Electronic Circuit Diode Wave Shaping

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Analog Electronics Circuit
Mohamed Albanna
1. p-n Junction diode
1) Normal and ideal diode.
A diode is a passive electronic device that has a positive anode
terminal and a negative cathode terminal and a nonlinear voltage-current
characteristic. It conducts only in one direction, from anode to cathode,
when it is forward biased or ‘‘on’’ (a diode is connected with its anode to
positive (vD ≥ 0); the positive current that flows (iD ≥ 0). When a diode is
connected in the reverse direction, with its cathode to positive vD < 0, we
say that it is reverse biased or ‘‘off, (not conduct).
Figure 1: Diode symbol and work principle
The ideal diode exhibits zero impedance when forward-biased and
infinite impedance when reverse-biased (Figure 2). Note that since either
current or voltage is zero, no power is dissipated by an ideal diode.
Figure 2: The ideal diode: (a) i–v characteristic; (c) equivalent circuit in the
forward direction; (d) equivalent circuit in the reverse direction.


A practical application of diode
Diode Wave Shaping
There are two types of diode wave shaping circuits: clippers and
clampers.
Clipper: Separate an input signal at a particular dc level (clip off)
and pass to the output, without distortion, the remaining upper or lower
portion of the input waveform. This property can be used to limit the
amplitude of a signal or to eliminate noise from an existing signal.
A diode will not conduct until the forward bias voltage is at least as
the forward break over voltage, (Ge diode 0.3V; Si diode 0.6V).
For the ideal diode the transition between states will occur at Vd = 0
V and Id = 0 A. Applying the Id = 0 A at Vd = 0 V to the network of Figure 3
will result in the configuration of the next figure, where it is recognized
that the level of vi that will cause a transition in state is For an input
voltage greater than VB volts the diode is in the short-circuit state, while
for input voltages less than VB volts it is in the open-circuit or “off” state.
Figure 3: Determining the transition and Vo.
When the diode is in the short-circuit state, the output voltage Vo can
be determined by applying Kirchhoff’s voltage law in the clockwise
direction Vi – VB – Vo = 0.
1- Parallel clipper
Figure 4 shows a parallel clipper circuit with a 5 Vpp triangular input
waveform.
Figure 4: Parallel Clipper Circuit and waveforms
During the positive half cycle, diode Cr1 is forward biased, the output
voltage equals the forward voltage drop of the diode VF=0.7 V for Si diode.
Any voltage greater than 0.7 V will shunt by diode Cr1, as a result, only a
portion of the positive half cycle is passed through to the output. With
diode Cr1 reverse biased, the entire input is passed through to the output.


By adding a dc voltage bias (battery) in series with the diode, the bias
of the diode enables control over the portion of the input signal passed to
the output. This effectively shifts the limiting point of the circuit.
In each part of the figures, we can write Kirchhoff’s voltage law
around the loop to determine the value of input voltage Vin that is
necessary to forward bias the diode. Assuming that the diodes are ideal,
we determine the value Vin necessary to forward bias each diode by
determining the value Vin necessary to make VD > 0. When Vin reaches the
voltage necessary to make VD > 0, the diode becomes forward biased and
the signal source Vin is forced to, or held at, the dc voltage VB. If the
forward voltage drop across the diode is not neglected, the clipping level is
found by determining the value of Vin necessary to make VD greater than
that forward drop (e.g., VD > 0.7 V for a Si diode).
In positive clipping Since Vin = VD+ VB, then VD = Vin - VB,
Since VD > 0 then Vin – VB > 0, and Vin > VB
Figure 5 illustrates the concept of applying a 1 V bias to a parallel
clipping circuit.
Figure 5: Positive biased Positive parallel Clipper Circuit and waveforms
During the positive half cycle, diode Cr1 is only forward biased when
the input signal is greater than +1.7 V (Vin > VB +VCr1 > 1+ 0.7 > 1.7). Any
portion of the input signal above +1.7 V will shunt by diode Cr1. When the
input signal decreases below +1.7 V, diode Cr1 is reverse biased and the
entire input signal is passed to the output.
Figure 6 shows the clipping effect for a bias voltage of -2 V.
Figure 6: Negative biased Positive parallel Clipper Circuit and waveforms
Diode Cr1 will remain forward biased until the potential at the anode
is (-2 + 0.7 = -1.3 V). As the input decreases below -1.3 V, (Vin < -VB + VCr1)


