1.1
Special Diodes
PN junction diodes are the most common type of devices. They are
used in power supplies to convert ac voltage to dc voltage. But this process
called rectification is not all that a diode can do. Now we will discuss diodes
used in other applications. This session begins with the Zener diode, which is
optimized for its breakdown properties.Zener diodes are very important
because they are ket to voltage regulation. This session also covers
optoelectronic diodes, Schottky diodes, varactors and other diodes.
1.1.1 The Zener Diode
Small-signal and rectfier diodes are never intentionally operated in
the breakdown region because this may damege them. A Zener diode is
different;it is a silicon diode that the manufacturerhas optimized for
operation in the breakdown region. The Zener diode is the backbone of
voltage regulators, circuits that hold the load voltage almost constant
despite large changes in line voltage and load resistance.
Figure. 1.46 Zener diode. (a) Schamatic symbol (b) alternative symbol; (c) graph of current versus voltage
I-V Graph
Figure 1.46(a) shows a schematic symbol of a Zener diode; 1.46(b) is an
alternative symbol. In either symbol, the lines resemble a Z, which stands for
“Zener”. By varying the doping level of silicon diodes, a manufacturer can
produce Zener diodes which breakdown voltages from about 2 to over 1000 V.
These diodes can operate in any of three regions: forward, leakage and
breakdown.
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Figure 1.46(c) shows the I-V graph of a Zener diode. In the forward
region, it starts conducting around 0.7 V, just like an ordinary silicon diode. In
the leakage region between zero and breakdown, it has only a small reverse
current. In a Zener diode, the breakdown has a very sharp knee, followed by an
almost vertical increase in current. Note that the voltage is almost constant,
approximately equal to Vz over most of the breakdown region. Data sheets
usually specify the value of Vz at a particular test current IZT.
Figure 1.46(c) also shows the maximum reverse current IZM. As long as
the reverse current is less than IZM, the diode is operating within its safe range.
If the current is greater than IZM, the diode will be destroyed. To prevent
excessive reverse current, a current-limiting resistor must be used.
Ideal Voltage Reference Circuit
1.47 shows a Zener voltage regulator circuit.
For this circuit, the output voltage should
remain constant, even when the output load
resistance varies over a fairly wide range, and
when the input voltage varies over a specific
range. The variation in VPS may be the ripple
voltage from a rectifier circuit.
We determine, initially, the proper input
resistance Ri. The resistance Ri limits the
current through the Zener diode and drops
the “excess” voltage between VPS and VZ. We
can write
Figure 1.47 A Zener diode voltage regulator
circuit
(1.36)
which assumes that the Zener resistance is zero for the ideal diode. Solving
this equation for the diode current, IZ, we get
(1.37)
where IL = VZ /RL , and the variables are the input voltage source VPS and the
load current IL.
For proper operation of this circuit, the diode must remain in the breakdown
region and the power dissipation in the diode must not exceed its rated value.
In other words:
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1. The current in the diode is a minimum, IZ(min), when the load current is
a maximum, IL(max), and the source voltage is a minimum, VPS(min).
2. The current in the diode is a maximum, IZ(max), when the load current is
a minimum, IL(min), and the source voltage is a maximum, VPS(max).
Inserting these two specifications into Equation (1.36), we obtain
(1.38.a)
and
(1.38.b)
Zener Resistance
In the ideal Zener diode, the Zener resistance is zero. In actual Zener
diodes, however, this is not the case. The result is that the output voltage will
fluctuate slightly with a fluctuation in the input voltage, and will fluctuate with
changes in the output load resistance.
Figure 1.48 shows the equivalent
circuit of the voltage regulator including the
Zener resistance. Because of the Zener
resistance, the output voltage will change
with a change in the Zener diode current.
Two figures of merit can be defined
for a voltage regulator. The first is the
source regulation and is a measure of the
change in output voltage with a change in
Figure 1.48 A Zener diode voltage regulator
circuit with a nonzero Zener resistance
source voltage.
The second is the load regulation and is a measure of the change in output
voltage with a change in load current.
