Download Bipolar Junction Transistors

Bipolar Junction Transistors
Conotinuing the study of semiconductor devices, we now turn to a more complicated structure of three alternating doped semiconductor regions, better known as
the bipolar junction transistor (BJT). The BJT, like all other transistors,1 is capable of being used as a large-signal (DC) switch or as a small-signal (AC) amplifiers.
In this chapter, we will build on our analysis of the diode to characterize the BJT
as an ON/OFF DC switch and as a physical amplifier that produces a gain greater
than one. In later chapters, we will consider the fundamentals of logic operations
using these DC switches. Just as an introduction to the lingo, we will be considering DC bias calculations to obtain the bulk characteristics of the BJT as well
as a relatively small superimposed small-signal or AC component that contains
information.
Physical Structure of the BJT
The semiconductor structure of a BJT is that of three alternating semiconductor
regions, either npn or pnp. We will perform our analyses for the npn-BJT shown
in Figure X, which can be modified easily for the pnp.
Figure 1. Bipolar Junction Transistor
The three regions of the BJT are labeled base (B), collector (C), and emitter
(E) to reflect what their basic operation is in terms of physical charge carriers. You
should immediately notice that the BJT’s structure is approximately the same as
two reversed diodes shown in Figure X; the analysis will use this observation as a
starting point.
1
Besides the BJT, electrical engineers also use (Metal-Oxide-Semiconductor) Field Effect Transistors (MOSFETs or just FETs) and Junction Field Effect Transistors (JFETs), each of which
may be used as DC switches or AC amplifiers. The physics and DC bias equations are more
complicated, so we will focus on the BJT as the “prototypical” transistor.
1
2
Figure 2
For diodes, we generally considered two possible cases of the diode voltage: V D <
0 Volts (0.7 Volts for the more practical diode) and VD > 0 Volts (again, 0.7 Volts
for practical diodes) corresponding to the diode either being ON or OFF. With three
semiconductor regions (or two diodes), there are now four possible combinations of
voltages between the terminals. We will only consider the forward-active mode,
which is when the collector-base voltage, VCB , and the base-emitter voltage, VBE ,
are both positive (consequently, VCE will be positive as well).
Figure 3. BJT Operating in Forward-Active Biasing Region.
We will most often treat the base terminal as the input terminal and the collector
as the output terminal. Delving into the internal operation of the BJT, we do indeed
see diode characteristics from base to emitter (with the same polarity as V BE )
and another diode from base to collector (this time opposite to the polarity of a
positive VCB ). The relative magic of the BJT is caused by a physical process called
avalanche multiplication across the p-type semiconductor region. When operated
with DC voltages in the forward-active mode, electrons are excited along the edge of
the emitter region with enough energy to overcome the base-emitter electric field, 2
and hop into the base region. Once in the base region, the electrons are quickly
propelled through the base due to the reverse-biased collector base junction; the
overall effect is a flow of electrons being emitted by the Emitter and collected by
the Collector.
If you are still having trouble picturing these descriptions, consider the quasimechanical diagram in Figure X.
2Recall that the position of an individual electron is a question for probability rather than
deterministic equations.
3
Figure 4
We draw loosely-held marbles in an otherwise flat board to represent the electrons
in a the emitter region, a flat base region filled with holes where the marbles can
fall into and become trapped, and then a strongly sloping collector region (due
to the reverse-biased base-collector junction). When enough energy is given to the
electrons in the emitter, they will be propelled towards the base region with enough
momentum to pass straight through the base (although a small number will become
trapped in the holes) and fly through to the collector. Also, you an imagine these
mobile marbles (electrons) colliding with other electrons, creating an avalanche of
carriers over the junction between regions. You should notice that the voltage
polarities are seemingly labeled backwards: gravity makes objects fall down, while
an electron in an electric field actually falls up.
A more careful analysis3 will show that the number of electrons crossing into
the base region and then directly into the collector is a constant factor β, known
as the BJT’s performance parameter, of the number of electrons exiting through
the base, measured as IB . Typical β’s of good BJTs range from about 50 to 200,
so the positive current from collector to emitter (flow of electrons from emitter to
collector) will largely dominate the base current. The nearly-completed model for
the BJT is shown in Figure X.
