4/27/2011
section 5_9 Frequency Response of the CE Amp
1/1
5.9 Frequency Response of
the Common-Emitter Amp
Reading Assignment: 491-503
Amplifiers made with BJTs are similar to amplifiers made with opamps—the both exhibit finite bandwidth.
HO: AMPLIFIER BANDWIDTH
The gain within the bandwidth is usually constant with respect to
frequency—we call this value the mid-band gain.
HO: MID-BAND GAIN
Large capacitors (e.g., COUS) determine the low-frequency limit
of amplifier bandwidth. We can explicitly determine this value be
analyzing the low-frequency small-signal circuit.
HO: THE LOW-FREQUENCY RESPONSE
Parasitic capacitors (e.g., Cπ) determine the high-frequency limit
of amplifier bandwidth. We can explicitly determine this value be
analyzing the high-frequency small-signal circuit.
HO: THE HIGH-FREQUENCY RESPONSE
Jim Stiles
The Univ. of Kansas
Dept. of EECS
4/20/2011
Amplifier Bandwidth
1/16
Amplifier Bandwidth
BJT amplifiers are band-limited devices—in other words, they
exhibit a finite bandwidth.
Q: ???
A: Say the input to a BJT small-signal amplifier is the eigen
function of linear, time-invariant system:
Vin cos ωt = Vin Re { e − jωt }
Since the small-signal BJT amp is (approximately) a linear
system, the output will likewise be the eigen function—an
undistorted sinusoidal function of precisely the same
frequency ω as the input!
+
+
Vin cos ωt
I
Vout cos ωt
−
−
Amplifier
Q: Of course that’s true! We know that:
Jim Stiles
The Univ. of Kansas
Dept. of EECS
4/20/2011
Amplifier Bandwidth
2/16
vout (t ) = Avo vin (t )
Therefore the magnitudes of the input and output sinusoids
are related as:
Vout = Avo Vin
Right?
A: Not necessarily!
The small-signal, open-circuit voltage gain of a BJT amplifier
depends on the frequency ω of the input signal!
Q: Huh!?! We determined earlier that the small-signal voltage
gain of this amplifier:
15 V
15 V
1K
3.7 K
vO (t )
COUS
β = 100
vi (t )
Jim Stiles
+
-
2.3K
1K
The Univ. of Kansas
COUS
Dept. of EECS
4/20/2011
Amplifier Bandwidth
was:
Avo =
3/16
vo
= −200
vi
So then if the small-signal input is:
vi (t ) = Vin cos ωt
isn't the small-signal output simply:
vo (t ) = −200Vin cos ωt
?????????
A: Maybe—or maybe not!
Again, the gain of the amplifier is frequency dependent. We
find that if ω is too high (i.e., large) or too low (i.e., small),
then the output might be much less than the 200 times larger
than the input (e.g., only 127.63 times larger than the input—
Doh!).
Now, the signal frequencies ω for which
vo (t ) = −200 Vin cos ωt
is an accurate statement, are frequencies that are said to lie
within the bandwidth of this amplifier (ω is just right!).
Conversely, frequencies ω for which:
vo (t ) ≠ −200 Vin cos ωt
Jim Stiles
The Univ. of Kansas
Dept. of EECS
4/20/2011
Amplifier Bandwidth
4/16
are frequencies ω that lie outside this amplifier’s bandwidth.
Fortunately, the frequencies that compose an amplifier’s
bandwidth typically form a continuum, such that the
frequencies outside this bandwidth are either higher or lower
than all frequencies within the bandwidth.
Perhaps a plot would help.
Avo ( ω )
200
ωL
ω
ωH
The frequencies between ωL and ωH thus lie within the
bandwidth of the amplifier. The gain within the bandwidth is
sometimes referred to as the midband gain.
For signals with frequencies less than ωL (fL ) , the amplifier
gain will be less than the midband gain—likewise for
frequencies greater than ωH (fH ) .
Jim Stiles
The Univ. of Kansas
Dept. of EECS
4/20/2011
Amplifier Bandwidth
5/16
Q: So what then is the value:
Avo =
vo
= −200
vi
determined for the example amplifier? It doesn’t seem to be
a function of frequency!
