TCC Curves: What about events below .001 seconds? 1

[buddy91082]

We have LED drivers with a rated inrush of 185 amps at 200 microseconds (.0002 seconds. We are trying to determine how many drivers we can place on a 20 amp, 120v circuit. The circuit breakers TCC curves start at .001 seconds. Also, the fuse TCC start at .01 seconds. What happens if the event is shorter then the published TCC curve and is off of the curve? Will the breaker or fuse trip or is the event short enough that the energy will not all the overcurrent protection device to trip?

Thanks.
b

 

[wbd]

I believe that the TCC curves have those time restrictions based on the fact that there is a definite time that it takes for a fuse to open or a breaker to operate. I think realities of the physical world prevent the fuse or breaker from operating any faster.

 

[cranky108]

At 0.06 cycles (in the US) it is not realistic to expect anything like a breaker or Fuse to operate that fast. Most such devices require a zero crossing of the current to interrupt, and in these short times that won't happen.

 

[krisys]

For your problem, you need transient switching study, rather than the protective device coordination study. Protective devices will not respond for such inrush and hence, there is no risk of mis-coordination due to the switching transient.

 

[davidbeach]

Current limiting fuses (or current limiting low voltage circuit breakers) must act in the first half cycle, before the first zero crossing, the inherent definition of current limiting. But most most MV fuses aren't current limiting and there are no MV (or higher) current limiting breakers. Fractions of a cycle is essentially noise.

 

[dpc]

I think that there is no way to know except by testing. It may vary between manufacturers and trip unit designs as well. In addition, I'm guessing the LED driver inrush is not linear or sinusoidal. Published TCC curves are based on pure 60 (or 50 Hz ) sine wave with no harmonics.

 

[bacon4life]

Since the minimum melt is essentially just I^2*t, would the physics of heating a fuse enough to damage it be any different for very short events? Around 50 drivers would damage one particular 20A fuse.
Your source impedance might also have an impact on how much current the LEDs draw.

 

[LionelHutz]

Since this is way below a 1/2 cycle of time, this would likely be dependent on the point of switching in the waveform. If you switch right at the peak of the voltage then you might trip or blow the fuse while if you switch near the zero crossing the increasing voltage would likely almost soft-charge the driver.

The specification sounds rather bogus. I would expect the instantaneous trip on a 20A breaker would likely be close to tripping with a single driver when you did manage to close the circuit at a voltage peak and try to draw 185A. Two drivers would likely randomly trip the breaker each time you closed the circuit at a voltage peak.

 

[VTer]

For a modeled case circuit breaker which most 120V 20A CBs are, your concern is the instantaneous portion of the TCC which is a magnetic element. It will essentially depend of the amount of force created by the current and the design of the electromagnet to force the trip mechanism open. If possible, I personally would try to size the number of drivers based on nominal rated current of the driver and CB rather then the transient characteristics and test to see what happens. Typically for a 20A CB, I do not exceed more than 12A of load. This allows for any potential concerns such as yours but also for future growth or changes that may required adding additional load to the circuit. Also, the amount of inrush current will depend on many different factors and may be restricted by source impedance. Another option is to discuss details with manufacturers of CB and driver if you have to be exact.

"Throughout space there is energy. Is this energy static or kinetic! If static our hopes are in vain; if kinetic ù and this we know it is, for certain ù then it is a mere question of time when men will succeed in attaching their machinery to the very wheelwork of nature". û Nikola Tesla

 

[HamburgerHelper]

If someone wanted something really fast, they could use an arc flash crowbar which GE claims needs at most 8ms to contain an arc.

 

[MatthewDB]

Things won't be so fast if there are more drivers per service. Current is limited to the available fault current. The source impedance will reduce the peak, making the inrush last longer.

Figure for residential, available fault current can be as low as 750A.

 

[itsmoked]

buddy91082; I feel for you. This has been a long running issue with LED drivers and their ridiculous inrush currents. Lots of people play LED driver roulette all the time with the caveat that when they lose they have to reset the breaker. You can also end up causing flicker issues with other loads on the same source.

From past experience I would say about 5 of those particularly high inrush units would be reliable on a 20A breaker.

One approach is to use a standard breaker and try it as dpc suggests. I'd wire the fixtures up as directly as possible and keep adding them until you get a trip. Knock off a few from that and go with that number in the installation. If after the install the different impedances involved you get a nuisance trip you then switch all the breakers to a C-curve. You won't have anymore trips. I'd keep that as a fall back tool though, as you'll be screwed if you start out with C-curves and then get nuisance trips.

I don't know what your situation is but if you can turn them all on with a zero voltage crossing switch then you will have no tripping issues up to the steady state circuit loading limit. This would be as simple as using a solid state zero crossing relay to turn on the circuits. That is one of the SSRs seen in this search depending, of course, on various details of your application.
DigiKey Search of 240V+/20A+ 120Vac triggered zero crossing SSRs

Keith Cress
kcress -

http://www.flaminsystems.com