Figure 5
The currents in each of the diodes and the controlled current source may be
written using the reverse saturation currents of each of the diodes.
3Refer to Pierret’s Semiconductor Device Fundamentals
4
IBE = I0BE · (e
VBE
VT
− 1)
IBC = I0BC · (e
−
VBC
VT
− 1) ≈ 0
ICE = β · I0BE
The final modification is to notice that the semiconductor junction (diode) between base and collector is reverse-biased and replace it by an open circuit (diode
turned OFF). The final model of the BJT operated in forward-active mode is shown
in Figure X.
Figure 6. Block equivalent of a BJT.
DC Model of the BJT
The first task we undertake whenever introducing any new electrical device is to
obtain the IV characteristics; for a BJT, we will be interested in how the collector
current, IC , changes with respect to the voltages VBE and VCE . The voltage VBE
sets the current into the base, IB , while the voltage VCE controls the width of the
semiconductor junctions and as a result contributes to the current from collector to
emitter. The classic IV curve for the BJT appears in Figure X (the multiple curves
are for different values of VBE ).
Figure 7
5
The characteristics to notice are the ratio between the collector current and the
base current, which we earlier termed the performance parameter β, and the small
degree of inclination in the forward active region (VCE > VBE ≥ 0.7V ). This slope
actually represents a non-ideality of the resistor corresponding to the AC output
resistance between collector and emitter. We quantify this slope by the imaginary
voltage intercept created by extrapolating the line far to the left. Calculation of
this Early voltage will later allow us to calculate AC characteristics of the device.
Figure 8. Calculation of the Early Voltage, VA
Example: Calculate the performance parameter and the Early voltage of a BJT
with the IV curve shown in Figure X.
Figure 9
The first and easiest to determine is the performance parameter, β. By estimating the average value of IC to be roughly 2.2 mA, we calculate β as the ratio of IC
to IB .
IC
2.2 mA
≈
= 63
IB
35 µA
To calculate the early voltage, we first determine the slope of IC versus VCE
β=
6
0.2 mA
mA
= 0.02
10 V
V
and then extrapolate the line to the left to find the imaginary intercept.
slope ≈
|∆IC | ≈ 2.2 mA
=⇒
VA ≈
(2.2 mA)
= 110 V
(0.02 mA/V )
A Simple DC BJT Circuit
In the last chapter, we saw that the diode could be treated as a type of voltageinduced switch; the BJT may also be used as a DC switch, but we will have the
added flexibility of a control voltage that is not in the same series voltage loop as
the switch. The simplest circuit in which we can see the BJTs response as a switch
is the BJT inverter in Figure X.
Figure 10. BJT Inverter
Do not be fooled by the way that the circuit is drawn in Figure X: the bottom
node has a ground symbol signifying 0 volts, while VDD at the top represents a
fixed DC potential (must be positive if in forward-active mode). The equivalent
method of drawing the BJT with obvious loops is shown in Figure X; you should
quickly realize how much easier Figure X is to draw than Figure X.
Figure 11. BJT Inverter
Testing the extreme values of input voltages will allow us to see that the current
flow in the BJT can be turned ON or OFF depending on the base voltage. When
VB is set to zero, the diode voltage from base to emitter is zero, which corresponds
to a flow of zero current. The controlled source between collector and emitter will
also produce zero current, forcing the current through, and therefore the voltage
dropped across, the resistor to be zero. As a result, Vout = VDD . When VB is set
to VDD , the BJT will try to transmit as much current as the rest of the system
will allow—the maximum current that may flow from collector to emitter (and
consequently through the resistor) is VDD
R . Such a large current flow forces the
entire voltage VDD to be dropped across the resistor, leaving an output voltage of
zero.
7
Vout =
(
VDD
0
Vin = 0
Vin = VDD
The output voltage (a real, physical quantity) must be continuous, so we require
that there be some input voltage between 0 and VDD where the output transitions
between VDD and 0, or simply high and low. These input/output relations also
show an interesting new tool for circuit design: a voltage-controlled switch. The
full input/output characteristics of the BJT inverter with VDD = 15 V appear in
Figure X.