diode Cr1 will be reverse biased allowing the input waveform to pass
through to the output. The limiting point of the circuit is now -1.3 V.
In negative clipping Since Vin = VD - VB or VD = Vin + VB,
Since VD > 0 then Vin + VB > 0, and Vin > - VB
Figure 7 shows a negative parallel clipper circuit with a 4 V bias
voltage.
Figure 7: Positive biased Negative parallel Clipper Circuit and waveforms
The negative half cycle has been clipped off in the output waveform.
Diode Cr1 remains in the forward biased condition until the input signal
increases above 3.3 V (Vin > VB - VCr1). The limiting point of the circuit is
set at 3.3 V.
Determine what value and polarity of total voltage Vin across the
combination is necessary to forward bias the diode.
Figure 8: Negative biased Negative parallel Clipper Circuit and waveforms
Double-ended limiting circuits: Both positive and negative peaks of
the output waveform are clipped. No more than one diode is forward
biased at any given time and that both diodes are reverse biased for –E1 <
vo < E2, the linear region.
Figure 9 Double-ended clipping,
If VB1 = 5 V, VCr1 = 0.7 V, then clip at 5.7 V if VB2 = 5 V, VCr2 = 0.7 V,
then clip at and -5.7 V

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The clipping takes place during the positive cycle only when the
input voltage is greater than battery voltage (i.e. Vin > VB). The clipping
level can be shifted up or down by varying the bias voltage (VB).
Figure 10-a. Series with positive bias
Clipping take place during the negative half cycle only when the
input voltage Vin > VB. The clipping level can be shifted up or down by
varying the voltage (-VB).
Figure 10-b. Series with negative bias
For these circuits, the direction of the diode suggests that the signal
must be positive to turn it on. The dc supply further requires that the
voltage be greater than VB volts to turn the diode on. The negative region
of the input signal is “pressuring” the diode into the “off” state, supported
further by the dc supply. In general, therefore, we can be quite sure that
the diode is an open circuit (“off” state) for the negative region of the input
signal.
Example 1. Figure 11 shows a negative series clipper circuit with
negative dc bias of 5 V.
Note that with the biasing voltage VB in series with the input voltage,
the resulting output signal has shifted in the dc level.
Figure 11: Negative Series Clipper with dc Bias


For example, in Figure 11 when the input voltage is +10 V, the
voltage at the anode of Cr1 is +15 V. This is because the biasing voltage VB
is summed with the input waveform voltage. Subtracting the forward
voltage drop of Cr1, results in a peak output voltage of 15 - 0.7 = +14.3 V.
When the anode voltage of Cr1 falls below 0.7 V, Cr1 is reverse biased
and the output voltage will be zero. This occurs when the input voltage
decreases past (Vin - VB ) 0.7 – 5 = -4.3 V.
A result of the biasing voltage being in series with the input signal is
a dc voltage shift of 5 V at the anode of Cr1 over the entire positive and
negative half cycles of the input. The forward voltage drop of the Cr1
results in a 4.3 V shifts at the output.
By reversing the orientation of the diode in a series clipper circuit,
the positive half cycle can be clipped from the input signal. This circuit is
called a positive series clipper.
Example 2. Figure 12 shows a positive series clipper circuit with a 5
V bias.
Figure 12: Positive Series Clipper with dc Bias
Note that Cr1 remains reverse biased until the input signal decreases
below (-10 + 5 - 0.7) -5.7 V. Therefore only the portion of the signal less
than -5.7 V will be passed through Cr1 to the output.
Example 3: Figure 13 shows a positive clipping circuit, which
removes any portion of the input signal vi that is greater than Vb and
passes as the output signal vo any portion of vi that is less than Vb.
Figure 13
As you can see, vD is negative when vi < Vb, causing the ideal diode to
act as an open circuit. With no path for current to flow through R, the


value of vi appears at the output terminals as vo. However, when vi ≥ Vb, the
diode conducts, acting as a short circuit and forcing vo=Vb.
The diode blocks for vi < 6V and conducts for vi ≥ 6V. Thus, vo = vi for
vi < 6V, and vo = 6V for vi ≥ 6V.
Example 4: In the positive clipping circuit of Figure 13, the diode is
ideal and vi is a 10 V triangular wave with period T. Sketch one cycle of
the output voltage vo if Vb = 6V. The diode blocks (acts as an open circuit)
for vi < 6V, giving vo = vi. For vi ≥ 6V, the diode is in forward conduction,
clipping vi to effect vo = 6V. The resulting output voltage waveform is
sketched in Figure 14.
Figure 14
Example 5: Reverse the diode in Figure 13 to create a negative
clipping network. (a) Let Vb = 6V, Sketch one cycle of the output
waveform if vS = 10 sinωt V.
(a) The diode conducts for vi ≤ 6 V and blocks for vi > 6V.
Consequently, vo = vi for vi > 6 V, and vo = 6 V for vi ≤ 6 V.
(b) With negative clipping, the output is made up of the positive
peaks of 10 sinωt above 6V and is 6V otherwise. Figure 18 displays the
output waveform.
Figure 18


