A SINLGE-SCR CIRCUIT WITH A FREE-WHEELING DIODE

 CIRCUIT DIAGRAM
CIRCUIT OPERATION
MATHEMATICAL ANALYSIS
SIMULATION
PSPICE SIMULATION
MATLAB SIMULATION
MATHCAD SIMULATION
SUMMARY 
CIRCUIT DIAGRAM
The circuit shown above differs from the circuit described in the previous page, which had only a single SCR.  This circuit has a free-wheeling diode in addition, marked D in the circuit shown above.  The circuit operation is described next.  The explanation is based on the assumption that both the diode and the SCR are ideal.  It means that the voltage drop across the device while in conduction is zero and the leakage current in the blocking state is zero.

CIRCUIT OPERATION

The source vs is an alternating sinusoidal source.  If vs = E * sin (wt), vs is positive when 0 < wt  < p,  and it is negative when p < wt <2p.  When vs starts becoming positive, the SCR is forward-biased but remains in the blocking state till it is triggered.  If the SCR is triggered at when wt = a, then a is called the firing angle.  When the SCR is triggered in the forward-bias state, it starts conducting and the positive source keeps the SCR in conduction till wt reaches  p radians.  At that instant, the current through the circuit is not zero and there is some energy stored in the inductor at wt =  p radians. In the absence of the free wheeling diode, the inductor would keep the SCR in conduction for part of the negative cycle till the energy stored in it is discharged. But when a free wheeling diode is present as shown in the circuit, the current has a path that offers almost zero resistance.  Hence the inductor discharges its energy during  p < wt < (2p + a)  through the free wheeling diode.  When there is a free wheeling diode, the current through the load tends to be continuous, at least under ideal conditions.  When the diode  conducts, the SCR remains reverse-biased, because the voltage vs is negative.
 
MATHEMATICAL ANALYSIS
An expression for the current through the load can be obtained as shown below.   It can be assumed that the load current flows all the time.  In other words, the load current is continuous.  When the SCR  conducts, the driving function for the differential equation is the sinusoidal function defining the source voltage.  During the period defined by  p < wt < (2p + a), the  SCR blocks current and acts as an open switch.  On the other hand, the free wheeling diode  conducts during this period, and the driving function can be set to be zero volts.  For  a < wt < p ,  equation (1) applies whereas equation (2) applies for the rest of the cycle.
 
As in the previous cases, the solution is obtained in two parts. The expressions for the complementary integral and the particular integral are the same.  The expression for the complementary integral is presented as equation (3). The particular solution is the steady-state response and is presented as equation (4).

The total solution is the sum of both the complimentary and the particular solution for a < wt < p .

The difference in solution is in how the constant A in complementary integral is evaluated. In the case of the circuit without free-wheeling diode, i(a) = 0, since the current starts building up from zero when the SCR is triggered during the positive half-cycle.  On the other hand, the current-flow is continuous, when there is a free wheeling diode.  Since the input to the RL circuit is a periodic function, we expect that the response of the circuit should also be periodic.  That means, the current through the load is periodic.  It means that

i(a) = i(2p + a).

Since the current through the load free-wheels during  < q < (2p + a) , we get that,

i(q) = i(p) *exp[-(q - p)/t] , where t  = wL/R.

We use ( q - p ) for the elapsed period in radians instead of q itself, since the free-wheeling action starts at  q = p .  From the total solution, we get i(p).

i(p) = A*exp(- p/t)   + (E/Z)*sin (p - a).

To obtain A, the following steps are necessary.  From the total solution, obtain an expression for i(a) by substituting a for q.  From the expression for the free-wheeling period,  obtain i(2p + a) by letting q = (2p + a) .  Since i(a) = i(2p + a), we can obtain A to be:


 

SIMULATION

The operation of the circuit can be simulated as shown below. During 0 < q < p , the expression for current is:
i(q) = A * exp [ -(q - a)/t) + (E/Z)*sin (q - b ),
where  t  = wL/R,   b = atan (t),   Z2 = R2 + (wL)2 = R2*(1 + t2and a is the firing angle.
During p < q < (2p + a) , the expression for current is:
i(q) = A*exp {- (q - a) /t}   + (E/Z)*sin (p - b)*exp {- (q - p) /t} .

The voltage across the inductor is obtained to be

vL(q ) = vs(q) - R*i(q) , for a < q < p and

           =  - R*i(q) , for  p < q < (2p + a) .

vs(q)  = E*sin (q).

