Download Section H5: High-Frequency Amplifier Response

Section H5: High-Frequency Amplifier Response
Now that we’ve got a high frequency models for the BJT, we can analyze the
high frequency response of our basic amplifier configurations.
Note: in the circuits that follow, the actual signal source (vS) and its
associated source resistance (RS) have been included. As discussed in the
low frequency response section of our studies, we always knew that this
source and resistance was there but we just started our investigations with
the input to the transistor (vin). Remember - the analysis process is the
same and the relationship between vS and vin is a voltage divider!
Again, in some instances in the following discussion, I will be using
slightly different notation and taking a different approach than your
author. As usual, if this results in confusion, or you are more
comfortable with his technique, let me know and we’ll work it out.
High Frequency Response of the CE and ER Amplifier
The generic common-emitter amplifier circuit of Section D2 is reproduced to
the left below and the small signal circuit using the high frequency BJT
model is given below right (based on Figures 10.17a and 10.17b of your
text). Note that all external capacitors are assumed to be short circuits at
high frequencies and are not present in the high frequency equivalent circuit
(since the external capacitors are large when compared to the internal
capacitances – recall that Zc=1/jωC gets small as the frequency or
capacitance gets large).
We can simplify the small signal circuit by making the following observations
and approximations:
¾ Cce is very small and may be neglected.
¾ rce >> (RC||RL), so rce may be neglected since rce||RC||RL is dominated by
RC||RL.
¾ rb’c is so much larger than all other resistances that it may be considered
“open” and removed from the circuit.
¾ rbb’ << rb’e so it may be neglected; i.e.,
rπ ≅ rb'e .
(Equation 10.62)
The above simplifications, along with the definition of Rin as
Rin = RB || rπ ,
(Equation 10.63)
and the reflection of Cb’c to the input and output circuits using Miller’s
theorem
C M1 = C b'c (1 − Av ) (input circuit )
,
⎛
1 ⎞
⎟⎟ (output circuit )
C M 2 = C b'c ⎜⎜1 −
Av ⎠
⎝
yields the simplified small
signal circuit presented to
the right. Note that this is a
corrected version of Figure
10.17c in your text. Using
the property that capacitors
in parallel add, we can
define a single capacitor in
the input circuit as
C in = C b'e + C M1 = C b'e + C b'c (1 − Av ) .
(Equation 10.64)
Putting all this together gives us the simplified, simplified circuit shown to
the left below. Note that, just like we did for the low frequency response, our
strategy will be to set all independent sources to zero (which, in this case,
also sets the dependent source to zero) as shown in the figure to the right
below. We can now calculate the individual time constants using the Method
of Open Circuit Time Constants.
Just as a sanity check – even if we had not neglected the capacitance Cce in
our first simplification, the parallel capacitance CM2 is so much larger that the
contribution of Cce would be negligible.
The Method of Open Circuit Time Constants is similar to the Method of Short
Circuit Time Constants, but now we set all capacitances to zero (open
circuit) except the one of interest. Equivalent resistances seen by the
capacitance of interest are then derived and the individual RC time constant
calculated. For the high frequency response using the poles of the transfer
function only, we may approximate the cutoff in radians per second (ω) or
Hertz (f) as
ωH
⎤
⎡
≅ ⎢∑ τ i ⎥
⎦
⎣ i
−1
=
1
,
∑ C i Reqi
fH =
ωH
.
2π
i
Using the figure above, we have
¾ Cin: with CM2 open, the resistance seen by Cin is equal to
RCin = RS || Rin = RS || RB || rπ ,
since the open current source separates the input and output circuits.
¾ CM2: with Cin open, the resistance seen by CM2 is equal to
RCM 2 = RC || RL
since, once again, the open current source separates the input an output
circuits.
Putting this all together, we can define the high frequency time constants for
the CE circuit as:
τ Cin = C in RCin ;
τ CM 2 = C M 2 RCM 2 ,
and an approximate value for the upper corner frequency, in radians per
second, is given by
ωH =
1
1
ω PCin
=
+
=
1
ω PCM 2
τ Cin
1
1
=
+ τ CM 2
C in RC in + C M 2 RC M 2
.
1
C in (RS || RB || rπ ) + C M 2 (RC || RL )
Since the capacitance on the input side is much larger than in the output
circuit, the high frequency cutoff is determined by dominant pole formed due
to the input resistance and capacitance. Expressing the high frequency cutoff
as a single dominant pole, we have
ωH =
1
;
Cin (RS || RB || rπ )
fH =
ωH
1
.
=
2π 2πCin (RS || RB || rπ )
(Equation 10.65)
Note that, if the source resistance, RS, becomes very small, the dominant
pole will switch from the input circuit to the output circuit. This results in a
higher cutoff frequency and a larger available bandwidth.
Since the expressions of Equations 10.64 and 10.65 are formulated in terms
of Rin and Av, the same equations may be used to calculate the high
frequency cutoff of the emitter resistor amplifier. The difference between the
common-emitter and emitter-resistor configurations is that Av is smaller and
Rin is larger for the ER, the cumulative effect of which results in a larger
bandwidth (higher cutoff frequency).
High Frequency Response of the CB Amplifier
The common base amplifier circuit is shown below and left, while the high
frequency small signal equivalent is given below and to the right (based on
Figures 10.21a and 10.21b of your text).
