John Larkin a écrit : > On Tue, 6 Jul 2010 19:06:37 -0500, "Tim Williams" > wrote: >
> Say you get 1A fault current from a CT (i.e., secondary referred), which throws your circuit into overload, so the voltage on the windings jumps to 5V (clamped by a perfect 5V supply, assuming ideal clamp diodes). The delivered power is evidently 1A * 5V = 5W, going into your supply. If the winding has a saturation flux of 1mWb, this fault current will flow for 1mWb / 5V = 0.2ms. The energy is 5W * 0.2ms =
1mJ, or 1A * 1mWb. >>
> Increase the fault to 10A. The winding is clamped at 5V, so 50W is delivered, and the fault again lasts for 0.2ms, because the flux is
1mWb. The energy is 10mJ. >>
> Increase the fault to 1kA. Now you get 5kW and 1J, beyond the capacity of a 1.5KE6. >>
>> Strike it with a bolt of lightning. 100kA gives 500kW peak and
100J, assuming the transformer doesn't fail first; if it has an internal resistance of 0.01 ohm, it will drop 1kV, probably breaking down the meager insulation in a CT. >>
> Tim >
> The usual practise in electronic metering is to have the CT secondary > drive a low-resistance wirewound or manganin strip shunt. The shunt > resistance is considerably less than the winding resistance. > Outrageous CT overloads don't damage the shunt... most of the power > dissipation is in the winding. The signal conditioning opamps or > whatever are protected by high value resistances between them and the > shunt. >
I'm designing an energy metering ammeter and am looking after a
0.1/0.2R, preferably SMT, shunt.
On the average all of them will sum up to half a billion euro energy, so it has to be accurate :-)
It'll work in some harsh environment and must :
- work @ 85°C Tamb, (100°C PCB temp)
- be low tempco (preferably lower than 20ppm/K)
- real low aging for less than yearly calibration
- preferably high initial accuracy to hopefully bypass one calibration step
None of the usual suspects fit the bill.
Any manufacturer / part series to suggest?