Download Multiple stage amplifiers

Multiple stage amplifiers
Aims:
• Examine a few common 2-transistor amplifiers:
-- Differential amplifiers
-- Cascode amplifiers
-- Darlington pairs
-- current mirrors
• Introduce formal methods for exactly analysing multiple stage amplifiers
L6
Autumn 2009
E2.2 Analogue Electronics
Imperial College London – EEE
1
Two stage BJT amplifiers
We study them separately because they very often appear as building blocks.
There are 9 possible cascades of 2 single stage transistor amplifiers.
We will study the shaded ones.
Comments
BJT
L6
Name
1st Stg 2nd Stg
(voltage amp)
CE
CE
High Voltage gain
cascode
CE
CB
High bandwidth
(op-amp)
CE
CC
High Zin low Zout
(current buffer) CB
CE
Higher Zout than CB/CG
(current buffer) CB
CB
Second stage to improve on CB/CG
(Not common)
CB
CC
Not common
(Not common)
CC
CE
Instead of CE, offers higher Zin
differential amp CC
CB
High voltage gain and bandwidth
darlington
CC
High current gain
Autumn 2009
CC
E2.2 Analogue Electronics
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2
Differential amplifier
•
•
•
L6
Half circuit (i.e. driven from one side) is CC followed by CB
Very wide frequency response
Extremely high voltage gain
Autumn 2009
E2.2 Analogue Electronics
Imperial College London – EEE
3
Cascode amplifier
•
•
•
•
L6
Wideband voltage amplifier
CE stage operates at gain=-1, minimising miller loading of input.
CB gives all the voltage gain, acting as transimpedance of value ZL
The cascode has a much higher output impedance (other than ZL) than the CE
amplifier (the common emitter Early resistance acts as series-series feedback
to the common base with loop gain =gmRCE)
Autumn 2009
E2.2 Analogue Electronics
Imperial College London – EEE
4
Darlington pair
•
•
•
•
•
L6
The darlington pair is a high gain power amplifier it has:
– Unity voltage gain
– High current gain equal to the product of the two transistor current gains
Often used as a single transistor for higher beta. But :
has high input DC voltage drop
Good frequency response due to the absence of shunt Miller feedback.
However, series Miller feedback introduces tendency for instability when
driving capacitive loads.
Autumn 2009
E2.2 Analogue Electronics
Imperial College London – EEE
5
Current mirrors
•
•
•
•
•
•
•
•
•
L6
Use one transistor with unity feedback as a transimpedance amplifier to measure
the VBE required for a given current.
Use a second transistor as transconductor to create a copy of the input current
Can make a current amplifier by using larger output transistor.
Current gain is in error due to base currents (i.e finite current gain)
No DC gain error in FET mirrors (remember the AC current gain of a FET scales as
the inverse of frequency!)
Main source of error transistor mismatch
– “VBE mismatch at a constant current” (BJT)
– VT mismatch in FET
AC analysis as in CE amplifier with extra source
admittance due to input transistor
Current mirrors are used for DC biasing multi-stage
amplifiers
Current mirrors often used load to a differential
amplifier to turn the differential amplifier into a
Simple current Mirror
differential transconductor.
Autumn 2009
E2.2 Analogue Electronics
Imperial College London – EEE
6
Improved current mirrors
The buffered mirror
The CC amplifier feeding the bases
reduces current gain error
The Wilson Mirror has high output Z, since output
stage is a cascode amplifier
L6
Autumn 2009
E2.2 Analogue Electronics
Imperial College London – EEE
7
Scaling Mirrors: The Widlar mirror
(a)
•
•
•
•
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(b)
(c)
The Widlar scaling mirror is often used as fixed scaling current source (a)
Can be made as a buffered or a Wilson source (c)
A feedback resistor can be added on the input side turning it into a
transconductor (b)
A base resistor as shown can provide “beta compensation” (i.e. introduce a
zero in the frequency response (c)
Autumn 2009
E2.2 Analogue Electronics
Imperial College London – EEE
8
Some more mirrors
(a)
(b)
(a) Buffered Widlar mirror
(b) The “gm-compensated” mirror
L6
Autumn 2009
E2.2 Analogue Electronics
Imperial College London – EEE
9
Current mirror as a differential amp load
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•
•
•
•
•
•
•
L6
The current mirror maps the left side current
differential into the right side.