Figure 12. .
You should notice that the calculated extreme points coincide well with the plot
and previous approximations: the output transition from high to low occurs at an
input voltage in the neighborhood of 0.7 volts (where the diode-like base-emitter
junction turns ON). Further, calculation of the base current will show that I B
DD
never exceeds the ratio VβR
which is typically on the order of a few µA. Taking
these results as a general set of assumptions for all BJTs, we will always use the
approximations VBE = 0.7 Volts and IB = 0 Amps (a BJT with 0 < VBE ¿ 0.7
V would not conduct much current and would therefore be considered as OFF–the
proper name is “cutoff” mode rather than forward-active).
8
Figure 13. .
AC Model of the BJT
Using the block-level model of the BJT in Figure XX, we may characterize an AC
response whenever the device is “ON” (operating in forward active) corresponding
to the addition of a small signal voltage to the DC base voltage. Take V B to be
the superposition of a DC voltage ( ≈ 0.7 Volts to turn the diode “ON”) and
a small sinusoidal signal, ∆VB (magnitude < 5 mV). The current into a diode,
and consequently the base of the transistor, can be written using the linear Taylor
approximation,
∆VB
∆VB
) = IB (VBDC ) + IB (VBDC )
Vt
Vt
from which we separate the AC portion of the base current
IB = IB (VBDC )(1 +
∆VB
Vt
The corresponding collector current in the controlled source will be the same AC
base current multiplied by the factor β. By transformming these characterizations
into a linear circuit, we obtain the circuit shown in Figure X as the AC model of
the transistor (comomnly known as the hybrid-π model).
∆IB = IB (VBDC )
Figure 14. .
The various parts of this AC model can be explained easily (all quantities ∆X
are the small signal perturbations of the DC quantities, X):
gm is the transconductance parameter that ultimately converts between changes
in base voltage and changes in collector current. Mathematically,
∆IC
β∆IB
βIB (VBDC )
IC
=
=
=
∆VB
∆VB
Vt
Vt
rπ is the input resistance that relates changes in base voltage to base current.
gm =
9
rπ =
β Vt
Vt
β
∆VB
=
=
=
∆IB
IB (VBDC )
IC
gm
ro is an output resistance that accounts for imperfections of the model in Figure
XX and for the physical limitations of semiconductor manufacturing. The formula
for this output resistance is
ro =
VA + VCE
,
IC
where VA is the Early voltage and VCE = VC − VE is the difference in DC
voltages between collector to emitter. The intuition underlying ro is grounded in
the fundamental device physics of the BJT,4 affected largely by the control of silicon
and doping impurities. The numerical values of VA can be anywhere from 50-200
Volts, corresponding to ro values on the order of 100 kΩ.
The non-idealities of a transistor can be explored far beyond those presented
above, but do not generally have a large effect. Hand analysis using the hybrid-π
model is the typical example of an engineering trade-off: simpler analyses are used
to obtain rough, but quick, solutions to hard problems. More precise answers are
then obtained by using a computer.
To summarize the basic AC parameters in terms of the DC bias points:
gm =
IC
Vt
rπ =
β
gm
ro =
VA + VCE
IC
Single-Stage BJT Amplifier
Now that we have characterized the transistor for both DC and AC voltages, we
are ready to build a single-stage BJT audio amplifier. The schematic in Figure XX
shows such an amplifier.
4A good reference is Grey & Meyer
10
Figure 15. .
We will assume that the values of the DC source voltage, VDD , Early voltage,
VA , the BJTs performance parameter, β, and all the resistors (typically measured
in kΩs) are all given. Further, we will generally disregard the specific values of the
capacitors C1 , C2 , and CE , since they are generally large (perhaps a few hundred
µF ) and used only to block the flow of DC currents while freely admitting the flow
of AC currents (within a certain range of frequencies). We can verify this property
mathematically by calculating the magnitudes of the capacitor impedances for the
extreme values of frequencies in the audio range (20 Hz to 20 kHz) and comparing
them to the impedances for DC signals (corresponding to ω = 0).
Example: Show that each of the capacitors in Figure X may be approximated
as DC open circuits and AC short circuits. Assume 330 µF capacitors.
