Electronic I Lecture 3

Diode Rectifiers

By

Asst. Prof Dr. Jassim K. Hmood

Diode Approximations 1- The Ideal Model

• When forward biased, act as a closed (on) switch

• When reverse biased, act as open (off) switch

Diode Approximations

This model neglects the effect of the barrier potential, the internal resistance, and other parameters

2. The Barrier Potential Model • The forward biased diode is represented as a closed

switch in series with a small ‘battery’ equal to the barrier potential VB (0.7 V for Si and 0.3 V for Ge)

• The positive end of the equivalent battery is toward the anode.

• This barrier potential cannot be measured by using a multimeter, but it has the effect of a battery when forward bias is applied.

• The reverse biased diode is represented by an open switch, because barrier potential does not affect reverse bias.

Diode Approximations

Diode Approximations

3. The Complete Diode Model

• The forward biased diode model with both the barrier potential and low forward (bulk) resistance ( r’d )

Diode Approximations

HALF-WAVE RECTIFIER CIRCUITS

• The basic rectifier circuit converts an ac voltage to a pulsating dc voltage.

• A filter is then added to eliminate the ac components of the waveform and produce a nearly constant dc voltage output.

HALF-WAVE RECTIFIER WITH RESISTOR LOAD

• The simplest single-phase diode rectifier is the single-phase half-wave rectifier.

• The circuit consists of only one diode that is usually fed with a transformer secondary.

• During the positive half-cycle of the transformer secondary voltage, diode conducts.

• During the negative half-cycle, diode stops conducting.

HALF-WAVE RECTIFIER WITH RESISTOR LOAD

• The average value of output voltage Vdc is defined as:

• The average current Idc is :

• The root-mean-square (rms) value of output voltage vo is Vo, which is defined as:

• And

• The rectification efficiency is

0.318p

dc p

VV V

2

p

o

VV

dc Pdc

V VI

R R

2

2

dcdc

in o d

I RP

P I r R

2

poo

VVI

R R

HALF-WAVE RECTIFIER WITH RESISTOR LOAD

By Substituting and simplifying, we get on:

If diode resistance rd is neglected, then

• PIV It is the maximum voltage across the diode in the reverse direction. PIV for diode in half wave rectifier is Vp

0.46 46%or

HALF-WAVE RECTIFIER WITH RESISTOR LOAD

0.46

1 dr

R

Half-wave rectifier output voltage with VP = 10 V and Von = 0.7 V.

1- For this case, the output voltage is one diode-drop smaller than the input voltage during the conduction interval:

2- The output voltage remains zero during the off-state interval. The input and output waveforms for the half-wave rectifier, including the effect of Von, are shown in the figure for VP = 10 V and Von = 0.7V.

HALF-WAVE RECTIFIER WITH RESISTOR LOAD

RECTIFIER with FILTER CAPACITOR

• The unfiltered output of the half-wave rectifier is not suitable for operation of most electronic circuits because constant power supply voltages are required to establish proper bias for the electronic devices.

• A filter capacitor can be added to filter the output of the rectifier circuit to remove the time-varying components from the waveform.

• A load must be connected to the circuit as represented by the resistor R.

• Now there is a path available to discharge the capacitor during the time the diode is not conducting.

RECTIFIER with FILTER CAPACITOR

• The output voltage is no longer constant as in the ideal peak-detector circuit but has a ripple voltage Vr . In addition, the diode only conducts for a short timeT during each cycle. This time DT is called the conduction interval, and its angular equivalent is the conduction angle qC where qC = ωT .

• the ripple voltage:

• The conduction angle and conduction interval are:

• The PIV of diode with capacitor filter is:

RECTIFIER with FILTER CAPACITOR

Half-wave diode rectifier

• Example: Find the value of the dc output voltage, dc output current, ripple voltage, conduction interval, and conduction angle for a half-wave rectifier driven from a transformer having a secondary voltage of 12.6 Vrms (60 Hz) with R = 15W and C = 25,000mF. Assume the diode on-voltage Von = 1 V.

• Solution

The ideal dc output voltage in the absence of ripple is

The nominal dc current delivered by the supply is

• The ripple voltage is

• The conduction angle is

• and the conduction interval is

Half-wave diode rectifier

Home work

• H.W4: Find the value of the dc output voltage, dc output current, ripple voltage, conduction interval, and conduction angle for a half-wave rectifier that is being supplied from a transformer having a secondary voltage of 6.3 Vrms (60 Hz) with R = 0.5W and C = 500,000mF. Assume the diode on voltage Von = 1 V.

• Answers: 7.91 V; 15.8 A; 0.527; 0.912 ms; 19.7◦