Single-Phase 2-Pulse Converters with Discontinuous Load Current

The study of single-phase 2-pulse converters has traditionally focused on continuous load current.In practical scenarios, however, the output current may become discontinuous under specific conditions, such as high firing angles or low load currents. Discontinuity refers to the state where the load current reaches zero during each half cycle before the next SCR in sequence is fired. Conversely, continuous operation implies that the load current consistently flows through SCR/diode or their combination without interruption.The performance of these converters degrades when the load current becomes discontinuous.To enhance efficiency, it is recommended to operate the DC load in continuous current mode, facilitated by incorporating freewheeling action and employing an external inductor in series with the load. This article delves into the operation of both single-phase full converters and semiconverters, particularly examining their behavior when the load current becomes discontinuous.

Discontinuous Current Operation in a Single-Phase Full Converter

The power circuit diagram for a single-phase full converter is illustrated in the figure.When the SCR pair T1 T2 is triggered at ωt = α, the load current initiates from zero and gradually increases.At the extinction angle β (where β > π), the load current diminishes to zero. Post ωt = π, T1 and T2, being reverse biased, undergo commutation at ωt = β when $i_0$ = 0. Between α and β, the output voltage $v_0$ mirrors the source voltage $v_s$. From β to (π + α), with no SCR conducting, the load voltage jumps from $V_m$sin β to E. At ωt = π + α, when the pair T3 T4 is triggered, the load current resumes its buildup, and $v_0$ follows $v_s$ waveform. At π + β, $i_0$ does not reach zero, causing $v_0$ to transition from $V_m$sin(π + β) to E as no SCR conducts.The figure also depicts the source current $i_s$.

Voltage and current waveforms for discontinuous load current for a single-phase full converter

In certain scenarios, the load current may reach zero at ωt = β, where β is less than π. This analysis assumes $V_m$sinβ < E. At β, $v_0$ undergoes a transition from $v_m$sinβ to E.The figure illustrates the waveforms for both load current $i_0$ and load voltage $v_0$.No SCR conducts from β to(π + α) and during this interval, therefore $v_0$ = E.

Based on the information provided earlier, the following observations can be noted:

1. Conduction period, α < ωt < β, T1, T2 conduct and $v_0 = v_s$.Also π + α < ωt < π+ β, T3, T4 conduct and  $v_0 = v_s$  and so on.
2. Idle period, β < ωt < π + α, no circuit element conducts and $v_0 = E$.

The voltage output in discontinuous current mode is lower than the value determined by the equation $V_0 = \frac{1}{\pi} \int_{\alpha}^{\alpha + \pi} V_m \sin \omega t \, d(\omega t) = \frac{2V_m}{\pi} \cos \alpha$. As mentioned earlier, load performance during discontinuous conduction is impaired.

Discontinuous Current Operation in a Single-Phase Semiconverter

The power circuit diagram for this converter is illustrated in the figure. In the controlled 2-pulse converter, when SCR T1 is triggered at $\omega t = \alpha$, the load current starts from zero, increases to a peak, and subsequently decreases to zero at $\beta > \pi$.

From $\alpha$ to $\pi$, T1 D conduct, and $v_0 = v_s$. At $\omega t = \pi$, as $v_s$ tends to become negative, FD becomes forward biased and starts conducting the load current. When FD conducts from $\pi$ to $\beta$, $v_0 = 0$.

From $\beta$ to $\pi + \alpha$, no circuit component  conducts, thus $v_0 = E$ as depicted in the figure below. During $\beta$ to $\pi + \alpha$, with zero load current, it becomes discontinuous. When T2 is triggered at $\pi + \alpha$, $i_0$ builds up as illustrated. At $2\pi$, FD becomes forward biased and conducts until $\pi + \beta$. While FD conducts, $v_0 = 0$. From $\pi + \beta$ to $2\pi + \alpha$, no circuit component is in conduction, thus $v_0 = E$. At $2\pi + \alpha$, T1 is triggered again, and the process repeats. The source current $i_s$ is also depicted in the figure(a).

Voltage and current waveforms for discontinuous conduction for a single-phase semiconverter

If the load current becomes zero before $\pi$, specifically for $\beta$ less than $\pi$, the current and voltage waveforms are depicted in figure(b).Here, $V_m\sin\beta$ is assumed to be less than $E$. During $\beta$ to $\pi + \alpha$, no circuit component conducts, hence $v_0 = E$.

Analyzing the waveforms of the single-phase semiconverter, the following observations are noted:

(a) When $\pi < \omega < \pi + \alpha$:

1. Conduction period, $\alpha < \omega t < \pi$, TID1 conduct and $v_0 = v_s$.Also for $\pi + \alpha < \omega t < 2\pi$,T2D2 conduct and  $v_0 = v_s$ and so on.
2. Freewheeling period, $\pi < \omega t < \beta$ ,FD conducts $i_{\text{fd}} = i_0$ and $v_0 = 0$. Also for $2\pi < \omega t < \pi + \beta$,  $i_{\text{fd}} = i_0$ and $v_0 = 0$ and so on. 
3. Idle period, $\beta < \omega t < \pi + \alpha$,no circuit component conducts, $i_0 = 0$ and $v_0 = E$.

(b)  when $\beta < \pi$ and $V_m \sin \beta < E$:

1. Conduction period, $$\alpha < \omega t < \beta$$ , $T_1$ $D_1$ conduct, $v_0 = v_s$. Also $π + \alpha < \omega t < π + \beta$,T2 D2 conduct,$v_0 = v_s$ and so no.
2. Freewheeling period, absent and $i_{\text{fd}} = 0$.
3. Idle period,$\beta < \omega t < π + \alpha$ and $π + \beta < \omega t < 2π + \alpha$,no circuit elements conducts,$i_0 = 0$ and $v_0 = E$.

The output voltage during discontinuous conduction is not determined by the equation $$V_0 = \frac{1}{\pi} \int_{\alpha}^{\pi} V_m \sin \omega t \, d(\omega t) = \frac{V_m}{\pi} (1+\cos \alpha)$$ . As mentioned earlier, the load performance with discontinuous load current degrades.

For a single-phase full converter when $\beta > \pi$ or $\beta < \pi$, and also for a single-phase semiconverter when $\beta < \pi$, the average load current is expressed as:

$I_{0} = \frac{1}{\pi R} \int_{\alpha}^{\beta} (V_{m}\sin \alpha t - E) \, d(\omega t) = \frac{V_{m}}{\pi R}(\cos \alpha - \cos \beta) - \frac{E}{\pi R}(\beta - \alpha)$

Average output voltage

$V_{0} = E + I_{0}R = \frac{V_{m}}{\pi}(\cos \alpha - \cos \beta) + E\left(1 - \frac{\gamma}{\pi}\right)$

Where,

$\gamma = conduction angle = \beta - \alpha$.

For a single-phase semiconverter with $\beta > \pi$, the average output current $I_{0}$ is expressed as

$$I_{0} = \frac{V_{m}}{\pi R}(\cos \alpha - \cos \beta) - \frac{E}{\pi R}(\beta - \alpha)$$ .

and average output voltage $V_0$ is

$$V_{0} = \frac{1}{\pi}\left[\int_{\alpha}^{\pi} V_{m} \sin \omega t \, d(\omega t) + E(\pi + \alpha - \beta)\right)$$

$= \frac{V_{m}}{\pi}(1+\cos \alpha) + E\left(1-\frac{\gamma}{\pi}\right)$.