A: The value -200 calculated for this amplifier is the
midband gain—it’s the gain exhibited for all signals that lie
within the amplifier bandwidth. Your book at times uses the
variable AM to denote this value:
Avo ( ω )
AM
fL
fH
ω
Q: So it’s actually the midband gain that we’ve been
determining from our small-signal circuit analysis (e.g.
AM = −200 )?
A: That’s exactly correct!
Jim Stiles
The Univ. of Kansas
Dept. of EECS
4/20/2011
Amplifier Bandwidth
6/16
Q: So how do we determine the frequency dependent gain
Avo ( ω ) ? More specifically, how do we determine midband gain
AM , along with fL and fH ?
A: The function Avo ( ω ) is simply the eigen value of the linear
operator relating the small-signal input and the small signal
output:
vo (t ) = L {vi (t )}
⇒
Vo (ω ) = Avo (ω ) Vi (ω )
Q: Yikes! How do we determine the eigen value of this linear
operator?
A: We simply analyze the small-signal circuit, determining
Vo (ω ) in terms of Vi (ω ) .
Specifically, we must explicitly consider the capacitance in
the small-signal amplifier—no longer can we make
approximations!
So, instead of vaguely labeling large capacitors as Capacitors
Of Unusual Size, let’s explicitly consider the exact values of
these large capacitors:
Jim Stiles
The Univ. of Kansas
Dept. of EECS
4/20/2011
Amplifier Bandwidth
7/16
15 V
15 V
1K
3.7 K
vO (t )
Ci
β = 100
vi (t )
+
-
2.3K
1K
CE
Likewise, we must consider the parasitic capacitances of the
BJT—specifically C μ and Cπ .
The small-signal circuit—when we explicitly consider these
capacitances—is thus:
Jim Stiles
The Univ. of Kansas
Dept. of EECS
4/20/2011
Amplifier Bandwidth
1
1
B
jωCi
8/16
Vo (ω )
C
jωC μ
+
+
-
0 .5 K
v be (ω )
2.3 K
3.7 K
1
Vi (ω )
jωCπ
1K
200 v be (ω )
−
E
1K
1
jωCE
We analyze this circuit to determine Vo (ω ) , and then the
eigen value—the small-signal gain—is determined as:
Avo (ω ) =
Vo (ω )
Vi (ω )
Q: So what again is the meaning of Vi (ω ) and Vo (ω ) ?
A: It’s the Fourier transform of vi (t ) and vo (t ) !
∞
Vi (ω ) = ∫ vi (t ) e
−∞
jωt
dt
and
∞
Vo (ω ) = ∫ vo (t ) e jωt dt
−∞
Q: So—I can’t recall—what’s the relationship between vi (t )
and vo (t ) ?
Jim Stiles
The Univ. of Kansas
Dept. of EECS
4/20/2011
A: If:
Amplifier Bandwidth
9/16
Vo (ω ) = Avo (ω ) Vi (ω )
Then in the time domain, we find that the input and output are
related by the always enjoyable convolution integral!!!
vo (t ) =
∞
∫ g (t − t ′) vi (t ′) dt ′
−∞
where the impulse response of the amplifier is of course:
g (t ) =
∞
− jωt
(
)
A
ω
e
dω
∫ vo
−∞
Q: What the heck? What happened to solutions like:
vo (t ) = −200 vi (t ) ??
A: This result implies that the impulse response of the
amplifier is:
g (t ) = −200 δ (t )
Such that:
vo (t ) =
∞
∫ g (t − t ′) vi (t ′) dt ′
−∞
∞
= −200 ∫ δ (t − t ′ ) vi (t ′ ) dt ′
−∞
= −200 vi (t )
Jim Stiles
The Univ. of Kansas
Dept. of EECS
4/20/2011
Amplifier Bandwidth
10/16
Q: You say that the result:
vo (t ) = −200 vi (t )
“implies” that the impulse response of the amplifier is:
g (t ) = −200 δ (t )
Are you saying the impulse response of the common-emitter
example is not this function?
A: It is definitely not that function. The impulse response
g (t ) = −200 δ (t )
is ideal—the impulse response of an amplifier with an infinite
bandwidth!
Q: So all our small-signal analysis up to this point has been
incorrect and useless???
A: Not at all! The small-signal gain we have been evaluating
up to this point (e.g., -200) is the amplifier midband gain AM .