Note that, unlike the CE/ER amplifier, no internal feedback capacitance
exists for the common base configuration. This is the most important
characteristic of the CB stage and means that there is no Miller effect, which
means that the capacitances are smaller, which means that the cutoff
frequency will be higher. Cool, huh?
Using the method of open circuit time constants, we can define equivalent
resistances for each of the remaining capacitances. With vS=0 and,
therefore, gmvbe=0, we have
¾ Cb’e: with Cb’c set to zero, the resistance seen by Cb’e is equal to
RCb'e =
rπ
β
|| RE || RS = re || RE || RS ,
since we have to reflect rπ to the emitter circuit and assuming β >>1 so
that β+1 ≈ β.
¾ Cb’c: with Cb’e set to zero, the resistance seen by Cb’c is equal to
RCb'c = RC || RL .
Putting this all together, we can define the high frequency time constants for
the CB circuit as:
τ Cb'e = C b'e RCb'e ;
τ Cb'c = C b'c RCb'c ,
and an approximate value for the upper corner frequency, in radians per
second, is given by
ωH =
1
1
ω PCb'e
=
+
1
ω PCb'c
=
τ Cb'e
1
1
=
+ τ Cb'c
C b'e RC b'e + C b'c RC b'c
.
1
C b'e (re || RE || RS ) + C b'c (RC || RL )
If the source resistance is not extremely small, the dominant pole will
normally be at the input circuit, so that the high frequency cutoff is
ωH =
1
;
C b'e (re || RE || RS )
fH =
ωH
1
.
=
2π 2πC b'e (re || RE || RS )
(Equation 10.72)
Note that the frequency defined in Equation 10.72 is quite high since Cb’e
and re are quite small. At such high frequencies, it is often necessary to take
into account effects that may generally be considered negligible. To more
accurately determine the high frequency cutoff of a CB configuration, a more
elaborate transistor model is necessary and it is generally a good idea to
verify design using a computer simulation package such as Spice.
High Frequency Response of the EF (CC) Amplifier
The generic circuit for the emitter follower (common collector) amplifier is
given to the left below and the high frequency small signal circuit is shown
below and to the right (Figures 10.22a and 10.22b of your text).
Your author states that the load for an EF amplifier is small and often
capacitive, so he has included the load capacitance, CL. Also, since the
voltage gain for an EF stage is approximately equal to positive one, the
Miller capacitances are both close to zero.
The equivalent circuit
using Miller’s theorem is
given in Figure 10.22c in
your
text
and
is
reproduced to the right,
where
C M1 = C b'e (1 − Av )
⎛
1 ⎞ and Rin = RB || [rπ + β (RE || RL )] .
⎟
C M 2 = C b'e ⎜⎜1 −
Av ⎟⎠
⎝
Note: It looks like your author is making β(RE||RL) >> rπ so that
Rin=RB||β(RE||RL). This may be true, and probably will be true, but I’m more
comfortable leaving rπ in for now. So…as has happened so many times
before…some of my equations won’t exactly match the text.
We can again define an equivalent capacitance for the input side and an
equivalent capacitance for the output side as
C in = C b'c + C M1 ≅ C b'c
C out = C L + C M 2 ≅ C L
,
and, by the method of open circuit time constants, the resistances seen by
each capacitor as
¾ Cin: with Cout open, the resistance seen by Cin is equal to
RCin = RS || Rin .
¾ Cout: with Cin open, the resistance seen by Cout is equal to
RCout = RE || RL .
The high frequency time constants for the EF circuit as:
τ Cin = C in RCin ;
τ Cout = C out RCout ,
and an approximate value for the upper corner frequency, in radians per
second, is given by
1
ωH =
1
ω PCin
+
1
=
ω PCout
τ Cin
1
1
1
=
=
+ τ Cout
C in RC in + C out RC out
C in (RS || Rin ) + C out (RE || RL )
.
Often, τCin >> τCout, so that the high frequency cutoff is limited by the input
circuit. If this is the case, the high frequency cutoff may be expressed as
ωH =
1
;
C b'c (RS || Rin )
Cascode Amplifier
fH =
ωH
1
.
=
2π 2πC b'c (RS || Rin )
(Equation 10.76, Modified)
The
cascode
configuration
was
introduced in Section D7 of the WebCT
notes and Chapter 5 of your text. This
configuration consists of a CE stage
driving a CB stage as shown in Figure
10.24 and to the right. As we discussed
earlier, this configuration combines the
advantages of the common-emitter and
common-base circuits. Specifically, this
configuration allows us to get the high
input impedance of the CE, while also
getting a much higher cutoff frequency
than would be achievable with either
the CE or ER alone.
Since the CE amplifier (Q1) sees the low input resistance of the CB, the gain
of the CE stage is approximately equal to –1, which reduces the effect of the
capacitance Cb’c. This capacitance is the source of the limiting high frequency
pole of the CE configuration due to the Miller effect and, by reducing its
effect, the pole occurs at a higher frequency. As we discussed earlier, the CB
configuration (Q2) is not affected by the Miller effect and already has a wide
bandwidth. With RB bypassed with CB1, the CB stage can provide a high
voltage gain and compensates for the low gain of the CE amplifier.
The results are: high voltage gain, wide bandwidth (high cutoff frequency),
high input resistance and high output resistance. Pretty good deal!