A single - phase, full bridge, fully controlled thyristor rectifier feeds a load comprising a 10  resistance in series with a very large inductance. The rectifier is fed from an ideal 230 V, 50 Hz sinusoidal source through cables which have negligible internal resistance and a total inductance of 2.28 mH. If the thyristors are triggered at an angle α=45 , the commutation overlap angle in degree (rounded off to 2 decimal places) is _________.

The correct answer is 4.8.

Step 1: Understand the Given Parameters
From the given single-phase, full-bridge, fully-controlled thyristor rectifier problem, we have:
RMS source voltage, $V_s=230\text{V}$
Source frequency, $f=50\text{Hz}$
Angular frequency, $\omega =2\pi f=2\pi \times 50=100\pi \text{rad/s}$
Source inductance, $L_s=2.28\text{mH}=2.28\times {10}^{-3}\text{H}$
Load resistance, $R=10\Omega$
Firing angle, $\alpha ={45}^{\circ }$
The load inductance is very large, which implies that the DC load current $I_d$ is continuous and ripple-free.

Step 2: Source Reactance Calculation
The source reactance $X_s$ is given by:

$X_s=\omega L_s=100\pi \times 2.28\times {10}^{-3}\approx 0.7163\Omega$

Step 3: Average Output Voltage Equation
For a single-phase full-bridge controlled converter considering the effect of source inductance (commutation overlap):

$V_{do}=\frac{2\sqrt{2}{V}_s}{\pi }\cos \alpha -\frac{2\omega L_s}{\pi }{I}_d$

Also, since the average load voltage across the resistance in steady-state is equal to $I_{d}R$ (the inductance absorbs zero average voltage):

$V_{do}=I_{d}R$

Equating both expressions for $V_{do}$:

$I_{d}R=\frac{2\sqrt{2}{V}_s}{\pi }\cos \alpha -\frac{2X_s}{\pi }{I}_d$

Rearranging to solve for $I_d$:

$I_d(R+\frac{2X_s}{\pi })=\frac{2\sqrt{2}{V}_s}{\pi }\cos \alpha$

Step 4: Compute Load Current $I_d$
Substitute the values into the formula:

$\frac{2\sqrt{2}{V}_s}{\pi }\cos \left({45}^{\circ }\right)=\frac{2\sqrt{2}\times 230}{\pi }\times \frac{1}{\sqrt{2}}=\frac{460}{\pi }\approx 146.423\text{V}$

$R+\frac{2X_s}{\pi }=10+\frac{2\times 0.7163}{\pi }=10+0.456=10.456\Omega$

$I_d=\frac{146.423}{10.456}\approx 14.004\text{A}$

Step 5: Calculate Commutation Overlap Angle ($\mu$)
The fundamental relation during overlap angle $\mu$ is given by:

$\cos \alpha -\cos (\alpha +\mu )=\frac{2\omega L_s{I}_d}{\sqrt{2}{V}_s}$

Substitute the calculated current $I_d$ into the formula:

$\cos \left({45}^{\circ }\right)-\cos ({45}^{\circ }+\mu )=\frac{2\times 0.7163\times 14.004}{\sqrt{2}\times 230}=\frac{20.062}{325.269}\approx 0.06168$

Now, calculate $\cos ({45}^{\circ }+\mu )$:

$\cos ({45}^{\circ }+\mu )=\cos \left({45}^{\circ }\right)-0.06168=0.70711-0.06168=0.64543$

Taking the inverse cosine:

${45}^{\circ }+\mu ={\cos }^{-1}\left(0.64543\right)\approx {49.80}^{\circ }$

$\mu ={49.80}^{\circ }-{45}^{\circ }={4.80}^{\circ }$

Thus, the commutation overlap angle rounded off to two decimal places is 4.8° (or 4.8).