A DIODE 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 one diode, labelled D1.  This circuit has another diode, marked D2 in the circuit shown above.  This diode is called the free-wheeling diode.   The circuit operation is described next.  The explanation is based on the assumption that the reader knows how the circuit without a free-wheeling diode operates.

 
CIRCUIT OPERATION
 
Let the source voltage vs be defined to be E*sin (wt).   The source voltage is positive when 0 < wt < p  radians and it is negative when p < wt < 2p  radians.  When vs is positive, diode D1 conducts and the voltage vc is positive. This in turn leads to  diode D2 being reverse-biased during this period.  During p < wt < 2p, the voltage vc would be negative if diode D1 tends to conduct. This means that D2 would be forward-biased and would conduct.  When diode D2 conducts, the voltage vc would be zero volts, assuming that the diode drop is negligible.  Additionally when diode D2 conducts, diode D1 remains reverse-biased, because the voltage vs is negative.
When the current through the inductor tends to fall, it starts acting as a source. When the inductor acts as a source, its voltage tends to forward bias diode D2 if the source voltage vs is negative and forward bias diode D1 if the source voltage vs is positive.  Even when the source voltage vs is positive, the inductor current would tend to fall if the source voltage is less than the voltage drop across the load resistor.
During the negative half-cycle of source voltage, diode D1 blocks conduction and diode D2 is forced to conduct.  Since diode D2 allows the inductor current circulate through L, R and D2, diode D2 is called the free-wheeling diode.  We can say that the current free-wheels through D2.
 
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 diode D1 conducts, the driving function for the differential equation is the sinusoidal function defining the source voltage.  During the period defined by  p < wt < 2p, diode D1 blocks current and acts as an open switch.  On the other hand, diode D2 conducts during this period, the driving function can be set to be zero volts.  For  0 < wt < p , the equation (1) shown below applies.  For the negative half-cycle of the source, equation (2) applies. It is the practice to obtain  the solution to a differential equation in two parts. The expressions for the complementary integral and the particular integral are the same, as outlined in the previous page.  The expression for the complementary integral is presented by equation (3).  The particular solution to the equation (1) 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 0 < q  < p, where wt = q,  the total solution is presented as equation (5).

The difference in solution for the circuit with a free-wheeling diode is in how the constant A in complementary integral is evaluated. In the case of the circuit without free-wheeling diode, i(0) = 0, since the current starts building up from zero at the start of every positive half-cycle.  On the other hand, the current-flow is continuous when the circuit contains a free-wheeling diode.  Since the input to the RL circuit is a periodic half-sinusoid 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(0) = i(2p).

Since the current through the load free-wheels during  p < q < 2p , we get equation (6).  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 can get i(p) from equation (7) by substituing q = p. To obtain A, the following steps are necessary.  From the total solution, obtain an expression for i(0) by substituting 0 for q.  Also obtain an expression for i(p) by substituting p for q in equation (7).  Using  this expression for i(p) in equation (6),  obtain i(2p) by letting q = 2p .  Since i(0) = i(2p), we can obtain A from equation (8).  In equation (8), the terms containing constant A are grouped on the left-hand side of equation and the other terms on the right-hand side.

SIMULATION

The operation of the circuit can be simulated as shown below. During 0 < q < p , the expression for current is presented as equation (9).  During p < q < 2p , the expression for current is shown as equation (10).
The voltage across the inductor is obtained to be

vL(q ) = vs(q) - R*i(q) , for 0 < q < p and

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

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 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 and then click on the button next to it.  Do not key-in a NaN.

PSPICE SIMULATION

The circuit that is used for Pspice simulation is shown below.  The nodes are numbered and the components have been labeled.

A Pspice program to simulate the circuit shown above is presented now.

* Half-wave Rectifier with free-wheeling diode and with RL Load
* A problem to find the diode current
VIN 1 0 SIN(0 340V 50Hz)
D1 1 2 DNAME
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)
.TRAN 10US 60.0MS 20.0MS 10US
.PROBE
.OPTIONS(ABSTOL=1N RELTOL=.01 VNTOL=1MV)
.END
 The waveforms obtained are presented now.  Since the program specifies that the waveforms be displayed from the second cycle, there is no output for the first 20 ms.   The waveform of voltage at the cathode of both diodes is shown below.

The waveform of current through diode D1 is presented next.

The waveform of current through diode D2 is shown below.

The waveform of current through load resistor is shown below.  It is the sum of both diode currents.

 

The waveform of voltage across the inductor is shown below.

The advantage with Pspice is the simplicity of the program.  In addition, the devices used are also simulated using the spice models.

MATLAB SIMULATION

A matlab program for simulating the half-wave rectifier with a free-wheeling diode is presented below.
 

% Program to simulate the half-wave 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>');

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);
A1=peakV/Z*sin(loadAng);
A2=peakV/Z*sin(pi-loadAng)*exp(-pi*tauInv);
A3=(A1+A2)/(1-exp(-2.0*pi*tauInv));
Ampavg=0;
AmpRMS=0;

for n=1:360;
  theta=n/180.0*pi;
  X(n)=n;
  if (n<180)
   cur=peakV/Z*sin(theta-loadAng)+A3*exp(-tauInv*theta);
   Ampavg=Ampavg+cur*1/360;
   AmpRMS=AmpRMS+cur*cur*1/360;
  else
   A4=peakV/Z*sin(pi-loadAng)*exp(-(theta-pi)*tauInv);
   cur=A4+A3*exp(-tauInv*theta);
   Ampavg=Ampavg+cur*1/360;
   AmpRMS=AmpRMS+cur*cur*1/360;
  end; 
  
  if (n<180)
    Vind(n)=peakV*sin(theta)-R*cur;
    Vout(n)=peakV*sin(theta); 
        diode2cur(n)=0;
        diode1cur(n)=cur;
  else
     Vind(n)=-R*cur;
         Vout(n)=0; 
         diode2cur(n)=cur;
         diode1cur(n)=0;
  end
  
  iLoad(n)=cur;
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,diode1cur)
title('Diode 1 current')
xlabel('degrees')
ylabel('Amps') 
grid
pause

plot(X,diode2cur)
title('Diode 2 current')
xlabel('degrees')
ylabel('Amps') 
grid

AmpRMS=sqrt(AmpRMS);
[A,message]=fopen('outhfr2.dat','w'); 
fprintf(A,'Avg Load Cur=\t%d\tRMS Load Cur=\t%f\n',Ampavg,AmpRMS);
fclose(A)
The responses obtained for the typical specified values are shown below.
The average and rms values of load current are presented below.
Avg Load Cur=   1.082254e+001   RMS Load Cur=   13.954542
MATHCAD SIMULATION

This circuit can be simulated using MathCad as shown next.  Click on DOWNLOAD to download the file containing the program and then open it using MathCad.   Click on VIEW ONLY to view this file in HTML format.


 

SUMMARY

It has been shown how the half-wave rectifier with a free-wheeling diode can be simulated using different
software packages.  The next page shows how a half-wave controlled rectifier circuit operates.

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