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The commonemitter transistor circuit is commonly used for voltage
amplification, as shown in the example below. Here we assume
and
, and .
The input voltage and current
Assuming the transistor is properly biased so that , we get
the input voltage
as the superposition of DC component
and a small AC inpupt
:
and find the base current by the input characteristic plot:
Due to the small dynamic range of the input voltage, the nonlinear
(approximately exponential) input characteristic can be linearized
locally as a resistance, the reciprocal of the slope of the input
characteristic curve around :
The base current can be approximated to be
which is also a superposition of a DC component and an AC
component with an amplitude . (Why can't we get the base current
as
?)
The output voltage and current
The output characteristic plot is shown below:
Here the value is
.
The load line is the plot of equation
, a straight line
that goes through the two points:
The collector current and the output voltage can be obtained
either algebraically or graphically:
 The output current:
 The output voltage:
Note:
 The AC sinusoidal component of the input voltage is amplified
times.
 The sinusoidal component of the output voltage (
)
is out of phase compared to that of the input signal
(
), i.e., the CE transistor circuit is a reverse amplifier.
The waveform of the output may be distorted if the DC component of
the input voltage (and thereby, the base current ) is either too
low or too high, causing the positive or negative peaks of the sinusoidal
component to exceed the linear range of the output characteristic plot, as
illustrated below:
We see that severe distortion in output will be caused if a transistor
amplification circuit is working near either the cutoff or the saturation
region. It is therefore desirable to properly set the DC operating point
around the middle of the linear range along the load line, to avoid to be
too close to either the saturation or cutoff region.
Example
Assume ,
, . Given the input voltage
or , find the corresponding output voltage
.

, the forward bias of BE PNjunction is insufficient
for it to conduct current, we have ,
,
. The transistor is cutoff (the switch is
open or opencircuit).
 , the BE PNjunction is forward biased, we can find
from the input characteristics, here assumed to be , and get
and
.
The transistor is in linear region.

, the BE junction is forward biased, we can find
from the input characteristics, here assumed to be . If the linear
relationship were to hold, we would get
and
.
This result is obviously wrong, indicating that the transistor is actually
in the saturation region (the switch is closed or shortcircuit),
i.e., the linear relation does not hold. In fact, it is
impossible for the transistor to draw from the voltage source,
as the maximum current is
when
. In this case, the actual can be approximated on the output
characteristics to be about , the intersection of load line and the
curve corresponding to ), and
.
Conclusion: a change in input from 0.2V to 0.8V switches the output
current from 0 to about 10 mA, and the output voltage from 15V to 0.2V,
and the transistor is in cutoff, linear, and saturation region, respectively.
is only valid when the transistor is in the linear region.
Next: DC Biasing
Up: ch4
Previous: Bipolar Junction Transistor (BJT)
Ruye Wang
20140802