Clamping
A circuit that places either the positive or negative peak of a signal at
a desired dc level is known as a clamping. A clamping adds a dc
component to the signal and does not change the shape or amplitude of the
input signal. Clamping circuits consist of a capacitor, a diode and a
resistor. An independent dc biasing voltage can also be employed to
provide an additional voltage shift. For clamping circuits, the total swing
in the output is equal to the total swing of the input. Only the dc reference
level has changed. Adding a dc bias to a clamping circuit will provide
additional shift in the dc reference level of the output.
When a signal drives an open-ended capacitor the average voltage
level on the output terminal of the capacitor is determined by the initial
charge on that terminal and may therefore be quite unpredictable. Thus it
is necessary to connect the output to ground or some other reference
voltage via a large resistor. This action drains any excess charge and
results in an average or dc output voltage of zero.
A simple alternative method of establishing a dc reference for the
output voltage is by using a diode clamp as shown in figure 19. By
conducting whenever the voltage at the output terminal of the capacitor
goes negative, this circuit builds up an average charge on the terminal that
is sufficient to prevent the output from ever going negative. Positive
charge on this terminal is effectively trapped."
Figure 19
The steps involved in analyzing clamper circuits are:
• always determine the forward biasing cycle first,
• treat the diodes initially as ideal diodes and
• Adjust the ideal voltage output for the practical diode forward
voltage drop.
1- Positive Clamper
Clamper circuit simply consists of a diode D and capacitor C as
shown in below Figure.
During the negative half cycle of the input voltage Vin the diode is
forward biased and the current flows through the circuit. As a result of
this, the capacitor C is charged to a voltage equal to the negative peak
value i.e., -Vm. Once the capacitor is fully charged to -Vm it cannot

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important to note that the capacitor acts essentially as a battery in series
with the input.
In analyzing clamper circuits with bias, it is easier to first examine
the cycle in which the diode is forward biased. For the circuit shown in
Figure 21, diode Cr1 is forward biased in the negative half cycle of the
input signal.
Figure 21. Clamping Circuit and Waveforms
As Cr1 is an ideal diode, when the input voltage decreases to -4 V, it
will conduct. This will charge C1 to a voltage of -6 V (the remaining
voltage of the input signal). The output voltage is equal to the biasing
voltage of VB (-4V). Examining the positive half cycle, diode Cr1 is reverse
biased. The 6V charge stored in capacitor C1 is summed with the input
voltage. The resulting output voltage is therefore +16 V. Note that the
voltage swing is still 20 V; only the reference level has changed.
If the diode in Figure 21 was a silicon diode, the forward voltage
drop would have to be taken into account. The voltage output waveform
for a silicon diode would have the positive half cycle clamped to +15.3 V
and the negative half cycle clamped to -4.7 V.
Figure 22 illustrates a summary of the configurations and expected
outputs for clamper circuits.
Figure 22. Clamping Circuits


Voltage regulation (Zener diode)
Voltage regulation is the measure of circuit ability to maintain a
constant output even when input voltage or load current varies. When
forward-biased, Zener diodes behave much the same as standard
rectifying diodes. In reverse-bias mode, they do not conduct until the
applied voltage reaches or exceeds the so-called Zener voltage, at which
point the diode is able to conduct substantial current, and in doing so will
try to limit the voltage dropped across it to that Zener voltage point.
Figure 1. Zener diode characteristics
Zener and Avalanche Diodes
Zener diodes (ZD) and avalanche diodes are pn-diodes specially built
to operate in reverse breakdown. Avalanche and Zener diodes are used as
voltage regulators and as overvoltage protection devices.
TC of Zener diodes is negative at VZ ≤ 3.5 to 4.5 V and is equal to
zero at about VZ ≈ 5 V.
TC of a Zener diode operating above 5 V is in general positive. Above
10 V the pn-diodes operate as avalanche diodes with a strong positive
temperature coefficient.
The motivation for using a Zener diode rather than a regular diode is
the high voltage switching applications. While a regular diode provides
switching behavior, the voltage involved is fairly small. With Zerner diode,
voltage switching occurs on the order of magnitude of 100 V. Moreover, a
very small change in the voltage induces a large change in current;
therefore, the resistance is very small by Ohm’s Law, as the slope of the
Zener diode characteristic is fairly large.
The behavior will be described by two effects.
1. Zener effect: Increasing the reverse current VS causes an increase
of the electric field in the depletion zone. The position of the valence band
and of the conducting band will be relatively shifted in the energy band
model. If the upper part of the valence band in the p-type is lying
energetically higher than the lower part of the conducting band in the n-
type, then electrons from the valence band in the p-type may tunnel