As long as the small-signal input vi (t ) resides completely
within the amplifier bandwidth, then the output will be:
vo (t ) = −200 vi (t )
Jim Stiles
The Univ. of Kansas
Dept. of EECS
4/20/2011
Amplifier Bandwidth
11/16
The problem occurs when the input signal lies—at least
partially—outside the amplifiers bandwidth.
In that case, we find that:
vo (t ) ≠ −200 vi (t )
And instead:
vo (t ) =
∞
∫ g (t − t ′) vi (t ′) dt ′
−∞
where:
∞
g (t ) =
∫ Avo (ω ) e
− jωt
dω
−∞
and the eigen value Avo (ω ) = Vo (ω ) Vi (ω ) is determined by
evaluating this small-signal circuit:
1
1
B
jωCi
C
jωC μ
Vo (ω )
+
+
-
3.7 K
Vi (ω )
v be (ω )
2.3 K
1
jωCπ
−
0 .5 K
1K
200 v be (ω )
E
1K
Jim Stiles
The Univ. of Kansas
1
jωCE
Dept. of EECS
4/20/2011
Amplifier Bandwidth
12/16
Q: What do you mean when you say that a signal lies “within
the amplifier bandwidth” or “outside the amplifier bandwidth?
How can we tell?
A: Use the Fourier Transform!
If we plot the magnitude of the Fourier Transform Vi (ω ) of
the input signal vi (t ) , we can see the spectrum of the input
signal:
Vi ( ω )
ω
For example, if you are attempting to amplify a signal
representing the audio of symphonic music, the spectrum
Vi (ω ) will include low-frequency signals (e.g., from the tubas),
mid-range frequency signals (e.g., from the trumpets), and
high-frequency signals (e.g., from the flutes).
Jim Stiles
The Univ. of Kansas
Dept. of EECS
4/20/2011
Amplifier Bandwidth
13/16
Now, if it is your desire to reproduce exactly this music at
the output of your amplifier, then the amplifier bandwidth
must be wide enough to include all these spectral components
!
Vo ( ω ) = Avo ( ω ) Vi ( ω )
f
fH
fL
For the case above, the input signal resides completely within
the bandwidth of the amplifier (i.e., between fL and fH ), and
so we find (for AM = −200 ) that:
Vo ( ω ) = ( 200 ) Vi ( ω )
and:
vo (t ) = −200 vi (t )
However, if the input spectrum resides (at least partially)
outside the amplifier bandwidth, e.g.:
Jim Stiles
The Univ. of Kansas
Dept. of EECS
4/20/2011
Amplifier Bandwidth
14/16
Vo ( ω ) = Avo ( ω ) Vi ( ω )
fL
fH
f
then we find that:
Vo ( ω ) ≠ ( 200 ) Vi ( ω )
and:
vo (t ) ≠ −200 vi (t ) !!!!
Instead, we find the more general (and more difficult!)
expressions:
Vo (ω ) = Avo (ω ) Vi (ω )
and:
vo (t ) =
Jim Stiles
∞
∫ g (t − t ′) vi (t ′) dt ′
−∞
The Univ. of Kansas
Dept. of EECS
4/20/2011
Amplifier Bandwidth
15/16
where the impulse response of the amplifier is:
g (t ) =
∞
− jωt
(
)
A
ω
e
dω
∫ vo
−∞
Q: So just what causes this amplifier to have a finite
bandwith?
A: For mid-band frequencies fm (i.e., between fL < fm < fH ),
we will find that the Capacitors Of Unusual Size exhibit an
impedance that is pretty small—approximately an AC short
circuit:
ZCOUS (ωm ) =
1
ωmCous
≅0
Likewise, the tiny parasitic capacitances C μ and Cπ exhibit
an impedance that is very large for mid-band frequencies—
approximately an open circuit:
Z C (ω m ) =
π
1
ω m Cπ
≅∞
However, when the signal frequency ω drops too low, the
COUS will no longer be a small-signal short.
The result is that the amplifier gain is reduced—the values of
the COUS determine the low-end amplifier bandwidth fL .
Jim Stiles
The Univ. of Kansas
Dept. of EECS
4/20/2011
Amplifier Bandwidth
16/16
Likewise, when the signal frequency ω is too high, the
parasitic caps will no longer be a small-signal open.
The result is that the amplifier gain is reduced—the values of
the parasitic capacitors determine the high-end amplifier
bandwidth fH .
