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. They
are:
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 .
LoadVolt1n := if (n<deg, VP1n, 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.
Next the ripple factor in output voltage of the bridge is computed. This ripple factor
should be much greater than the ripple factor for the load current, since the
load inductance acts as a filter for harmonic components.
Function for line current during overlap
Verification
Next THD in line current is found out. Find the Fundamental trignometric Fourier
Series components.
The THD in line current is obtained as shown above.
