Lecture 4c
Determination of Poles & Zeros of CS Amplifiers (2)
Objectives
„ To calculate different poles and zeros which
determine the frequency response of CS
amplifiers
„ To calculate the dominant poles and the
amplifier bandwidth
Direct Determination of High-Frequency Poles and Zeros:
The high frequency poles and zeros may be determined by direct
analysis.
In this case all the coupling and bypass capacitors may be
considered short circuit.
However, the high frequency model of the transistor should be
used since the transistor internal capacitances can't considered
open circuit at high frequencies.
The following example will illustrate this method for common
source amplifier.
Example 1:
Drive an expression for the high frequency response AvH(s) of the
common source amplifier shown in Figure 1. Hence, determine the
midband gain Amid, high frequency poles and zeros, and the high
cutoff frequency.
Figure 1: Common source amplifier
Solution:
The CS amplifier is redrawn for small signal analysis as shown in
Figure 2 by grounding the DC voltage source VDD and replacing
the transistor by its high frequency small signal model.
The biasing resistors R1 and R2 are combined into RG.
∴
RG
= R1 || R 2
All the coupling and bypass capacitors are considered short circuit
at high frequency.
Figure 2: The CS amplifier in Figure 1 redrawn for small signal analysis with the
transistor replaced by its high frequency small signal model
The small-signal model can be simplified by using Thevinin
Theorem as shown in Figure 3.
v th = v sig
RG
R sig + RG
and Rth =
R sig RG
R sig + RG
Also, R L′ = R L || R D
Figure 3: Simplified circuit to calculate the frequency response of the CS amplifier
Writing the two nodal equations at nodes A and B in frequency
domain,
vo
+ g mv gs + (v o −v gs )sC GD = 0
R L′
v gs sC GD +
v gs −v th
R th
+ (v gs −v o )sC GD = 0
Solving the two nodal equations found in the preceding slide by
eliminating vgs yields,
V o (s) =
v th (s) (sC GD - g m )
Rth
Δ
⎛C
⎛
1
1 ⎞⎞
1
Δ = s 2 (C GS C GD ) + s ⎜ GS + C GD ⎜ g m +
+
⎟ ⎟⎟ +
⎜ R′
R th R L′ ⎠ ⎠ R L′ .R th
⎝
⎝ L
∴
AvH (s ) =
v o (s )
v sig (s )
∴
AvH (s ) =
1 (sC GD - g m )
R sig
Δ
Let us define
CT
⎛
R′ ⎞
= C GS + C GD ⎜ 1 + g m R L′ + L ⎟
R th ⎠
⎝
∴
Δ = s 2C GS C GD +
∴
AvH (s ) =
sCT
1
+
R L′ R ′.R th
1
R sig C GS .C GD s 2 +
(sC GD - g m )
sC T
1
+
R L′ .C GS .C GD R ′.R th .C GS .C GD
vo
(s) =
v th (s) (sC GD - g m )
Δ
R th
⎛C
⎛
1
1
Δ = s 2 (C GS C GD ) + s ⎜ GS + C GD ⎜ g m +
+
⎜ R′
R th R L′
⎝
⎝ L
v o (s )
1 (sC GD - g m )
AvH (s ) =
=
v sig (s )
R sig
Δ
∴
⎞⎞
1
⎟ ⎟⎟ +
⎠ ⎠ R L′ .R th
⎛
R′ ⎞
= C GS + C GD ⎜1 + g m R L′ + L ⎟
R th ⎠
⎝
sC
1
Δ = s 2C GS C GD + T +
R L′ R ′.R th
Let us define C T
∴
∴
1
AvH (s ) =
R sig C GS .C GD s 2 +
(sC GD - g m )
sC T
1
+
R L′ .C GS .C GD R ′.R th .C GS .C GD
Amid may be found from the last expression by assuming s → 0 .
∴
A mid
=
- g m R L′ .RG
R sig + RG
High-frequency response is given by 2 poles, one finite zero and
one zero at infinity. Finite right-half plane zero, ωZ = + gm/CGD > ωT
can easily be neglected.
For a polynomial s2 + sA1 + A0 with roots a and b, if one root is
much larger than the other one we can assume that a = A1 and b =
A0/A1.
