Finding the Harmonic Input Current of a Dc Converter A highly inductive load is supplied by a converter as shown in Figure 5.37a. The average load current is Ia = 100 A and the load ripple current can be considered negligible (ΔI = 0) . A simple LC input filter with Le = 0.3 mH and Ce = 4500 µF

Question

Finding the Harmonic Input Current of a Dc Converter

A highly inductive load is supplied by a converter as shown in Figure 5.37a. The average load current is [latex]I_a = 100 A[/latex] and the load ripple current can be considered negligible [latex](ΔI = 0)[/latex] . A simple LC input filter with [latex]L_e = 0.3 mH[/latex] and [latex]C_e = 4500 µF[/latex] is used. If the converter is operated at a frequency of 350 Hz and a duty cycle of 0.5, determine the maximum rms value of the fundamental component of converter-generated harmonic current in the supply line.

Accepted Answer

For [latex]I_a = 100 A, f = 350 Hz, k = 0.50, C_e = 4500 µF[/latex], and [latex]L_e = 0.3 mH[/latex], [latex]f_0=1/(2\pi \sqrt{C_eL_e} )=136.98 Hz[/latex] Equation (5.156)

[latex]i_{1h}(t)=\frac{I_a}{\pi } sin 2\pi k cos 2\pi ft+\frac{I_a}{\pi } (1 -cos 2\pi k)sin 2\pi ft[/latex] (5.156)

can be written as

[latex]I_{1h}(t) = A_1 cos 2\pi ft + B_1 sin 2\pi ft[/latex]

where [latex]A_1 = (I_a/\pi) sin 2\pi k[/latex] and [latex]B_1 = (I_a/\pi) (1 - cos 2\pi k)[/latex]. The peak magnitude of this current is calculated from

[latex]I_{ph}=(A_1^2 +B_1^2)^{1/2}=\frac{\sqrt{2}I_a }{\pi }(1 -cos \pi k)^{1/2}[/latex] (5.159)

The rms value of this current is

[latex]I_{1h}=\frac{I_a }{\pi }(1 -cos 2\pi k)^{1/2}=45.02 A[/latex]

and this becomes maximum at k = 0.5. The fundamental component of converter-generated harmonic current in the supply can be calculated from Eq. (5.157)

[latex]I_{ns}=\frac{1}{1 + (2n\pi f)^2L_eC_e}I_{nh}=\frac{1}{1 + (nf/f_0)^2}I_{nh}[/latex] (5.157)

and is given by

[latex]I_{1s}=\frac{1}{1 + (nf/f_0)^2}I_{1h} =\frac{45.02}{1+(350/136.98)^2}=5.98 A[/latex]

If [latex]f/f_0 \gg 1[/latex], the harmonic current in the supply becomes approximately

[latex]I_{1s}=I_{1h}\left(\frac{f_0}{f} \right)^2[/latex]

Question 5.9: Finding the Harmonic Input Current of a Dc Converter A highl...

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