The large signal response of this circuit is
Vout=tanh(V+-V-)
This circuit (a 3 stage amplifier! Why?)
This circuit It has extremely high voltage gain: AV is
of the order of VA/Vth
This circuit is also used for mixers if a
transconductor is used in the place of the tail
current source.
There is no Miller effect on the left half circuit
If this circuit drives a current sink at the output
there is no Miller effect on the right half circuit
either!
The diff-amp has an extremely wide frequency
response. This is partly a consequence of the
resistive impedance match between the output of
the first stage (emitter of Q1)and input of the
second stage (emitter of Q2).
Autumn 2009
E2.2 Analogue Electronics
Q1
Q2
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10
Two stage FET amplifiers
• The analogy we observed between single stage BJT and FET amplifiers applies,
to two stage amplifiers. The correspondence is, as before, EÆS, BÆG, CÆD.
• The behaviour of BJT and FET configurations is very similar, except for the
difference on the input side of the small signal equivalent circuit.
• A very useful possibility opens up: Use a FET for one stage and a BJT for the
other. Mixed bipolar-FET two-stage combinations try to exploit the smaller input
admittance of FETs and the better frequency response and power handling
capability of bipolars at the same time.
• This approach gives rise to the “BiCMOS” manufacturing technologies which
use FETs for input stages and BJTs for output stages, especially line drivers.
FET
L6
C o m m e n ts
N am e
1 s t S tg
2 n d S tg
(vo lta g e a m p )
CS
CS
H ig h V o lta g e g a in
cascod e
CS
CG
H ig h b a n d w id th
(o p -a m p )
CS
CD
H ig h Z in lo w Z o u t
(c u rre n t b u ffe r)
CG
CS
H ig h e r Z o u t th a n C B /C G
(c u rre n t b u ffe r)
CG
CG
S e c o n d s ta g e to im p ro ve o n C B /C G
(N o t c o m m o n )
CG
CD
N ot com m on
(N o t c o m m o n )
CD
CS
In s te a d o f C E , o ffe rs h ig h e r Z in
d iffe re n tia l a m p
CD
CG
H ig h vo l ta g e g a in a n d b a n d w id th
d a rlin g to n
CD
CD
H ig h c u rre n t g a in
Autumn 2009
E2.2 Analogue Electronics
Imperial College London – EEE
11
Multistage amplifiers
• Multistage amplifiers are difficult to compute if the components are not unilateral.
• For unilateral amplifiers things are simple. We multiply gains with appropriate
voltage dividers.
V1
V2
VL
Rs
ROUT1
RO2
+
A1 V1
-
Amp1
A2 V2
RL
-
Source
RIN1
-
Vs
+
RIN2
Amp2
Load
VL
1
1
1
1
, Yx =
, x ∈ {in1, in 2, L}
=
A1
A2
Vs 1 + RsYin1 1 + Rout1Yin 2 1 + Ro 2YL
Rx
• For non-unilateral amplifiers:
• The input impedance of each stage depends on the input impedance of the
next stage
• The output impedance of each stage depends on the output impedance of
the preceding stage.
• This problem has a solution but involves the solution of sets of simultaneous
quadratic equations.
