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

 

hi M,
Are any of these wire tables helpful.?
E
https://www.google.com/search?sxsrf=ALiCzsZha16Lui3HH9rsXNFMirndwG7RkA:1664286463169&source=univ&tbm=isch&q=copper+wire+current+table&client=firefox-b-d&fir=VX4BLifTlqXkJM%2CmixeaqnP1TK5uM%2C_%3BLedU4UIU5YbflM%2CXS1REax0pacE7M%2C_%3BsT9Wj0pftscaKM%2CHJZEKZh59ow3SM%2C_%3BDkfFGOXf6WHG4M%2CjH6Yb2Owdx8BvM%2C_%3Bz_P2Aj7fMyI1DM%2CF4m6FLFwIMg9JM%2C_%3Bpf8T3PYCUVwQfM%2CmgPMGtv5P8KzIM%2C_%3BJhM4rnON6iSGFM%2C9JcbA29quHs5hM%2C_%3BVZOhiCZKzD0fbM%2CkRP2mI0FguZpaM%2C_%3BWflbpAiVDQwGxM%2C-qdTAFor9kcEyM%2C_%3BQAdK0EYjtzD6xM%2CmgPMGtv5P8KzIM%2C_&usg=AI4_-kQx9sUqKczTHQTNuqRS34b2Hf41ng&sa=X&ved=2ahUKEwjb9MObjrX6AhUjSkEAHUEvBfAQjJkEegQIAhAC&biw=1604&bih=891


Added this link, shows maximum ranges.
https://www.engineeringtoolbox.com/wire-gauges-d_419.html

 

Lets say the pulse is 100A for 2ms repeating every 2s. That's a 0.1% duty cycle. So the average current is 100mA.
So the wire will not melt! But the resistance (and inductance) of the wire is still there and may effect the operation of your device - whatever that is.

 

hi M,
Are any of these wire tables helpful.?
E
https://www.google.com/search?sxsrf=ALiCzsZha16Lui3HH9rsXNFMirndwG7RkA:1664286463169&source=univ&tbm=isch&q=copper+wire+current+table&client=firefox-b-d&fir=VX4BLifTlqXkJM%2CmixeaqnP1TK5uM%2C_%3BLedU4UIU5YbflM%2CXS1REax0pacE7M%2C_%3BsT9Wj0pftscaKM%2CHJZEKZh59ow3SM%2C_%3BDkfFGOXf6WHG4M%2CjH6Yb2Owdx8BvM%2C_%3Bz_P2Aj7fMyI1DM%2CF4m6FLFwIMg9JM%2C_%3Bpf8T3PYCUVwQfM%2CmgPMGtv5P8KzIM%2C_%3BJhM4rnON6iSGFM%2C9JcbA29quHs5hM%2C_%3BVZOhiCZKzD0fbM%2CkRP2mI0FguZpaM%2C_%3BWflbpAiVDQwGxM%2C-qdTAFor9kcEyM%2C_%3BQAdK0EYjtzD6xM%2CmgPMGtv5P8KzIM%2C_&usg=AI4_-kQx9sUqKczTHQTNuqRS34b2Hf41ng&sa=X&ved=2ahUKEwjb9MObjrX6AhUjSkEAHUEvBfAQjJkEegQIAhAC&biw=1604&bih=891


Added this link, shows maximum ranges.
https://www.engineeringtoolbox.com/wire-gauges-d_419.html

I check this before but it gives the current range that wire can stand with for full duty cycle

But
What if the wire got much more pulsed current than its range

 

Lets say the pulse is 100A for 2ms repeating every 2s. That's a 0.1% duty cycle. So the average current is 100mA.
So the wire will not melt! But the resistance (and inductance) of the wire is still there and may effect the operation of your device - whatever that is.

Sorry did not catch what you mean by average current.
Is it mean the current will be 100mA through the wire instead of 100A?

 

Sorry did not catch what you mean by average current.
Is it mean the current will be 100mA through the wire instead of 100A?

It means that we get the average current since the full current is not applied 100% of the time.

But
What if the wire got much more pulsed current than its range

You still calculate the average current.

Lets say the pulse is 100A for 2ms repeating every 2s. That's a 0.1% duty cycle. So the average current is 100mA.
So the wire will not melt! But the resistance (and inductance) of the wire is still there and may effect the operation of your device - whatever that is.

In this very good example we have 100 amps with an On Time of 0.002 Sec. The duty cycle (On Time) is expressed as a ratio between pulse width (0.002 Sec) and period time (2.0 Sec). So 0.002 / 2.000 = 0.001 and convert to a percentage so *100 = 0.1%. and 0.1% of 100 amps becomes 100 mA.

Is there a rule to select wire gauge for a pulses current

Yes and that is the rule. Remember the Ampacity is the maximum current that a conductor can carry continuously under the conditions of use without exceeding its temperature rating. Current is measured in amperes or “amps. Since pulsed is not continuous. So we get the average current. Make sense?

Ron

 

You all seem to be forgetting the fundamentals.
The wire current carrying capacity is determined by the heat generated due to the current, which is proportional to the RMS value not the average value.
For that you need to use the squared value of the current, since power is proportional to I²R.
Thus, a pulse of 100A for 2ms with a 2s rep rate would generate power in the wire equal to the average power value that a 100² x 2ms/2 = 10A continuous current would generate.
So the wire gauge should be selected to be able to carry 10A continuous current (which the 18 AWG wire can readily carry).
But one problem might be build up of heat if the wire is wound into a small coil, since the average dissipated power will be 10²A * 3.2Ω = 32W.

