selecting wire gauge for pulses current

Momentory

Joined Sep 4, 2022
67
What is the ESR of that capacitor?

If you charge it to 320 V, then as soon as you start dumping it into that coil, the voltage is going to drop and you will never get anywhere near your 100 A target since you need the full 320 V available when you get there.

If it is 3.5 mH and 3.2 Ω, the time constant is about 1.1 ms. That means that if you apply a constant 320 V to it, after 1.1 ms the current would only be up to about 63 A.

Now let's consider how much charge you have available in that cap.

At 320 V, a 470 uF cap will have 0.15 coulombs of charge. Even if you could pull it out at a constant 100 A, that's only 1.5 ms worth of charge. But, again, that voltage is going to be dropping precipitously while this is happening.
Unfortunately the datasheet of the capacitor has not named capacitor ESR

So the capacitor will not delivery 100A!!

How to prevent this massive voltage and will this voltage drop generate heat across the wire?

Danko

Joined Nov 22, 2017
1,489
With capacitor 470 μF and voltage 320 V average current through coil will 14.9 A, and RMS current 27.4 A.
For RMS current 100A you should use voltage 980 V during 2 ms.
Pulse duration is more than 2 ms, so diagrams and calculations are for 10 ms pulse.

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Momentory

Joined Sep 4, 2022
67
With capacitor 470 μF and voltage 320 V average current through coil will 14.9 A, and RMS current 27.4 A.
For RMS current 100A you should use voltage 980 V during 2 ms.
I can not understand why all this lost of current!!

WBahn

Joined Mar 31, 2012
27,862
I can not understand why all this lost of current!!
It's because a capacitor's voltage is proportional to the charge stored on it. As that charge is turned into current, the voltage drops. The more current, the faster the drop. The initial voltage on the capacitors isn't going to driving current through the coil's resistance, it is going to increasing the current in the coil's inductance. As that current is building up to a peak, the charge is being consumed faster and fast. The simulation shows that the peak current is reached at about 1.5 ms (so pretty close to the 1.1 ms L/R time constant) at which point the voltage on the capacitor is down to about 180 V, which is only enough to support about 56 A in the coil's resistance (since the current is, momentarily, not changing, there is no voltage associated with the inductance and it's all due to resistance at that moment). After that, the capacitor voltage continues to fall and now the coil current starts dropping accordingly.

I'm actually a bit surprised that the pulse is both as short as it is and as high as it is.

You're going to have a very hard time getting your current pulses to be any shorter than 3 ms to 5 ms given that the time constant of your coil is over 1 ms.

Momentory

Joined Sep 4, 2022
67
It's because a capacitor's voltage is proportional to the charge stored on it. As that charge is turned into current, the voltage drops. The more current, the faster the drop. The initial voltage on the capacitors isn't going to driving current through the coil's resistance, it is going to increasing the current in the coil's inductance. As that current is building up to a peak, the charge is being consumed faster and fast. The simulation shows that the peak current is reached at about 1.5 ms (so pretty close to the 1.1 ms L/R time constant) at which point the voltage on the capacitor is down to about 180 V, which is only enough to support about 56 A in the coil's resistance (since the current is, momentarily, not changing, there is no voltage associated with the inductance and it's all due to resistance at that moment). After that, the capacitor voltage continues to fall and now the coil current starts dropping accordingly.

I'm actually a bit surprised that the pulse is both as short as it is and as high as it is.

You're going to have a very hard time getting your current pulses to be any shorter than 3 ms to 5 ms given that the time constant of your coil is over 1 ms.
Thanks for this great explanation

But if so, and according to the schematic data the pulse duration ~ 5ms and the Max current is 56A

(If that current is continued to 5ms but it wont as it will decrease)
While between pulse and next pulse is 3s

So the duty cycle should be

5/ 3000 = 0.00168

So according to our formula

A√D

D= duty cycle

56√0.00168 = 2.29

So it is lower than the nominal current of the 18 Awg magnet wire

So it will not got much heat

Right?

