Mathematical Analysis

When there is an inductor in series with each input line, it is necessary to find out its effect. We need to find out:

a. The reduction in output voltage.
b. The duration of commutation overlap.
c. The relationship between the firing angle and the commutation overlap.

I. REDUCTION IN OUTPUT VOLTAGE

Calculations by hand are carried out assuming that the dc link current remains steady without any ripple. The source voltages at its terminals and the output voltage appear as shown below, assuming that the inductances belong to the source.

It is seen that there are six notches in one input cycle. The reduction in average output voltage can be found out as follows. Let SCR S₁ be in conduction and let S₃ be triggered. Let the current through the dc link be I_DC. Then current through Y-phase has to rise from zero to I_DC, whereas current through R-phase has to fall from - I_DC to zero. On the other hand, loop current i_LOOP marked in the sketch below has to rise from zero to I_DC. This means that during commutation current through Y-phase would rise from zero to I_DC and the volt-second area the output misses out is L₂I_DC, that absorbed by the inductor in the Y-phase.

From the volt-seconds lost per commutation, we can find the total volt-seconds lost in one input cycle. Since there are six commutations per cycle, the total volt-seconds lost per cycle is expressed as shown by equation (1). Dividing this area by the time corresponding to one cycle, we get the average voltage reduction in output. The time corresponding to one input cycle is 1/f, where f is the line frequency. Then the average reduction in output voltage is obtained as shown in equation (2). In equation (2), we make use of the relation that the angular frequency, w = 2pf. It is to be noted that commutation overlap occurs only when there is continuous conduction through the load and the average output voltage is expressed by equation (3). In equation (3), U is the peak line-to-line voltage and a is the firing angle.

II. COMMUTATION OVERLAP ANGLE

The commutation of commutation overlap depends on:

a. the firing angle,
b. the dc link current and
c. the source inductance or the inductance in series with each phase.

To find out the commutation overlap, it is sufficient to analyse one commutation. Let SCR S₅ be in conduction and let SCR S₁ be triggered at a firing angle of a. As seen earlier, the loop current i_LOOP builds up from zero to I_DC. This means that the current in the inductance in R-phase builds up to I_DC, whereas it decreases to zero in the inductance in the B-phase. Let the commutation last for an angle m. Then during commutation, the voltage across the source inductance is expressed as shown in equation (4). In equation (4), the loop current is denoted as ‘i’. Since q = wt, the equation for commutation overlap can be represented as shown in equation (5). During this interval, the loop current changes from zero to I_DC. Hence equation (6) defines how current in R phase changes.

The solution of equation (6) is presented in equation (7). Equation (7) can be re-arranged and presented as shown in equation (8). Solving for m, we obtain equation (9).

It is seen that overlap angle m increases

a. as the firing angle moves closer to either 0^o or 180^o ,
b. as the dc link current becomes larger, and
c. as the link inductance gets larger.

The above equation has been obtained based on the assumption that the dc link current remains steady, which happens only when the dc link inductance is relatively large. In practice, it is not true and hence the above equation yields only an approximate result.

The rest of mathematical analysis follows the familiar route. The items of interest are:

a. RMS output voltage of bridge circuit,
b. Average output voltage of bridge circuit,
c. Ripple Factor of output voltage of the bridge,
d. RMS output voltage/voltage across load resistor,
e. Average output voltage (across load resistor),
f. Ripple factor of output voltage (across load resistor),
g. RMS line current,
h. RMS value of fundamental component in line current,
i. THD in line current,
j. Displacement power factor,
k. Apparent power factor and
l. Harmonic analysis.

In order to simulate it is necessary to have an expression for line current and load current. Let us consider one output cycle, starting from the instant SCR S₁ is triggered till SCR S₂ is triggered. During this period, the R-phase is defined as shown in equation (10). When SCR S₁ is triggered, we have commutation overlap till the load current is transferred from SCR S₅ to SCR S₁ and let the commutation overlap angle be m. During commutation overlap, the current through SCR S₁ rises from zero to load current. At the end of commutation overlap, line current is obtained from the expression in equation (11). During the period of commuation overlap, the voltage source that is seen by the dc link circuit is described as shown in equation (12), where v_OL represents the source during overlap period.

On simplifying equation (12), we get equation (13). During this period, the source appears to have a source inductance equalling 1.5L₂ . We get this value because the path through SCR S₆ contains L₂ whereas the path through both S₁ and S₅ appears to have an equivalent inductance of value equal to 0.5L₂. The current through the load at the instant when SCR S₁ is triggered can be expressed as shown in equation (14).. The values of Z and f in equation (14) are expressed in equation (15).

The constant A in equation (14) is to be evaluated and t₂ = tan (f). At the end of commutation, this current would be equal to the line current. Hence we obtain equation (16). The terms used in equation (17) are defined in equation (18).

Equation (16) is the first of the three equations we need in order to obtain a solution and this equation equates the line current to the load current at the end of commutation. Equation (17) expresses the load current at the end of commutation, using another constant B and B is also to be evaluated. Another equation can be formed as shown in equation (19), which equates the line current at the end of commuation to the load current. Unlike equation (16), equation (19) makes use of constant B. Equations (16) and (19) are two of the three equations required to solve for A, B and d. The third equation is obtained as follows. The load current at the instant when q = p /3 can be computed in two ways, one from equation (16) and the other from equation (19) and these two expressions can be equated to yield equation (20).

From these three equations, the three unknowns, A, B and m, can be obtained. These equations have been used in the program written for simulation.