Finding the Currents of a Dc Converter with an RL Load A converter is feeding an RL load as shown in Figure 5.4 with Vs = 220 V, R = 5 , L = 7.5 mH, f = 1 kHz, k = 0.5, and E = 0 V. Calculate (a) the minimum instantaneous load current I1, (b) the peak instantaneous load current I2,

Question

Finding the Currents of a Dc Converter with an RL Load

A converter is feeding an RL load as shown in Figure 5.4 with [latex]V_s = 220 V, R = 5 Ω , L = 7.5 mH, f = 1 kHz, k = 0.5[/latex], and E = 0 V. Calculate (a) the minimum instantaneous load current [latex]I_1[/latex], (b) the peak instantaneous load current [latex]I_2[/latex], (c) the maximum peak-to-peak load ripple current, (d) the average value of load current [latex]I_a[/latex], (e) the rms load current [latex]I_o[/latex], (f) the effective input resistance
[latex]R_i[/latex] seen by the source, (g) the rms chopper current [latex]I_R[/latex], and (h) the critical value of the load inductance for continuous load current. Use PSpice to plot the load current, the supply current, and the freewheeling diode current.

Accepted Answer

[latex]V_s = 220 V, R = 5 Ω , L = 7.5 mH, E = 0 V, k = 0.5[/latex], and [latex]f = 1000 Hz[/latex]. From Eq. (5.23),

[latex]I_2=I_1e^{-kTR/L}+\frac{V_s - E}{R}(1 - e^{-kTR/L})[/latex] (5.23)

[latex]I_2 = 0.7165 I_1 + 12.473[/latex] and from Eq. (5.24),

[latex]I_3 = I_1=I_2e^{-(1-k)TR/L}-\frac{E}{R} (1 -e^{-(1-k)TR/L})[/latex] (5.24)

[latex]I_1 = 0.7165 I_2 + 0.[/latex]

a. Solving these two equations yields [latex]I_1 = 18.37 A[/latex].

b. [latex]I_2 = 25.63 A[/latex].

c.[latex] ΔI = I_2 - I_1 = 25.63 - 18.37 = 7.26 A[/latex]. From Eq. (5.29),

[latex]\Delta I_{max}=\frac{V_s}{R}tan h\frac{R}{4fL}[/latex] (5.29)

[latex]I _{max} = 7.26 A[/latex] and Eq. (5.30)

[latex]\Delta I_{max}=\frac{V_s}{4fL}[/latex] (5.30)

gives the approximate value, [latex]I_{ max} = 7.33 A[/latex].

d. The average load current is, approximately,

[latex]I_a=\frac{I_2 -I_1}{2}=\frac{25.63 + 18.37}{2}=22 A[/latex]

e. Assuming that the load current rises linearly from [latex]I_1[/latex] to [latex]I_2[/latex], the instantaneous load current can be expressed as

[latex]i_1=I_1+\frac{\Delta It}{kT}[/latex] for 0 < t < kT

The rms value of load current can be found from

[latex]I_0=\left(\frac{1}{kT}\int_{0}^{kT}{i_1^ 2 dt} ight)^{1/2}=\left[I_1^2\frac{(I_2 -I_1)^2}{3}+I_1(I_2-I_1) ight]^{1/2}[/latex] (5.32)

= 22.1 A

f. The average source current

[latex]I_s = kI_a = 0.5 × 22 = 11 A[/latex]

and the effective input resistance [latex]R_i = V_s/I_s = 220/11 = 20 Ω[/latex] .

g. The rms converter current can be found from

[latex]I_R=\left(\frac{1}{T}\int_{0}^{kT}{i_1^ 2 dt} ight)^{1/2}=\sqrt{k}\left[I_1^2+\frac{(I_2 - I_1)^2}{3}+I_1(I_2-I_1) ight]^{1/2}[/latex] (5.33)

[latex]=\sqrt{k}I_0=\sqrt{0.5} imes 22.1=15.63 A[/latex]

h. We can rewrite Eq. (5.31)

[latex]x=\frac{E}{V_s} \leq \frac{e^{kz}-1}{e^z-1}[/latex] (5.31)

as

[latex]V_S\left(\frac{e^{kz}-1}{e^z - 1} ight)=E[/latex]

which, after iteration, gives, [latex]z = TR/L = 52.5[/latex] and[latex] L = 1 ms× 5/52.5 = 0.096 mH[/latex]. The SPICE simulation results [32] are shown in Figure 5.6, which shows the load current I(R), the supply current - [latex]I(V_s)[/latex], and the diode current [latex]I(D_m)[/latex]. We get [latex]I_1 = 17.96[/latex] A and [latex]I_2 = 25.46 A[/latex].

Question 5.2: Finding the Currents of a Dc Converter with an RL Load A con...

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