through the now very small forbidden energy gap into the conduction
band in the n-type where they will be accelerated toward the positive pole.
Fig. 2: Zener effect in the energy band model.
2. Avalanche effect: Increasing the reverse voltage VS causes an
increase of the kinetic energies of electrons, which are responsible for the
reverse current IS. If the energy of the electrons is sufficiently large, then
these can release bound electrons of the valence band and may reach the
conduction band, where they will be accelerated toward the positive pole.
Then these electrons, which are already present in the conduction band,
will gain kinetic energy, which is high enough to release further electrons
from the valence or conduction band (avalanche effect).
The size of the breakdown voltage depends on doping. In an ordinary
diode the breakdown voltage is between 50 Volts and more than 1000
Volts. A Zener diode is highly doped such that the breakdown voltage is
below 5 V depending on the amount of doping. Z-diodes are often used for
stabilizing voltages, where the current can steeply increased by a slight
varying voltage at the so called breakdown point. Z-diodes are used in
power supplies of almost all electric apparatuses, as safety diode in circuits
and in order to limit voltages.
Zener Diode Regulator Model
In breakdown, the diode is modeled with a voltage source, VZ, and a
series resistance, RZ. RZ models the slope of the I-V characteristic.
Figure 3, basic zener diode regulator model
Important considerations:
1: Zener diode must remain in the breakdown region.

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25.2 mA of current. In order for the 1 kΩ series "dropping" resistor to
drop 32.4 volts (reducing the power supply's voltage of 45 volts down to
12.6 across the zener), it still must conduct 32.4 mA of current. This leaves
7.2 mA of current through the zener diode.
Example: Find the voltage vZ across the Zener diode if iZ =10 mA and
it is known that VZ =5.6 V, IZ =25 mA, and RZ =10 Ω.
Since 0.1 IZ ≤ iZ ≤ IZ, operation is along the safe and predictable
region of Zener operation. Consequently,
vZ = VZ + iZRZ = 5.6 + (10*10-3
)(10) = 5.7V
Problem
The Zener diode regulator circuit shown in Figure 4 has an input
voltage of that varies between 10 and 14 V, and a load resistance that
varies between RL = 20 and 100 Ω. Assume a 5.6 V zener is used and
assume IZ (min) = 0.1 IZ (max). Find the value of Ri required and the
maximum rating of the diode.
Figure 4 Zener diode regulator circuit
Problem
A Zener diode is connected in a voltage regulator circuit as shown in
Figure 4. The Zener voltage is Vz = 10 V and the Zener resistance is
assumed to be rz = 0. (a) Determine the value of Ri such that the Zener
diode remains in the breakdown region if the load current varies from IL =
50 to 500 mA and the input voltage varies from VI = 15 to 20 V. Assume IZ
(min) = 0.1 IZ (max). (b) Determine the power rating required for the
Zener diode and the load resistance
Problem
The secondary voltage in the circuit shown in Figure 5 is vs = 12
sin(at) V. The Zener diode parameters Vz = 8 V at Iz = 100 mA and rz = 0.5
Ω. Let Vv = 0, Ri = 3Ω. Determine the percent of regulation for load
currents between IL = 0.2 and 1 A. Find the C such that the ripple voltage
is no larger than 0.8 V.


Figure 5
Power Supply:
Most of the electronic device and circuits needs direct current (dc) to
work. Batteries produce dc, but there is a limit to how much energy and
how much voltage a battery can provide. The electricity from the utility
company is alternating current (ac) with a frequency of 60 Hz. The energy
from a wall outlet is practically unlimited, but it must be converted from
ac to dc, to be suitable for electronic equipment. Most power supplies
consist of several stages, always in the same order (Fig. 6).
Fig. 6, Block diagram of a power supply. Sometimes a regulator is not
needed.
First, the ac encounters a transformer that steps the voltage either
down or up, depending on the exact needs of the electronic circuits.
Second, the ac is rectified, so that it becomes pulsating dc with a
frequency of 60 Hz. This is almost always done by one or more
semiconductor diodes.
Third, the pulsating dc is filtered, or smoothed out, so that it becomes
a continuous voltage having either positive or negative polarity with
respect to ground.
Finally, the dc voltage might need to be regulated. Some equipment is
finicky, insisting on just the right amount of voltage all the time. Other
devices can put up with some voltage changes.
Power supplies that provide more than a few volts must have features
that protect the user from receiving a dangerous electrical shock. All
power supplies need fuses and/or circuit breakers to minimize the fire
hazard in case the equipment shorts out.
Rectification ("frequency shifting")
Most rectifier diodes are made of silicon and are therefore known as
silicon rectifiers. Two important features of a power-supply diode are, the