To get an idea of the magnitude of the capacitor impedances, we simply plug in
a frequency of 0 Hz for DC signals and a range of values from 20 Hz to 20 kHz for
AC signals.
When ω = 0,
When ω = 20 · 2π,
When ω = 20000 · 2π,
1
|=
jωC
1
|ZC | = |
|=
jωC
1
|=
|ZC | = |
jωC
|ZC | = |
1
=∞
0
1
= 24 Ω
0.0415
1
= 0.024 Ω
41.469
The approximation that the capacitors are DC open circuits is extremely good,
while the largest resistance value within the audio range (20 Hz - 20 kHz) is no more
than ≈ 25Ω; compared to resistor values in the kΩ range, this capacitor impedance
is near enough to being a short circuit.
11
Figure 16
The other important trick that will come into play is that DC sources are AC
zeros; that is, the AC part, or time-varying component, of a DC source (constant)
is always zero. This observation suggests that the DC source voltage, V DD , will be
a ground node for the AC analysis.
DC Analysis
The massive amplifier circuit can be broken into three separate circuits (separated by C1 and C3 ) for DC analysis.
Figure 17. DC Equivalent of the Single-Stage BJT Amplifier.
We will calculate:
1. The DC input voltage
2. The DC output voltage
3. The DC bias points of the transistor (all the voltages and currents at the
transistor terminals)
DC Input voltage
The capacitor C1 blocks DC current from flowing out of the input source Vin into
the amplifier. Therefore, the DC portion of the input voltage is negligible to other
parts of the circuit and will often be taken as zero. If we consider multiple-stage
amplifiers, the DC input voltage may not be zero, but will not directly affect the
12
gain.
DC Output Voltage
The DC output voltage will not affect the actual amplifier block due to C3 , but
it may be useful in its own right. We will calculate VoutDC as a voltage divider
between Rout1 and Rout2 .
Figure 18
VoutDC =
Rout2
VDD
Rout1 + Rout2
BJT Bias points
We will use the approximations learned from the BJT inverter, namely that
IB = 0 and VBE = 0.7, to solve for the DC values of voltages and currents in the
amplifier circuit, called the DC bias. The order of solution will always be V B , VE ,
IE , IC , and VC . We can then re-approximate IB using the preceding solutions.
Base Voltage:
Based on the initial assumption that IB = 0 and the fact that the DC current
left through C1 is also zero, any current flowing downward in R1 must also flow
through R2 , which means that they are in series for the DC source voltage VDD .
The voltage VB is thus obtained by a routine voltage divider between R1 and R2 .
13
Figure 19. .
VB =
R2
VDD
R1 + R 2
Emitter Voltage:
To find VE , we need only apply the condition that VBE , the potential difference
between the base and emitter, is a constant 700 mV.
Figure 20. .
VBE = VB − VE = 700 mV
=⇒
VE = VB − 0.7 V
Emitter Current:
IE depends only on the resistor RE and the voltage across it, VE . We see
this because any current flowing out of the emitter must flow downwards through
14
RE (since no DC current may flow through the capacitor, CE ) creating a voltage
VE = I E RE .
Figure 21
IE =
VE
RE
Collector Current:
We obtain IC by considering the BJT as a generic three-terminal device and
applying KCL. The assumption that IB = 0 forces the collector and emitter currents
to be identical.
Figure 22
IE = I B + I C = I C
Collector Voltage:
The entire current entering the collector must go through RC in a similar fashion
to the relationship between IE and RE ; this current creates a voltage of IC RC across
15
the resistor RC . The collector voltage is then equal to the source voltage, VDD ,
minus the voltage drop across RC .
Figure 23
VC = VDD − −IC RC
Base Current revisited:
The initial approximation for the base current was zero since it is a (large) factor
of β less than the emitter and collector currents. The error introduced is only a
few percent, which is certainly acceptable for hand calculations. We may improve
the initial approximation by setting
’
IB
=
IC
β
AC Analysis
Now that we have calculated all of the DC voltages and currents in the circuit,
we will move forward to calculating the AC gain–this is where the real magic, or at
least practical concern, of the BJT as an amplifier becomes evident!
