Effect of Source Inductance

• The source does possess internal impedance.

• If the source impedance is resistive, then there will be a voltage drop across the resistance and the average voltage output of a converter gets reduced by an amount equal to:

  • lo Rs for a single-phase converter and by

  • 2lo Rs for a 3-phase converter

Here lo is the constant dc load current and Rs is the source resistance per phase.

• The source inductance causes the outgoing and incoming SCRs to conduct together is called the commutation period or the overlap period.

• The output voltage during this time is equal to the average value of the conducting-phase voltages.

• For a single-phase converter, the load voltage will be zero and for a 3-phase converter, the load voltage is $(V a+V b) / 2$ The angular-period, during which both the incoming ar outgoing SCRs are conducting, is known as commutati angle or overlap angle ' $\mu$ ' in degrees or radians.

The effect of source inductance is:

(i) To lower the mean output voltage,

(ii) To distort the output voltage and current waveforms

(iii) To modify the performance parameters of the converter

• More predominant in full converters than semi-converters.

During the commutation of $\mathrm{T}_{1}, \mathrm{~T}_{2}$ and $\mathrm{T}_{3}, \mathrm{~T}_{4}$; i.e. during the overlap angle, ' $\mu$ ' $K V L$ - for the loop abcda of $$ \begin{aligned} &v_{1}-L_{s} \cdot \frac{d i_{2}}{d t}=v_{2}-L_{s} \frac{d i_{2}}{d t} \\ &v_{1}-v_{2}=L_{s}\left(\frac{d i_{1}}{d t}-\frac{d i_{2}}{d t}\right) \end{aligned} $$ $V_{1}=V_{m} \sin \omega t$, then $V_{2}=-V_{m} \sin \omega t$ $L_{s}\left(\frac{d i_{1}}{d t}-\frac{d i_{2}}{d t}\right)=2 V_{m} \sin \omega t$

As the load current is assumed constant throughout, $i_{1}+i_{2} \diamond i_{0}$, Differentiating this with respect to $t$, we get $$ \frac{d i_{1}}{d t}+\frac{d i_{2}}{d t}=0 $$ From (1) $\frac{d i_{1}}{d t}-\frac{d i_{2}}{d t}=\frac{2 V_{m}}{L_{s}} \sin \omega t$ Adding eqn. (2) \& (3) gives $$ \frac{d i_{1}}{d t}=\frac{V_{m}}{L_{s}} \sin \omega t $$ Load current $\dot{i}_{1}$ through thyristor pair $T_{1}, T_{2}$ builds up from zero to $i_{1}=i_{0}$, during the overlap angle; i.e. at $\omega t=\alpha, i_{l}=0$ and at $\omega t=(\alpha+\mu), \dot{l}_{l}=I_{0 .}$. From eqn. (4) $$ \int_{0}^{I_{0}} d i_{1}=\frac{V_{m}}{L_{s}} \int_{\alpha / \omega}^{(\alpha+\mu) / \omega} \sin \omega t \cdot d t $$ $$ I_{0}=\frac{V_{m}}{\omega L_{s}}[\cos \alpha-\cos (\alpha+\mu)] $$

It is seen from the figure that output voltage Vo is zero from $\alpha$ to $(\alpha+\mu)$. Thus the average output voltage Vox is given by $$ \begin{aligned} V_{\alpha x} &=\frac{V_{m}}{\pi} \int_{(\alpha+\mu)}^{(\alpha+\pi)} \sin \omega t \cdot d(\omega t)=\frac{V_{m}}{\pi}[\cos (\alpha+\mu)-\cos (\alpha+\pi)] \\ &=\frac{V_{m}}{\pi}[\cos \alpha+\cos (\alpha+\mu)] \end{aligned} $$ Average value of output voltage at no load, $V_{0}=2 \mathrm{Vm} / \pi \operatorname{Cos} \alpha$ Maximum mean output voltage, $\mathrm{Vom}=2 \mathrm{Vm} / \pi$ Eqn. (6) can be expressed as $$ \begin{aligned} V_{o x} &=\frac{\text { Maximum mean output voltage at no load }}{2}[\cos \alpha+\cos (\alpha+\mu)] \\ &=\frac{V_{o m}}{2}[\cos \alpha+\cos (\alpha+\mu)] \end{aligned} $$ From eqn. (5) ... $\quad \cos (\alpha+\mu)=\cos \alpha-\frac{\omega L_{s}}{V_{m}} I_{0}$

Substituting this value of $\operatorname{Cos}(\alpha+\mu)$ in Eq. (6) gives $$ V_{a x}=\frac{2 V_{m}}{\pi} \cos \alpha-\frac{\omega L_{s}}{\pi} I_{0}=\frac{2 V_{m}}{\pi} \cos \alpha-2 f L_{s}, I_{0} $$ From above eqn., a DC equivalent circuit can be drawn as shown

The effect of source inductance $L s$ is to present an equivalent resistance of $\omega L s / \pi$ ohm, in series with internal voltage of rectifier, $(2 \mathrm{Vm} / \pi) \cos \alpha$. The voltage drop due to $L$ is proportional to $I 0$ and $L$. As the load current (or Ls) increases, the commutation interval ( $\mu)$ increases and as a consequence, the average output voltage decreases. As long as $\mu<\pi$, the output voltage is given by Eq. (8). When $\mu=\pi$, the load will be permanently short circuited by SCRs and the output voltage will be zero because during the overlap angle, all SCRs will be conducting. If $\alpha=0$, then the mean output voltage can be controlled over $\mu<\alpha<180^{\circ}$. Also, the maximum value of firing angle can be $(180-\mu)$. Practically, thyristor takes some time to regain its forward blocking Capability, therefore, the maximum possible firing angle can be $180-\mu-\delta=180-(\mu+\delta)$ where $\delta / \omega$ is thyristor turn-off time including the factor of safety. Here $\delta$ is called the recovery angle.