L6
Autumn 2009
E2.2 Analogue Electronics
Imperial College London – EEE
12
Input - output impedance of a loaded amplifier
•
We calculate the input impedance of a voltage amplifier driving a load ZL :
i1 = g11v1 + g12i2 ⎫
⎪ i1 = g11v1 − g12YL v2 ⎫
v2 = g 21v1 + g 22i2 ⎬ ⇒
⎬⇒
v
g
v
g
Y
v
=
−
2
21 1
22 L 2 ⎭
⎪
i2 = −v2YL
⎭
g12 v2YL = g11v1 − i1 ⎫⎪
v1 1 + g 22YL
⎬ ⇒ Z in = =
g
Y
v
g
v
1
+
=
i1 g11 + Δ g YL
( 22 L ) 2 21 1 ⎪⎭
•
A similar calculation for the output impedance of a voltage amplifier driven by a
finite impedance Thevenin source ZS gives:
i1 = g11v1 + g12i2 ⎫
⎪ i1 = g11 ( vs − i1Z S ) + g12i2 ⎪⎫
v2 = g 21v1 + g 22i2 ⎬ ⇒
⎬⇒
⎪ v2 = g 21 ( vs − i1Z S ) + g 22i2 ⎭⎪
v1 = vs − i1Z S
⎭
g 22 + Δ g Z S
Z out = dv2 / di2 =
1 + g11Z S
L6
Autumn 2009
E2.2 Analogue Electronics
Imperial College London – EEE
13
Gain of a fully loaded voltage amplifier
ZS
VS
i2
i1
v1
G11
G22
G21V1
v2
YL
G12i2
We start with the amplifier definition, plus the source-load boundary conditions:
i1 = g11v1 + g12i2
v2 = g 21v1 + g 22i2
v1 = vs − i1Z s
i2 = −v2YL
After some algebra we conclude that:
v2
g 21
g 21
=
=
, Δ g = g11 g 22 − g 21 g12
vs (1 + g11Z s )(1 + g 22YL ) − g12 g 21Z sYL 1 + g11Z S + g 22YL + Δ g YL Z s
L6
Autumn 2009
E2.2 Analogue Electronics
Imperial College London – EEE
14
Cascade connection: Transmission Parameters
In a cascade connection,
• V1 of network X2= V2 of network X1
• I1 of network X2 = -I2 of network X1
We can define a new set of parameters so that we have a simple way to
calculate the response of cascades of amplifiers.
A suitable definition is:
⎡v1 ⎤ ⎡ A B ⎤ ⎡ v2 ⎤
⎢ i ⎥ = ⎢ C D ⎥ ⎢ −i ⎥
⎦⎣ 2⎦
⎣ 1⎦ ⎣
With this definition, the ABCD parameters of a cascade of two networks are
found from the matrix product of the individual ABCD matrices ports labelled for
clarity):
3
3
1
2
1
X1
X2
⎡v1 ⎤ ⎡ A1 B1 ⎤ ⎡ v2 ⎤ ⎫
⎢ i ⎥ = ⎢ C D ⎥ ⎢ −i ⎥ ⎪
⎡v1 ⎤ ⎡ A1
1⎦ ⎣ 2 ⎦ ⎪
⎣ 1⎦ ⎣ 1
⇒
⎬ ⎢ ⎥=⎢
⎡ v2 ⎤ ⎡ A2 B2 ⎤ ⎡ v3 ⎤ ⎪ ⎣ i1 ⎦ ⎣C1
⎢ −i ⎥ = ⎢C D ⎥ ⎢ −i ⎥ ⎪
2 ⎦ ⎣ 3 ⎦⎭
⎣ 2⎦ ⎣ 2
L6
Autumn 2009
X3=X1X2
B1 ⎤ ⎡ A2
D1 ⎥⎦ ⎢⎣C2
E2.2 Analogue Electronics
B2 ⎤ ⎡ v3 ⎤ ⎡ v1 ⎤ ⎡ A3
⇒⎢ ⎥=⎢
⎢
⎥
⎥
D2 ⎦ ⎣ −i3 ⎦ ⎣ i1 ⎦ ⎣C3
B3 ⎤ ⎡ v3 ⎤
D3 ⎥⎦ ⎢⎣ −i3 ⎥⎦
Imperial College London – EEE
15
Transmission (or ABCD) parameters (2)
The transmission matrix elements are related to the 4 gains:
A=
⎡v1 ⎤ ⎡ A B ⎤ ⎡ v2 ⎤
⎢ i ⎥ = ⎢ C D ⎥ ⎢ −i ⎥
⎦⎣ 2⎦
⎣ 1⎦ ⎣
∂v1
∂v2
∂i1
C=
∂v2
A cascade of 2 amplifiers has gains:
i2 = 0
1
G21
i2 = 0
1
=
Z 21
=
B=
−∂v1
∂i2
−∂i1
D=
∂i2
=
v2 = 0
v2 = 0
−1
Y21
−1
=
H 21
⎡ A B ⎤ ⎡ A1 B1 ⎤ ⎡ A2 B2 ⎤ ⎡ A1 A2 + B1C2 A1 B2 + B1 D2 ⎤
⎢C D ⎥ = ⎢C D ⎥ ⎢C D ⎥ = ⎢C A + D C C B + D D ⎥ ⇒
⎣
⎦ ⎣ 1
1⎦ ⎣ 2
2⎦
1 2
1 2
1 2⎦
⎣ 1 2
g f 1g f 2 y f 1z f 2
g f 1 y f 2 y f 1h f 2
1
1
gf =
yf =
=
=
1
1
1
1
y
z
g
g
y f 1h f 2 − g f 1 y f 2
−
f1 f 2
f1 f 2
−
−
