# Power Electronics (EE 203)

A single-phase half-wave controlled rectifier with a resistive load 10 Ω is connected to a 120-V, 60-H source. If the DC output voltage is 25% of the maximum possible average output voltage. Calculate:

(a) The firing angle ………………………………….. (0.5 mark)

(b) The DC voltage across the load ….….………….. (0.5 mark)

(c) The RMS load Current ………………………….. (0.5 mark)

(d) The average current of the SCR ………………… (0.5 mark)

(e) The commutation angle of the SCR …….……….. (0.5 mark)

(f) The conduction angle of the SCR ……..………… (0.5 mark)

1. The maximum possible average output voltage is

$Vmax = Vac/\pi * \int_0^\pi (sin(x)dx) = -(Vac/\pi)*cos(x)|_0^\pi =2/\pi * Vac$

For simplicity at this question I will consider Vac =1 V

The DC output voltage is

$0.25*Vmax =1/\pi*\int_\alpha^\pi (sin(x)dx) =-(1/\pi)*cos(x)|_\alpha^\pi =$

$=1/pi(1-cos(\alpha))$

$0.25*Vmax -1/\pi = -(1/\pi)*cos(\alpha)$

$0.25*2-1 =-cos(\alpha)$

$cos(\alpha) =0.5$

$\alpha = 60 degree$

The firing angle is 60 degree

2.The maximum possible average voltage is

$V_{max} =(2/\pi)* Vac =2/\pi *120 = 76.39 Volts$

The DC volatge on the load is

$Vdc = 0.25*Vmax =0.25*76.39 =19.1 Volts$

3. The RMS load Volatge for Vac =1 is

$VRMS = \sqrt{(1/(\pi-\alpha)}*\int_\alpha^\pi (sin^2(x)dx) =$

$=\sqrt{3/(2 \pi)} *(1/2*x-1/4*sin(2 x))|_{\pi/3}^\pi =$

$=\sqrt{3/(2 \pi)} *(-\pi/6 +\pi/2 +1/2*sin(2\pi/3) -1/2sin(2\pi)) =$

$=\sqrt{3/(2\pi)}*(2\pi/6 +0.5*0.866 -0.5) =\sqrt{3/(2\pi)}*0.98) = 0.684$

For $Vac =120 Volts$

$VRMS = 0,684*120 =82.08 Volts$

The RMS current is

$IRMS = VRMS/R = 82.08/10 =8.208 A$

4. The commutation angle of the SCR is simply the firing angle. This angle is 60 degree or pi/3

5. The conduction angle is the equal to the maximum conduction angle minus the commutation angle

$\alpha = 180 -60 =120 degree$ or $2*\pi/3$