Recall-Lecture 4

DC Analysis
 Representation of diode into three models



Ideal case – model 1 with V = 0
Piecewise linear model 2 with V has a given value
Piecewise linear model 3 with V and forward resistance, rf
• Diode AC equivalent model
– During analysis, must perform DC analysis first to
calculate ID in order to obtain rd
DC ANALYSIS
DIODE = MODEL 1 ,2
OR 3
CALCULATE DC
CURRENT, ID
AC ANALYSIS
CALCULATE
rd
DIODE = RESISTOR,
rd
CALCULATE AC
CURRENT, id
BREAKDOWN VOLTAGE
The
breakdown voltage is a function of the doping concentrations
in the n- and p-region of the pn junction.
Large doping concentrations result in smaller break-down voltage.
Reverse biased voltage – ET 
The electric field may become large enough for the covalent bond
to break, causing electron-hole pairs to be created.
So, electrons from p-type are swept to n-region by the electric
field and holes from the n-type are swept to the p-region
The movement will create reverse biased current known as the
Zener Effect.
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ECE 1231
Zener Effect and Zener Diode

The applied reverse biased voltage cannot increase without limit since at
some point breakdown occurs causing current to increase rapidly.

The voltage at that point is known as the breakdown voltage, VZ

Diodes are fabricated with a specifically design breakdown voltage and are
designed to operate in the breakdown region are called Zener diodes. Circuit
symbol of the Zener diode:
NOTE: When a Zener diode is reversebiased, it acts at the breakdown region,
when it is forward biased, it acts like a
normal PN junction diode

Such a diode can be used as a constant-voltage
reference in a circuit.

The large current that may exist at breakdown
can cause heating effects and catastrophic failure
of the diode due to the large power dissipated in
the device.

Diodes can be operated in the breakdown region
by limiting the current to a value within the
capacities of the device.
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ECE 1231

Avalanche Effect


While these carriers crossing the space-charge
region, they also gain enough kinetic energy.
Hence, during collision with other atoms, covalent
bond is broken and more electron-holes pairs are
created, and they contribute to the collision process
as well. Refer to figure below
Electron with
high kinetic
energy
e
© Electronics
e
atom
h
e
atom
h
eh
atom
ECE 1231
Other Types of Diodes
Photodiode
The term ‘photo’ means light. Hence, a photodiode converts optical energy
into electrical energy. The photon energy breaks covalent bond inside the
crystal and generate electron and hole pairs
Solar Cell
Solar cell converts visible light into
electrical energy. The working
principle is the same as photodiode
but it is more towards PROVIDING the
power supply for external uses
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ECE 1231
Light Emitting Diode
An LED is opposite of photodiode this time, it converts electrical energy into light
energy – Normally GaAs is used as the material for LED. During diffusion of
carriers – some of them recombines and the recombination emits light waves.
Schottky Barrier Diode
A Schottky Barrier diode is a metal
semiconductor junction diode. The
metal side is the anode while the ntype is the cathode. But the turn on
voltage for Schottky is normally
smaller than normal pn junction
diode
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ECE 1231
Zener Diode
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ECE 1231
Chapter 3
Diode Circuits
Voltage Regulator
Voltage Regulator - Zener Diode
A voltage regulator supplies constant voltage to a load.

The breakdown voltage of a Zener
diode is nearly constant over a wide
range of reverse-bias currents.

This make the Zener diode useful in
a voltage regulator, or a constantvoltage reference circuit.
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
Example
A Zener diode is connected in a voltage regulator circuit. It is given that VPS = 20V, the Zener
voltage, VZ = 10V, Ri = 222  and PZ(max) = 400 mW.
a. Determine the values of IL, IZ and II if RL = 380 .
b. Determine the value of RL that will establish PZ(max) = 400 mW in the diode.
For proper function the circuit the following conditions must be 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
For general thumb of rule for design this circuit is, so from the last Equation
Maximum power dispassion in the Zener diode is
EXAMPLE 1 (Example 3.3 from textbook)
Design a Zener diode voltage regulator circuit as shown
in Figure . 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).
simple Zener diode voltage
regulator circuit
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
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,
Vo = Vs – V
i
Vp
V
V
Equation of VO and current when diode is conducting
vD
• Vs< V, diode off, open circuit, no current flow,Vo = 0V
• Vs> V, diode conducts,current flows and Vo = Vs – V
Vs = Vpsin t
Vp
Notice that the
peak voltage of Vo
is lower
V
Vs >V