E2.2 Analogue electronics
Problem sheet 5 (Week 8)
An NPN bipolar transistor is given with C BE = 5 pF , CBC = 1 pF , CCE = 0.1 pF , VA = 100V , β = 200 .
For the purposes of this problem sheet the transistor is biased at I C = 1mA
QUESTION 1: Use the second form of the Miller Theorem to derive an expression for the input
impedance of a common collector amplifier which drives an inductive load LE . Derive an
expression for the frequency at which the input impedance of this amplifier becomes real, and a
value for the input impedance at that frequency. Neglect the effect of CBC .
The calculation will be greatly simplified if you use the following substitutions:
R = Rπ , C = CBE , L = ( β + 1) LE ,τ = RC ,τ 1 =
L
R
ANSWER
The input impedance of the CE amplifier will be :
Z in = ( β + 1) jω LE +
Rπ
1 − ω 2ττ 1 + jωτ 1
R − ω 2 RLC + jω L
=
=R
1 + Rπ jωCBE
1 + RjωC
1 + jωτ
1 − ω 2ττ 1 + jωτ 1 − jωτ (1 − ω 2ττ 1 + jωτ 1 )
1 + jω (τ 1 − τ + ω 2τ 2τ 1 )
1 − ω 2ττ 1 + jωτ 1
Z in = R
=R
=R
1 + jωτ
1 + ω 2τ 2
1 + ω 2τ 2
This becomes real when Re Z in = 0 ⇒ τ 1 − τ + ω 2τ 2τ 1 = 0 ⇒ ω 2 =
τ − τ1 1 ⎛ τ1 ⎞
=
1−
τ 2τ 1 ττ 1 ⎜⎝ τ ⎟⎠
The value of the real resistance is:
Z in =
R
1 + ω 2τ 2
=
Rπ2CBE
R
g
τ
R
, =
π , with ωT = m
τ ⎛ τ ⎞ τ
β + 1) LE LEωT
CBE
1 + ⎜1 − 1 ⎟ 1 (
τ1 ⎝ τ ⎠
E22 Analogue Electronics
PS5 Answers
p 1/4
QUESTION 2: Calculate the value of the load resistance RL which results in the maximum
voltage gain of a Common Emitter amplifier employing Miller Cancellation as a function of the
frequency at which Miller cancellation is implemented. You may assume that the source
impedance satisfies Rs << Rπ . Sohow that in the limit of very high Miller cancellation frequency
the maximum obtainable gain varies as ω −2 .
ANSWER
If Rs << Rπ , we can neglect Rπ .Miller cancelation implies that at the frequency of interest
ω 2CBC LM = 1 and CBC does not play a role at all.
The voltage gain is:
− g m ( RA / / RL )
vL
1
1
=
=
vS
1 + sRs CBE 1 + s ( RA / / RL ) CCE
1 + sRs CBE
=
1
2
1 + ω 2 Rs2CBE
⎛ R R ⎞
− gm ⎜ A L ⎟
⎝ RA + RL ⎠ =
⎛ R R ⎞
1 + s ⎜ A L ⎟ CCE
⎝ RA + RL ⎠
⎛ R R ⎞
gm ⎜ A L ⎟
⎝ RA + RL ⎠
2
⎛ R R ⎞ 2
1 + ω ⎜ A L ⎟ CCE
⎝ RA + RL ⎠
2
We need to maximize the quantity
x
1 + x2
with x =
RA RL
CCE , since the only thing we can vary
RA + RL
is RL
⎛
x2 ⎞
2
1
+
x
−
⎜
2 ⎟ d
d
x
dx d
x
RA RL
1
+
x
⎜
⎟
=
=
=
2
2
2
⎜
⎟ dRL RA + RL
1+ x
dRL 1 + x
dRL dx 1 + x
⎜
⎟
⎝
⎠
2
( RA + RL ) RA − RA RL =
RA
1
=
>0
2
2
2
2
2
2
+
R
R
(
)
+
+
+
x
x
R
R
1
1
( )
( ) ( A L)
A
L
This is a monotonically increasing with RL function and the maximum gain is therefore:
max
g m RA
1
1
vL
=
=
2
1 + sRs CBE 1 + sRACCE
vS
1 + ω 2 Rs2CBE
E22 Analogue Electronics
PS5 Answers
g m RA
1+ ω R C
2
2
A
2
CE
→
gm
ω Rs CBE CCE
2
p 2/4
QUESTION 3: A negative feedback amplifier is built with a dominant pole forward amplifier and
feedback gain constant with frequency. Prove that the closed loop amplifier is also a dominant
pole amplifier whose gain-bandwidth product is constant and equal to the gain-bandwidth product
of the forward amplifier.
ANSWER
The forward gain is G ( s ) =
G0
. The reverse gain is H ( s ) . The closed loop gain is:
1 + sτ
G0
G (s)
(1 + sτ ) =
G0
G0
1
=
=
=
A( s) =
1 + G ( s ) H 1 + G0 H 1 + G0 H + sτ 1 + G0 H 1 + sτ / (1 + G0 H )
(1 + sτ )
=
G0
τ
G′
, G′ =
,τ ′ =
⇒ G′ / τ ′ = G0 / τ QED
1 + sτ ′
1 + G0 H
1 + G0 H
QUESTION 4: (Tutorial Question Week 8-9)
Show that the admittance matrix of the “Pi” network shown on the left below is:
⎡Y + Y
Y =⎢ 1 C
⎣ −YC
−YC ⎤
Y2 + YC ⎥⎦
Connect an admittance YC between the input and output of a network with the admittance
y12 ⎤
⎡y
matrix: Y0 = ⎢ 11
⎥ . Use KCL to show that the resulting circuit (shown in the figure below)
⎣ y21 y22 ⎦
⎡y +Y
has an admittance matrix equal to: Y ' = ⎢ 11 C
⎣ y21 − YC
y12 − YC ⎤
y22 + YC ⎥⎦
The admittance matrix of a 2-port is defined as: Yij =
∂ii
∂v j
i.e. the current in port “i” is
δ vk ≠ j = 0
measured by an ideal ammeter as a voltage differential is applied on port “j”.
YC
Y1
Yc
E22 Analogue Electronics
Y0
Y2
PS5 Answers
p 3/4
ANSWER
By definition,
y11 =
∂i1
∂v1
y21 =
= Y1 + YC
y12 =
v2 = 0
∂i2
∂v1
= −YC
y22 =
v2 = 0
∂i1
∂v2
∂i2
∂v2
= −YC
v1 = 0
= Y2 + YC
v1 = 0
−YC ⎤
⎡Y
A network with only Yc between ports 1 and 2 has an admittance matrix Y ' = ⎢ C
⎥ . The
⎣ −YC YC ⎦
network on the right consists of two networks connected in parallel, so its admittance matrix is the
sum of the individual admittance matrices.(You can see this by taking the derivatives)
⎡y +Y
It follows that the total admittance matrix is Y ' = ⎢ 11 C
⎣ y21 − YC
E22 Analogue Electronics
PS5 Answers
y12 − YC ⎤
y22 + YC ⎥⎦
p 4/4