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the cycle. For this reason, the bridge circuit makes more efficient use of
the transformer.
The diodes act to route the current from both halves of the ac wave
through the load resistor in the same direction, and the voltage developed
across the load resistor becomes the rectified output signal. The diode
bridge is a commonly used circuit and is available as a four-terminal
component in a number of different power and voltage ratings.
Power supply filtering
Electronic equipment doesn’t like the pulsating dc that comes
straight from a rectifier. The ripple in the waveform must be smoothed
out, so that pure, battery-like dc is supplied. The filter does this.
The simplest filter is one or more large-value capacitors, connected in
parallel with the rectifier output. Electrolytic capacitors are almost always
used. They are polarized; they must be hooked up in the right direction.
Typical values range in the hundreds or thousands of microfarads.
A simple filter. The capacitor, C, should have a large capacitance.
The more current drawn, the more capacitance is needed for good
filtering. This is because the load resistance decreases as the current
increases. The lower the load resistance, the faster the filter capacitors will
discharge. Larger capacitances hold charge for a longer time with a given
load.
Filter capacitors work by “trying” to keep the dc voltage at its peak
level (Fig. 8). This is easier to do with the output of a full-wave rectifier
(shown at A) as compared with a half-wave circuit (at B). The remaining
waveform bumps are the ripple. With a half-wave rectifier, this ripple has
the same frequency as the ac, or 60 Hz. With a full-wave supply, the ripple
is 120 Hz. The capacitor gets recharged twice as often with a full-wave
rectifier, as compared with a half-wave rectifier. This is why the ripple is
less severe, for a given capacitance, with full-wave circuits.


Figure 8 Filtered output for full-wave rectification (A) and half-wave
rectification (B).
The rectified wave is dc only in the sense that it does not change
polarity. But it does not have constant value and has plenty of ripples i.e.
small waves. It has to be smoothed out in order to obtain authentic direct
current. This can be done by means of a low-pass filter, which is shown in
Figure below.
Full-wave bridge with RC filter.
The full-wave bridge diodes prevent flow of current back out of
capacitor. The capacitor is an energy storage element. The energy stored
in a capacitor is 2
CU
2
1
E = . For C in F (farads) and U in V (Volts), E
comes out in J (jouls) and J = Watt/sec. The capacitor value is chosen so
that f/1Rload >> , where f is the ripple frequency.
For power line sine wave it is 2*50 Hz = 100 Hz. It allows ensuring
small ripples, by making the time constant for discharge much longer than
the time between recharging (the capacitor is charging very quickly, while
discharging is very slow). It is quite easy to calculate the approximate
ripple voltage. Let us assume that the load current stays constant (it will,
for small ripples). The load causes the capacitor to discharge somewhat
between cycles. The capacitor will lose some voltage, let us say UΔ .
In this case, we have: t
C
I
U ΔΔ = , (from
dt
dU
CI = )
Instead of tΔ we use f/1 or f2/1 respectively for half-wave
rectification and for full-wave rectification. Finally we obtain approximate
ripple voltage:
For halve-wave
fC
I
U load=Δ , for full-wave
fC2
I
U load=Δ .
If one wanted to do exact calculation with no approximation, one
would use the exact exponential formula.


2. Bipolar Junction Transistors (BJT)
The Transistor as a Switch
Both the npn and pnp type bipolar transistors can be made to operate
as "ON/OFF" type solid state switches. The areas of operation for a
transistor switch are known as the Saturation Region and the Cut-off
Region. This means then that we can ignore the operating Q-point biasing
and voltage circuitry required for amplification, and uses the transistor as
a switch by driving it back and forth between its fully-OFF (cut-off) and
fully-ON (saturation) regions.
The pink shaded area at the bottom of the curves represents the
"Cut-off" region while the blue area to the left represents the
"Saturation" region of the transistor. Both these transistor regions are
defined as:
1. Cut-off Region
Here the operating conditions of the transistor are zero input base
current (IB), zero output collector current (IC) and maximum collector
voltage (VCE) which results in no current flowing through the device.
Therefore the transistor is switched Fully-OFF.
Cut-off Characteristics
• The input and Base are grounded (0v)
• Base-Emitter voltage VBE < 0.7v
• Base-Emitter junction is reverse biased
• Base-Collector junction is reverse biased
• Transistor is "fully-OFF" (Cut-off region)
• No Collector current flows (IC = 0)
• VOUT = VCE = VCC = "1"
• Transistor operates as an "open switch"
Then we can define the "cut-off region" or "OFF mode" when using
a bipolar transistor as a switch as being, both junctions reverse biased,
VB < 0.7v and IC = 0. For a pnp transistor, the Emitter potential must be
negative with respect to the Base.
2. Saturation Region


