Single Phase Fully Controlled Converters with Source Inductance

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teamques10 ★ 69k

The single-phase fully-controlled bridge converter is shown in Fig.1(a). $L_{s}$ is the source inductance. The load current is assumed constant. The a.c. supply may be represented by its Thevenin's equivalent circuit, each phase being a voltage source in series with its inductance. The major contributor to the supply impedance is the transformer leakage reactance. The equivalent circuit of fully controlled single converter is shown in Fig.1(b). When the terminal L of the source voltage, $e_{s},$ is positive, then the current flows through the path $L-L_{s}-T_{1}-R-L-T_{2}-N .$ This is shown as $e_{1}, L_{s}, T_{1}, T_{2},$ and load in Fig.1(b). Similarly, when terminal N of source voltage, $e_{s},$ is positive, load current flows through the path $N-T_{3}-$ load $-T_{4}-L_{s}-L .$ This is shown as $e_{2},L_{s}, T_{3}, T_{4}$ and load in Fig.1(b). The related circuit voltage and current waveforms are shown in Fig.2.

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When $T_{1}, T_{2}$ are triggered at a firing angle $\alpha,$ the commutation of already conducting thyristors $T_{3}, T_{4}$ begins. Because of the presence of source inductance $L_{s},$ the current through outgoing thyristors $T_{3}, T_{4}$ decreases gradually to zero from its initial value of $I_{d},$ whereas in case of incoming thyristors $T_{1}, T_{2},$ the current builds up gradually from zero to full value of load current, $I_{d} .$ During the commutation of $T_{1}, T_{2}$ and $T_{3}, T_{4},$ i.e. during the overlap angle $\mu,$ KVL for the loop $\mathrm{OPQRO}$ of Fig.2 ignoring SCR drops, gives

$e_{1}-L_{s} \frac{\mathrm{d}_{i T_{1}}}{\mathrm{d} t}=e_{2}-L_{s} \frac{\mathrm{d}_{i_{3}}}{\mathrm{d} t}$

$e_{1}-e_{2}=L_{S}\left(\frac{\mathrm{d}_{i T_{1}}}{\mathrm{d} t}-\frac{\mathrm{d}_{i_{3}}}{\mathrm{d} t}\right)-----(1)$

But from Fig.2, we have the relations

$e_{1}=E_{m} \sin \omega t \ and \ e_{2}=-E_{m} \cdot \sin \omega t$

Substituting in Eq.1 we get

$L_{s}\left(\frac{\mathrm{d}_{i T_{1}}}{\mathrm{d} t}-\frac{\mathrm{d}_{i_{3}}}{\mathrm{d} t}\right)=2 E_{m} \times \sin \omega t-----(2)$

Since the load current is assumed constant, we can write

$i_{T_{1}}+i_{T_{3}}=I_{d}-----(3)$

Differentiating Eq.3, with respect to t

$\frac{\mathrm{d}_{i T_{1}}}{\mathrm{d} t}=\frac{\mathrm{d}_{i T_{3}}}{\mathrm{d} t}-----(4)$

Substitute Eq.4 in Eq.3, we get

$L_{s}\left(2 \frac{\mathrm{d}_{i T_{1}}}{\mathrm{d} t}\right)=2 E_{m} \cdot \sin \omega t$

$\therefore \quad \frac{\mathrm{d}_{i T_{1}}}{\mathrm{d} t}=\frac{E_{m}}{L_{s}} \sin \omega t-----(5)$

If the overlap angle is $\mu,$ then the current through thyristor pair $T_{1}, T_{2}$ builds up from zero to $I_{d}$ during this interval.

Therefore, at $\omega t=\alpha, i_{T_{1}}=0 \quad$ and at $\omega t=(\alpha+\mu), i_{T_{1}}=I_{d}$

$\therefore$ Therefore, from Eq.5 we can write

$\int_{0}^{I_{d}} d_{i T_{1}}=\frac{E_{m}}{L_{s}} \int_{\alpha / \omega}^{(\alpha+\mu) / \omega} \sin \omega t \cdot \mathrm{d}(\omega t)$

$\therefore \qquad I_{d}=\frac{E_{m}}{\omega L_{s}}[\cos \alpha-\cos (\alpha+\mu)]-----(6)$

It can be observed from Fig.2 that the output voltage is zero during the interval $\mu .$ There are two commutations in each cycle. Thus, the average output voltage is given by

$E_{\mathrm{dc}}=\frac{E_{m}}{\pi} \int_{\alpha+\mu}^{\pi+\alpha} \sin \omega t \cdot \mathrm{d}(\omega t)=\frac{E_{m}}{\pi}[-\cos \omega t]_{\alpha+\mu}^{\pi+\alpha}$

$=\frac{E_{m}}{\pi}[\cos (\alpha+\mu)-\cos (\alpha+\pi)]$

$\therefore \quad E_{\mathrm{dc}}=\frac{E_{m}}{\pi}[\cos \alpha+\cos (\alpha+\mu)]-----(7)$

Now, from Eq. 5, we have

$\cos (\alpha+\mu)=\cos \alpha-\frac{\omega L_{s}}{E_{m}} I_{d}$

Substituting this value of $\cos (\alpha+\mu)$ in eq. 7, we get

$E_{\mathrm{dc}}=\frac{E_{m}}{\pi}\left[\cos \alpha+\cos \alpha-\frac{\omega L_{s}}{E_{m}} I_{d}\right] E_{\mathrm{dc}}=\frac{2 E_{m}}{\pi} \cos \alpha-\frac{\omega L_{s}}{E_{m}} I_{d}-----(8)$

Also, from Eq. 6, we have, $\cos \alpha=\frac{\omega L_{s}}{E_{m}} I_{d}+\cos (\alpha+\mu)$

Substituting this value of $\cos \alpha$ in Eq. 7, we get

$E_{\mathrm{dc}}=\frac{2 E_{m}}{\pi} \cos (\alpha+\mu)+\frac{\omega L_{s}}{\pi} I_{d}-----(9)$

With the help of Eq.8 a d.c. equivalent circuit for a two-pulse single-phase fully-controlled converter can be drawn, as shown in Fig.3.

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Diode D in Fig.3 indicates that load current is unidirectional. This equivalent circuit shows that the effect of source inductance is to present an equivalent resistance of magnitude $\frac{\omega L_{s}}{\pi}$ in series with interval voltage of rectifier $\frac{2 E_{m}}{\pi} \cos \alpha$ . With the load current $I_{d}$ the votage drop is $\frac{\omega L_{s} I_{d}}{\pi} .$ Hence it becomes clear that with source inductance, the output voltage of a converter is reduced by $\frac{\omega L_{s} I_{d}}{\pi}$ value. The variation of output-voltage with output-current is shown in Fig.4. From this Fig.4, it is also clear that as load current $I_{d}$ (or source inductance) increases, the commutation interval or the overlap angle increases and as a consequence, the average output voltage decreases.

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Thus, in single-phase fully-controlled converter, as long as commutation angle $\mu$ is less than $\pi$ the output voltage is given by Eq.7 and when $\mu$ is equal to $\pi$ the load will be permanently short-circuited by SCRs and the output voltage will be zero because during the overlap period, all SCRs will be conducting.

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