OPERATION WITH RL LOAD 
CIRCUIT OPERATION
MATHEMATICAL ANALYSIS
SIMULATION

PSPICE SIMULATION
MATHCAD SIMULATION
SUMMARY

CIRCUIT OPERATION
The circuit of the three-phase fully-controlled bridge rectifier circuit with an RL load and a voltage source has been shown above.  The purpose of providing a dc source in the dc link is to illustrate how two-quadrant operation can take place.  When the firing angle is held between 0o and 90o, the circuit operates in the rectifier region.   When the firing angle is held between 90o and 180o, the circuit can operate in the inverter region if the source E is sufficiently negative.  In the inverter region, the power flow is from the dc link to the ac 3-phase source. The inductor reduces the ripple in dc link current.
MATHEMATICAL ANALYSIS

When the current flow in the dc link is continuous, the average output voltage can be calculated as outlined in the previous page.  It is difficult to estimate what the average output would be if  there is a source present in the dc link and the conduction is discontinuous.  If there be no source in the dc link, the average output voltage for discontinuous conduction is expressed by equation (1).  The analysis of the circuit is along the lines described for the single-phase controlled rectifier circuit.  In order to get an expression for the line/phase current, it is necessary to get an expression for the load current.  The expression for load current is obtained from the expression for output voltage.  The output voltage is described by equation (2).
 
The differential equation that describes the load current is expressed by equation (3).  The solution is of the form expressed by equation (4).  The impedance of load is Z and the load angle is f.  They are defined as shown in equation (5).

In the equation above, w is the angular frequency in radians/second corresponding to  the source frequency. When the load current is continuous, equation (6) is valid. Using equation (6),   equation (7) is obtained.  Solving for A, we get equation (8). When the conduction is discontinuous, iL(0) = 0 and then A is evaluated as shown in equation (9).
 


Once the value of A is known, the load current can be found out.  From the load current, an expression for the line current can be obtained.  When SCR S1 is ON,  the line current equals the load current.  The line current is the negative of load current when SCR S4 is ON, and it is zero when neither S1 nor S4 is ON.   From the expression of the load current, the rms value of line current, the rms value of the fundamental component of line current, the THD in line current, the harmonic spectrum of line current, the DPF and the apparent power factor can be determined as outlined in the earlier pages. 
 

SIMULATION

The parameters to be keyed in are the firing angle, the value of source in the dc link (between 1.0 and -1.0 preferably) and the ratio of load reactance to load resistance.  The load reactance is wL, where w is the angular frequency in rad/s corresponding to the ac source frequency. Click on Start Button for the program to respond.

PSPICE SIMULATION

* Three-phase Full-wave Fully-Controlled Bridge Rectifier
VA 1 0 SIN(0 340V 50Hz)
VB 2 0 SIN(0 340V 50Hz 0 0 -120)
VC 3 0 SIN(0 340V 50Hz 0 0 -240)
XT1 1 4 11 4 SCR
XT3 2 4 13 4 SCR
XT5 3 4 15 4 SCR
XT4 5 1 14 1 SCR
XT6 5 2 16 2 SCR
XT2 5 3 12 3 SCR
RP 4 0 100K
RN 5 0 100K
R1 4 6 10
L1 6 5 31.8M

VP1 21 0 PULSE(0 10 3333.3U 1N 1N 100U 20M)
VP2 22 0 PULSE(0 10 6666.7U 1N 1N 100U 20M) 
VP3 23 0 PULSE(0 10 10M 1N 1N 100U 20M) 
VP4 24 0 PULSE(0 10 13333.3U 1N 1N 100U 20M) 
VP5 25 0 PULSE(0 10 16666.7U 1N 1N 100U 20M) 
VP6 26 0 PULSE(0 10 0M 1N 1N 100U 20M) 
RP1 21 0 100K
RP2 22 0 100K
RP3 23 0 100K
RP4 24 0 100K
RP5 25 0 100K
RP6 26 0 100K
EP1 11 4 poly(2) (21,0) (22,0) 0 1 1
EP2 12 3 poly(2) (22,0) (23,0) 0 1 1
EP3 13 4 poly(2) (23,0) (24,0) 0 1 1
EP4 14 1 poly(2) (24,0) (25,0) 0 1 1
EP5 15 4 poly(2) (25,0) (26,0) 0 1 1
EP6 16 2 poly(2) (26,0) (21,0) 0 1 1


* Subcircuit for SCR
.SUBCKT SCR 101 102 103 102
S1 101 105 106 102 SMOD
RG 103 104 50
VX 104 102 DC 0
VY 105 107 DC 0
DT 107 102 DMOD
RT 106 102 1
CT 106 102 10U
F1 102 106 POLY(2) VX VY 0 50 11
.MODEL SMOD VSWITCH(RON=0.0105 ROFF=10E+5 VON=0.5 VOFF=0)
.MODEL DMOD D((IS=2.2E-15 BV=1200 TT=0 CJO=0)
.ENDS SCR


.TRAN 10US 60.0MS 0.0MS 10US
.PROBE
.FOUR 50 V(4,5) I(VA) I(L1)
.OPTIONS(ABSTOL=1N RELTOL=.01 VNTOL=1MV)
.END
The results of the Fourier series analysis are presented below.

The Fourier Series Spectrum of the output voltage



The Fourier Series Spectrum of the Line Current (Phase A)



The Fourier Series Spectrum of the Load Current

It can be seen that the lowest harmonic frequency in the output voltage is at 300 Hz, the sixth harmonic, whereas there is hardly any harmonic present in the load current because of the relatively large inductance in the load circuit.  On the other hand, the 5-th and the 7-th harmonics are visibly high in the line current.

MATHCAD SIMULATION

The Matlab file can be downloaded by clicking on the image below.

SUMMARY

This page has described the operation of the three-phase fully-controlled bridge rectifier circuit with an RL load.  The next page shows the effect of adding source inductance.



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