Download Section F5: Darlington Circuit

Section F5: Darlington Circuit
To provide improved performance and input/output
characteristics, single transistors may be combined
to form compound devices. A commonly used
compound device is known as the Darlington
configuration and is shown to the right (a modified
version of Figure 8.16 in your text). In this
representation, two npn BJTs are cascaded and are
behaviorally equivalent to a single npn transistor.
This single compound device possesses desirable
characteristics such as high input impedance, low
output impedance and high current gain; but does have the disadvantages
of an almost doubled VBE (overall VBE for the pair is 1.2V to 1.4V instead of
the 0.6V to 0.7V for single silicon BJTs) and the fact that any leakage
current from the first transistor is amplified by the second transistor. A
Darlington pair may also be created using two pnp devices, particularly in
discrete circuit design, or through the use of an npn and a pnp. The resulting
compound device may be considered a single transistor and, in the following
discussion, will be used in either the CE or EF (CC) configuration.
Assuming rO1 and rO2 are very
large so that they may be
neglected, and that β1 and β2 are
both much greater than one (i.e.,
β1≈ β1+1 and β2≈ β2+1), the ac
small signal model of the npn
version of the Darlington pair is
shown to the right. By making
the assumption that emitter
currents are approximately equal
to
collector
currents
(i.e.,
α1=α2≈1):
i e = i e2 = β 2 i b2 = β 2 i e1 = β 2 β 1 i b1 ,
we can see that the combination looks like a single high β (β=β1β2)
transistor. Note that although the effective collector current iC calculated by
adding iC1 and iC2, the multiplier of β2 makes iC2 the dominant contributor to
the sum. Also note that the Q-point for the first transistor may be different
from the second transistor so, in general, rπ1≠rπ2.
Darlington EF (CC) Amplifier
Figure 8.17, reproduced to the left below, illustrates the Darlington pair used
in an EF (CC) amplifier configuration. Figure 8.18, corrected and presented
to the right below, shows the small signal model for this EF (CC) amplifier,
using ideal capacitors, with the approximations that iC=iC2 (since the output
is taken at the emitter and the contribution of Q2 dominates the total
collector current) and that the impedance from Q1, when reflected to the
emitter circuit of Q2, is negligible; i.e.
rπ 1 / β 1 + rπ 2 ≈ rπ 2 = β 2 re2 .
From the small signal circuit in Figure 8.18 (above right), we can see that
the load on the second transistor (at E2) is RL||RE. The equivalent load on
the
first
transistor
(using
impedance
reflection
at
E1)
is
β2(RL||RE)+rπ2≈β2(RL||RE), which is also the input resistance of the second
transistor. Your author states that, in practice, the first transistor can be of
lower power rating than the second. This is due to the β1 multiplication
factor of currents through the second transistor.
Using the impedance reflection technique to reflect all resistances to the
base circuit of Q1, we can define the input resistance of the Darlington EF
(CC)amplifier as:
Rin = RB || [β 1 rπ 2 + β 1 β 2 (RE || RL )]
= RB || [β 1 β 2 (re2 + (RE || RL )]
.
(Equation 8.40, Corrected)
Just for information, if it happens that rπ1 cannot be neglected, the complete
expression for the input resistance is:
Rin = RB || [rπ 1 + β 1 rπ 2 + β 1 β 2 (RE || RL )] = RB || [β 1 re1 + β 1 β 2 (re2 + (RE || RL )] .
Using our standard definition for current gain as iL/iin, we can derive the
required currents by current division at the input and output:
iL =
i b1
RE β 1 β 2 i b1
RE + RL
R + β 1 β 2 [re2 + (RE || RL )]i b1
RB i in
=
; or i in = B
RB + β 1 β 2 [re2 + (RE || RL )]
RB
.
Taking the ratio of currents and dividing by β1β2, we get the final expression
for the current gain to be
Ai =
RB
RE
.
RB / β 1 β 2 + re2 + (RE || RL ) RE + RL
(Equation 8.41, Corrected)
The voltage gain may be expressed as
AV =
re2
RE || RL
,
+ (RE || RL )
which is approximately equal to one, as expected for the EF (CC) amplifier
configuration.
Comparing the expressions above for input resistance and current gain with
those of a single transistor EF (CC) amplifier (please refer to Section D4), we
can see that both of these characteristics are much larger for circuits
employing the Darlington pair than for single transistor circuits.
Darlington CE Amplifier
Figure 8.19a, reproduced to the left below, illustrates the Darlington pair
used in a CE amplifier configuration. Figure 8.19b, given on the right below,
shows the ac small signal equivalent. Note that in this case, we include both
dependent current sources, as well as rπ1 since the output is taken at the
collector of the Darlington pair.
Before we get into an investigation of the ac characteristics, let’s look at the
relationship between the emitter resistances re1 and re2. Recall that we
calculate the emitter resistance through the ratio of the thermal voltage VT
(=26mV at room temperature) and the collector current by assuming that
the collector and emitter currents are approximately equal. Using this
definition, as well as the current relationships previously defined for the
Darlington pair, and assuming that β1 and β2 are both much greater than
one, we get:
re2 =
VT
IC2
rπ 2 = β 2 re2
re1
V
V
VT
β V
= T = T =
= 2 T = β 2 re2
I C1
I B2
IC 2 / β 2
IC 2
.
(Equation 8.43, Modified)
rπ 1 = β 1 re1 = β 1 β 2 re2 = β 1 rπ 2
We may now define the input resistance as the equivalent resistance seen at
the base of Q1:
Rin = RB || [rπ 1 + β 1 rπ 2 ] = RB || 2rπ 2 = RB || 2β 1 β 2 re2 .
(Eqns 8.42 & 8.44)
To calculate the current gain, we need expressions for iL and iin. Using
current division, we get
iL =
i b1 =
RC i C
; where i C = −(β 1 i b1 + β 2 i b2 ) = −(β 1 i b1 + β 1 β 2 i b1 ) ≅ − β 1 β 2 i b1
RC + RL
(RB + 2β 1 βre2 )i b1
RB i in
RB i in
=
; so i in =
RB
RB + rπ 1 + β 1 rπ 2
RB + 2β 1 β 2 re2
Taking the ratio of output (load) current to input current and dividing by
β1β2, we get an expression for the current gain of the Darlington pair CE
amplifier as
Ai =
− RB RC
.
(RB / β 1 β 2 + 2re2 )(RC + RL )
(Equation 8.46)
The voltage gain may be found by taking the ratio of output (load) voltage
to input voltage, or by using the gain impedance formula:
Av =
− RC || RL
.
2re2
(Equation 8.47)
Using a Darlington pair in the CE configuration still allows for a large voltage
gain, but the major improvements are in the input resistance and current
gain. Refer to Section D2 for a comparison of single transistor CE
characteristics.
As we have seen, the Darlington pair may be considered a single transistor
when used in amplifier circuits – in fact, some manufacturers package this
compound transistor circuit into a single package with only three external
leads (base, collector and emitter). However, although we have been
concentrating on the single transistor characteristics of the compound
transistor, there are some important differences. In addition to the
previously mentioned increase in the total VBE drop, primary among the
potential difficulties is the achievable speed of operation. Changing the
voltage across any junction requires a finite amount of time, since charges
must be moved and electrons and holes move at a finite speed within a
material. Since the Darlington transistor pair has two base emitter junctions
in series, this combination operates more slowly than a single transistor. To
increase the speed of operation, a resistor may be placed between the
emitter of the first transistor and the base of the second transistor.