EECS 105 Spring 2004, Lecture 27
Lecture 27:
Frequency response
Prof J. S. Smith
Department of EECS
University of California, Berkeley
EECS 105 Spring 2004, Lecture 27
Prof. J. S. Smith
Context
Today, we will continue the discussion
of single transistor amplifiers by
looking at common source amplifiers
with source degeneration (also
common Emitter amplifiers with
emitter degeneration.
We will then start discussing the
frequency response of single stage
amplifier, the frequency response of
CE amps, and the Miller
approximation.
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University of California, Berkeley
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EECS 105 Spring 2004, Lecture 27
Prof. J. S. Smith
Lecture Outline
z
z
z
z
z
Emitter Degeneration
Frequency response of the CE and
CS current amplifiers
Unity-gain frequency ωT
Frequency response of the CE as
voltage amp
The Miller approximation
Department of EECS
University of California, Berkeley
EECS 105 Spring 2004, Lecture 27
Prof. J. S. Smith
Bias sensitivity
z
z
z
When a transistor biasing circuit is designed, it is
important to realize that the characteristics of the
transistor can vary widely, and that passive
components vary significantly also.
Biasing circuits, must therefore be designed to
produce a usable bias without counting on specific
values for these components.
One example is a BJT base bias in a CE amp. A
slight change in the base-emitter voltage makes a
very large difference in the quiescent point. The
insertion of a resistor will improve sensitivity.
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University of California, Berkeley
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EECS 105 Spring 2004, Lecture 27
Prof. J. S. Smith
Typical “Discrete” Biasing
VCC
z
RC
R1
In this scheme, the base bias
voltage is given by a voltage
divider:
R1
VB = VCC
z
R2
R1 + R2
The emitter will approximately
follow the base voltage, so the
emitter current is:
RE
⎛
⎞
R1
I E ≈ ⎜ VCC
− VBE ,on ⎟ / RE
R1 + R2
⎝
⎠
Department of EECS
University of California, Berkeley
EECS 105 Spring 2004, Lecture 27
Prof. J. S. Smith
Gain for “Discrete” Design
Rin ≈ rπ + ( β + 1) RE
Rin ≈ ( β + 1) RE
vs
RC
R1
~ vs RC / RE
R2
vs / RE
~ vs
RE
Rin 2 ≈ ( β + 1) RE || R1 || R2
Can be made large to couple
All of source to input (even with RS)
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University of California, Berkeley
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EECS 105 Spring 2004, Lecture 27
Prof. J. S. Smith
Insensitivity to transistor parameters
z
Most of the circuit parameters are independent of
variation of the transistor parameters, and depend
only on resistance ratios. That is often a design
goal, but in integrated circuits we will not want to
use so many resistors.
Department of EECS
University of California, Berkeley
EECS 105 Spring 2004, Lecture 27
Prof. J. S. Smith
Emitter Degeneration
z
The addition of the resistor at the emitter will have
several additional effects:
–
–
–
z
The transconductance will be reduced
The output resistance will be increased
The input resistance will be increased
Each of these are a result of the negative feedback
from the presence of the emitter resistor.
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University of California, Berkeley
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EECS 105 Spring 2004, Lecture 27
Prof. J. S. Smith
Source Degeneration
z
z
Source degeneration for a FET is not as common as
emitter degeneration is for a BJT, because the gain
of a FET is already lower than for an BJT, and its
input impedance is already ∞.