We begin the AC analysis by substituting short circuits for the capacitors and
redrawing DC voltage sources as AC zeroes (ground nodes). The resulting circuit
is shown in Figure X.
16
Figure 24. .
We can simplify this circuit slightly by redrawing it with all resistors collected
in recognizable forms as shown in Figure X.
Figure 25
You should notice that the emitter is now directly connected to ground (R E goes
away since it is in parallel with a short circuit). We then make the next substitution,
that of the AC model of the BJT, making sure to match up the correct nodes at
base, collector, and emitter.
Figure 26. .
We then collect the resistors outside the BJT and create a single circuit. Notice
that R1 and R2 are in parallel since they each go from base to emitter (ground
node) for the AC circuit, as are Rout1 and Rout2 . Also notice that the output
voltage, Vout is measured from collector to ground (emitter). The resulting overall
AC circuit is shown in Figure X.
17
Figure 27. .
Solving this circuit is much easier than it may appear. We break up the total
gain, A = VVout
, into a pair of voltage ratios that represent the effect of each side of
in
the circuit.
A=
Vout
Vπ Vout
=
·
Vin
Vin Vπ
π
Starting with the first voltage ratio, VVin
represents a voltage divider between the
parallel combination of R1 , R2 , and rπ and the input resistor, Rs .
Vπ
R1 kR2 krπ
=
Vin
R1 kR2 krπ + Rs
The second voltage ratio works out to just be an Ohm’s law calculation between
the controlled current source of magnitude gm Vπ and the parallel set of resistors
RC , ro , Rout1 , and Rout2 .
Vout
= −gm (ro kRC kRout1 kRout2 )
Vπ
The negative sign in the last term shows up because the flow of current in the
resistors is opposite the polarity of the output voltage. (Many students mistake
a negative gain as the signal getting smaller, thus defeating the purpose of an
amplifier. The actual effect is to boost the signal by a factor of |A| and induce a
180◦ phase shift.) Combining the two terms, we obtain the overall AC gain of the
BJT amplifier.
A = −gm (ro kRC kRout1 kRout2 ) ·
R1 kR2 krπ
R1 kR2 krπ + Rs
Example: Plot the output voltage of the single-stage BJT amplifier shown in
Figure X. You may assume that the device is operating in the forward active region,
and that the BJT’s parameters are: VA = 50 V and β = 200.
Figure 28
Step 1: Apply assumptions
VBE = 0.7 V, IB = 0 A, while the capacitors act as DC open circuits and AC
short circuits.
Step 2: Solve for the DC bias points. We will repeat the same order as before:
VB , VE , IE , IC , VC , andIB .
18
20kΩ
(20V ) = 4 V
20kΩ + 80kΩ
3.3V
IE =
= 1 mA
3.3kΩ
VB =
VE = 4V − 0.7V = 3.3 V
IC = 1 mA
1mA
= 5 µA
200
Other DC terms: Along with the DC bias of the transistors, we are also interested
in the DC output voltage and the difference of collector and emitter voltages (to
calculate ro .
VC = 20V − 8kΩ · 1mA = 12 V
IB =
20 kΩ
(20 V ) = 10 V
20 kΩ + 20 kΩ
Step 3: We calculate the BJT’s AC parameters using the DC bias points.
VCE = 12 V − 3.3 V = 8.7 V
1 mA
= 38.7 mΩ−1
25.86 mV
Step 4: AC Model
gm =
VoutDC =
rπ =
200
= 5.17 kΩ
38.7 mΩ−1
ro =
100 V + 8.7 V
= 108.7 kΩ
1 mA
We begin by replacing the capacitors with short circuits, and all DC sources by
AC zeros (ground nodes). The second step is to substitute the AC model of the
transistor, collecting resistors where possible.
Figure 29
Solving for the intermediate gains,
Vπ
20kk80kk5.17k
=
= 0.987
Vin
20kk80kk5.17k + 50
Vout
= (−38.7 m0) · (8kk20kk20kk108.7k) = −165
Vπ
we can obtain the overall AC gain, A.
Vout Vπ
Vout
=
·
= −163
Vin
Vπ Vin
The most common error in any of these calculations is to neglect the 10 3 or 10−3
implied in “kΩ” or “mA.”