1.52. Photo Diode and LED Circuits
A photodiode converts an optical signal into an electrical current, and a light
emitting diode (LED) transforms an electrical current into an optical signal.
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Photodiode Circuit
Figure 1.49 shows a typical photodiode
circuit in which a reverse-bias voltage is applied
to the photodiode. If the photon intensity is
zero, the only current through the diode is the
reverse-saturation current, which is normally
very small. Photons striking the diode create
excess electrons and holes in the space-charge
region. The electric field quickly separates these
excess carriers and sweeps them out of the
space-charge
region,
thus
creating
a
photocurrent in the reverse-bias direction. The
photocurrent is
Figure 1.49 A photodiode circuit. The
diode is reverse biased
(1.39)
where η is the quantum efficiency, e is the electronic charge,
is the photon
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flux density (#/cm −s), and A is the junction area. This linear relationship
between photocurrent and photon flux is based on the assumption that the
reverse-bias voltage across the diode is constant. This in turn means that the
voltage drop across R induced by the photocurrent must be small, or that the
resistance R is small.
Example 1.13
Calculate the photocurrent generated in a photodiode.
For the photodiode shown in Figure 1.49 assume the quantum efficiency is 1,
the junction area is 10−2 cm2, and the incident photon flux is 5 × 1017cm−2 − s−1
From Equation (1.39), the photocurrent is
Comment: The incident photon flux is normally given in terms of light intensity,
in lumens, foot-candles, or W/cm2. The light intensity includes the energy of
the photons, as well as the photon flux.
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LED Circuit
A light-emitting diode (LED) is the inverse of a photodiode; that is, a current is
converted into an optical signal. If the diode is forward biased, electrons and
holes are injected across the space-charge region, where they become excess
minority carriers. These excess minority carriers diffuse into the neutral n- and
p-regions, where they recombine with majority carriers, and the recombination
can result in the emission of a photon.
LEDs are fabricated from compound semiconductor materials, such as gallium
arsenide or gallium arsenide phosphide. These materials are direct-band gap
semiconductors. Because these materials have higher band gap energies than
silicon, the forward-bias junction voltage is larger than that in silicon-based
diodes.
It is common practice to use a seven-segment
LED for the numeric readout of digital instruments,
such as a digital voltmeter. The seven-segment
display is sketched in Figure 1.50. Each segment is
an LED normally controlled by IC logic gates.
Figure 1.51 shows one possible
circuit connection, known as a
common-anode display. In this circuit,
the anodes of all LEDs are connected to
a 5 V source and the inputs are
controlled by logic gates. If VI 1 is
“high,” for example, D1 is off and there
is no light output. When VI 1 goes “low,”
D1 becomes forward biased and
produces a light output.
Figure 1.50 Seven-segment LED display
Example 1.14
Determine the value of R required to
limit the current in the circuit in Figure
1.51 when the input is in the low state.
Assume that a diode current of 10 mA
produces the desired light output, and
that the corresponding forward-bias
voltage drop is 1.7 V.
Figure 1.51 Control circuit for the sevensegment LED display
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If VI = 0.2 V in the “low” state, then the diode current is
The resistance R is then determined as
Comment: Typical LED current-limiting resistor values are in the range of 300
to 350 Ω.
Figure 1.52 Optoisolator using an LED and a photodiode
One application of LEDs and photodiodes is in optoisolators, in which
the input signal is electrically decoupled from the output (Figure 1.52). An
input signal applied to the LED generates light, which is subsequently detected
by the photodiode. The photodiode then converts the light back to an electrical
signal. There is no electrical feedback or interaction between the output and
input portions of the circuit.
1.53. The Schottky-Barrier Diode (SBD)
The Schottky-barrier diode (SBD) is formed by bringing metal into
contact with a moderately doped n-type semiconductor material. The resulting
metal- semiconductor junction behaves like a diode, conducting current in one
direction (from the metal anode to the semiconductor cathode) and acting as an
open-circuit in the other, and is known as the Schottky-barrier diode or simply
the Schottky diode. In fact, the current-voltage characteristic of the SBD is
remarkably similar to that of a pn-junction diode, with two important
exceptions:
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1. In the SBD, current is conducted by majority carriers (electrons). Thus
the SBD does not exhibit the minority-carrier charge-storage effects
found in forward-biased pn junctions. As a result, Schottky diodes can
be switched from on to off, and vice versa, much faster than is possible
with pn-junction diodes.