It is preferable to normalize vs with respect to E and the current with respect to E/R and then the only value that is required to be known for solving for the current is  t.  The applet shown below simulates this circuit.  You have to key-in the ratio t in the text-field to the left of the button and the firing angle in degrees in the text-field to the right of the button and  then click on the button next to it.  Do not key-in a NaN.


 
 

PSPICE SIMULATION
The circuit used for Pspice simulation is shown below.
The Pspice program is presented next.
* Half-wave Rectifier with a free-wheeling diode
* A problem to find the SCR current
VIN 1 0 SIN(0 340V 50Hz)
XT1 1 2 5 2 SCR
VP 5 2 PULSE(0 10 1667U 1N 1N 100U 20M)
L1 2 3 31.8MH
R1 3 0 10
D2 0 2 DNAME
.MODEL DNAME D(IS=10N N=1 BV=1200 IBV=10E-3 VJ=0.6)
* 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 20.0MS 10US
.PROBE
.OPTIONS(ABSTOL=1N RELTOL=.01 VNTOL=1MV)
.END

The waveform of voltage at the cathode of SCR



The waveform of current through the load



The waveform of voltage across the SCR


The waveform of current through the SCR


The waveform of current through the free-wheeling diode


The waveform of voltage across the inductor



 
MATLAB SIMULATION
The Matlab program is presented below.
% Program to simulate the half-wave controlled rectifier circuit
% The circuit has a free-wheeling diode
% 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=exp(R*(pi+fangRad)/X);
k2=exp(R*(fangRad-pi)/X);
A=peakV/Z*(sin(pi-loadAng)+sin(loadAng-fangRad)*k1)/(k1-k2);

Ampavg=0;
AmpRMS=0;

Cur180=peakV/Z*sin(pi-loadAng)+A*k2;

for n=1:360;
  theta=n/180.0*pi;
  X(n)=n;
  if (n<fangDeg)
    cur=Cur180*exp(-(pi+theta)*tauInv);
        Vind(n)=-R*cur;
        iLoad(n)=cur;
    Vout(n)=0;
        VSCR(n)=peakV*sin(theta);
        curSCR(n)=0;
        diodecur(n)=cur;
        Ampavg=Ampavg+cur*1/360; 
        AmpRMS=AmpRMS+cur*cur*1/360;
  elseif ((n>=fangDeg) & (n<180)) 
    cur=peakV/Z*sin(theta-loadAng)+A*exp(-tauInv*(theta-fangRad));
    Ampavg=Ampavg+cur*1/360; 
        AmpRMS=AmpRMS+cur*cur*1/360;
        Vind(n)=peakV*sin(theta)-R*cur;
        iLoad(n)=cur;
    Vout(n)=peakV*sin(theta);   
    VSCR(n)=0;
    curSCR(n)=cur;
        diodecur(n)=0;
  else
    cur=Cur180*exp((pi-theta)*tauInv);
    Vind(n)=-R*cur;
        iLoad(n)=cur;
        Vout(n)=0;
    VSCR(n)=peakV*sin(theta);
    curSCR(n)=0;
        diodecur(n)=cur;
        Ampavg=Ampavg+cur*1/360; 
        AmpRMS=AmpRMS+cur*cur*1/360;
  end;
end;  
  
  
plot(X,iLoad)
title('The Load current')
xlabel('degrees')
ylabel('Amps') 
grid
pause

plot(X,Vout)
title('Voltage at cathode')
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
pause

plot(X,curSCR)
title('SCR Current')


xlabel('degrees')
ylabel('Amps') 
grid
pause

plot(X,diodecur)
title('diode Current')
xlabel('degrees')
ylabel('Amps') 
grid

AmpRMS=sqrt(AmpRMS);
[C,message]=fopen('hwavec2.dat','w'); 
fprintf(C,'Avg Load Cur=\t%d\tRMS Load Cur=\t%f\n',Ampavg,AmpRMS);
fclose(C)
The output file is re-produced below.
Avg Load Cur=   1.009749e+001   RMS Load Cur=   13.337142
Next the plots obtained for the typical specified values are displayed.


 


 
 
 
 

MATHCAD SIMULATION
The Mathcad program can be either downloaded or viewed in HTML format.

 

 

SUMMARY

This page has described the operation of a half-wave rectifier wirh a free-wheeling diode.  Next we take up the study of single-phase, full-wave, fully-controlled rectifier.

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