Here the transistor will be biased so that the maximum amount of
base current is applied, resulting in maximum collector current resulting
in the minimum collector emitter voltage drop which results in the
depletion layer being as small as possible and maximum current flowing
through the transistor. Therefore the transistor is switched "Fully-ON".
Saturation Characteristics
• The input and Base are connected to VCC
• Base-Emitter voltage VBE > 0.7v
• Base-Emitter junction is forward biased
• Base-Collector junction is forward biased
• Transistor is "fully-ON" ( saturation region )
• Max Collector current flows (IC = Vcc/RL)
• VCE = 0 ( ideal saturation )
• VOUT = VCE = "0"
• Transistor operates as a "closed switch"
Then we can define the "saturation region" or "ON mode" when
using a bipolar transistor as a switch as being, both junctions forward
biased, VB > 0.7v and IC= Maximum. For a pnp transistor, the Emitter
potential must be positive with respect to the Base.
Then the transistor operates as a "single-pole single-throw" (SPST)
solid state switch. With a zero signal applied to the Base of the transistor it
turns "OFF" acting like an open switch and zero collector current flows.
With a positive signal applied to the Base of the transistor it turns "ON"
acting like a closed switch and maximum circuit current flows through the
device.
An example of an npn Transistor as a switch being used to operate a
relay is given below. With inductive loads such as relays or solenoids a
flywheel diode is placed across the load to dissipate the back EMF
generated by the inductive load when the transistor switches "OFF" and
so protect the transistor from damage. If the load is of a very high current
or voltage nature, such as motors, heaters etc, then the load current can be
controlled via a suitable relay as shown.


Basic npn Transistor Switching Circuit
To operate the transistor as a switch the transistor needs to be turned
either fully "OFF" (cut-off) or fully "ON" (saturated). An ideal transistor
switch would have infinite circuit resistance between the Collector and
Emitter when turned "fully-OFF" resulting in zero current flowing
through it and zero resistance between the Collector and Emitter when
turned "fully-ON", resulting in maximum current flow.
In practice when the transistor is turned "OFF", small leakage
currents flow through the transistor and when fully "ON" the device has a
low resistance value causing a small saturation voltage (VCE) across it.
Even though the transistor is not a perfect switch, in both the cut-off and
saturation regions the power dissipated by the transistor is at its
minimum.
In order for the Base current to flow, the Base input terminal must
be made more positive than the Emitter by increasing it above the 0.7 volts
needed for a silicon device. By varying this Base-Emitter voltage VBE, the
Base current is also altered and which in turn controls the amount of
Collector current flowing through the transistor. When maximum
Collector current flows the transistor is said to be saturated. The value of
the Base resistor determines how much input voltage is required and
corresponding Base current to switch the transistor fully "ON".
Inside the transistor, the small base current joins with the collector
current to become the emitter current (or the emitter current divides to
become the base and collector currents). This is expressed as
IE = IC + IB (1)
Where
IE = emitter current
IC = collector current
IB = base current


However, because the collector current is usually so much larger than
the base current, a useful approximation is to consider the collector current
to be equal to the emitter current:
IE  ~ IC (2)
The transistor dissipates power anytime there is a current flowing
through it and a voltage is across it, expressed as
PD
= IC
VCE
(3)
Where
PD
= power dissipated by the transistor
IC
= collector current
VCE
= voltage between the collector and emitter
Example 1:
Using the transistor values of: β = 200, Ic = 4 mA and IB = 20 uA,
find the value of the Base resistor (RB) required to switch the load fully
"ON" when the input terminal voltage exceeds 2.5 V.
. .
The next lowest preferred value is: 82 kΩ, this guarantees the
transistor switch is always saturated.
Example 2
Using the same values, find the minimum Base current required to
turn the transistor "fully-ON" (saturated) for a load that requires 200 mA
of current when the input voltage is increased to 5.0 V. Also calculate the
new value of RB.
Transistor Base current:
Transistor Base resistance:
. .
.
The behavior of the transistor may be illustrated by a load line on the
output characteristics of the transistor. A load line is a line whose
current-voltage (I-V) plot represents a load resistance, where the load
line cuts any of the IB lines.


Fig 1: Transistor input and output characteristics.
If the collector current IC is plotted against the collector-to-emitter
voltage VCE, a family of curves for various fixed values of VBE or IB results,
as in Figure 1. These curves show that as VCE increases, IC rises very
rapidly and then turns over as it is limited by IB. In the Common Emitter
circuit, if IB were reduced to zero, then IC would also be zero (apart from a
small leakage current ICE0). Hence there would be no voltage drop in
either RC or RE, and practically all of VCC would appear across the
transistor. That is, under cut-off conditions,
VCE → VCC for IB = 0 (4)
Conversely, if IB were large, IC would be very large, almost all of VCC
would be dropped across RC + RE and
B
EC
CC
C Iearglfor
RR
V
I
+
→ (5)
Actually, because the initial rise in IC for the transistor is not quite
vertical, there is always a small saturation voltage VCES across the
transistor under these conditions, where VCES means the voltage across the
transistor in the common-emitter mode when saturated. In this saturated
condition VCES ≈ 0.3 V for small silicon transistors.