 

There's one more thing to consider.
How much will the pulse voltage be reduced by the resistance of the wire?
If you're putting 100A through 0.75mm^2 wire, which has a resistance of 24mΩ/metre (48mΩ/metre there and back), then the voltage of the pulse is going to drop 4.8V for each metre of cable.

 

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
Resistance of wire 3.2 ohm
Gauge 18 awg

The length is 160 what? feet? miles? inches? cm?

Based on the charts here:
https://www.powerstream.com/Wire_Size.htm

The DC resistance of 18 AWG wire 6.385 Ω/ 1000 ft. That would imply that your length is right at 500 ft (which would be 167 yards, or 153 m). So are you measuring the length in yards???

How are you driving this pulse? With 3.2 Ω of resistance, your driver has to deliver 320 V, even assuming no inductance. There WILL be some inductance, and you are talking some pretty serious di/dt. Even if it ramps up from 0 to 100 A in the first millisecond and then back down to zero in the second millisecond, you are talking 100,000 A/s.

Also, that resistance is for frequencies in which the current penetrates the conductor more-or-less completely. High frequency currents tend to get crowded to the surface (the skin) of the conductor, resulting in a much higher effective resistance. As the chart shows, the frequency above which you need to really start considering this is 17 kHz (assuming your wire is solid copper). The pulses you are talking about should have most of their energy well below that, so you can probably ignore skin effects.

Assuming you can drive your pulse -- and we'll assume that you can ramp it up so quick that it looks like it instantly goes to 100 A for the full 2 ms -- there are two big thermal issues involved. The easy one to get a handle on is the average temperature rise. This is dictated by the RMS current through the wire. The worse case is the pulse lasting 2 ms and the repetition period being 2 s.

Irms = sqrt( (100 A)² * (2 ms) / (2 s) ) = 3.16 Arms

That is well below the current that would exceed the thermal limits for chassis wiring (which is 16 A), but it exceeds the limits for wire used for power transmission (which is 2.3 A). So this probably warrants some further investigation. One factor to consider is the limits due to the coating of the magnet wire you are using.

The other factor is how hot the wire gets during the time that the current is actually flowing. This has to do with the thermal capacity of the wire and the thermal conductivity to its surroundings and is much harder to get a handle on. But given that your average temperature appears to be getting up quite a bit, how hot must the wire be getting in that 2 ms pulse such that, even with cooling down for the remainder of 2 s, the average temperature is still elevated quite a bit?

Unless you investigate these issues further and unless you have good thermal coupling to the environment and the wire isn't tightly packed, I'd recommend using a larger wire.

 

Thus, a pulse of 100A for 2ms with a 2s rep rate would generate power in the wire equal to the average power value that a 100² x 2ms/2 = 10A continuous current would generate.

And this is why I harp on properly tracking units instead of just tacking on whatever units we hope and pray and would like the answer to have.

(100 A)² x (2ms)/(2s) = 10 A², not 10 A

So, if units had been tracked, the fact that this answer is wrong would have been glaringly obvious.

RMS is the ROOT of the MEAN of the SQUARE

√[(100 A)² x (2ms)/(2s)] = 3.16 A

 

Mea culpa.

 

The magnet wire lenght is ~160 meter
Wound as a coil
Has a resistance of 3.2 ohm

So if the current pulse is 2 millisecond and the rest time is 2 seconds

I want to ensure the wire will stand with this 100A or no

The best way to ensure that is get how much the wire temperature will be

To get this we need to estimate the average dissipated power

So how to calculate this?

My formula is that the heat generated by the wire does not exceed the heat generated from nominal operating current

A√(D) < A0

A ≡ pulse current (80A)

D ≡ duty cycle (pulse duration divided by total cycle duration)

A0 ≡ nominal current (2.7A according to datasheet of the wire)

If my pulse current multiplied by the square root of the duty cycle is less than the nominal rated current than the wire can handle the pulse current.

100×√(0.001) = 3.16A

Is that right?

If right so i need to low the Current to 80A so it be 2.5A which less than nominal current

 

That's in the ballpark. It is still ignoring how high the temperature gets in the transient. Trying to equate the average heat generated by the two scenarios is implicitly making the assumption that the thermal time constant is much longer than the period of the pulse train (the 2 s). But if the thermal time constant is too short, it will heat up to a high temperature during the pulse and come back down during the remainder. Yes, on average, the temperature might be fine, but the peak temp may not be. It's like sticking your head in an oven and your feet in ice water -- on average you are quite comfortable.

 

How much inductance does the coil have? If it's more than 6.4mH it won't get to 100A during the length of the pulse.

 

That's in the ballpark. It is still ignoring how high the temperature gets in the transient. Trying to equate the average heat generated by the two scenarios is implicitly making the assumption that the thermal time constant is much longer than the period of the pulse train (the 2 s). But if the thermal time constant is too short, it will heat up to a high temperature during the pulse and come back down during the remainder. Yes, on average, the temperature might be fine, but the peak temp may not be. It's like sticking your head in an oven and your feet in ice water -- on average you are quite comfortable.

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

 

How much inductance does the coil have? If it's more than 6.4mH it won't get to 100A during the length of the pulse.

It is 3.5 mh

 

It is 3.5 mh

And what is your supply voltage?

 

And what is your supply voltage?

470uf - 450v capacitor
Will charged to 320v

So i think the pulse time will be very short as the capacitor will discharge in very few milliseconds

And will take about 2-4 second to get charge again

So longer time for the next pulse
Adding to this a peltier and fan as cooling system directed to the coil

 

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.

 

The magnet wire lenght is ~160 meter
Wound as a coil
Has a resistance of 3.2 ohm

470uf - 450v capacitor
Will charged to 320v

It seems like this:


But it should be like that: