FULLY-CONTROLLED SINGLE-PHASE SCR BRIDGE RECTIFIER: RL LOAD

CIRCUIT OPERATION
MATHEMATICAL ANALYSIS
PERFORMANCE PARAMETERS
SIMULATION
PSPICE SIMULATION
MATHCAD SIMULATION
MATLAB SIMULATION
SUMMARY 
This chapter describes the operation of a single-phase fully- controlled bridge rectifier circuit with a load consisting of both a resistor and an inductor in series. The analysis in this page is based on the assumption that the SCRs are ideal controlled switches. The main purpose of the fully-controlled bridge rectifier circuit is to provide a variable dc voltage from an ac source.
 
CIRCUIT OPERATION

 

 

The circuit of a single-phase fully-controlled bridge rectifier  circuit is shown in the figure above. The circuit has four SCRs. It is preferable to state  that the circuit has two pairs of SCRs, with S1 and S forming one pair and, S2  and S4  the  other pair.  For this circuit,  vs  is a sinusoidal voltage source. When it is positive, the SCRs S1 and S3 can be  triggered and then current flows from vs through  SCR S1,  load inductor L, load resistor R, SCR S3 and back into the source. In the next half-cycle,   the other  pair of SCRs  conducts.  Even though the direction of current through the source alternates from one half-cycle to the other half-cycle,  the current through the load remains unidirectional.

The main purpose of this circuit is to provide a variable dc output voltage, which is brought about by varying the firing  angle. Let vs = E sin wt, with 0  < wt < 360o. If wt = 30o when S1 and S3 are triggered,  then the firing angle is said to be 30o.  In this instance, the other pair is triggered when wt= 210o.
 

 

When vs changes from a positive to a negative value,  current through the load does not fall to zero value at the instant  wt = p radians, since the load contains an inductor and the SCRs continue to conduct, with the inductor acting as a source. When the current through an inductor is falling, the voltage across it changes sign compared with the sign that exists when its current is rising. When the current through the inductor is falling, its voltage is such that the inductor delivers power to the load resistor, feeds back some power to the ac source under certain conditions and keeps the SCRs in conduction forward-biased.  If the firing angle is less than the load angle, the energy stored in the inductor is sufficient to maintain conduction till the next pair of SCRs is triggered.  When the firing angle is greater than the load angle, the current through the load becomes zero and the conduction through the load becomes discontinuous. Usually the description of this circuit is based on the assumption that the load inductance is sufficiently large to keep the load current continuous and ripple-free.

The operation of the circuit is illustrated by animating the functioning of this circuit. Key in the firing angle in degrees and click the button. The source voltage, and the bridge output voltage are also displayed. It is assumed here that the load inductance is quite large.  The animation is correct only if the firing angle is less than 90o. The programs under simulation section will run correctly for any firing angle.

MATHEMATICAL ANALYSIS

Analysis by hand is based on the assumption that the load inductance is sufficiently large to keep the load current ripple-free. Programs written for computer simulation are not based on this assumption and they simulate the operation based on the parameter keyed-in.

When the load current is continuous, the average value of the output voltage is obtained as follows. Let the supply voltage be vs = E*Sin (q ), where q varies from 0 to 2p radians. Since the output waveform repeats itself for every half-cycle, the average output voltage is expressed in equation (1) as a function of a, the firing angle.

If the load current is continuous, the r.m.s. value of output voltage is obtained as shown in equation (2).

The maximum average output voltage occurs at a firing angle of 0o. Let it be Vom.  Then the ripple factor RF(a) is defined as shown in equation (3).    Equations (4) and (5) apply when the conduction is discontinuous, and the extinction of current occurs b radians after zero-crossing of the source voltage in either direction.  That is, there is no current through in the regions defined by   b <   q   <  a and (p + b) <  (p + q )  < (p + a).
 

The average output voltage, the r.m.s. output voltage and the ripple factor vary with the firing angle as shown below, based on the assumption that the load inductance is large.The plots shown below have been normalized with respect to Vom. For example, when the firing angle is 60o, the average output is shown to be 0.5. It means that the actual average output voltage is 0.5Vom. It can also be seen that when the firing angle is 0o, the r.m.s. output voltage is about 1.1Vom and the ripple factor is about 0.48. The ripple factor increases as the firing angle increases.

If the firing angle is highly retarded or if the load inductance is not sufficiently large, conduction of load current is not continuous. Let us consider the positive half-cycle. When q equals the firing angle, the SCRs S1 and S3 are triggered. Let the load current start from zero at q = a and let the load current fall to zero when q = p + b , before the next pair of SCRs is triggered. It means that b < a . Then the average output voltage is computed as shown in equation (4),  whereas the r.m.s. output voltage is computed as shown in equation (5).
 

The value of b can be found out by iteration, as shown in earlier pages. For the case when the conduction is discontinuous, the plots of the average output voltage, the r.m.s. output voltage and the ripple factor are illustrated below. Key-in the ratio of wL/R, where L is the load inductance, R is the load resistance and w is the angular frequency of the source. In practice this ratio can vary from a low value to about 5. The load angle can be defined to be tan-1(wL/R). When the firing angle is less than the load angle, the conduction is continuous. As explained below, the load current is discontinuous if the firing angle is greater than the load angle.


 

PERFORMANCE PARAMETERS

In this section, the expressions for the load current are first derived. Then an expression for the line current can be obtained. The r.m.s line current, the r.m.s. value of the fundamental of line current, the THD in line current, the frequency spectrum of line current, the frequency spectrum and the ripple content of  bridge output voltage and the frequency spectrum and the ripple content of  voltage across the load resistor are also determined.

For the case of discontinuous conduction, let the load current in each half cycle flow from q = a till q = p + b , where b < a. During this period, the differential equation that applies to the load current is calculated according to equation (6), where a is the firing angle, (p +b) is the extinction angle and the conduction angle within a half-cycle is (p + b - a ). The solution to the above equation is obtained as given shown in equation (7). In equation (7),  t represents wL/R,the ratio of load reactance to load resistance.f is the load angle, defined to be tan-1(wL/R). When the conduction is discontinuous, the load current starts from zero value at q = a . Hence equation (8) is obtained.  When f < a , A has a negative value and we have discontinuous conduction when f < a . It means that the firing of the SCRs is retarded and the firing angle, a is greater than the load angle, f .

When the conduction is continuous, equation (9) applies. The solution to equation (9) is presented as equation (10). Since the signal applied to the load circuit is a periodic signal, the load current is also periodic, after a transient period. It means that equation (11) is valid for continuous conduction.

For continuous conduction, f > a and A is positive. When f = a, the conduction is about to become continuous and the value of A should satisfy both expressions for A. It is seen that when f = a, A = 0 according to both expressions for A. When f = a , the current drawn from the source is sinusoidal and equals (E/Z)*Sin (wt - f), where Z is defined as shown in equation (12).
 


Once the load current is defined for a half-cycle, the expression for line current can be obtained. It equals the load current when SCRs S1 and S3 are conducting and it is the negative of the load current when the other pair of SCRs is conducting. Further analysis can be carried out using Fourier series and the DPF and PF of the line current can be determined.

When the circuit is switched on initially, the load current may settle down to a periodic response after a few output cycles. The time constant of the load is L/R and a time period corresponding to five times the constant should elapse before the load current becomes periodic. For example if the load time constant is 20 ms, a time period of 100 ms should pass before the load current becomes periodic. This time period corresponds to five input voltage cycles at 50 Hz and six input cycles at 60 Hz.

SIMULATION

The plots for voltage have been normalized with respect to Vom, the maximum average output voltage and the plots for line current with respect to Irms,max, which is 0.707E/R. To see the periodic response, key-in firing angle in degrees in the box on the left-hand side and the ratio t in the box on the right-hand side and then click start button. The boxes have default values filled-in.
 

The statistical details related to the output voltage have been normalized with respect to Vom and those related to the source current with respect to Irms,max. The amplitude of each harmonic in the output voltage has been normalized with respect to E which is the amplitude of the source voltage , whereas the amplitude of each harmonics in the source current has been normalized with respect to E/R. The applet to follow displays the statistical details.
 

The applet below displays the transient response.  As the ratio gets larger and larger, the load current requires larger period to elapse before it becomes periodic.
 


 
 

PSPICE SIMULATION
The circuit used for simulation is shown below.

 

 

The program is presented below.

* Full-wave Bridge Rectifier with a resistive load
VIN 1 0 SIN(0 340V 50Hz)
XT1 1 2 5 2 SCR
XT2 0 2 6 2 SCR
XT3 4 0 7 0 SCR
XT4 4 1 8 1 SCR
VP1 5 2 PULSE(0 10 1667U 1N 1N 100U 20M)
VP2 6 2 PULSE(0 10 11667U 1N 1N 100U 20M)
VP3 7 0 PULSE(0 10 1667U 1N 1N 100U 20M)
VP4 8 1 PULSE(0 10 11667U 1N 1N 100U 20M)
L1 2 3 31.8M
R1 3 4 10
R2 1 0 1MEG
R3 2 0 1MEG
R4 4 0 1MEG

* 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
.FOUR 50 V(2,4) I(VIN)
.PROBE
.OPTIONS(ABSTOL=1N RELTOL=.01 VNTOL=1MV)
.END
The responses obtained are shown below.
 
The waveform of voltage at the output of bridge

The waveform of load current

The waveform of voltage across the SCR

The waveform of voltage across the inductor

The waveform of  line current

The average load current

The rms line current

Fourier frequency spectrum of  voltage at the output of bridge

Fourier frequency spectrum of  line current

MATHCAD SIMULATION
You can either download or view the Mathcad file by clicking on the respective image.
MATLAB SIMULATION

 

 

The Matlab program is presented below.

% Program to simulate the full-wave fully-controlled bridge rectifier
% Simulation at a specified firing angle 
% Enter the peak voltage, frequency, inductance L in mH and resistor R  
disp('Typical value for peak voltage is 340 V')
peakV=input('Enter Peak voltage in Volts>');
disp('Typical value for line frequency is 50 Hz')
freq=input('Enter line frequency in Hz>');
disp('Typical value for Load inductance is 31.8 mH')
L=input('Enter Load inductance in mH>');
disp('Typical value for Load Resistance is 10.0 Ohms')
R=input('Enter Load Resistance in Ohms>');
disp('Typical value for Firing angle is 30.0 degree')
fangDeg=input('Enter Firing angle within range 0 to 180 in deg>');
fangRad=fangDeg/180.0*pi;

w=2.0*pi*freq;
X=w*L/1000.0;
if (X<0.001) X=0.001; end;
Z=sqrt(R*R+X*X);
tauInv=R/X;
loadAng=atan(X/R);
k1=peakV/Z;
k2=2.0*k1*sin(loadAng-fangRad)/(1.0-exp(-pi*tauInv));
k3=k1*sin(loadAng-fangRad);
if (fangRad<loadAng)
  A=k2;
  sw=1;
else
  A=k3;
  sw=2;
end;  

Ampavg=0;
AmpRMS=0;

for n=1:360;
  theta=n/180.0*pi;
  X(n)=n;
  if (sw==1);
    if (n<fangDeg);
          cur=k1*sin(pi+theta-loadAng)+A*exp(-tauInv*(pi+theta-fangRad));
          vbr(n)=peakV*sin(theta+pi);
          vind(n)=vbr(n)-R*cur;
          iLoad(n)=cur;
          vSCR(n)=peakV*sin(theta);
          Ampavg=Ampavg+cur*1/360; 
          AmpRMS=AmpRMS+cur*cur*1/360;
        elseif ((n>=fangDeg) & (n<(180+fangDeg)));   
          cur=k1*sin(theta-loadAng)+A*exp(-tauInv*(theta-fangRad));
          vbr(n)=peakV*sin(theta);
          vind(n)=vbr(n)-R*cur;
          iLoad(n)=cur;
          vSCR(n)=0;
          Ampavg=Ampavg+cur*1/360; 
          AmpRMS=AmpRMS+cur*cur*1/360;
        else (n>=(180+fangDeg)); 
      cur=k1*sin(theta-pi-loadAng)+A*exp(-tauInv*(theta-pi-fangRad));
          vbr(n)=peakV*sin(theta-pi);
          vind(n)=vbr(n)-R*cur;
          iLoad(n)=cur;
          vSCR(n)=peakV*sin(theta);
          Ampavg=Ampavg+cur*1/360; 
          AmpRMS=AmpRMS+cur*cur*1/360;  
         end;
  
  else
    if (n<fangDeg);
          cur=k1*sin(pi+theta-loadAng)+A*exp(-tauInv*(pi+theta-fangRad));
          if (cur>0);
            vbr(n)=peakV*sin(theta+pi);
            vind(n)=vbr(n)-R*cur;
            iLoad(n)=cur;
            vSCR(n)=peakV*sin(theta);
            Ampavg=Ampavg+cur*1/360; 
            AmpRMS=AmpRMS+cur*cur*1/360;
          else;
            vbr(n)=0.0;
                vind(n)=0.0;
                iLoad(n)=0.0;
            vSCR(n)=peakV*sin(theta)/2.0;
          end;
          
        elseif ((n>=fangDeg) & (n<(180+fangDeg)));   
          cur=k1*sin(theta-loadAng)+A*exp(-tauInv*(theta-fangRad));
          if (cur>0);
            vbr(n)=peakV*sin(theta);
            vind(n)=vbr(n)-R*cur;
            iLoad(n)=cur;
            vSCR(n)=0;
            Ampavg=Ampavg+cur*1/360; 
            AmpRMS=AmpRMS+cur*cur*1/360;
      else;
            vbr(n)=0.0;
                vind(n)=0.0;
                iLoad(n)=0.0;
            vSCR(n)=peakV*sin(theta)/2.0;
          end;
                        
        else (n>=(180+fangDeg)); 
      cur=k1*sin(theta-pi-loadAng)+A*exp(-tauInv*(theta-pi-fangRad));  
          vbr(n)=peakV*sin(theta-pi);
          vind(n)=vbr(n)-R*cur;
          iLoad(n)=cur;
          vSCR(n)=peakV*sin(theta);
          Ampavg=Ampavg+cur*1/360; 
          AmpRMS=AmpRMS+cur*cur*1/360;  
        end;
  end;
end;  
  
  
plot(X,iLoad)
title('The Load current')
xlabel('degrees')
ylabel('Amps') 
grid
pause

plot(X,vbr)
title('Bridge Output volt')
xlabel('degrees')
ylabel('Volts') 
grid
pause

plot(X,vind)
title('Inductor Voltage')
xlabel('degrees')
ylabel('Volts') 
grid
pause

plot(X,vSCR)
title('SCR Voltage')
xlabel('degrees')
ylabel('Volts') 
grid


AmpRMS=sqrt(AmpRMS);
[C,message]=fopen('fwrl1ph1.dat','w'); 
fprintf(C,'Avg Load Cur=\t%d\tRMS Load Cur=\t%f\n',Ampavg,AmpRMS);
fclose(C);
The responses obtained for the typical specified values have been displayed below.

The results obtained with a firing angle of 60o  have been displayed below.  The other values are the same as for the previous set.  When the firing angle is retarded this much, the load current is discontinuous.


 
 

SUMMARY

This page has described the operation of a fully-controlled bridge rectifier circuit with an RL load.  Even though the circuit has been simulated using Pspice, Mathcad and Matlab,  it is difficult to reach the level of interactive programming that can be achieved with Java or C by using these packages.  Next page describes the operation of this circuit when the source inductance is also taken into account.



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