The rectifier in Figure 10.12 has the specifications given in Table 10.4.

Determine the harmonics when the SAG inductor of Example 10.5 is used at the input.

For simulation purposes, the circuit is represented in Figure 10.19; the input represents the rectifier output.

Question

The rectifier in Figure 10.12 has the specifications given in Table 10.4.

Determine the harmonics when the SAG inductor of Example 10.5 is used at the input.

For simulation purposes, the circuit is represented in Figure 10.19; the input represents the rectifier output.

Accepted Answer

PSPICE and other simulation packages are based on constant inductance. An alternative approach is to use discretized equations in the time domain. The method of simulation is based on the ElectroMagnetic Transient Program (EMTP) by Dommel [3], which can evaluate the variable inductance at each value of current.

The general solution for the three nodes in Figure 10.19, recognizing that there are no input currents at nodes 2 and 3, is:

[latex]\left[\begin{array}{c}i_1\left(t_n ight) \ 0 \ 0 \end{array} ight]=\left[\begin{array}{ccc}Y_{11} & Y_{12} & Y_{13} \ Y_{21} & Y_{22} & Y_{23} \ Y_{31} & Y_{32} & Y_{33}\end{array} ight] \cdot\left[\begin{array}{c}v_1\left(t_n ight) \v_2\left(t_n ight) \ v_3\left(t_n ight)\end{array} ight]+\left[\begin{array}{c}I_1\left(t_{n-1} ight) \ I_2\left(t_{n-1} ight) \ I_3\left(t_{n-1} ight)\end{array} ight][/latex]

The terms in the admittance matrix are:

[latex]Y_{11}=\frac{1}{R} \quad Y_{12}=-\frac{1}{R} \quad Y_{13}=0[/latex]

[latex]Y_{21}=-\frac{1}{R} \quad Y_{22}=\frac{1}{R}+\frac{\Delta t}{2 L_{ eff }} \quad Y_{23}=-\frac{\Delta t}{2 L_{ eff }}[/latex]

[latex]Y_{31}=0 \quad Y_{32}=-\frac{t}{2 L_{ eff }} \quad Y_{33}=\frac{\Delta t}{2 L_{ eff }}+\frac{2 C}{\Delta t}[/latex]

[latex]t_n=n \Delta t[/latex]

[latex]L_{eff}[/latex] must be calculated at [latex]i_{1}(t_{n - 1})[/latex] according to Equation 10.15.

[latex]L_{ eff }=L(i)+i \frac{d L(i)}{d i}=L(i)+i \frac{d L}{d \mu_r} \frac{d \mu_r}{d i}[/latex] (10.15)

The current sources [latex]I_{i}(t_{n - 1})[/latex] are known from past history:

[latex]I_1\left(t_{n-1} ight)=0[/latex]

[latex]I_2\left(t_{n-1} ight)=I_2\left(t_{n-2} ight)+\frac{\Delta t}{L_{ ext {eff }}}\left[v_2\left(t_{n-1} ight)-v_3\left(t_{n-1} ight) ight] \quad n=2,3 \ldots[/latex]

[latex]I_{3}(t_{n–1})[/latex] includes contributions from [latex]L_{eff}[/latex] and C and, since we are dealing with a constant power output rather than a load resistor, this effect may be included:

[latex]I_3\left(t_{n-1} ight)=I_{30}\left(t_{n-1} ight)-I_2\left(t_{n-1} ight)+\frac{P_0}{v_3\left(t_{n-1} ight)} \quad n=2,3 \ldots[/latex]

[latex]I_{30}\left(t_{n-1} ight)=-I_{30}\left(t_{n-2} ight)-\frac{4 C}{\Delta t} v_3\left(t_{n-1} ight) \quad n=2,3 \ldots[/latex]

The initial conditions are:

[latex]v_2(0)=v_3(0)=i_i(0)=0, I_2(0)=0, I_{30}(0)=0 I_3(0)=0 .[/latex]

If [latex]i_{1}[/latex] results in a negative value, it is reset to zero and both [latex]I_{2} ext{ and } I_{3}[/latex] are adjusted accordingly, since in reality the diodes block negative current at the output of the rectifier.
The results are shown in Figure 10.20 for one half cycle. The peak current in the circuit (3.0 A) is less than [latex]I_{max}[/latex] (3.5 A in Figure 10.11). The harmonic spectrum of the current waveform is obtained by the Fast Fourier Transform and is shown in Figure 10.21. The harmonic currents are normalized to the fundamental current in each case. The limits from Table 10.2 are also shown, and the harmonics are within the allowable limits

Table 10.4 Specifications
Input voltage (rms)	220 V
Output capacitance	330 μF
Inductor resistance	2.5 Ω
Frequency	50 Hz
Output power	200 W

Question 10.6: The rectifier in Figure 10.12 has the specifications given i...

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