Jim Stiles
The Univ. of Kansas
Dept. of EECS
4/22/2011
Midband Gain
1/4
Mid-band Gain
15 V
15 V
Q: So, to find the mid-band
gain of this amplifier:
1K
3.7 K
vO (t )
COUS
β = 100
vi (t )
+
-
2.3K
1K
COUS
we must find the analyze this small signal circuit:
1
1
B
jωCi
C
jωC μ
Vo (ω )
+
+
-
3.7 K
Vi (ω )
v be (ω )
2.3 K
1
jωCπ
0 .5 K
1K
200 v be (ω )
−
E
1K
Jim Stiles
The Univ. of Kansas
1
jωCE
Dept. of EECS
4/22/2011
Midband Gain
2/4
to determine:
Vo (ω )
(
)
Avo ω =
Vi (ω )
and then plotting the magnitude:
Avo ( ω )
AM
ωL
ω
ωH
we determine mid-band gain AM , right?
A: You could do all that, but there is an easier way.
Recall the midband gain is the value af Avo ( ω ) for
frequencies within the amplifier bandwidth. For those
frequencies, the AC coupling capacitors (i.e., COUS) are
approximate AC short-circuits (i.e., very low impedance) .
Jim Stiles
The Univ. of Kansas
Dept. of EECS
4/22/2011
Midband Gain
3/4
Likewise, for the signal frequencies within the amplifier
bandwidth, the parasitic BJT capacitances are approximate
AC open-circuits (i.e., very high impedance).
Thus, we can apply these approximations to the capacitors in
our small-signal circuit:
B
Vo (ω )
C
+
+
-
2.3 K
3.7 K
v be (ω )
Vi (ω )
0 .5 K
1K
200 v be (ω )
−
E
1K
Now simplifying this circuit (look, no capacitors!):
Vo (ω )
+
Vi (ω )
+
-
vbe
-
Jim Stiles
0.37 K
RC =1 K
200 v be
The Univ. of Kansas
Dept. of EECS
4/22/2011
Midband Gain
4/4
Q: Hey wait! Isn’t this the same small-signal circuit that we
analyzed earlier, where we found that:
vo (t ) = −200 vi (t ) ??
A: It is exactly!
All of the small-signal analysis that we performed previously
(i.e., the circuits with no capacitors!) actually provided us with
the mid-band amplifier gain.
Taking the Fourier transform of the equation above:
Vo ( ω ) = −200 Vi ( ω )
= e jπ 200 Vi ( ω )
Thus, the midband gain of this amplifier is:
AM = −200 = e jπ 200
Jim Stiles
The Univ. of Kansas
Dept. of EECS
4/22/2011
Low-Frequency Response
1/11
Low-Frequency Response
Q: OK, I see how to determine mid-band gain, but what about
determining amplifier bandwidth?
It seems like I have no alternative but to analyze the exact
small-signal circuit (explicitly considering all capacitances):
1
1
B
jωCi
C
jωC μ
Vo (ω )
+
+
-
3.7 K
v be (ω )
2.3 K
1
Vi (ω )
jωCπ
−
0 .5 K
1K
200 v be (ω )
E
1K
And then determine:
Avo (ω ) =
1
jωCE
Vo (ω )
Vi (ω )
and then plot the magnitude:
AM
Avo ( ω )
fH
fL
Jim Stiles
The Univ. of Kansas
f
Dept. of EECS
4/22/2011
Low-Frequency Response
2/11
And then from the plot determine the amplifier bandwidth
(i.e., determine fL and fH )?
A: You could do all that, but there is an easier way.
An amplifier frequency response Avo ( ω ) (i.e., its eigen value!)
can generally be expressed as the product of three distinct
terms:
Avo ( ω ) = FL ( ω ) AM FH ( ω )
The middle term is the of course the mid-band gain—a
number that is not frequency dependent.
The function FL ( ω ) describes the low-frequency response of
the amplifier—from it we can determine the lower cutoff
frequency fL .
Conversely, the function FH ( ω ) describes the high-frequency
response of the amplifier—from it we can determine the
upper cutoff frequency fH .
Q: So just how do we determine these functions FL ( ω ) and
FH ( ω ) ??
