Half-wave and Full-wave Diode Rectifiers

by Tory W. Vaughan III, BEE Georgia Institute of Technology

NOTE: The response is found for both a half-wave and full-wave rectifier circuit with a single capacitor filter.

Introduction

This worksheet calculates the response of both a half-wave and a full-wave diode rectifier circuit with a single capacitor filter, using the power of Maple to solve the non-linear differential equations.We will assume that the input to the circuits is a 60 Hz sine wave.

> restart:

The Ideal Diode

Define some constants and variables for the diode which we will assume to be silicon.

> degreesC := 23:

> T := degreesC+273: The absolute temperature in degrees Kelvin

> N := 2: A constant which for silicon is 2

> K := 1.38*10^(-23): Boltzmann's constant

> Q := 1.6*10^(-19): The charge on an electron

> vt := K*T/Q;

The ideal diode equation can be used to define the diode current.

>

We plot a graph of the diode current versus voltage.

> plot(id, V=0..1, thickness=2, labels=["V (volts)","I (amps)"],
title=" Ideal Silicon Diode Characteristics");

The Input Voltage

The input voltage to the circuit is the secondary voltage of a transformer.In this example we will use a value of 40*sin(377*t) volts.

We plot the input voltage.

> Vinput := 40*sin(377*t):

> plot(Vinput, t=0..5*10^(-2), labels=["Time (secs)","V (volts)"],
title=" Input Voltage vs. Time");

Half-Wave Rectifier

We will first calculate the output voltage of a half-wave rectifier circuit with a single capacitor filter using numerical techniques. We write the circuit equation using one node equation as follows, where i1 is the diode current.

>

We now replace the diode voltage V by its' circuit equivalent - the source voltage minus the output voltage.

> id:=subs(V=40*sin(377*t)-Vt(t),id);

We check to make sure id has its' new value.

> id;

Replace i1 in the circuit equation by id.

> eq1:=subs(i1=id, eq1);

This non-linear first order differential equation, eq1, describes the circuit.

Entering the value for C (farads) & RL (ohms)

> C:=2200*10^(-6): RL:=20:

We now solve the differential equation subject to the initial condition Vt=0 at t=0 and using a step size of 10 micro-seconds:

> ff := dsolve({eq1, Vt(0)=0}, Vt(t), type=numeric, method=classical, start=0, stepsize=.00001);

and plot the ouput voltage.

> odeplot(ff, [t,Vt(t)], 0..0.05, axes=boxed, labels=["Time","Voltage"],
title=" Output Voltage of Half-Wave Rectifier");

Full-Wave Rectifier

The full-wave rectifier circuit with a single capacitor filter is identical to the half-wave case except that the circuit equation is different, since there are two diodes in the equation. We begin the analysis by unassigning some of the variables.

> C:='C': RL:='RL': Vt(t):='Vt(t)': Vt:='Vt': i1:='i1': id:='id':

Since the input voltage is taken across a grounded center-tapped transformer, the circuit equation eq2 becomes:

>

The ideal diode equation is the same as before:

>

We replace each diode voltage by its' circuit equivalent:

> i1:=subs(V=40*sin(377*t)-Vt(t),id);

> i2:=subs(V=-40*sin(377*t)-Vt(t),id);

Check to make sure the circuit equation has the new values of diode voltages in it.

> eq2;

Enter the value of circuit components as before.

> C:=2200*10^(-6): RL:=20:

We now solve for the differential of output voltage with respect to time.

> ff2 := dsolve({eq2, Vt(0)=0}, Vt(t), type=numeric, method=classical, start=0, stepsize=.00001);

and plot the ouput voltage.

> odeplot(ff2, [t,Vt(t)], 0..0.05, axes=boxed, labels=["Time","Voltage"],
title=" Output Voltage of Full-Wave Rectifier");

Conclusion

In this worksheet we calculated the response of both a half-wave and a full-wave rectifier circuit with a single capacitor filter using the power of Maple's "dsolve" routines to solve non-linear differential equations. It is interesting to change the values of temperature, capacitance, and load resistance to see the effects on the output voltage.

Disclaimer: While every effort has been made to validate the solutions in this worksheet, Waterloo Maple Inc. and the contributors are not responsible for any errors contained and are not liable for any damages resulting from the use of this material.