Which Residential Voltage & Frequency Arrangement Is Best?

With a fast break switch and capacitor snubber, its even less.

True enough, but the situation is even worse on DC!

thanks, I suspected thats what it was.

NT

Not for motors. We use inductive loads on AC (with resultant spike problems) that would be simply impractical on DC.

AC switching is largely a problem of voltage (and getting an adequate break path), DC one of current (and arcing)

Even 240V can kill. Trouble is, it doesn't do so often enough.

---------------------------- "The Natural Philosopher" wrote in message news: snipped-for-privacy@lotis.uk.clara.net...

I don't see the purpose of a deliberately slow break on AC but not for DC. Inductive effects will occur on interruption of DC current. In fact they may be worse than for AC. With DC, the mechanism for interrupting the arc is to lengthen it and cool it so it becomes unstable. If the circuit is inductive then there will be a transient voltage. A large enough gap to do this is needed and a fast opening does reduce heating and arc erosion in the switch. With AC the problem is easier as the current naturally goes through 0 every half cycle so that all that is necessary is a gap large enough to prevent restriking. If it takes a couple of half cycles to fully open, this is generally not a problem and opening in less than 1/2 cycle may not be worth pursuing. This makes interruption of AC much easier than DC and large openings are not needed. In addition, most household circuits and loads are near unity power factor which reduces the recovery voltage at current zero. Hence, for lighting circuits, cheaper, smaller and quiet, easy to operate, switches can be used. For many years a mercury switch was common - similar to the switch in some thermostats - just a pair of contacts in a tube with a drop of mercury- tilt the tube to open or close the contacts. Also note that the peak transient voltage due to inductance twice the peak AC voltage or less. This is not a problem for the insulation and wiring.

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Don Kelly @shawcross.ca remove the X to answer

If I understand what you're saying, this isn't true. At the instant of trying to interrupt the current flow, the inductor will, by it's nature, try to maintain that current flow, and generate a voltage spike sufficient to do so across the switch gap. If the switch gap is small because it's a slow break, then only a small voltage spike is required. If the switch gap is large because it's a fast break, then a much larger voltage spike is required and generated.

The slow break on AC is a reflection on the much larger number of inductive loads (and much higher inductance) which you are going to find used on AC supplies than you are on DC supplies.

Fast break on DC is required to minimise power dissipated by the arc in the switch, and as one of the mechanisms to quench the arc. You don't generally have to worry about inductive loads, but where you do, a simple reverse diode across the load will handle allowing the current to die away gracefully without generating a back-EMF any more than the diode's forward voltage drop.

I disagree regarding the fault current. That would be true if all of the system impedances were the same. However, transformer impedance will be twice the Ohms at 480 volts as at 240 volts for the same percent impedance, and the wiring can have four times the impedance for the same losses and voltage drop %. Therefore, fault current would be the same or lower!

Ben Miller

Correction... Let me try this again now that I am awake. The transformer impedance will be four times, and available fault current will be one half, at twice the voltage.

Ben Miller

No, really ?

I was thinking of cross phase problems - e.g. if someone has a nice long extension lead ...

This also depends on the transformer size. In the UK, we often supply the housing in several roads from a single 1MVA transformer (partly because 240V carries further more easily than 120V). In the US, it's more common to find only a few houses running from a very much smaller transformer, which is going to have a higher impedance.

Absolutely right. I assumed the same kVA rating for purposes of discussion. The OP stated that fault current would be higher at the higher voltage, apparently based on a discussion in another newsgroup, which is incorrect. It sounds like whoever said that assumed that all of the system impedences stay the same at the either voltage.

Ben Miller

Fault current is not normally determined by the supply, but by the fault and the voltage applied to it. Unless the fault is a user very firmly wiring L to N, the supply capacity doesnt come into it.

Carbon and arc faults are non linear, twice the v causes more than twice the i.

NT

the trouble is you cant safely rely on every inductive appliance having a diode fitted and working, so you still must use arc breaking switches.

NT

Please note that in an AC circuit, any arc actually goes unstable and breaks at or close to current zero. There is no attempt to actually "break" the current as with DC. Ideally the object is to try to maintain 0 current from the time that the current naturally goes to 0. That is, after the "break, di/dt=i=0. If the switch does break the current before current zero, then there will be a spike which may cause the gap to restrike and this chopping may be repeated leading to a series of the kind of spikes that you indicate. This was once a problem with air blast breakers on interruption of relatively small transformer magnetising currents.In a good switch, with interruption at current zero, the object is to have the gap strength rise faster than the recovery voltage (which could have a high frequency component. This recovery transient is determined by the circuit parameters on each side of the switch. Another factor is that typical household switches are intended for such things as lighting loads, not motors. Motor switches such as those on my furnace, are designed for interrupting inductive circuits. would be the case with AA at the same switch opening and opening speed.

I note that the problem of interrrupting the current will present the same problem -possibly worse- with DC in circuits which have inductance such as motors or relay coils.

However, I will try to go over some references to check this out as my comment is based on memory of material dealing with HV circuit breakers. I know that high transients can occur but these are generally due to wave propagation and reflections rather than an inductive kick and I could have faulty memory.

Good point

In UK terminology, a "Fault current" is a current limited mainly by the supply impedance (L-N fault) or the earth fault loop impedance (L-E fault), with the fault itself having a very low impedance.

If the fault itself has any appreciable impedance, then it generates an "overload current", not a "fault current" (at least, in UK terminology).

Arc impedance, if ballasted only by the supply impedance, rapidly collapses towards zero as the current rises to high levels.

The microswitches in our Triton shower are the worst of both worlds..... fast opening and a small gap. :(

One of them switches the 20A element and the other 10A. Doing some ball-park sums shows the sort of problem.

Suppose the 100ft of T&E cabling has an inductance of 100uH, and the stray capacitance is effectively 1000pF. This will form a resonant circuit that will ring when the switch is opened. The energy stored in the charged inductor kicks off the ring, and then the inductor and capacitor exchange energy as they resonate. To get the peak transient voltage we need only consider the first exchange of energy, from the charged L into the C.

Inductor energy= 0.5*L*I^2. Capacitor energy= 0.5*C*V^2.

Suppose the switch opens at the peak of the 50Hz sine, when a full 20*1.414A is flowing through the inductor.

The resonant frequency of 100uH+1000pF is about 503KHz, and the Vpk is reached at the peak of the first quarter sine. ie, The voltage spike goes from roughly zero to 9KV in about 0.5uS..... quite a nasty dV/dT for a switch that has only just started to open.

Ball park numbers only (and many approximations), but it does give a flavour of the problem. The saving grace for Triton is that the switch only opens rarely at the full 28A, and the 50Hz sine means that the arc is only sustained for between 5 to 10mS. Still, it shows why those tiny microswitch contacts have such a short life. :(