Figure 2 universal bias arrangement
Note that because the transistor is non-ohmic (that is, it does not obey
Ohm’s law), the voltage across it may only be determined by using the
(ohmic) voltage drop across the resistors RC and RE according to Equation
(6). At the quiescent point this is: VQ = VCC – IQ (RC + RE)
From the circuit of Fig. 2 a,
)6()RR(IVV ECCCCCE +−= , which may be rewritten as
)7()VV(
RR
1
RR
V
RR
V
I CCCE
ECEC
CE
EC
CC
C −
+
−=
+
−
+
+
=
Figure 3. An example of a BJT constant current source
It is simple to plot the load line for the circuit in figure 3. The open-
circuit voltage for the x-axis value of the load line is simple the voltage
across CE when the transistor is removed from the circuit. This value is:
VCE (Sat) indicate the maximum possible value of VCE in the circuit.
Next we will calculate the short-circuit current for the y-axis value of
the load line. This value is:


This is the straight-line equation to the dc load-line (compare y = mx
+ c), showing that its slope is –1/ (RC + RE) and that it crosses the IC axis at
VCC / (RC + RE) as expected. The actual position of a point is determined by
where this load line crosses the output characteristic in use, that is, by
what value of VBE or IB is chosen. For example, the quiescent point for the
transistor is where the load line crosses the output curve defined by VBE =
VBEQ (or IB = IBQ) to give VCE = VQ and IC = IQ.
When the value of collector resistor has been chosen, bias is applied
by passing current into the base so that the collector voltage drops to the
desired value of around 0.5 VCC where VCC is the supply voltage. For any
bias system, the desired base current must be equal to: 0.5 VCC/RLhFE with
VCC in volts, RL (the load) in kΩ, hFE as a ratio.
The amplification of a transistor can be recognized with the so called
current-control characteristic. The collector current IC is plotted with
respect to IB we can see that a small base current causes a large collector
current.
Fig. 4: Linear amplification range of a transistor (current-control
characteristic).
Under the proper operating conditions, IC will be some multiple of IB;
in other words, the transistor is a current amplifier. The forward current
gain (hFE or β) is expressed as:
Where
hFE = forward current gain
IC = collector current
IB = base current
Example 1
A power transistor with an hFE
of 50 is operating with a load current
(IC
) of 3 A. What is the base current?


Solution:
Recall that a transistor is basically a current amplifier where the IC
is
hFE
times larger than the base current (IB
). By rearranging Equation 8, we
can solve for IB
:
Let us see how the voltage-divider bias helps us set a Q-point
independent of β. The term “voltage-divider bias” comes from the voltage
divider formed by R1 and R2 in figure 3. The voltage across R2 should
forward bias the emitter diode, VBE is forward-biased and VCE is reverse-
biased. If we Thevenize the circuit as shown below, we get the circuit
shown in figure 5 below.
Figure 5: Circuit of figure 3 Thevenized at the base terminal, KVL loop
shown.
Since the transistor is in active region, IC*IE = ICE. Then
,
We can write a loop equation as shown in figure 5:
Now, hence, the equation above becomes:
If we want to swamp out the effects of β, a good rule of thumb is: R4
is 100 times greater than Rth/ β. Thus we want: Rth ≤ 0.01*β*R4. Thus, in


our circuit: Rth = 250 Ω < 0.01*220*150 = 330 Ω.
Notice that the BJT will provide an amplified constant ICE as long as
the Rth condition holds.
Method of creating the bias voltage
Figure 6 is a voltage-divider circuit shows that resistors R1 and R2
create a base bias voltage of 0.8 V (ignoring any loading effects).
Consulting the input curve of Figure 1, we see that 0.8 V would cause
a base current IB
of 2 mA. Notice the load line shows that if 0.1 mA ≤ IB <
0.2 mA, then the BJT is guaranteed to be in active region.
Figure 6: A voltage-divider bias circuit.
The hFE
of the transistor is given as 100, so rearranging Equation 8
we can calculate the collector current: IC
= hFE
IB
= 100*2 mA = 200 mA
Notice that the load is in series with the collector so that all 200 mA is
going through the transistor and the load.
Now consider an input signal voltage of ±0.1 V, which is
superimposed onto the base through a capacitor. In other words, the input is
added to the 0.8 Vdc bias voltage to form an ac base voltage that cycles from
0.7 to 0.9 V, as shown in Figure 7. Referring again to the base curve (Figure
1), we see that a base voltage of 0.7-0.9 V translates into a base current of 1-3
mA, which the transistor multiplies into a collector current of 100-300 mA,
and all of this current goes through the load.


