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# The Transistor Common Emitter Amplifier

A basic transistor amplifier which has the emitter lead in common with the
input and output circuits is shown below. The input voltage
is between base and emitter and the output voltage between collector and emitter.
The voltage *V*_{BB} is introduced into the base circuit so that
the emitter-base junction is forward-biased for control of the
emitter-collector current. The base-to-emitter voltage is referred to as
*v*_{be} and the base current as *i*_{b}.
The power supply voltage *V*_{CC} is necessary to the circuit
to make the collector positive with respect to the emitter and to provide a
source for the output current. The collector-to-emitter voltage and collector
current are *v*_{ce} and *i*_{c}, respectively.
The resistance *R*_{L} is the load. The base current of
the transistor controls the current through the collector and through the load
resistor.

## Output Characteristic Curves

Unfortunately, the resistance of the emitter and collector junctions of transistors are not always constant. Hence, Ohm's Law cannot always be used to express the relationship between the various currents and voltages in a transistor amplifier. For this and other reasons, this information is usually supplied by the manufacturer in graphical form. The figure below shows a graph of the collector (output) characteristics for the common emitter configuration of a transistor. The graph consists of a series of curves. Each curve shows, with the base current held constant, the collector current variation as the collector-to-emitter voltage is changed.

There are particular values for *i*_{b}, *v*_{be},
*i*_{c}, and *v*_{ce} even when no signal is applied
to the input. The no-signal values of *i*_{b}, *v*_{be},
*i*_{c}, and *v*_{ce} are called the **quiescent**,
or average, values and are given the symbols *I*_{B}, *V*_{BE},
*I*_{C}, and *V*_{CE}. The point defined by
*I*_{C} and
*V*_{CE} on the output characteristic curve is called the quiescent point (Q).
The location of the quiescent point on the characteristic curve is an important
part of the analysis of an amplifier circuit. The procedure is as follows:
First a "load line" is determined for the values of supply voltage
*V*_{CC} and load resistance *R*_{L} which are to be
used in the amplifier circuit. Assume that *i*_{b} is so small
that the collector current *i*_{c} = 0. If this were true, the
voltage drop across *R*_{L} would be zero and
*v*_{ce} would equal *V*_{CC}. This point 1 is plotted
on the output characteristic curve (figure above). Then it is assumed that the
base current is so large that the transistor becomes a perfect conductor so that
*v*_{ce} = 0 and
*i*_{c} = *V*_{CC}/*R*_{L}. Plot this
point 2 on the characteristic curve, and join points 1 and 2 with a straight
line as in the figure above. This line is called the load line because it is
determined only by the values of the load and *V*_{CC}.

For the selected values of *V*_{CC} and *R*_{L}
the resulting values of *v*_{ce} and *i*_{c}
must fall along the load line. The figure above, point Q, shows the no-signal values
for *V*_{CE} and *I*_{C} as determined by the load
line and the output characteristic for the selected average base current
*I*_{B} = 100 µA. The instantaneous base
current *i*_{b} depends on the sum of base bias current and input
signal, and, by moving along the load line, the instantaneous values of
*v*_{ce} and *i*_{c} can be read from
the graph for any given value of *i*_{b}.

Note that the load line should fall well within the maximum
collector-dissipation line. The maximum collector dissipation
*P*_{max} is a characteristic of the transistor and is usually
given in the transistor manuals or catalogue descriptions. The maximum
collector-dissipation line is drawn by connecting all the points which satisfy
the equation *I*_{C}*V*_{CE} = *P*_{max}.
The quiescent base current can be estimated by assuming that the base-to-emitter
junction is simply the transistor input resistance *h*_{ie}.
If *v*_{s} = 0 (quiescent state),
*I*_{B} = *V*_{BB}/(*R*_{S}
+*h*_{ie}). Often *R*_{S} will be very much greater
than *h*_{ie} so that
*I*_{B} ≈ *V*_{BB}/*R*_{S}.
Thus the choice of bias voltage will depend on the nature of the signal source.
Other methods of biasing will be discussed later.

## Transistor Amplifier Characteristics

The change in *v*_{ce} or *i*_{c} as a result of an
input signal (a change in *i*_{b}) could be determined by a
graphical analysis of the characteristic curves as above for each value of
*i*_{b}. However, it is much more convenient to analyze the
circuit mathematically after substituting an equivalent circuit for the
transistor. It is not possible to derive a simple circuit which would replace
the transistor for its entire operating range. This is because the
characteristic curves of the transistor are not linear, as shown e.g. in the figure
above. A linear equivalent circuit is only a valid approximation over small
portions of the operating range. In other words, the equivalent circuit
will not describe the relationship of the total quantities *i*_{b},
*v*_{be}, *i*_{c}, and *v*_{ce}, but it
will describe the relationship of the small changes in these quantities.
Equations for the various amplifier characteristics may be derived from the
equivalent circuit and the mathematical relations used to obtain that circuit.
Some of the characteristics (e.g. the voltage and current gains) of the common-emitter
amplifier are given below. The actual derivations are not shown here.

