The Miller Theorem
and the
Frequency Response of the
Common Emitter Amplifier
L2
Autumn 2009
E2.2 Analogue Electronics
Imperial College London – EEE
1
A useful approximation: The Miller Theorem
•
Consider a shunt admittance connected between the input and output of an
inverting voltage amplifier of gain G.
VIn
Iin
•
Y
IF
- IF VOut
In
Out
-G
I1
-G
I2
Iout
(1+G)Y
(1+1/G)Y
Assuming that Y does not affect the gain, KCL at the input says that:
iin = i1 + iF = i1 + ( vin − vout ) Y = i1 + ⎡⎣(1 + G ) Y ⎤⎦ vin
iout = i2 − iF = i2 − ( vin − vout ) Y = i2 + ⎡⎣(1 + 1/ G ) Y ⎤⎦ vout
•
•
L2
This is the same current as we would get if we connected Y(1+G) in parallel to
the input and Y(1+1/G) in parallel to the output (on the right)
Note that this is an approximation; Y always changes the gain,
unless Z out // Z L = 0. Nonetheless, very often it is a good approximation.
Autumn 2009
E2.2 Analogue Electronics
Imperial College London – EEE
2
An application of the Miller theorem
Input impedance of the inverting amplifier (if op-amp has zero Yin):
R2
R1
vin
G
Vout
Z in = R1 +
R2
(1 + G )
We can even use the Miller theorem to calculate the gain of this circuit
for finite op-amp gain G:
Av =
R2 / ( G + 1) )
vout
−GR2
− R2
= −G
=
, lim Av =
vin
R1 + R 2 / ( G + 1) R1G + R1 + R2 G →∞
R1
This is the correct answer for the closed loop gain as we shall later see!
SF1.6p34
L2
Autumn 2009
E2.2 Analogue Electronics
Imperial College London – EEE
3
The small signal model the of BJT
The model below is useful to frequencies of up to 100s of MHz
The model is strictly valid for B-E voltage variations smaller than VT=25mV.
To analyse a circuit:
– Each transistor in a circuit is replaced by this model
– The “base spreading resistance” RBB is lumped with the thevenin
impedance of the signal source. It is not shown explicitly in these
notes.
What does RBB do:
• RBB is responsible for most of the noise of transistor amplifiers
• RBB is responsible for the power gain rolling-off at high frequencies
• RBB is often ignored at “low” (f < MHz) frequencies
•
•
•
L2
Autumn 2009
E2.2 Analogue Electronics
Imperial College London – EEE
4
BJT model: impedances and transconductance
•
The large signal BJT equation gives the transconductance:
⎛ V ⎞
I C = I 0 eVbe / Vth − 1 ⎜1 + CE ⎟ I 0 eVbe / Vth
VA ⎠
⎝
kT
Vth = B 26 mV at room temperature
e
∂I
I
gm = C C
∂Vbe Vth
(
•
However, since IE=(1+1/β)IC ,
Rπ =
•
)
∂VBE
= β VT / I C = β / g m
∂I B
Same equation gives output conductance:
GCE = 1/ RCE =
L2
Autumn 2009
∂I C
I
= C
∂VCE VA
E2.2 Analogue Electronics
Imperial College London – EEE
5
BJT model Capacitances
BC junction is reverse biased:
– depletion capacitance
– overlap capacitance
•
BE junction is forward biased:
– depletion capacitance
– storage (diffusion capacitance)
– overlap capacitance
CBE = CBE 0 + C j + CD
•
CE capacitance is small
– overlap capacitance
CCE CCE 0
•
Fringing/Overlap capacitances:
CBC 0 , CBE 0 , CCE 0
•
Junction capacitance:
– Reverse biased diode
– M=1/2-1/3, from doping
– Φ approx 0.8V
– M, Φ usually determined from two
bias points
•
L2
CBC = CBC 0 + C j
•
Diffusion capacitance
– from transit time
Autumn 2009
Cj =
C0
⎛ VBE ⎞
⎜1 −
⎟
Φ ⎠
⎝
CD = g mτ T
E2.2 Analogue Electronics
M
,
τ T = 1/ 2π fT
Imperial College London – EEE
6
An important figure of merit:
the BJT Current gain and the transit frequency fT
The transistor is connected as a current amplifier:
• Driven by a current source
• Driving a current meter
fT is the frequency at which the current gain drops to unity.
fT is usually specified by the transistor manufacturer as a figure of merit.
fT is a rough estimate of the highest frequency at which the transistor
can be used as an amplifier. Typically fT / 10 is this highest frequency.