∴
ωP 1 =
ωP 1 ≅
1
R thC T
A0
A1
Assume that the midband gain is very large
⎛ R′ ⎞
( i.e. g m R L′ ⎜1 + L ⎟ , and g m R L′C GD C GS )
⎝ Rth ⎠
gm ⎛
1
1 ⎞
+
ωP 2 ≅
⎜1 +
⎟
C GS ⎝ g m Rth g m R L′ ⎠
gm
ωP 2 ≅
C GS
∴
∴
∴
ωP 1 =
A0
1
≅
A1
R thC T
⎛ R′
Assume that the midband gain is very large ( i.e. g m R L′ ⎜ 1 + L
⎝ R th
gm ⎛
gm
1
1 ⎞
ωP 2 ≅
+
⎜1 +
⎟ ≅
C GS ⎝ g m R th g m R L′ ⎠
C GS
⎞
⎟ , and
⎠
g m R L′C GD C GS )
∴
Smallest root that gives first pole limits frequency response and
determines ωH. Second pole is important in frequency
compensation as it can degrade phase margin of feedback
amplifiers.
Assuming ωp1 ωp2
(ωp1 is the dominanat pole)
∴
Assuming ωp1
∴
1
RthCT
ωp2 (ωp1 is the dominanat pole)
ωH ≅ ωP 1 =
ωH ≅ ωP 1 =
1
R thC T
Dominant pole model at high frequencies for CS amplifier is shown
in Figure 4.
Figure 4: Equivalent circuit assuming dominant pole model
Miller’s Theorem
Another way to find the high frequency poles is by calculating the
equivalent capacitance for CGD at the amplifier input. This may be
done using Miller's Theorem as shown next.
For the circuit shown we can state the following.
Figure 5: Miller's Theorem
v o (s ) = − Av 1 (s )
i s (s ) = sC [v 1 (s ) − v o (s ) ]
i s (s )
= sC (1 + A )
v 1 (s )
v o (s ) = − Av 1 (s )
Y (s ) =
i s (s ) = sC [v 1 (s ) −v o (s ) ]
Y (s ) =
i s (s )
= sC (1 + A )
v 1 (s )
The total input capacitance = C (1+A) because total voltage across
C is vc = vi (1+A) due to the inverting voltage gain of amplifier.
Applying Miller's Theorem to the CS Amplifier
For the common source amplifier referring to Figure 6 we may
write.
CT
CT
= C GS + C GD (1 + A )
= C GS + C GD (1 + g m R L )
CT
= C GS + C GD (1 + A ) = C GS + C GD (1 + g m R L )
Figure 16a: Applying Miller's theorem to the CS amplifier
Figure 6b: Resulting circuit from applying Miller's theorem to the CS amplifier
Open-Circuit Time Constant Method to Determine ωH
At high frequencies, impedances of the coupling and bypass
capacitors are small enough to be considered short circuits.
Open-circuit time constants associated with the impedances of
device capacitances are considered instead.
ωH
≅
1
m
∑ R ioC i
i =1
where Rio is the resistance at the terminals of the ith capacitor Ci
with all other capacitors open-circuited.
Example 2:
Derive an expression for the high cutoff frequency for the circuit in
Example 1 using the OCTC method.
Solution:
Using OCTC method for the circuit in Figure 3.
For CGS the resistance seen across the capacitor terminals may be
calculated from Figure 7.
RGSO = R th
So the time constant associated with CGS equals CGS. Rth
Figure 7: Circuit used to calculate the open-circuit time constant associated with CGS
For CGD the resistance seen across the capacitor terminals may be
calculated from Figure 8.
vx
ix
RGDO =
RGDO = Rth ⋅ (1 + g m R L′ +
R L′
)
Rth
vx
R′
= R th .(1 + g m R L′ + L )
ix
R th
RGDO =
So the time constant associated with CGD equals CGD · Rth (1 +
gmR`L + R`L/Rth)
Figure 8: Circuit used to calculate the open-circuit time constant associated with CGD
The high cutoff frequency is calculated using the SCTC as:
ωH
≅
1
2
∑ R ioC i
i =1
1
+ RGDOC GD
ωH ≅
ωH
ωH
RGSOC GS
1
≅
R thCT
1
1
≅ 2
=
∑ R ioC i RGSO C GS + RGDO C GD
i =1
=
1
R thC T