The **voltage gain** could be approximated by the equation

where *β* is the transistor current gain
and *h*_{ie} is the transistor input resistance in the
common-emitter circuit. The minus sign indicates that the
output voltage is 180° out of phase from the input voltage.

The **current gain** *A*_{i} is approximately

Again the minus sign is an indication of the phase reversal between the input and output signals.

The **power gain** *A*_{p} is simply the product of the current and
voltage gains

The power gain does not include power losses in the signal source or in
transferring the output signal to a load other than *R*_{L}.
It is simply the ratio of the signal power dissipated in *R*_{L}
to the signal power dissipated in the transistor input resistance
*h*_{ie}.

The **input resistance** *R*_{in} is

The **output resistance** *R*_{out} could be approximated by
the equation

where *h*_{oe} is the conductance between the collector and emitter.
As this equation indicates, the output resistance *R*_{out} is in
this case the resistance that the transistor as a power source presents to the
load *R*_{L}. The output resistance of the entire amplifier circuit
as seen by a device attached to the output terminals *v*_{ce}
would actually be *R*_{out} in parallel with *R*_{L}.

The common-emitter amplifier is thus shown to possess the properties of voltage gain, current gain, phase reversal of the input signal, low input resistance, and high output resistance.

## Transistor Biasing

The use of a separate power source for the input and output circuits of
the transistor amplifier is costly and inconvenient. Methods of providing
the proper bias voltages from a single power source have, therefore, been
devised. One of the most widely used bias systems is the voltage-divider
type shown in the figure below (see the
Types of Bias section for more info).
Fixed bias is provided in this circuit by the
voltage-divider network consisting of *R*_{B1}, *R*_{B2},
and the collector supply voltage (*V*_{CC}).
Resistor *R*_{E}, which is connected in series with the emitter,
provides the emitter with self-bias. A large capacitor
bypassing *R*_{E} will alleviate the loss in gain for AC signals.

**Quiescent Point Determination**

After it has been concluded that widely used transistor amplifiers could
have a resistance in the emitter circuit and a voltage divider in the base,
it is suitable to learn how to determine the quiescent point for such circuits.
First, the load line is drawn in the usual way (see the graph of output
characteristics above), care being taken to use *R*_{E} +
*R*_{L} in calculating the current for *V*_{CE} = 0.
Now the circuit of figure above is redrawn considering the voltage divider as
a series battery *V*_{BB} and resistor *R*_{B}
(see
Thevenin's theorem).
It is first approximated that the emitter and the base
have very nearly the same potential (*V*_{E} = *V*_{B}
≈ *V*_{BB}) and that *I*_{E} is very nearly
*I*_{C}. Therefore,
*I*_{C} ≈ *V*_{BB}/*R*_{E}.
Looking at the load line to find the value of *I*_{B} corresponding
to this value of *I*_{C}, one can improve on the approximation
by correcting the base voltage to
*V*_{BB} - *I*_{B}*R*_{B}. Use this
value of the base voltage to calculate a better value for *I*_{C}.
Usually this second approximation is sufficiently accurate.

## Example of Transistor Bias Design

An example of how to go about determining reasonable resistance values for
a transistor-amplifier circuit is given here. First, choose the desired quiescent
point from the transistor's collector-characteristic curve. Evaluate from the
characteristics or look up values for *β*, *h*_{ie},
and *h*_{oe} in datasheets. Assume that, at the desired quiescent point
(*I*_{C} = 1 mA, *V*_{CE} = 5 V),
*h*_{oe} = 55, *h*_{ie} = 2720 Ω, and
*h*_{oe} = 14 μS. Choose *R*_{L} according
to gain requirements and impedance matching, high values for current gain,
moderate values for voltage gain with moderate output impedance. For
this example *R*_{L} = 25 kΩ. *R*_{B}
should be larger than *h*_{ie} to prevent
excessive signal loss through the bias divider. Therefore, choose
*R*_{B} = 25 kΩ. For good stability choose
*R*_{E} = *R*_{B}/5 = 5 kΩ.
The required DC supply voltage is then
(assuming *I*_{E} ≐ *I*_{C})

The bias voltage *V*_{BB} should equal the base
voltage plus the voltage drop
*I*_{B}*R*_{B}. *V*_{BB} is thus

where *V*_{BE} = 0.7 for silicon transistors.
*V*_{BB} can also be calculated as

Parallel combination of *R*_{B1} and
*R*_{B2} is *R*_{B}

Solving the two previous simultaneous equations for the bias resistors,
*R*_{B1} = 142 kΩ and
*R*_{B2} = 30.3 kΩ. The formulas derived previously for
the voltage- and current-gain of the transistor amplifier apply also to this circuit.
The amplifier output resistance is 1/*h*_{oe} in parallel with
*R*_{L}. The amplifier input resistance is *h*_{ie}
in parallel with *R*_{B}.

Note: The current through the voltage divider *R*_{B1},
*R*_{B2} shoud be at least 10 × *I*_{B}.
In our case the current through *R*_{B2}