L2
Autumn 2009
E2.2 Analogue Electronics
Imperial College London – EEE
7
An important figure of merit:
the BJT Current gain and the transit frequency fT
iB = vBE ( s ( CBE + CBC ) + 1/ Rπ ) ⎫⎪
⎬⇒
iC = g m vBE
⎪⎭
β0
iC
gm
g m Rπ
= " h fe " =
=
=
iB
s ( CBE + CBC ) + 1/ Rπ 1 + sRπ ( CBE + CBC ) 1 + s β 0 / 2π fT
where fT =
β0
2π Rπ ( CBE + CBC )
the current gain is h fe ( f ) =
=
β
gm
gm
, since: Rπ = 0
2π ( CBE + CBC ) 2π CBE
gm
β0
⇒
1 + j β 0 f / fT
lim h fe =
f > fT / β 0
fT
1
with fT =
2πτ T
f
fT allows the easy calculation of CBE + CBC since gm only depends on the
collector bias current! Note that CBE>>CBC so knowing fT defines CBE
L2
Autumn 2009
E2.2 Analogue Electronics
Imperial College London – EEE
8
The Common Emitter amplifier
Load
Source
Amplifier
Model parameters:
L2
Autumn 2009
⎧
∂I C
IC
=
=
g
⎪ m
∂VBE VT
⎪
−1
−1
−1
⎪
⎛ ∂ ( IC / β ) ⎞
⎛ ∂I B ⎞
⎛ ∂I C ⎞
β
⎪
R
β
=
=
=
=
⎨ π ⎜
⎜
⎟
⎟
⎜
⎟
∂
∂
∂
V
V
V
gm
⎝ BE ⎠
⎝ BE ⎠
BE
⎝
⎠
⎪
−1
⎪
⎛
⎞
I
∂
VA
⎪R =
C
⎪ CE ⎜⎝ ∂VCE ⎟⎠
IC
⎩
VT = kT / q = 25mV at 290K=17C
E2.2 Analogue Electronics
Imperial College London – EEE
9
The Common Emitter amplifier (2)
Combine VS , Z S , Rπ into a Thevenin circuit:
Rπ
VS
=
VS 1 = VS
Rπ + Z S 1 + Z S Yin
Rπ Z S
ZS
=
RS 1 = Rπ / / Z S =
Rπ + Z S 1 + Z S Yin
Consider RCE and Z L together: R2 = RCE / / Z L =
RCE Z L
ZL
=
RCE + Z L 1 + Z LYout
and note that Z S = RS + RBB
L2
Autumn 2009
E2.2 Analogue Electronics
Imperial College London – EEE
10
The Common Emitter amplifier (3)
Gain:
vout ∂V2
Vs1
− gm Z L
1
Av =
g m R2 =
≡
=−
vs
Vs
∂Vs
(1 + RsYin ) (1 + Z LYout )
If Yin= 0 and Yout = 0 this is the familiar:
vout
Av =
≡ − gm Z L
vs
Note that lowercase letters represent small signal variations, i.e. differentials
L2
Autumn 2009
E2.2 Analogue Electronics
Imperial College London – EEE
11
The Common Emitter amplifier (4):
Input and output impedance
−1
Input Impedance:
∂VBE ⎛ ∂I B ⎞
=⎜
Z in =
⎟ = Rπ
∂I B ⎝ ∂VBE ⎠
−1
Output Impedance:
L2
Autumn 2009
Z out
∂VC ⎛ ∂I C ⎞
=
=⎜
⎟ = RCE
∂I out ⎝ ∂VCE ⎠
E2.2 Analogue Electronics
Imperial College London – EEE
12
Frequency response of
the Common Emitter amplifier
If we momentarily neglect CBC this would be the same as before but with:
Input Impedance:
Rπ
1
Z in =
= Rπ / / CBE =
1 + jωCBE Rπ
Yin
Output Impedance:
Z out =
We still have: A V =
L2
Autumn 2009
RCE
1
= RCE / / CCE =
1 + jωCCE RCE
Yout
vout
− gm Z L
1
v
− gm Z L
=
and out = G =
vs
(1 + RsYin ) (1 + Z LYout )
vB
(1 + Z LYout )
E2.2 Analogue Electronics
Imperial College London – EEE
13
Dealing with CBC
ICBC
