Can someone help me understand how the circuit breaker will operate in this scenario?

Question

For some reason I can't wrap my head around how inverse time overcurrent relays work.

Here are some trip currents and times from this curve for the Schneider Electric, Square D, QO 15A miniature circuit breaker (click image for larger size):

100 ms: 90 to 150 A

1.0 s: 67 to 120 A

10 s: 27 to 45 A

22.5 A: 25 to 125 s

30 A: 9 to 40 s

45 A: 2.7 to 12 s

Here is what I understand so far. If 90 to 150 A of current flows through the circuit for 100 ms than the circuit breaker will trip. If it flows for less than 100 ms and returns to normal current then the circuit breaker won't trip.

But if 90 to 150 A of current flow for 50 ms and then 67 to 120 A flow for 0.95 s and then 30 A for 30 s and then returns to normal current. When will the circuit breaker trip and why?

I feel like I'm missing some crucial piece information and that's the reason why I can't answer this simple question but I have no clue what it is.

Answer 1

I feel like I'm missing some crucial piece information and that's the reason why I can't answer this simple question but I have no clue what it is.

The breaker has two parts.

1. The magnetic part acts on instantaneous current - that is the surge rating of the breaker, say 10x the nominal current.

2. The thermal part acts on the time integral of current. The integrator is leaky, and has some time constant. Due to the leakage, the running integral of the rated current is just below the threshold. In the short term, if the time integral of the current is above the threshold, the breaker trips.

   Thus, you'll notice that the I-t curve follows roughly the constant \$I\cdot t\$ product - since what trips the thermal part is the time integral. But, thanks to the leakage, as \$t \to \infty,\$ the curve transitions from \$I \cdot t\$ to \$I \to I_{rated}^-\$, where \$I^-\$ means "just below of".

Answer 2

That is a thermal/magnetic breaker.

In the very large overload case(Here somewhere between 7 and 15 * rated current and up) the magnetic field trips the breaker with basically zero delay, and the springs then force the contacts apart clearing the fault within a single cycle.

For smaller overloads the thermal side takes over and has a delay, there is a bimetallic strip heated by the current passing thru it which if it gets too hot bends and activates the trip mechanism, in that sense it is RMS current responding (And yes the things behaviour varies with ambient temperature).

In normal operation you stay to the left of the breaker rating, with possibly the occasional foray into a multiple of the rating during startup surges and such, but still staying to the left of the grey area.

Answer 3

The circuit breaker is actually an integral of the transferred energy. The integral of i²t is what the circuit breaker graph is showing; if you know your current profile, you can calculate your cumulative energy and compare it with the circuit breaker energy graph.

Answer 4

The power in the bimetal strip is actually proportional to the square of the current. At a point where the heat loss is not very significant, the approximate trip time is inversely proportional to the current squared. For 10 sec., it takes about about 2.25 x rated current. For 1 sec., it takes a about 2.25√10 x rated current (actually bit less (~6.9x), because at that current, the heat loss is even less significant). At 7-15 x rated current (that big vertical area), it can trip from either the bimetal strip or the magnet. The magnet is only just pulling hard enough to trip it and can take up to about 1/4 sec. Above 15x, it trips as fast as the magnet can trip the mechanism (1 cycle max.).