2. The forward voltage drop of a conducting SBD is lower than that of a pnjunction diode. For example, an SBD made of silicon exhibits a forward
voltage drop of 0.3 V to 0.5 V, compared to the 0.6 V to 0.8 V found in
silicon /wz-junction diodes. SBDs can also be made of gallium arsenide
(GaAs) and, in fact, play an important role in the design of GaAs circuits.
Gallium-arsenide SBDs exhibit forward voltage drops of about 0.7 V.
Apart from GaAs circuits, Schottky diodes find application in the
design of a special form of bipolar-transistor logic circuits, known as SchottkyTTL, where TTL stands for transistor- transistor logic.
Before leaving the subject of Schottky-barrier diodes, it is important to
note that not every metal-semiconductor contact is a diode. In fact, metal is
commonly deposited on the semiconductor surface in order to make terminals
for the semiconductor devices and to connect different devices in an integratedcircuit chip. Such metal-semiconductor contacts are known as ohmic contacts
to distinguish them from the rectifying contacts that result in SBDs. Ohmic
contacts are usually made by depositing metal on very heavily doped (and thus
low-resistivity) semiconductor regions.
1.54 Varactor Diode
The Varactor (also called the voltage-variable capacitance, varicap,
epicap and tuning diode) is widely used in television receivers, FM receivers
and other communications equipment because it can be used for electronic
tuning.
Basic Idea
In Figure 1.53(a) the depletion layer is
between the p region and n region. The
p and n regions are like the plates of a
capacitor and the depletion layer is like
the dielectric. When a diode is reverse
biased, the width of the depletion layer
increases with the reverse voltage.
Since the depletion layer gets wider
with more reverse voltage, the
Figure.1.53 Varactor. (a) Doped regions arelike capacitor
plates separated by a dielectric (b) ac equivalent circuit
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Capacitance becomes smaller. It’s as though you moved apart the plate of a
capacitor. The key idea is that capacitance is controlled by reverse voltage.
Equivalent Circuit and symbol
Figure 1.53(b) shows the ac
equivalent circuit for a reverse-biased
diode. In other words, as far as an ac
concerned, the varactor acts the same
as a variable capacitance. Figure
1.53(c) shows the schematic symbol for
a varactor. The inclusion of a capacitor
in series with the diode is a remainder
that a varactor is a device that has
been optimized for its variablecapacitance properties.
Capacitance Decreases
Reverse Voltage
at
Higher
Figure. 1.53 (c) Schematic symbol (d) capacitance
versus reverse voltage
Figure 1.53(d) shows how the capacitance varies with reverse voltage.
This graph shows that the capacitance gets smaller when the reverse voltage
gets larger.The really important idea here is that reverse dc voltage control
capacitance.
How is abvaractor used? It is connected in parallel with an inductor to
form a oarallel resonant circuit. This circuit has only one frequency at which
maximum impedance occurs. This frequency is called the resonant frequency.
If the dc reverse voltage to the varactor is changed, the resonant frequency is
also changed. This is the principle behind electronic tuning of a radio station, a
TV channeland so on.
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Exercise problems
1. (a)Calculate Vbi for a GaAs pn junction at T = 300 K for Na = 1016 cm−3
and Nd = 1017 cm−3. (b) Repeat part (a) for a Germanium pn junction
with the same doping concentrations. (Ans. (a) Vbi = 1.23 V, (b) Vbi =
0.374 V)
2. Calculate the junction capacitance of a pn junction. Consider a silicon
pn junction at T = 300 K, with doping concentrations of Na = 1016 cm−3
and Nd = 1015 cm−3. Assume that ni = 1.5 × 1010 cm−3 and let Cjo = 0.5
pF. Calculate the junction capacitance at VR = 1V and VR = 5V.