WBahn

Joined Mar 31, 2012
27,862
Thanks for this great explanation

But if so, and according to the schematic data the pulse duration ~ 5ms and the Max current is 56A

(If that current is continued to 5ms but it wont as it will decrease)
While between pulse and next pulse is 3s

So the duty cycle should be

5/ 3000 = 0.00168

So according to our formula

A√D

D= duty cycle

56√0.00168 = 2.29

So it is lower than the nominal current of the 18 Awg magnet wire

So it will not got much heat

Right?
The RMS current is A√D for a rectangular pulse. Your RMS current is going to be much lower. You are probably okay, but keep in mind that comparing the RMS current to the constant current limit of the wire only addresses average heating, not transient heating. That limit is dictated by the thermal mass and thermal time constant, which we don't know (and which you'd need to measure).

Danko

Joined Nov 22, 2017
1,489
@Momentory
I guess it is exactly what you want:
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WBahn

Joined Mar 31, 2012
27,862
Why the current goes here to 100A in 1 ms? Does it because discharging of 4 caps together?
Yep. Notice that the time to peak is still almost the same at about 1.5 ms. Not much you can to about that time constant unless you REALLY slam it with very high voltage and then actively kill it at the other side to get it to fall fast, too.

Danko

Joined Nov 22, 2017
1,489
One more variant:
____

BobaMosfet

Joined Jul 1, 2009
2,053
Is there a rule to select wire gauge for a pulses current

Lets say i have magnet copper wire of 18 Awg, and i want to give it a pulse current 100A for 1-2 milliseconds and 2-4 seconds before the next pulse for 10 mins

So it be like

1-2 milliseconds pulse , 2-4 seconds no pulse , 1-2 milliseconds pulse, ....etc for 10 mins

Iam not sure the wire can stand or no, but is there any rule to select a gauge?

Wire lenght ~160 meter
Resistance of wire 3.2 ohm
Gauge 18 awg
I did a deep dive myself on this, as there was nobody out there who had actually calculated max amperage per wire gauge (it devolves into a chemistry question, rather than an electronics question). Understand that all the tables out their for 'standards' are safety based. They are concerned with overheated wire setting fire to things around it- not the maximum a wire can handle. Goal being to prevent fires.

There are many factors to consider- specific resistivity, dimensions, material, ambient temperature, wire temperature (thermal dissipation factors), etc.

NEC (National Electric Code) is going to rate an 18AWG wire as capable of safely handling up to 2.3A. But it could theoretically handle up to 27A just before it slags to molten- again THEORETICALLY because that assumes wire temp is 30C unchanging, and ambient is 25C, which is unrealistic. Wire heats, so, the melt temp is going to be far less than 27A.

Pulsed will allow you to go higher than the rated 2.3A the first time, and less with each successive pulse unless the wire is allowed to cool completely.

drjohsmith

Joined Dec 13, 2021
541
A peltier

Add a cooling system a peltier and a fan directed to the coil will makes it cool down more fast even during the pulse
Two things to consider

heating, which is the resistance of the / current / ambient temperature / ability to dissipate

voltage drop. A high speed pulse , has a very big amount of EMI , and the "AC" resistance of the wire will drop the voltage

Danko

Joined Nov 22, 2017
1,489
Pulsed will allow you to go higher than the rated 2.3A the first time, and less with each successive pulse unless the wire is allowed to cool completely.
For every pulse temperature increased:
Δt [°C] = (P_ave [W] * T_period ) / ( 381 [W * s) / (kg * °C)] * M_copper [kg]),
so for circuit #31
Δt = (15 * 2) / (381 *1.172) = 0.067 °C
and for circuit #28
Δt = (40 * 2) / (381 *1.172) = 0.179 °C

Irving

Joined Jan 30, 2016
3,186
How big is this coil? Length, inside and outside diameter? Air core or iron?

