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Recall-Lecture 6
• Zener effect and Zener diode
– When a Zener diode is reverse-biased, it acts at
the breakdown region, when it is forward
biased, it acts like a normal PN junction diode
• Avalanche Effect
– Gain kinetic energy – hit another atom –
produce electron and hole pair
• Voltage Regulator using Zener Diode
3. The remainder of VPS
drops across Ri
1. The Zener diode holds the
voltage constant regardless of
the current
2. The load
resistor sees
a constant
voltage
regardless of
the current
Voltage Regulator
For proper function the circuit must satisfied the following conditions.
1. The power dissipation in the Zener diode is less than the rated value
2. When the power supply is a minimum, VPS(min), there must be minimum
current in the Zener diode IZ(min), hence the load current is a maximum,
IL(max),
3. When the power supply is a maximum, VPS(max), the current in the diode is a
maximum, IZ(max), hence the load current is a minimum, IL(min)
AND
Or, we can write
Considering designing this circuit by substituting IZ(min) = 0.1 IZ(max),
now the last Equation becomes:
Maximum power dispassion in the Zener diode is
EXAMPLE 1
Consider voltage regulator is used to power
the cell phone at 2.5 V from the lithium ion
battery, which voltage may vary between 3
and 3.6 V. The current in the phone will vary 0
(off) to 100 mA(when talking). Calculate the
value of Ri and the Zener diode power
dissipation
Solution:
The stabilized voltage VL = 2.5 V, so the Zener diode voltage must be VZ = 2.5 V.
The maximum Zener diode current is
The maximum power dispassion in the Zener diode is
The value of the current limiting resistance is
• Example 2
Range of VPS : 10V– 14V
RL = 20 – 100 
VZ = 5.6V
Find value of Ri and calculate the maximum power rating of the diode
Rectifier
Rectifier Circuits

A DC power supply is required to bias all electronic circuits.

A diode rectifier forms the first stage of a dc power supply.
Diagram of an Electronic Power Supply

Rectification is the process of converting an alternating (ac) voltage
into one that is limited to one polarity.

Rectification is classified as half-wave or full-wave rectifier.
Rectifier Parameters
Relationship between the number of turns of a
step-down transformer and the input/output
voltages
𝑉𝑃
𝑉𝑆
=
𝑁1
𝑁2
The peak inverse voltage (PIV) of the diode is the peak value of the voltage
that a diode can withstand when it is reversed biased
Duty Cycle: The fraction of the wave cycle over which the diode is
conducting.
•
vs< V, diode off, open circuit, no
current flow, Vo = 0V
vs > V, diode conducts, current flows,
v = vs – V
•
o
V
Equation of VO and current when diode is conducting
vs < V, diode off, open circuit, no current flow, vo = 0V
• vs > V, diode conducts, current flows and vo = vs – V
•
Consider a sine wave where
vs = v
m sin
t and
v
m
is the
peak value
v
m
Notice that the
peak voltage of Vo
is lower
V
vs > V
Example
Consider the rectifier circuit in the figure below. Let R = 1 k, and the diode
has the properties of V = 0.6 V and rf = 20 .
Assume
i.
ii.
v = 10 sin t (V)
s
Determine the peak value of the diode current
Sketch vO versus time, t. Label the peak value of vO.
v
s
Solution
vO
vs > V
FULL WAVE RECTIFIER
• Center-Tapped
• Bridge
Full-Wave Rectification – circuit
with center-tapped transformer

Positive cycle, D2 off, D1 conducts;
vo– vs + V = 0
vo = vs - V
 Negative cycle, D1 off, D2 conducts;
vo– vs + V = 0
vo = vs - V

Since a rectified output voltage occurs during
both positive and negative cycles of the input
signal, this circuit is called a full-wave rectifier.

Also notice that the polarity of the output
voltage for both cycles is the same
v =v
s
m
sin t
vm
V
-V
Notice again that the peak voltage of Vo is lower
since vo = vs - V
• vs < V, diode off, open circuit, no current flow,
vo = 0V
Full-Wave Rectification –Bridge Rectifier

Positive cycle, D1 and D2 conducts, D3 and D4 off;
V + vo + V – vs = 0
vo = vs - 2V

Negative cycle, D3 and D4 conducts, D1 and D2 off
V + vo + V – vs = 0
vo = vs - 2V
Also notice that the polarity of the output voltage for both cycles is the same
• A full-wave center-tapped rectifier circuit is shown in the figure below.
Assume that for each diode, the cut-in voltage, V = 0.6V and the diode
forward resistance, rf is 15. The load resistor, R = 95 . Determine:
i.
peak output voltage, vo across the load, R
ii. Sketch the output voltage, vo and label its peak value.
25: 1
125 V (peak
voltage)
( sine wave )
• SOLUTION
i.
peak output voltage, Vo
vs (peak) = 125 / 25 = 5V
V +ID(15) + ID (95) - vs (peak) = 0
ID = (5 – 0.6) / 110 = 0.04 A
vo (peak) = 95 x 0.04 = 3.8V
ii.
3.8V
V
-V