Half - Wave Rectifier with RL Load

 E = 340 V

 R = 10 W

 wL = 2 x p x 50 x 31.8 x 10^-3   W

 a = atan (wL/R)

 a = 0.785    rad

The value of L is set to be 31.8 mH. The frequency is set to be 50 Hz.  The load angle is calculated for the specified values of R, L and f.

 
 Z = 14.135   W

 A = 17 Amp

The value of the coefficient A has been computed. Now the plots of the source voltage, the current thr ough the diode and the voltage at the  cathode of diode are obtained. A range variable, called q, is created first and it is varied from 0 deg to 360 deg. At each degree,   the current through the diode is computed.
 
The expression for current, Curq , shown above would yield positive values if q is less than the conduction angle and would yield negative values if  q is higher than the conduction angle. The statement below creates an array called i_q and it is set equal to Cur_q , if Cur_q is positive. If Cur_q   is negative, i_q is set equal to zero. The expression within the brackets is similar to arithmetic IF operator in C language. Given

x :=if (expr1,expr2, expr3 ),

x is assigned the value of expr2, if expr1 evalautes to Boolean TRUE value  and x is assigned the value of expr3, if expr1 evalautes to Boolean FALSE value. This arithmetic IF operation is carried out for each value of array  i_q taking into account the corresponding value of Cur_q . That is, both i_q and Cur_q  have the same index, q when evaluation is performed.
 
 

The voltage across the resistor can be computed as shown below.  Next the voltage across the inductor is computed. Again the arithemtic IF operation is carried out. When the diode current is positve, the voltage  across the inductor is the difference between the source voltage and the voltage across the resistor. When the diode current is zero, the voltage across the inductor is also zero. Next the conduction angle is determined. MathCad allows a particular construct.  Assign a guess value for conduction angle. Here it is called b and is assigned a value of p radians. Next form a block staring with Given statement. The equality that should exist is presented below. The program then evaluates the correct value of b for which the equality is true. Then that value of b is assigned to another variable called d in this program.

b := p

Given

d := Find(b)

d  = 3.94  rad

When wt = d, the current through the load becomes zero. The dc value of this current is found out as shown below. Here this current is  expressed as a function of x and this function is integrated over
the period of conduction.

  I_DC = 9.187 Amp
 

The ac source shown in Fig. 1.1 has to supply this dc current. An ac source should not normally be  required to supply a dc current. The  mains ac supply is distributed using transformers and a transformer is  not designed to supply a dc current. It is preferable to avoid using half-wave rectifier circuits.