Finding the Performance Parameters of an AC Power Supply with a PWM Control The load resistance of the ac power supply in Figure 13.13 is R = 2.5 Ω. The dc input voltage is Vs = 100 V. The input inverter operates at a frequency of 20 kHz with one pulse per half-cycle. The on-state voltage drops

Question

Finding the Performance Parameters of an AC Power Supply with a PWM Control

The load resistance of the ac power supply in Figure 13.13 is [latex]R = 2.5 Ω[/latex]. The [latex]d_c[/latex] input voltage is [latex]V_s = 100 V[/latex]. The input inverter operates at a frequency of 20 kHz with one pulse per half-cycle. The on-state voltage drops of transistor switches and diodes are negligible. The turns ratio of the transformer is [latex]a = N_s/N_p = 0.5[/latex]. The output inverter operates with a uniform PWM of four pulses per half-cycle. The width of each pulse is [latex]δ = 18°[/latex]. Determine the rms load current. The ripple voltage on the output of the rectifier is negligible. Neglect the losses in the transformer, and the effect of the load on the resonant frequency is negligible.

Accepted Answer

The rms output voltage of the input inverter is [latex]V_1 = V_s = 100 V[/latex]. The rms transformer secondary voltage is [latex]V_2 = aV_1 = 0.5 × 100 = 50 V[/latex]. The dc voltage of the rectifier is [latex]V_o = V_2 = 50 V[/latex]. With the pulse width of δ = 18, Eq. (6.31)

[latex]V_o=\left[\frac{2p}{2\pi } \int_{(\pi /p-\delta )/2}^{(\pi /p+\delta )/2}{V^2_sd(\omega t)} ight]^{1/2} =V_s\sqrt{\frac{P\delta }{\pi } }[/latex] (6.31)

gives the rms load voltage, [latex]V_L=V_o\sqrt{p\delta /\pi }=50\sqrt{4 imes 18/180}=31.6 V[/latex]. The rms load current is [latex]I_L = V_L/R = 31.6/2.5 = 12.64 A[/latex].

Question 13.5: Finding the Performance Parameters of an AC Power Supply wit...

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