High current/low duty cycle?

Depends what losses you are prepared for, but in general, no.

Power loss in the conductor will be proportional to I². So a steady 10 times the current will dissipate 10²=100 times the power. For a 10% duty cycle, this will reduce to

10%x100=10 times the power loss you get at 10amps. It depends if your application can handle this loss and if the cable can safely dissipate 10 times the power without damage or fire risk.

In many respects, this is the same problem you get with a low power factor load. In your case, the power factor is effectively 0.1 (10%). Conductors have to be oversized relative to the load's power rating.

Increasing the current 10 times increases the amount of heat 100 times. In the simplest case, I think you can use the square root of the duty cycle. If the current is only flowing 10% of the time, it is equivalent to a continuous current of root 0.10 = 32% of the current.

/Clas-Henrik

Smaller wire will not perform very well. When I was testing circuit breakers with a high current test set we used 4/0 welding cable for up to an 400 amp setting. Then 2- 4/0's for 800 then 3-4 for 1600. If we were just doing one test 3 was ok. If we were doing all of the tests for certification had to use 4 cables. The insulation would blister if we did not use 4 cables. Also our cables were 3 feet long.

No. Size for peak, and it depends on what is connected at both ends. You can't get 100 amps down a 10 amp wire, unless it is really, really short. Frequency does not matter. V=I*R look up R for wire in a book. How much voltage can you loose? Same thing applies to PCBs Go back to cutting metal, and leave the electrical stuff alone, you'll start a fire, or get sombody zapped. (seriously)

It seems to me that a duty cycle of 10 ms on, 90 ms off would have a vastly different effect than 10 minutes on and 90 minutes off. Think about what you're saying.

The wires are very short. I'm worried about heating, not voltage loss (and it's spelled "lose", not "loose"). The other posters pointed out adeptly that heating varies as the square of current, so sqrt(10%) =

32% is the appropriate figure for sizing the wire.

Hey, thank's for the snide reply (seriously). Now run along, kid.

Don

The problem is not described sufficiently enough to yield a specific answer. Missing are:

-current amplitude

-duty cycle range

-Length of wire

-Allowable Voltage drop (what is connected at both ends)

-Failure modes (either end) (don't assume that either end will never fail)

-startup surge

-loads (phase and magnitude)

-flammable materials

-compliance with safety standards (you may not have the latitude in choosing a wire size)

- rule of thumb is to not let the temp of the wire rise more than 10 degree C above ambient. But that applies to average power (what your proposing) and not peak, peak can melt the wire too, or cause enough voltage drop so it is unusable at the other end.

are you dealing with 100 amps over 30 feet using 10 g copper to a AC motor duty cycled at 10mson/90msoff? or 110V 1 amp over 1 foot using 18 g copper to a light bulb with 10/90 ms on/off? or 3.1 V 1.2 amp over 1/2 foot using 24 g copper to a positional control valve

your example of duty cycles illustrates why one sizes to peak current. If your duty cycle generator varies by frequency or pulse width, or stays stuck on....

On 4 Jun 2004 07:04:26 -0700 eromlignod wrote: |> No. Size for peak, and it depends on what is connected at both ends. |> You can't get 100 amps down a 10 amp wire, |> unless it is really, really short. Frequency does not matter. | | It seems to me that a duty cycle of 10 ms on, 90 ms off would have a | vastly different effect than 10 minutes on and 90 minutes off. Think | about what you're saying.

It also depends on your thermodynamics, inductance, capacitance, etc.

|> V=I*R look up R for wire in a book. How much voltage can you loose? |> Same thing applies to PCBs | | The wires are very short. I'm worried about heating, not voltage loss | (and it's spelled "lose", not "loose"). The other posters pointed out | adeptly that heating varies as the square of current, so sqrt(10%) = | 32% is the appropriate figure for sizing the wire.

1. 100 amps at 100%.

1. 1000 amps at 100% is 10 times the current, but 100 times the heating as number 1 (due to "I square times R") with the same size of wire.

2. 1000 amps at 10% is still the 100 times the heating, but for only 1/10th the time, so the net average is 10 times the heating as number 1, plus the pulsing effects depending on frequency.

1. 1000 amps at 100% with 10 times the wire (R is 1/10th) is only 10 times the heating as number 1.

2. 1000 amps at 10% with 10 times the wire is 10 times the heating for 1/10th the time, so the net average is the same heating as number 1.

Note that the same heating on 10 times the wire will have 3 times the surface area. So while the BTU release will be the same the heat rise and other thermodynamics will be different.