A=
Step 5: Plot Vout
Vout = VoutDC + Voutac = VoutDC + A · Vin = 10 − 0.489 · cos(1000t)V
Figure 30
19
Exercise: Determine the values of RE and β such that the single-stage amplifier
pictured in Figure X has a transconductance of 50 m0 and a total AC gain of 150.
Assume that the Early voltage is infinite, VA = ∞.
Figure 31
Comments on the Analysis
The hardest part of analyzing BJT amplifiers is to gain an intuition for whether
or not the various calculations result in reasonable values. As rules of thumb, you
should calculate all DC voltages to be between the upper and lower supply voltages, usually zero and VDD , values of gm measured in m0, rπ measured in kΩ, and
ro in hundreds of kΩ, resulting in gains of roughly 50-200 (remember that gains
are negative). These values are intentionally vague, but relatively consistent for
practical single-stage audio amplifiers. You should also notice that the value of DC
bias voltages, and consequently AC parameters and gain, can change by varying
component values within the circuit, which is where we turn next.
Effects of Varying Component Values
Scenario: Your company has completed the manufacture of a major system, but
one portion fails because a single-stage BJT amplifier has a gain that is 10% too
low. The design was contracted out to another company who is unavailable on the
weekend, so nobody can help. To make things worse, the amplifier is on a single
chip, so you cannot change any of the component values (resistors or capacitors).
What do you do?
The only parameter that you still control is the power supply, so you pull a piece
of scrap paper and begin playing with equations: when you increase the source voltage, the base voltage goes up, with a corresponding increase in the emitter voltage.
That causes the emitter current (and collector current) to also go up, resulting in
a larger transconductance. You know that gain is linearly related to gm (and that
DC sources are AC zeros independent of value), so you suggest adding a couple of
AA batteries out of your walkman in series with the source voltage. Bingo: the
system works, and your two AA batteries earn you a nice bonus.
Yes, it’s a little farfetched. But it is by no means inconceivable: as an engineer
of any form, people will expect you to have the answer to everything mechanical,
electrical, industrial, etc, and will not always choose to be civil in their requests!
The rest of the chapter on BJTs is a fairly simple analysis of “what if;” that is,
what happens to the (gain/power/etc.) when we vary (R1 , RC , etc.). This is more
common than it might seem since most practical resistors are only accurate within
5% of their nominal values, and other components may vary widely.
Following the example, we saw that an increase in VDD caused increases in VB ,
VE , IE , IC , and gm . Had we checked, there would have been a decrease in rπ and
ro , as well as an increase in VoutDC . Rather than explain these in words, it is much
20
easier to tabulate them as a positive, negative, or no change. The table below shows
the effects of increasing VDD and that of reducing R1 on the other circuit values.
table example
You will notice that the gain, A, depends on many terms, some certainly having
more significance than others. The table above shows that the overall gain (magnitude) increased despite the fact that rπ decreases, causing a marginal decrease
π
. The overall gain equation depends more heavily on the
in the voltage ratio VVin
increase in the transconductance than the decrease in input resistance (test small
changes to prove this to yourself).
The process above works well for incremental changes in component values, but
what if we drastically change a value? Let’s consider RC → ∞. Plugging and
chugging through equations will work until you obtain a collector voltage of −∞.
Clearly, DC voltages should be between the source and ground, so something is
wrong! RC → ∞ means that the collector resistor is being replaced by an open
circuit, which precludes current from entering the collector, effectively turning the
transistor off as an amplifier. While we will not explore all the possible cases of
extremes in component values, you should realize that the equations derived only
hold when the BJT is biased in the forward active mode.
STUFF TO ADD: One more exercise of one-stage amp, variation on R2 exercise, variation on Rout2 example with percentages, variation on beta exercise,
complete table for changes,
Questions:
1. For the BJT inverter shown in Figure X, calculate how much power is dissipated in the circuit when the input voltage is 0 V and when it is VDD .
2. Power dissipation is a major concern in battery operated electronics, many of
which use an Inverter similar to that of the BJT inverter. What condition should
be placed on the series resistance, R, to minimize the power dissipation?