A: The low-frequency response FL ( ω ) is dependent only on
the large capacitors (COUS) in the amplifier circuit. In other
Jim Stiles
The Univ. of Kansas
Dept. of EECS
4/22/2011
Low-Frequency Response
3/11
words the parasitic capacitances have no affect on the lowfrequency response.
Thus, we simply “ignore” the parasitic capacitances when
determining FL ( ω ) !
For example, say we include the COUS in our common-emitter
example, but ignore C μ and Cπ . The resulting small-signal
circuit is:
1
jωCi
B
C
Vo (ω )
+
+
-
0.37 K
Vi (ω )
Vbe (ω )
−
0 .5 K
1K
200 Vbe (ω )
E
1K
1
jωCE
To simplify this analysis, we first determine the Thevenin’s
equivalent circuit of the portion of the circuit connected to
the base.
We start by finding the open-circuit voltage:
Jim Stiles
The Univ. of Kansas
Dept. of EECS
4/22/2011
Low-Frequency Response
4/11
1
jωCi
+
-
0.37 K
Vi (ω )
Vo oc (ω )
Vooc ( ω
⎛
⎞⎟
0.37
⎜
⎟
) = Vi ( ω )⎜⎜
1 ⎟
⎜⎝ 0.37 + jωCi ⎠⎟
⎛ jω ( 0.37 )Ci
= Vi ( ω )⎜⎜
⎜⎝ 1 + jω ( 0.37 )Ci
⎞⎟
⎟
⎠⎟
And the short-circuit output current is:
1
jωCi
+
-
0.37 K
Vi (ω )
Iosc (ω )
Iosc (ω ) =
Vi (ω )
1
jωCi
=Vi (ω ) ( jωCi )
And thus the Thevenin’s equivalent source is:
Zth ( ω )
+
-
Jim Stiles
Vth (ω )
Vth ( ω ) = Vooc ( ω )
⎛ jω ( 0.37 )Ci
= Vi ( ω )⎜⎜
⎜⎝ 1 + jω ( 0.37 )Ci
The Univ. of Kansas
⎞⎟
⎟
⎠⎟
Dept. of EECS
4/22/2011
Low-Frequency Response
5/11
Vooc ( ω )
Zth ( ω ) = sc
Io ( ω )
⎛ jω ( 0.37 )Ci
= ⎜⎜
⎜⎝ 1 + jω ( 0.37 )Ci
=
⎞⎛
⎟⎟⎜ 1
⎜
⎠⎟⎜⎝ jωCi
⎞⎟
⎟
⎠⎟
( 0.37 )
1 + jω ( 0.37 )Ci
Likewise, the two parallel elements on the emitter terminal
can be combined:
Z E (ω ) =
1
jωCE
1+
1
jωCE
=
1
1 + jωCE
Thus, the small-signal circuit is now:
Zth ( ω )
B
C
Vo (ω )
+
+
-
Vth (ω )
v be (ω )
−
0 .5 K
1K
200 v be (ω )
E
Z E (ω )
From KVL:
Jim Stiles
The Univ. of Kansas
Dept. of EECS
4/22/2011
Low-Frequency Response
6/11
0 +Vth − Ib ( Zth + 0.5 ) − ( β + 1 ) Ib ZE = 0
⇒
Ib =
Vth
Zth + 0.5 + 101ZE
From Ohm’s Law:
Vbe = 0.5Ib
Therefore:
Vo ( ω ) = −200Vbe ( 1 )
= −200 ( 0.5 )
Vth ( ω )
Zth + 0.5 + 101ZE
⎛
−100
= Vth ( ω )⎜⎜
⎜⎝ Zth + 0.5 + 101ZE
⎞⎟
⎟⎟
⎠
Inserting the expressions for the Thevenin’s equivalent
source, as well as ZE .