Figure 7: A simple transistor amplifier.
Notice that the output voltage VC
is out of phase with the output
current IC
. This occurs because the transistor must lower its resistance to
increase the current, and a lower resistance creates a lower voltage drop.
When we add a sinusoidal source with amplitude of ΔVBB in series
with VBB, in response to this additional source, the base current will
become iB + ΔiB leading to the collector current of iC + ΔiC and CE voltage
of VCE + ΔVCE. These changes are shown in figure 7.
Example 2
Find the load current in the transistor circuit shown in Figure 8. Use
the input curves in Figure 1 and assume that the hFE
of the transistor is 70.
Solution
Figure 8: A transistor circuit
We want to find the output current, and because we know the
current gain is 70, the problem comes down to finding out what the base


current is. In most cases, such as this one, it is the base voltage that we can
calculate (or measure), so we will use the input curves to find the
corresponding base current. First, calculate the base bias voltage from R1
and R2 using the voltage-divider rule:
.
.
Now, knowing that the base voltage is 0.77 V, we use the input curve of
Figure 1 to determine that IB
is about 1.7 mA. The load current IC
can now
be calculated by rearranging Equation 8:
IC
= hFE
IB
= 70*1.7 mA = 119 mA
3- Field-Effect Transistors FET
Transfer Conductance
The slope of the transfer curve, dIDS /dVGS, is the dynamic forward
transfer conductance, or mutual transfer conductance, gm. We see that gm
starts off at zero when VGS = VP and increases as IDS increases, reaching a
maximum when IDS = IDSS. Since IDS = IDSS [1 – (VGS /VP)]
2
, gm can be
obtained as
P
P
GS
DSS
GS
DS
m
V
V
V
1
I2
dV
dI
g
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
−
== (1)
Since
DSS
DS
P
GS
I
I
V
V
1 =−
we have that
P
DSSDS
P
DSSDS
DSSm
V
I.I
2
V
I/I
I2g ==
The maximum value of gm is obtained when VGS = 0 (IDS = IDSS) and
is given by gm (VGS = 0) = gm0 = 2IDS / (VP).


Figure 1: A metal oxide semiconductor field effect transistor
Example 1
For the power MOSFET whose characteristic curves are shown in
Figure 1 (b), find the gain gm.
Solution
From Equation 1, we see that the gain gm is the change in output
current due to the corresponding change in input gate voltage. Thus, we
pick two convenient gate voltages—say, 5 V and 6 V (which represents a
change of 1 V) — and divide that into the corresponding change in the load
current IDS
:
)mho1or(S1
V1
A1
)V5V6(
)A1A2(
V
I
g
GS
DS
m ==
−
−
==
Δ
Δ
Some important characteristics of Motorola MTW24N40E power
MOSFET are as follows:
1. Maximum load current (IDS
) is 24 A.
2. The gate threshold voltage (VGS(th)), the gate voltage that just
causes the FET to conduct, is between 2 and 4 V.
3. the minimum FET gain, called the forward transconductance
(gFS
), is 11 mho. (Note: gFS
is the same as gm.)
4. Total power dissipation for the FET (PD
) should be less than 250 W.
Example 2
A particular power FET (MTW24N40E) was found to just start
conducting when VGS
= 2.5 V. Also, when VGS
= 3 V, the load current (ID
)
was measured to be 6 A. Is this FET operating within specifications?
Solution
The spec sheet shows that the gate-threshold voltage VGS (th) can be
between 2 V and 4 V. Thus if this FET starts conducting with a gate


voltage of 2.5 V, it meets this spec.
The second set of data gives one operating point. Specifically, we
are told that an input voltage (VGS
= 3 V) gives a corresponding output
current (ID
= 6 A). These numbers allow us to calculate the gain
(transconductance)
12
V5.0
A6
)V5.2V3(
)A0A6(
V
I
gg
GS
DS
FSm ==
−
−
===
Δ
Δ
The operating value of gFS
= 12 is above the minimum spec sheet value
of 11, so the FET is OK.
EXERCISES
1. A transistor has a current gain of 60 and a collector current of 5 A.
Find the base current.
2. A transistor has a current gain of 40 and a base current of 25 mA.
Find the collector current.
3. Calculate the exact and approximate emitter currents for the
transistor described in Exercise 12.
4. A transistor with a gain of 35 has an input curve similar to that
shown in Figure. The base voltage (VBE
) is measured to be 0.8 V.
a. Find the base current.
b. Find the collector current.
5. For the transistor circuit shown in Figure, calculate IC if the
resistor R1 is changed to 1.3 kΩ. (Assume input curve Figure applies.)
6. For the transistor circuit shown in Figure 4.19, calculate the IC if
the resistor R2 is changed to 120 kΩ. (Assume input curve Figure applies.)
7. For the FET circuit shown in Figure, find the load current if the
gate voltage is 6.5 V.
8. A circuit uses the MTW24N40E MOSFET (Figure) and has a gate
voltage (VGS
) of 5 V. Find the minimum and maximum drain current that
could be expected in this circuit. (Assume gm = 12 mho.)

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