It is widely used, however, to raise the output
impedance of CS amplifiers
Department of EECS
University of California, Berkeley
EECS 105 Spring 2004, Lecture 27
Prof. J. S. Smith
Source degeneration in a CS amp
z
z
The input impedance of
the amplifier is still ∞
Assume that the body is
still connected to ground
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University of California, Berkeley
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EECS 105 Spring 2004, Lecture 27
Prof. J. S. Smith
Small signal model
z
CS amp with source degeneration:
i0
+
vin
−
+
+
vs
RD
v0
RS
−
−
Department of EECS
University of California, Berkeley
EECS 105 Spring 2004, Lecture 27
Prof. J. S. Smith
Circuit gain calculation
z
z
z
From KCL at the source,
vs v s
+ = g m ( vi − vs ) + g mb (0 − vs )
RS r0
At the Drain:
v
i0 + s = g m ( vi − vs ) + g mb (0 − vs )
r0
The circuit’s transconductance:
Gm =
z
i0
gm
=
vi 1 + ( g + g ) R + RS
m
mb
S
r0
Because of the body effect, the output is still
dependant on the transistor characteristics
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University of California, Berkeley
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EECS 105 Spring 2004, Lecture 27
Prof. J. S. Smith
Output impedance
z
To calculate the output impedance, we put in a test
current and calculate the voltage:
i0
+
+
vin
+
vs
−
it
vt
RS
−
−
Department of EECS
University of California, Berkeley
EECS 105 Spring 2004, Lecture 27
Prof. J. S. Smith
Output impedance
z
Since all of the test current must go through RS, if
we take i1 to be the current through ro, we have:
vt = vs + i1r0 = vs + r0 [1 + g m + g mb ) RS ]
z
From which we get the output resistance:
R0 =
Department of EECS
vt
= RS + r0 [1 + ( g m + g mb ) RS ]
it
University of California, Berkeley
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EECS 105 Spring 2004, Lecture 27
Prof. J. S. Smith
Frequency response
z
z
So far, we have modeled the small signal response
of the stages for low frequencies (but not so low
that we couldn’t neglect the DC blocking coupling
capacitors!)
Now we will put in the parasitic capacitances, and
analyze the changes in the transfer functions of the
circuits at higher frequencies.
Department of EECS
University of California, Berkeley
EECS 105 Spring 2004, Lecture 27
Prof. J. S. Smith
Parasitic Capacitances
z
If we look at the small signal model of the FET that
we developed a few weeks ago, we have
capacitances between the gate and the source, and
the gate and the drain
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University of California, Berkeley
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EECS 105 Spring 2004, Lecture 27
Prof. J. S. Smith
CS with parasitics
z
When we take into account a finite source
impedance in a common source amplifier, the
capacitances will reduce the voltage swing at the
gate at high frequencies.
Department of EECS
University of California, Berkeley
EECS 105 Spring 2004, Lecture 27
Prof. J. S. Smith
Parasitic Capacitances
The transfer function will be a low pass filter, with a
pole at the frequency determined by the source
resistance and the capacitance.
rs
vs
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University of California, Berkeley
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EECS 105 Spring 2004, Lecture 27
Prof. J. S. Smith
Miller approximation
z
z
z
z
However, the capacitance from the gate to the drain
has a large effect, because the voltage on the drain
is amplified, with a negative gain coefficient.
This means that the contribution of the capacitance
from the gate to the drain is comparable to the same
capacitance to ground, multiplied by the gain of the
transistor.
This is called the Miller effect.
We can approximate the effect, by simply putting
in a capacitance to ground, multiplied by the low
frequency gain. This is called the Miller
approximation
Department of EECS
University of California, Berkeley
EECS 105 Spring 2004, Lecture 27
Prof. J. S. Smith
High frequency zero
At very high frequencies, the gain flattens out again,
because the capacitor couples from the gate to the
drain directly, as a passive circuit
rs
vs
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University of California, Berkeley
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EECS 105 Spring 2004, Lecture 27
Prof. J. S. Smith
Magnitude Bode Plot
pole
Low frequency
βo
gain
Unity current gain
0 dB
zero
ωp
ωT
ωz
Department of EECS
University of California, Berkeley
EECS 105 Spring 2004, Lecture 27
Prof. J. S. Smith
CE Amplifier with Current Input
Find intrinsic current gain by driving with infinite source impedance and
Zero load impedance…
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University of California, Berkeley
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EECS 105 Spring 2004, Lecture 27
Prof. J. S. Smith
Short-Circuit Current Gain
Pure input current
(RS = 0 Ω)
Small-signal
short circuit
(could be a DC
voltage source)
Substitute equivalent circuit model of transistor and do the “math”
Department of EECS
University of California, Berkeley
EECS 105 Spring 2004, Lecture 27
Prof. J. S. Smith
Small-Signal Model: Ai
Note that ro, Ccs play no role (shorted out)
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University of California, Berkeley
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EECS 105 Spring 2004, Lecture 27
Prof. J. S. Smith
Phasor Analysis: Find Ai
KCL at the output node:
I out = g mVπ + (0 − Vπ ) / Z µ
KCL at the input node:
I in = Vπ / Zπ + (Vπ − 0 ) / Z µ
Solve for Vπ :
Vπ = (1/ Zπ + 1/ Z µ ) −1 I in
Department of EECS
University of California, Berkeley
EECS 105 Spring 2004, Lecture 27
Prof. J. S. Smith
Phasor Analysis for Ai (cont.)
