Hi guys:
I'm an ME working on a project where it would be advantageous to use
high currents (and low voltages) pulsed on a short PWM duty cycle. If
the cycle frequency is fairly high, do I really have to size my
conductors for the peak current, or can I size them for the average
current?
In other words, if I'm pulsing 100 amps for 10 ms on a 100 ms cycle
(10%), can I size the wire for 10 amps?
Thanks for all relplies.
Don
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.
/ClasHenrik
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 34
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....
> 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%.
2. 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.
3. 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.
4. 1000 amps at 100% with 10 times the wire (R is 1/10th) is only 10
times the heating as number 1.
5. 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.
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