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 is1mWb. The energy is 10mJ. >>
> Increase the fault to 1kA. Now you get 5kW and 1J, beyond the capacity of a 1.5KE6. >>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. >>
>> Strike it with a bolt of lightning. 100kA gives 500kW peak and
> 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?