MrAl

Joined Jun 17, 2014
9,623
Is there a rule to select wire gauge for a pulses current

Lets say i have magnet copper wire of 18 Awg, and i want to give it a pulse current 100A for 1-2 milliseconds and 2-4 seconds before the next pulse for 10 mins

So it be like

1-2 milliseconds pulse , 2-4 seconds no pulse , 1-2 milliseconds pulse, ....etc for 10 mins

Iam not sure the wire can stand or no, but is there any rule to select a gauge?

Wire lenght ~160 meter
Resistance of wire 3.2 ohm
Gauge 18 awg

Hello there,

I hate to say this, but there is a bit more to it than just average current or average power or average RMS current. This is not as obvious because of the way the problem was stated. The problem was stated with something like a 2ms pulse and a long time off adding up to a duty cycles of 0.1 percent or a factor of 0.001. The current was stated as being 100 amps, and #18AWG wire has 3.35 Ohms for 160 meters (160 meters as stated).
Now lets state it a little differently and see what happens.

We pulse the same wire with 100 amps at a duty cycle of 0.1 percent or a factor of 0.001. Now lets choose the time periods. Let's say 100000 seconds 'off' and 100 seconds 'on'. So what is the average current or average power?
The answer is, with the wire being 'on' for 100 seconds it's going to melt and break apart with power heating of over 30kw. If not with 100 seconds, then lets go with 1000 seconds 'on' and the corresponding time off. What is the average current or power?
They are both going to be zero because surely the wire melts and breaks apart.

Another scenario...
We pulse the wire with 1 amp and the wire has ideal insulation. What is the average current or power? Again it doesnt matter because the wire inside builds up so much heat that it's going to melt eventually, there's no escape for the heat.

So what is missing.
We would need the shape of the wire and consider the dimensions probably round cross section. We would then need to apply the heat equation to the physical dimensions and maybe assume there is no insulation to make it easier.
More simply, we dont know the rate of cooling.

What is a similar situation.
A fuse.
Since i think someone already did this more extensive calculation you can look up the fusing current of copper wire, with a melting point (if i rem right) of 1083 degrees C (double check temperature though). That will tell you the time it takes to melt a wire with a given current and wire diameter or radius or whatever.

If you want to calculate a better result then you have to use the heat equation and specify all the physical aspects of the system, which in this case is the wire.
If you dont want to do that then do what many an engineer has done in the past: test it. Maybe start with 4 meters of wire and test it for temperature rise and go from there. You can make a table of pulse time and current and temperature rise.

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Danko

Joined Nov 22, 2017
1,489
@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.

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Momentory

Joined Sep 4, 2022
67
How big is this coil? Length, inside and outside diameter? Air core or iron?
- About 160 meter 18 Awg
- Dc resistance about 3.3 ohm
- Air core
- inside = 1.5 cm
- outside ~ 4.7 cm

Momentory

Joined Sep 4, 2022
67
i am searching for best a cooling system to keep it cool and long its lifespan.

I have some options to keep the coil cool and prevent to reach to maximum heat and melting

Option 1

Use a peltier with a fan directed to the coil to minimize the heat
( actually i do not think this will work significantly but the wire won't melted at least)

Option 2,3 need house or tank for the coil

Option 2

Use paraffin wax
during phase change it has a the ability to absorb a lot of heat.

The heat of fusion of paraffin wax is about 90 - 95 Btu / lb and the heat capacity is about 0.64 Btu / lb / *F.

They used it with the Lunar Rovers batteries.

Option 3

Mineral oil ( transformer oil)

Mineral oil can dissipate heat out of the coil and also it is very great insulator, they even use it with PC ( water tank PC).

It should be exchange once a year

But i do not know
will option 2,3 need cooling circle or no?!

WBahn

Joined Mar 31, 2012
27,862
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.