
1000V 100UF DC-link Film Capacitor

View attachment 277756
Yes i think the only way to determine the temp is test the coil and measure its temp

For capacitor, i will use diode across the coil i still can use electrolytic capacitor
 
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