(Ans. For VR = 1V, Cj=0.312 pF, For VR = 5V, Cj= 0.168 pF)
3. A silicon pn junction at T = 300 K is doped at Nd = 1016 cm−3 and Na =
1017 cm−3. The junction capacitance is to be Cj = 0.8pF when a reverse
bias voltage of VR = 5V is applied. Find the zero-biased junction
capacitance Cjo. (Ans. Cjo = 2.21 pF)
4. Consider the circuit in Figure shown below. Let VPS = 4 V, R = 4 kΩ, and
IS = 10−12A. Determine VD and ID, using the ideal diode equation and the
iteration method. (Ans. VD = 0.535 V, ID = 0.866 mA)
Figure for problem 4. A simple diode circuit
5. Determine the diode voltage and current in the circuit shown in Figure
for problem 4 using a piecewise linear model. Also determine the power
dissipated in the diode. (Ans. VD=0.622V, ID=2.19mA , PD=1.36mW)
6. Consider the diode and circuit in problem 4, Determine VD and ID, using
the graphical technique. (Ans. VD 0.54V, ID 0.87 mA)
7. Determine the diode voltage and current in the circuit shown in figure for
problem 4 using a piecewise linear model. Assume piecewise linear diode
parameters of Vγ=0.6 V and rf = 10Ω.(Ans. ID=2.19 mA, VD=0.622 V)
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8. The power supply (input) voltage in the circuit of Figure for problem 4 is
VPS = 10 V and the diode cut-in voltage is Vγ = 0.7 V (assume rf = 0). The
power dissipated in the diode is to be no more than 1.05 mW. Determine
the maximum diode current and the minimum value of R to meet the
power specification. (Ans. ID = 1.5 mA, R = 6.2 kΩ)
9. Determine diode voltages. The reverse saturation currents of a pn
junction diode and a Schottky diode are IS = 10−12 A and 10−8 A,
respectively. Determine the forward-bias voltages required to produce 1
mA in each diode. (Ans.0.539v for PN junction diode and 0.299V for
Schottky diode)
10. A pn junction diode and a Schottky diode both have forward-bias
currents of 1.2 mA. The reverse-saturation current of the pn junction
diode is IS = 4 × 10−15 A. The difference in forward-bias voltages is 0.265
V. Determine the reverse-saturation current of the Schottky diode. (Ans.
IS = 1.07 × 10−10 A)
11. Consider a simple constant-voltage reference circuit and design the
value of resistance required to limit the current in this circuit. Consider
the circuit shown in Figure below. Assume that the Zener diode
breakdown voltage is VZ = 5.6 V and the Zener resistance is rz = 0. The
current in the diode is to be limited to 3 mA. (Ans. R=1.47 KΩ)
Figure for problem 11 Simple circuit containing a Zener diode in which the Zener diode is
biased in the breakdown region
12. Consider the circuit shown in Figure for problem 11. Determine the
value of resistance R required to limit the power dissipated in the Zener
diode to 10 mW. (Ans. R = 2.46 kΩ)
13. A Zener diode has an equivalent series resistance of 20Ω. If the voltage
across the Zener diode is 5.20 V at IZ = 1 mA, determine the voltage
across the diode at IZ = 10 mA. (Ans. VZ = 5.38 V)
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14. Determine the value of resistance R required to limit the current in the
circuit shown in Figure below to I = 15 mA. Assume Vγ = 1.7 V, rf = 15 Ω,
and VI = 0.2 V in the “low” state. (Ans. R = 192 Ω)
Figure for problem 14 Control circuit for the sevensegment LED display
15. Consider the circuit in Figure below. Let Vγ = 0. (a) Plot vO versus vI over
the range −10 ≤ vI ≤ +10 V. (b) Plot i1 over the same input voltage range
as part (a).
Figure for problem 15
16. For the circuit in Figure below, (a) plot vO versus vI for 0 ≤ vI ≤ 15 V.
Assume Vγ = 0.7V. Indicate all breakpoints. (b) Plot iD over the same
range of input voltage.
Figure for problem 16
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