Biasing Bipolar Junction Transistors (BJTs)
Introduction
We now consider a
BJT
amplifier circuit that is especially useful because it
maintains the operating (quiescent) point selected by the designer pretty much
independent of the transistor parameters. T
his characteristic of the circuit is
important in manufacturing because the values of parameters for individual
transistors of a specified type can differ dramatically from the values typical of
that type. The parameters even for a particular transistor ca
n change
substantially as the temperature changes.
This **** miracle circuit is actually very simple:
The voltage
is the power supply voltage. The capacitors
and
are
coupling (or d

c blocking) capacitors. They are chosen large enough to have a
very low (more on how

low

is

low later) impedance for all signal frequencies.
They therefore permit the signals to flow into and then out of the amplifier (and
perhaps
into subsequent amplifier stages for more amplification) practically
unimpeded while blocking the passage of any bias (d

c) currents. The operating
point under quiescent conditions (no input signal) is therefore determined solely
by the circuit between
and
: that is, solely by
,
,
,
and
. (In
principle, the operating point dep
ends also on various parameters of the
transistor,
, but that they do not have much influence is a special feature of this
miracle circuit. More on this later.)
How does this circuit work? The key point is that the voltage divider
formed by
and
sets a reference voltage that (ideally) maintains the base voltage of
the transistor,
, at a constant value. Now let's consider what happens in this
circuit if t
he operating point tries to change. Suppose the current through the
transistor,
, tries to increase (because of a change in temperature, for
example). This increase increases the voltage across the emitter resistor,
But
since the base voltage of the transistor is fixed by the voltage divider, an increase
in
causes the base

to

emitter voltage,
, to decrease. But, according to the
characteristic curves fo
r a
BJT
, a decrease in
leads to a decrease in the base
current,
, that in turn leads to a decrease in the collector current,
. The circuit
therefore automatically acts to counte
ract any increase in the transistor current at
the operating point. In fact, it is easy to see from a similar argument that the
circuit also counteracts any decrease in the collector current. In summary, then,
the circuit automatically tries to keep the op
erating current through the transistor
at a constant value. (This effect is a particular example of
negative feedback
, a
process that we will consider in more detail later. A thermostat is another
example of negative feedback.) Because, all other things be
ing equal, any
change in collector

to

emitter voltage for the transistor,
, produces a
corresponding change in collector current, the circuit also attempts to maintain a
constant operating voltage,
, acro
ss the transistor. In short, the circuit tries
very hard to maintain the operating point for the transistor that the designer
intended.
We now outline a procedure for choosing values of
,
,
,
and
that
will set the operating point of the transistor where we want it, or at least near
there. Although we will write down equations to document the procedure and to
refer to during lat
er analysis,
your objective should be to learn to carry out the
procedure without memorizing a single equation
.
Procedure
1. For present purposes, assume that
(the small control current in a
BJT
) is
negligible and that
. (
depends mainly on what material the transistor
is made of, not on the size or geometry of the transistor. For silicon transistors,
the most common type,
.)
2. Choose
(or
) and
(
by
KCL
since
is negligible) to specify the
quiescent operating point you want. (If you choose the operating voltage and
current t
oo large, the transistor can burn up. The data sheet for the transistor
may contain a limit for the power dissipated by the transistor,
. The
product of
and
must be less than t
his value.)
3. Choose
. (Later, we'll see that we need to choose
for good
bias stability. For silicon transistors, therefore, we can achieve good bias stability
without using up too much power supply vol
tage by choosing
.)
4. From the quiescent operating point, we can determine
from Ohm's law:
5. Choose
. For maximum symmetric swing (operating point a
t the center of
the load line under quiescent conditions), choose
Thus, under quiescent conditions, we can determine
from Ohm's law:
(Later, we'll see that the value of
can be chosen to set the gain of the
amplifier, at the cost of sacrificing maximum symmetric swing. For small signals,
maximum symmetric swing may be unimportant. Maximum symmetric swing is
most important for large signals, such as tho
se in power amplifiers.)
6. Calculate
from KVL:
7. We have determined
,
and
at this point. We s
till must choose
and
. We now develop two simultaneous equations to determine them. As a
preliminary event, we calculate the base reference voltage:
8. To get the first equation for
and
, we impose the condition that
be
divided down to
.
9. To g
et the second equation, set the current in the base voltage divider to 10%
of
. (This choice, as we'll see later, gives good bias stability yet increases the
current drain on the power supply by only 10%.)
(This calculation neglects
, which we know is small. Later, we'll justify this
assumption in more detail.)
10. We can solve for
and
by substituting the second equation into
the
first:
11. Now that we know
, we can solve for
.
The design of the amplifier is now complete
:
,
,
,
and
are all
determined. Remember that we have chosen
for maximum symmetric swing.
Later,
we'll see how to choose
to obtain a specified voltage gain from the
amplifier.
Example:
Choose a quiescent point for a transistor in the circuit above and use
the 2N2222 NPN BJT model in PSpice to see how close the quiescent poin
t in
the simulation is to your design values.
For the desired operating point, we choose, arbitrarily:
Thus,
For maximum symmetric swing, choose
so
that
and
Now we must choose
and
. Set the current through the divider to 10% of
, or
2 mA
. Thus,
and
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