Power Electronics Interactive Text

Chapter 4
Fully Controlled 1 - PH SCR Bridge Rectifier

Section 3
Operation with Source Inductance

A Single - Phase Fully Controlled Bridge Rectifier Circuit With Source Inductance

This program describes the operation of a single-phase fully-controlled rectifier circuit with a source inductance. The presence of a source inductance introduces commutation overlap before the load current gets switched from one pair of SCRs to the other. The effect of source inductance can be seen in the output voltage waveform , given that the load current is continuous. The load current is continuous if the firing angle is less than the load angle. Let L = L1 + L2. The load angle f is then tan -1 (wL/R). The program given below works when a < f. Let t = (wL1)/R and a the firing angle, a. When a > f , the load current is discontinuous and the behaviour of the circuit is similar to the circuit without source inductance operating in the discontinuous mode. The peak source voltage is assumed to be unity and the current E/R is also assumed to be unity.

The solution is slightly difficult. Four unknown values are to be computed. Theyare:

i. the load current at wt = a, called A ,

ii. the instant b when the commutation overlap ends,

iii. the load current at wt = b, called B , and

iv. the coefficient for the exponential term, K.

The equations are formed as follows.

Let the load current be i(a) at the instant of triggering and let it be i(b) at wt = b.

Then the supply current changes from i(a) to -i(b) when wt varies from a to b.

During this period, the load current varies from i(a) to i(b) exponentially, with

the time constant in radians being t. The third equation is based on the periodic

nature of the load current. Since the load current repeats itself every p radians,

i(a) = i(p+a). Another expression can be formed for the load current i(b) using

the source voltage and the coefficient for the exponential term.

For solving the problem, the SOLVE BLOCK facility within MathCad is used. First, the guess values of the variables to be solved for are assigned. Then in the block below, the four constraints for solving are stated in the form of equations, with the equality sign created using by pressing CONTROL and = keys simultaneously. The program yields the solution as an array. The solution technique is elegant. The first equation equates the change in load current due to source voltage being applied across it directly from a to b . The second equation finds the load current at b from its value at a, based on the exponential decay. From the equation that would define the load current from b till (p + a), two equations can be obtained. At wt = b , the exponential part would equal K. At wt = p + a, the current would have decayed.

Now the plots of load current, line current and output voltage of the bridge can be obtained. Define a range variable, n, to correspond to the degrees within a cycle of source voltage. Obtain the angle qn in radians.

The equation presented below computes the load current when wt < a. During this part, the solution is obtained by equating wt = p + qn, and the elapsed angle for exponential decay is (p + qn - b)

When a < wt < b, the load current decays exponentially. Note that the load current at wt = a is A.

The expression below computes the load current when b < wt < ( p + a).

Now a single expression for the load current for half-a-cycle can be obtained.

From the expression for load current over half-a-cycle, an expression for the load current over a whole cycle is obtained.

Next an expression for load current is developed. At instants outside the overlap region, the line current has the same amplitude as the load current. It has the same sign when SCRs S1 and S4 conduct and has the opposite sign when SCRs S2 and S3 conduct. During overlap period, its value is different from that of the load current.

The Plot of Load Current

The Plot of Line Current

Getting an expression for the Load Voltage

At instants outside the overlap period, the bridge output voltage is the drop across the load resistor plus the voltage across the load inductance. Now two expressions are needed, one for wt < a and another for b < wt < p .

LoadVolt1_n := if (n<deg, VP1_n, 0.0)

The Plot of Load Voltage

Calculating the reduction in output voltage due to commutation overlap

During commutation over lap, the output voltage is zero. All the four SCRs are in conduction, with the current through the incoming SCRs rising from zero to load current level and the current through the outgoing SCRs falling from the load current level to zero. There is a slight reduction in the output voltage. Let the commutation overlap in degrees be m. Let the average voltage with no source inductance be VavNoOL and the average voltage with overlap be VAvgWOL.

Commutation overlap angle is obtained as follows.

deg

Next the output voltage without overlap is obtained. It is assumed that there is no source inductance.

To obtain, the actual output voltage given a peak voltage, multiply the above value by the peak value of the source.

From the bridge output voltage during the period when there is no overlap, the average value of bridge voltage can be obtained. During the overlap period, the bridge output voltage is zero.

To obtain, the actual output voltage given a peak voltage, multiply the above value by the peak value of the source.

Fractional reduction in output Voltage, FRed, is

The unit value corresponds to E, the peak value of source voltage.

Verification

The average load current should have the same value as the average bridge output voltage, since the average voltage across the load inductor is zero.

To obtain, the actual current given a peak voltage, multiply the above value by E/R, where E is the peak value of the source and R the load resistance.