I out = ( g m − jωCµ )Vπ
Substituting for Vπ
Ai ( jω ) =
( g m − j ωC µ )
(1 / Zπ ) + jωCµ
Substituting for Zπ = rπ || (1/jωCπ)
Ai ( jω ) =
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Zπ =
β 0 (1 − jωCµ / g m )
1 + jω rπ (Cπ + Cµ )
rπ
1 + jω rπ Cπ
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EECS 105 Spring 2004, Lecture 27
Prof. J. S. Smith
Short-Circuit Current Gain Transfer Function
Transfer function has one pole and one zero:
Ai ( jω ) =
β o (1 − jω [C µ / g m ])
1 + jω [rπ (Cπ + C µ )]
Ai ( jω ) =
β o (1 − jω / ω z )
(1 + jω / ω p )
Note: Zero Frequency much larger than pole:
ωz =
gm β 1
1
=
>β
= βω p
Cµ rπ Cµ
rπ (Cµ + Cπ )
Department of EECS
University of California, Berkeley
EECS 105 Spring 2004, Lecture 27
Prof. J. S. Smith
Magnitude Bode Plot
pole
Ai ( jω ) ≈
βo
βo
1
=
jω / ω p jω / ωT
Unity current gain
0 dB
zero
ωp
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ωT
ωz
University of California, Berkeley
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EECS 105 Spring 2004, Lecture 27
Prof. J. S. Smith
Transition Frequency ωT
ω T = β oω p =
gm
Cπ + C µ
Dependence on DC collector current:
Cπ = C je + Cdiff
Base Region
Transit Time!
Cdiff = g mτ F = τ F qI / kT
Limiting case: f = ωT →
T
2π
1
2πτ F
Department of EECS
University of California, Berkeley
EECS 105 Spring 2004, Lecture 27
Prof. J. S. Smith
Common Source Amplifer: Ai(jω)
DC Bias is problematic: what sets VGS?
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University of California, Berkeley
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EECS 105 Spring 2004, Lecture 27
Prof. J. S. Smith
CS Short-Circuit Current Gain
Transfer function: Ai ( jω ) =
g m (1 − jωC gd / g m )
jω (C gs + C gd )
Department of EECS
University of California, Berkeley
EECS 105 Spring 2004, Lecture 27
Prof. J. S. Smith
MOS Unity Gain Frequency
z
Since the zero occurs at a higher frequency than
pole, assume it has negligible effect:
Ai ≈
gm
=1
jω (C gs + C gd )
g
ωT ≈ m =
C gs
µ Cox
ωT =
gm
(C gs + Cgd )
W
(VGS − VT )
3 µ (VGS − VT )
L
=
2
L2
2
WLCox
3
Performance improves like L^2 for long channel devices!
For short channel devices the dependence is like ~ L^1
3 µ (VGS − VT )
ωT ≈
~
L2
2
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µ
VGS − VT
µ Eeff v
L
=
= =τL
L
L
L
Time to
cross
channel
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EECS 105 Spring 2004, Lecture 27
Prof. J. S. Smith
Miller Impedance
z
Consider the current flowing through an impedance
Z hooked up to a “black-box” where the voltage
gain from terminal to the other is fixed (as you can
see, it depends on Z)
Av =
v2
v1
I
Z
v1
v2
I=
1 − Av
v1 − v2 v1 − Av v1
=
= v1
Z
Z
Z
Department of EECS
University of California, Berkeley
EECS 105 Spring 2004, Lecture 27
Prof. J. S. Smith
Miller Impedance
z
Notice that the current flowing into Z from terminal
1 looks like an equivalent current to ground where
Z is transformed down by the Miller factor:
I = v1
z
1 − Av
Z
→ Z M ,1 =
1 − Av
Z
From terminal 2, the situation is reciprocal
−I =
1 − Av−1
v2 − v1 v2 − Av−1v2
=
= v2
Z
Z
Z
Z M ,2 =
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Z
1 − Av−1
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EECS 105 Spring 2004, Lecture 27
Prof. J. S. Smith
Miller Equivalent Circuit
Z M ,1 + Z M ,2 = Z
Note:
Z M ,1 =
z
z
z
Z
1 − Av
Z M ,1 =
Z
1 − Av−1
We can “de-couple” these terminals if we can
calculate the gain Av across the impedance Z
Often the gain Av is weakly dependent on Z
The approximation is to ignore Z, calculate A, and
then use the decoupled miller caps
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