Post on 27-Jan-2021
transcript
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◦