⎛
⎞⎟
−100
⎟
⎜⎝ Zth + 0.5 + 101ZE ⎠⎟
⎛
−200
⎛ jω ( 0.37 )Ci ⎞⎟⎜⎜
= Vi ( ω )⎜⎜
⎟⎜
2 ( 0.37 )
202
⎜⎝ 1 + jω ( 0.37 )Ci ⎠⎟⎜⎜
+
+
1
⎜⎝ 1 + jω ( 0.37 )C
1 + jωCE
i
Vo ( ω ) = Vth ( ω )⎜⎜
Jim Stiles
The Univ. of Kansas
⎞⎟
⎟
⎟⎟⎟
⎟
⎠⎟
Dept. of EECS
4/22/2011
Low-Frequency Response
7/11
Now, it can be shown that:
1
2 ( 0.37 )
1 + jω ( 0.37 )Ci
+1+
202
1 + jωCE
≅
jω ( CE 203 )
1 + jω ( CE 203 )
Therefore:
⎛ jω ( 0.37 )Ci
Vo ( ω ) = Vi ( ω ) ⎜⎜
⎜⎝ 1 + jω ( 0.37 )Ci
⎞⎟⎜⎛ jω ( CE 203 ) ⎞⎟
⎟( −200 )
⎟⎟⎜⎜
⎠⎝ 1 + jω ( CE 203 ) ⎟⎠⎟
And so:
⎛ jω ( 0.37 )Ci
⎜⎝ 1 + jω ( 0.37 )Ci
Avo ( ω ) = ⎜⎜
⎞⎟⎜⎛ jω ( CE 203 ) ⎞⎟
⎟( −200 )
⎟⎟⎜⎜
⎟
⎟
+
1
jω
⎠⎝
(CE 203 ) ⎠
Now, since we are ignoring the parasitic capacitances, the
function FH ( ω ) that describes the high frequency response
is:
And so:
FH ( ω ) = 1
Avo ( ω ) = FL ( ω ) AM FH ( ω ) = FL ( ω ) AM
By inspection, we see for this example:
AM = −200
Jim Stiles
Å We knew this already!
The Univ. of Kansas
Dept. of EECS
4/22/2011
And:
Low-Frequency Response
⎛ jω ( 0.37 )Ci
FL ( ω ) = ⎜⎜
⎜⎝ 1 + jω ( 0.37 )Ci
8/11
⎞⎟⎜⎛ jω ( CE 203 ) ⎞⎟
⎟
⎟⎟⎜⎜
⎠⎝ 1 + jω ( CE 203 ) ⎠⎟⎟
Now, let’s define:
ωP 1 =
1
2.7
=
Ci
0.37Ci
and
ωP 2 =
203
CE
Thus,
⎛ j ( ω ωP 1 )
⎜⎝ 1 + j ( ω ω
FL ( ω ) = ⎜⎜
P1
⎞⎛
⎟⎟⎜⎜ j ( ω ωP 2 )
⎟⎟⎜ 1 + j ( ω ω
) ⎠⎝
⎞⎟
⎟⎟⎟
)
P2 ⎠
Now, functions of the type:
⎛ j ( ω ωP )
⎜⎜
⎜⎝ 1 + j ( ω ω
P
⎞⎟
⎟
) ⎠⎟⎟
are high-pass functions:
2
2
j ( ω ωP )
( ω ωP )
=
2
1 + j ( ω ωP )
1 + ( ω ωP )
with a 3dB break frequency of ωP .
Jim Stiles
The Univ. of Kansas
Dept. of EECS
4/22/2011
Low-Frequency Response
2
( ω ωP )
1 +( ω ωP
0
−3
2
)
9/11
[ dB ]
log ωP
log ω
Thus:
⎧
for ω > ωP
≅ 1.0
⎪
⎪
⎪
⎪
2
⎪
⎪
( ω ωP )
⎪
=
for ω = ωP
⎨ 0.5
2
⎪
1 + ( ω ωP )
⎪
⎪
⎪
⎪
⎪
for ω = 0
⎪
⎩0
As a result, we find that the transfer function:
Avo ( ω ) = FL ( ω ) AM
⎛ j ( ω ωP 1 )
= ⎜⎜
⎜⎝ 1 + j ( ω ω
P1
⎞⎛
⎟⎟⎜ j ( ω ωP 2 )
⎟⎟⎜⎜ 1 + j ( ω ω
) ⎠⎝
⎞⎟
⎟⎟( −200 )
⎟
P 2 )⎠
will be approximately equal to the midband gain AM = −200
for all frequencies ω that are greater than both ωP 1 and ωP 2 .
Jim Stiles
The Univ. of Kansas
Dept. of EECS
4/22/2011
Low-Frequency Response
10/11
I.E.,:
Avo ( ω ) ≅ AM = −200
if ω > ωP 1 and ω > ωP 2
Hopefully, it is now apparent (please tell me it is!) that the
lower end of the amplifier bandwidth—specified by frequency
ωL —is the determined by the larger of the two frequencies
ωP 1 and ωP 2 !
