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 V 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");

Warning, expecting only range variable V in expression id to be plotted but found name id

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);

Error, (in dsolve/numeric/process_input) system must be entered as a set/list of expressions/equations

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);

Error, (in dsolve/numeric/process_input) system must be entered as a set/list of expressions/equations

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");