Vendors for Semiconductor fuses normally may supply let-through charts as well as melting and clearing I2t figures. When used for dc application where the short circiut waveform is known to have the following equation I(t)=Isc(1-exp(-at)), what method will be more accurate:



Method 1: Use the Let-thru chart, with Iprospective= Isc to determine the let through current, Ic, and then use the equation above to find the melting time. Note in this method the published melting I2t is not used.



Methopd 2: Use the melting I2t data and equate it the square of the above equation and solve iteratively for the melting time, and the let-thru current Ic. note this method does not use the let-through charts



Both of the above methods provide significantly different answers.

Hello guruxyz

Welcome to the forum.

Semiconductor fuses are designed for the protection of semiconductors, primarily SCRs.

To select a semiconductor fuse to protect an SCR, first refer to the SCR data sheet and determine the maximum I²t of the SCR. This is always quoted in the data sheet.
Next, dtermine the maximum operating voltage of the circuit and selct the fuse to have a total clearing I²t at that voltage of less than the i²t of the SCR.

Best regards,
Mark.

Thank you Mark for your quick reply

In addition to protecting semiconductor devices, semiconductor fuses are also used to ensure fast fault clearing times as well as to reduce the fault duty below the interrupt rating of some protection devices.

In my application, I am trying to evaluate the point (Let through amps and melting time) an internal fuse, inside a rectifier will blow, and thus limit its contribution to a fault on the main DC bus.

Hello guruxyz

Yes, the major advantage of the semiconductor fuse is that they essentially limit the energy let through under fault conditions whereas a standard HRC fuse limits the time of the fault current.

I would be concerned only with the clearing I²t of the fuse in determineing the prospective fault energy. The melting I²t determines that the fuse is going to fail, so use that if you want to find out how much fault current you can tolerate before the fuse is damaged. i.e. a safe operating fault or impulse current that will not damage the fuse.

Under fault conditions, fractions of a second, you can use the integral of the current squared to give you the maximum let through (clearing I2t) for mast waveforms. The I²t does not apply to longer lower level faults however. I have seen people try to apply the I²t to 30 second overload currents. I would calculate the integral of the I2t of your fault current to detemine the time taken for the fault to clear.

Best regards,
Mark.

Hello Mark

But which method is more accurrate. There is large descrepencies in th eresults between the two methods?

Any body would like to volunteer an answer?