The larger of the two frequencies is called the dominant pole
of the transfer function FL ( ω ) .
For our example—comparing the two frequencies ωP 1 and ωP 2 :
ωP 1 =
1
2.7
=
0.37Ci
Ci
and
ωP 2 =
203
CE
it is apparent that the larger of the two (the dominant pole!)
is likely ωP 2 —that darn emitter capacitor is the key!
Say we want the common-emitter amplifier in this circuit to
have a bandwidth that extends down to fL = 100 Hz
The emitter capacitor must therefore be:
Jim Stiles
The Univ. of Kansas
Dept. of EECS
4/22/2011
Low-Frequency Response
2πfL > ωP 2 =
⇒ CE >
11/11
203
CE
203
203
=
= 32,300μF
(
)
2πfL
2π 1000
!!!!
This certainly is a Capacitor Of Unusual Size !
Jim Stiles
The Univ. of Kansas
Dept. of EECS
4/22/2011
High Frequency Response
1/4
High-Frequency Response
To determine the high-frequency response of our example
common-emitter amp, we simply consider explicitly the
parasitic capacitances in the small-signal model, while
approximating the COUS as small-signal short-circuits:
1
B
Vo (ω )
C
jωC μ
+
+
-
3.7 K
Vi (ω )
v be (ω )
2.3 K
1
jωCπ
−
0 .5 K
1K
200 v be (ω )
E
Now, since we are ignoring the COUS, the function FL ( ω )
that describes the low- frequency response is:
FL ( ω ) = 1
And so:
Avo ( ω ) = FL ( ω ) AM FH ( ω ) = AM FH ( ω )
Jim Stiles
The Univ. of Kansas
Dept. of EECS
4/22/2011
High Frequency Response
2/4
We will find that the high-frequency response will
(approximately) have the form.
⎛
⎞⎛
⎞⎟
1
1
⎟
⎜
⎜
FH ( ω ) = ⎜
⎟⎟⎜
⎟
⎜⎝ 1 + j ( ω ωP 3 ) ⎟⎜
1 + j ( ω ωP 4 ) ⎠⎟
⎠⎝
Now, functions of the type:
⎛
1
⎜⎜
⎜⎝ 1 + j ( ω ωP
⎞⎟
⎟
) ⎠⎟
are low-pass functions:
1
2
1 + j ( ω ωP )
=
1
2
1 + ( ω ωP )
with a 3dB break frequency of ωP .
1
1 +( ω ωP
0
−3
Jim Stiles
2
)
[ dB ]
log ωP
The Univ. of Kansas
log ω
Dept. of EECS
4/22/2011
High Frequency Response
3/4
Thus:
for ω < ωP
⎪⎧⎪ ≅ 1.0
⎪⎪
⎪⎪
1
= ⎪⎨ 0.5
for ω = ωP
2
⎪
1 + ( ω ωP )
⎪⎪
⎪⎪
⎪⎪⎩ 0
for ω → ∞
As a result, we find that the transfer function:
Avo ( ω ) = AM FH ( ω )
⎛
⎞⎛
⎞⎟
1
1
⎟⎟⎜
= ( −200 )⎜⎜
⎟⎠⎝⎜ 1 + j ( ω ω ) ⎠⎟⎟
⎜⎝ 1 + j ( ω ωP 3 ) ⎟⎜
P3
will be approximately equal to the midband gain AM = −200
for all frequencies ω that are less than both ωP 3 and ωP 4 .
I.E.,:
Avo ( ω ) ≅ AM = −200
if ω < ωP 3 and ω > ωP 4
Hopefully, it is now apparent (please tell me it is!) that the
higher end of the amplifier bandwidth—specified by
frequency ωH —is the determined by the smaller of the two
frequencies ωP 3 and ωP 3 !
The smaller of the two frequencies is called the dominant
pole of the transfer function FH ( ω ) .
Jim Stiles
The Univ. of Kansas
Dept. of EECS
4/22/2011
High Frequency Response
4/4
Generally speaking, to increase the value ωH , we must
increase the DC bias currents of our amplifier design!
Jim Stiles
The Univ. of Kansas
Dept. of EECS