Phase Controlled Rectifiers Practice Questions

Average output voltage

Average output current $=I_0$ (constant) RMS value of supply current $=I_s=I_0$ RMS value of supply voltage $=V_s=\frac{V_m}{\sqrt{2}}$ Input power factor,

Load current $=I_0=10A$ RMS value of total source current $I_s=I_0\sqrt{\frac{2}{3}}=10\sqrt{\frac{2}{3}}=8.165A$ Supply current $i_s$ can be expressed by fourier series, $i_s(t)=\sum_{n=1,3,5}^{\infty }\frac{4I_0}{n \pi}\sin \frac{n \pi }{3}\sin (n\omega t-n\alpha )$ RMS value of fundamental current,

Input to the converter, $V_s=\left ( \frac{1}{4} \right )230=57.5V$ Diode conducts when $V_s\geq E$

Charging current flows during $\theta _1 \leq \omega t \leq \pi-\theta _1$ and can be expressed as,

Let DC link current $=I_d$ DC voltage applied to the inverter $V_d=420V$ Power fed to the inverter

Current through each thyristor flows for period of $2 \pi/3$ . So, rms current of thyristor.

$1-\phi$ full wave half controlled bridge converter without free wheeling diode is shown in the figure. At $\omega t=\alpha , T_1$ is fired and $T_1$ starts conducting load current flows through $T_1D_1$ for $\alpha \lt \omega t \lt \pi$ . At $\omega t=\pi, V_s$ becomes negative and At $D_1$ gets reverse biased and $D_2$ is forward biased. So during $\pi \lt \omega t \lt \pi+\alpha$ , free wheeling action takes place through $T_1$ and $D_2$ and output voltage becomes zeros. At $\omega t=\pi+\alpha$ , Load $T_2$ is triggered and load current is transferred from $T_1$ to $T_2$ . So, during $\pi +\alpha \lt \omega t \lt 2\pi$ , $T_2D_2$ conducts. At $\omega t=\pi+\alpha$ , It may be possible that load current is not transferred completely from $T_1$ to $T_2$ , and $T_1$ and $T_2$ may be conducting simultaneously which results in short circuit of the supply for short direction.

$T_1$ and $T_4$ gets forward biased, when $V_m \sin \theta _1 \leq E$

Without considering source inductance i.e. (Transformer inductance) After considering transformer leakage inductance.

Load current is constant $I_0=10A$ In $3-\phi$ full converter with free wheeling diode, input displacement factor (IDF) $= \cos \alpha = \cos 60^{\circ}=0.5$ RMS value of source current $I_s=\sqrt{I_0^2 \times \frac{2 \pi}{3} \times \frac{1}{\pi}}=I_0\sqrt{\frac{2}{3}}$ RMS value of $n^{th}$ harmonic $I_{sh}=\frac{4I_0}{\sqrt{2}n \pi}\sin \frac{n \pi}{3}$ RMS value of fundamental current $I_{s1}=\frac{2\sqrt{2}}{\pi}I_0 \sin 60^{\circ}=\frac{\sqrt{6}}{\pi}I_0$ Current distortion factor (CDF) $=\frac{I_{s1}}{I_s}=\frac{\sqrt{6}}{\pi}I_0 \times \left ( \frac{1}{I_0}\sqrt{\frac{3}{2}} \right )$ $=\frac{3}{\pi}=0.955$ Input power factor $(IPF) =CDF \times IDF = 0.955 \times 0.5\approx 0.478$

Maximum value of inout voltage $V-m=400\sqrt{2}\;V$ Since load is purely resistive, therefore peak instantaneous output voltage $V-m=400\sqrt{2}\;V$

Supply voltage waveform is shown in figure (a). It has peak value of $230\sqrt{2}\; V$ at $\omega t=\frac{\pi}{2}$ . If the thyristor is triggered at diring angle $\alpha \lt \frac{\pi}{2}$ , output voltage will also have peak value of $230\sqrt{2}\; V$ The supply voltage at $\alpha =135^{\circ}$ $V_s|_{\alpha =135^{\circ}}=V_m \sin 135^{\circ}=230\sqrt{2}\times \frac{1}{2}=230V$ So, when thyristor is triggered at $\alpha =135^{\circ}$ 230 V will appear across the load, it is also the peak value of the output voltage.

The thyristor must turn on even for minimum value of $V_b$ . Gate current for gauranteed turn-on $=I_b=10mA$ , neglecting voltage between gate and cathode. Therefore, required value of $R=\frac{V_{b(min)}}{I_g}=\frac{8}{10mA}=800\Omega$

When 'a' is positive with resoect to 'b' diode $D_1$ conducts and diode $D_2$ is reverse baised. Applying KVL, $V_{D2}=100V$ (rms) So, diode $D_2$ is subjected to reverse voltage of 100 V (rms). Similarly, during negative half cycle, $D_1$ is subjected to reverse voltage of 100 V (rms). Thus, for diode $D_1$ and $D_2$ peak reverse voltage is $100\sqrt{2}V$ (maximum value of the voltage across the diodes.)

By drawing the output voltage. Each diode conducted for $120^{\circ}$ only. there are three pulses of output voltage during one cycle of inout voltage. So, frequency of ripple =3f=3 x 400=1200 Hz

Due to free wheeling action, negative voltage is eliminated and power factor is improved in single-phase half controlled rectifier.

$\frac{\text{Peak to peak ripple voltage}}{Peak output dc voltage}=\frac{V_{ML}-V_{ML} \sin 150^{\circ}}{V_{ML}}=0.5$

$V_{rms}=\frac{V_m}{\sqrt{2}}$ $P=\frac{V^2_{rms}}{R}=\frac{V^2_{m}}{2R}$ $\;\;=\frac{200^2}{2 \times 50}=400W$

Currently no explanation available

Lowest harmonic present is fifth harmonic. Its frequency = 50x5=250 Hz.