g f 1g f 2 y f 1z f 2
g f 1 y f 2 y f 1h f 2
zf =
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1
1
1
−
z f 1g f 2 hf 1z f 2
Autumn 2009
=
z f 1g f 2hf 1z f 2
hf 1z f 2 − z f 1g f 2
hf =
1
1
1
−
h f 1h f 2 z f 1 y f 2
E2.2 Analogue Electronics
=
h f 1h f 2 z f 1 y f 2
z f 1 y f 2 − h f 1h f 2
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16
Transmission (or ABCD) parameters (3)
⎡v1 ⎤ ⎡ A B ⎤ ⎡ v2 ⎤
⎢ i ⎥ = ⎢C D ⎥ ⎢ −i ⎥
⎦⎣ 2⎦
⎣ 1⎦ ⎣
• Note the sign of i2 and also the reverse sense of signal flow. The sign is chosen so
the ABCD matrix of a cascade of two networks is just the matrix product of the
individual ABCD matrices (compare this to the messy loading calculation before!)
• The reverse sense of signal flow is to keep the matrix finite if an amplifier is
unilateral.
• The conversion from, say, Y to ABCD follows the same logic as the Y(H) calculation:
⎡v1 ⎤ ⎡ A B ⎤ ⎡ v2 ⎤ ⎡ 1 0 ⎤ ⎡ v1 ⎤ ⎡ A B ⎤ ⎡ 0
⎢ i ⎥ = ⎢C D ⎥ ⎢ −i ⎥ ⇒ ⎢Y Y ⎥ ⎢ v ⎥ = ⎢C D ⎥ ⎢ −Y
⎦ ⎣ 2 ⎦ ⎣ 11 12 ⎦ ⎣ 2 ⎦ ⎣
⎦ ⎣ 21
⎣ 1⎦ ⎣
⎡ A B ⎤ −1 ⎡Y22 1 ⎤
⎢C D ⎥ = Y ⎢ Δ Y ⎥ , Δ y = Y11Y22 − Y21Y12
⎣
⎦
11 ⎦
21 ⎣ Y
1 ⎤ ⎡ v1 ⎤
⇒
−Y22 ⎥⎦ ⎢⎣ v2 ⎥⎦
Remember that all ABCD parameters are inversely proportional to the gains. This
is the reason for formally choosing port 2 as the input port.
The intuitive choice of input at port 1 would make all parameters inversely
proportional to the reverse gains, which are small, and usually not very accurately
determined.
L6
Autumn 2009
E2.2 Analogue Electronics
Imperial College London – EEE
17
Composition rules summary
For the exact calculation of circuit interconnections we can use 2-port matrix algebra:
G1
Y1
G1+G2
Y1+Y2
G2
Y2
shunt-series: add G matrices
shunt-shunt: add Y matrices
H1
Z1
H1+H2
Z1+Z2
H2
Z2
series-shunt: add H matrices
series-series: add Z matrices
X1
X2
X1X2
cascade connection: multiply ABCD matrices
L6
Autumn 2009
E2.2 Analogue Electronics
Imperial College London – EEE
18
Multistage amplifiers: summary
• Calculation of the response of unilateral multi-stage amplifiers is simple:
• Product of gains and voltage dividers.
• Calculation of the response of non-unilateral multistage amplifiers involves
determining for each stage:
• the effect of source impedance on gain and output impedance
• the effect of load impedance of input impedance and gain
• This typically leads to a set of simultaneous quadratic equations.
• A conceptually simpler analysis method involves the transmission or “ABCD”
parameters which allow to describe all the loading effects of a non-unilateral
cascade through a matrix product.
• With the introduction of ABCD parameters we have introduced simple ways to
describe any connection between 2-port circuits.
L6
Autumn 2009
E2.2 Analogue Electronics
Imperial College London – EEE
19