Mathematical Analysis

The analysis of this circuit is slightly more complex. The differential equations that describe the operation of the circuit are presented below. Because of the filter capacitor, the current in the inductor can become discontinuous for a light load, even if the firing angle is not high. Hence the differential equations are described for the two cases separately.

I. DISCONTINUOUS CONDUCTION

When the current through the load is discontinuous, the load current starts building up from zero value when one of the pair of SCRs is triggered and it falls to zero before the next pair of SCRs is triggered. Let the firing angle be a and let the current through the inductor become 0 when wt = p + b, where b < a. That is, there is current flow during a < wt < (p + b) SCRs S₁ and S₃ in conduction and during (p + a) < wt < (2p + b) SCRs S₂ and S₄ in conduction. This means that ther will no current flow during (p + b) < wt < (p + a) and b) < wt < a. Let the supply voltage v_s be E*Sin (q), where q = wt. Let the voltage across the capacitor be v_C(q) and the current through the inductor be i_L. Then equations (1) and (3) are for the periods when there would no current flow. When SCRs S₁ and S₃ are in conduction, both the line current and the load current have the same magntitude and polarity and equation (2) applies. When SCRs S₂ and S₄ are in conduction, the line current is the negative of the load current and equation (4) is to be used. The current through the capacitor is the difference of the dc link inductor current and the current through the load resistor, as shown in equation (5).

II. CONTINUOUS CONDUCTION

For continuous conduction, it is sufficient if the equations are described for half-a-cycle only. In the other half-cycle, the only difference is in the source current waveform. Since the source current has half-wave symmetry, it is sufficient if it is described over half-a-cycle. Let the firing angle be a. Let the commutation overlap angle be d. Then it means the SCRs are triggered when q = a or when q = p + a, the process of commutation ends d radians later. Let the source current be i_s. Then during a < q < ( a + d ), the entire source voltage is applied across the source inductor, as shown in equation (6). During this period, the output voltage of the bridge is zero and the voltage across the inductor is then the voltage across the output capacitor. Since the voltage across the output capacitor tends to reverse the current through the inductor, equation (7) describes the current-voltage relationship in the inductor for this period. During ( a + d ) < q < (p + a) , the voltage across the source inductor and the dc link inductor is defined by equation (8). During this period, the line current and the load current have both the same magnitude and polarity, as shown by equation (9). Equation (10) defines the current-voltage relationship of the output capacitor. As shown in equation (10), capacitor current is the difference between the inductor current and the current through the load resistor.

III. SOLUTION

The solution of the equations for both the cases is carried out using numerical technique.