Selecting a Fast-Acting Fuse for Protecting a Thyristor A fuse is connected in series with each thyristor of type S30EF in the single-phase full converter, as shown in Figure 10.1a. The input voltage is 240 V, 60 Hz, and the average current of each thyristor is Ia = 400 A. The ratings of thyristors

Question

Selecting a Fast-Acting Fuse for Protecting a Thyristor

A fuse is connected in series with each thyristor of type S30EF in the single-phase full converter, as shown in Figure 10.1a. The input voltage is 240 V, 60 Hz, and the average current of each thyristor is [latex]I_a = 400 A[/latex]. The ratings of thyristors are [latex]I_{T(AV)} = 540 A, I_{T(RMS)} = 850 A, I^2t = 300 kA^2s at 8.33 ms, i^2\sqrt{t} = 4650 kA^2\sqrt{s} [/latex] , and [latex] I_{TSM} = 10 kA [/latex] with reapplied [latex]V_{RRM} = 0 [/latex], which may be the case if the fuse opens within a half-cycle. A 540-A fuse has the ratings of a maximum current [latex]I_{max} = 8500 A[/latex], a fuse [latex]i^2t = 280 KA^2[/latex], and the total clearing time [latex]t_c = 8 ms[/latex]. The resistance of fault circuit is negligible and inductance is L = 0.07 mH. Determine the suitability of the fuse to protect the devices.

Accepted Answer

[latex]V_s = 240 V, f_s = 60 Hz[/latex]. The short-circuit current, also known as the prospective rms symmetric fault current, is

[latex]I_{sc}=\frac{V_s}{Z}=240 imes \frac{1000}{2\pi imes 60 imes 0.07}=9094 A[/latex]

For the 540-A fuse and [latex]I_{sc} = 9094 A[/latex], the maximum peak fuse current is 8500 A, which is less than the peak thyristor current of [latex]I_{TSM }= 10 kA[/latex]. The fuse [latex]i^2t[/latex] is [latex]280 kA^2s[/latex] and total clearing time is [latex]t_c = 8 ms[/latex]. Because [latex]t_c [/latex] is less than 8.33 ms, the [latex]i^2\sqrt{t} [/latex] rating of the thyristor must be used. If thyristor [latex]i^2\sqrt{t} = 4650 imes 103 kA^2\sqrt{s} [/latex] , then at [latex]t_c = 8 ms[/latex], thyristor [latex]i^2t = 4650 imes 10^3\sqrt{0.008} = 416 kA^2s[/latex], which is 48.6% higher than the [latex]i^2t[/latex] rating of the fuse ([latex]280 kA^2s[/latex]). The [latex]i^2t[/latex] and peak surge current ratings of the thyristor are higher than those of the fuse. Thus, the thyristor should be protected by the fuse.

Question 17.6: Selecting a Fast-Acting Fuse for Protecting a Thyristor A fu...

This Question has been Answered.

Already have an account?

Login

Related Answered Questions