In the absence of CBC we know that:
the current through CBC is:
vout
− gm Z L
=G=
vB
(1 + Z LYout )
I CBC = jωCBC (VB − VC ) = jωCBC (1 − G ) VB = jωCBC (1/ G − 1) VC
We can account for CBC by introducing an additional input “Miller” admittance:
YM ,in = jωCBC (1 − G )
and an additional output “Miller” admittance:
YM ,out = jωCBC (1 − 1/ G )
L2
Autumn 2009
E2.2 Analogue Electronics
Imperial College London – EEE
14
Frequency response of the CE amplifier
VS1 and RS1 is the Τhevenin equivalent of RS VS and Rπ
Use the Miller Theorem to get:
with C2 = CBE + (1 + G ) CBC (1 + G ) CBC and C3 = CCE + (1 + 1/ G ) CBC CBC
The low requency gain from V1 to V2 is: G = − g m R2 − g m Z L
L2
Autumn 2009
E2.2 Analogue Electronics
Imperial College London – EEE
15
Frequency response of the CE amplifier (2)
the gain of this amplifier is:
− gm
dV2
R2
G
=
=
dVS 1 1 + sRS 1C2 1 + sR2C3 (1 + sRS 1C2 )(1 + sR2C3 )
input pole
output pole
If the gain is large the input pole can be at a much lower frequency than the output pole:
τ IN = RS 1C2 = ( Rπ / / RS ) ( CBE + (1 + G ) CBC ) RS (1 + G ) CBC ( since RS Rπ )
τ OUT = R2C3 = ( CBC (1 + 1/ G ) + C CE ) ( RCE / / Z L ) CBC Z L (since CBC CCE )
The gain-bandwidth product of this amplifier is:
G = g m R2 g m Z L
⎫
g m R2
1
G
⎪
. lim GBW .
1
1 ⎬ ⇒ GBW =
G →∞
BW =
2
2
(1
)
2
π
π
π
R
C
R
C
+
G
R
C
S 2
S BC
S BC
2π RS 1C2 2π RS C2 ⎪⎭
This is independent of G!
if τ in > τ out
The tiny CBC and the source impedance can define the gain-bandwidth product of the CE amp.
This suggests that feedback is at work!
Note that the input pole will be at a lower frequency than the output pole if
G RS 1 > R2 ⇒ g m RS 1 R2 > R2 ⇒ g m RS = β RS / Rπ > 1
L2
Autumn 2009
E2.2 Analogue Electronics
Imperial College London – EEE
16
Importance of the gain-bandwidth product
the gain of the CE amplifier is:
− gm
dV2
R2
G
=
=
dVS 1 1 + sRS 1C2 1 + sR2C3 (1 + sRS 1C2 )(1 + sR2C3 )
input pole
output pole
One of the two pole frequencies is usually much lower than the other.
We then say that we have a dominant pole amplifier whose frequency response is:
Av ( s ) =
G0
G0
=
for a DC gain G0 and a bandwidth
1 + sτ 1 + jωτ
ωB = 1/ τ
We have already seen that in the limit of large G0 the Gain-Bandwidth product:
GB = G0ωB = G0 / τ
is independent of the amplifier gain.
We will see later in the course that this is a very general result which applies to
all one pole feedback amplifiers regardless of how they are constructed.
L2
Autumn 2009
E2.2 Analogue Electronics
Imperial College London – EEE
17