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TSEK03 Integrated Radio Frequency Circuits 2017/Ted Johansson 1/6
Tutorial 1: Noise, solutions
Problem 1
In the amplifier schematic shown below, determine the total input-referred noise
voltage. Consider only the thermal noise sources and ignore the gate noise of the
transistors. Neglect channel-length modulation and body effect.
Vin
Vdd
RL
M1
Vout
Solution:
There are two thermal noise sources as shown below.
RL
M1
Vout
Vin
Since:
,
then because of uncorrelation we can add the noise currents to get:
To get the input-referred noise voltage we should divide the total output noise by the
gain square as:
TSEK03 Integrated Radio Frequency Circuits 2017/Ted Johansson 2/6
Problem 2 (5.4 Course book)
Determine the noise figure of the stages below, with respect to a source impedance of
Rs. Neglect channel-length modulation and body effect.
Vin
Vout
VDD
M1
M2
I1
+
–
RS
Solution:
Gain:
gm2Vout
Vin
+
–
RS
gm1Vx
Vout
Vx
There is no “ac” path from M1 => Vx = 0
m
S
in
out
v
g
R
V
V
A
1
=
=
Noise of M2:
gm2Vout
RS
gm1Vx
Vout
Vx
In,M2
TSEK03 Integrated Radio Frequency Circuits 2017/Ted Johansson 3/6
In,M 2 = −gm2Vn,M 2 out
Vn,M 2 out
2
=
In,M 2
2
gm2
2
=
4kTgm2γ
gm2
2
=
4kTγ
gm2
Noise of M1:
gm2Vout
RS
gm1Vx
Vout
Vx
In,M1
There is no “ac” path from M1 => 0
1
,
1 =
+ M
n
x
m I
V
g (1)
Also,
x
S
out
m V
R
V
g =
2 (2)
From (1) and (2):
Vout =
In,M1
gm1
⋅
1
gm2RS
⇒
V
n,M1 out
2
=
In,M1
2
gm1
2
⋅
1
gm2
2
RS
2
=
4kTγgm1
gm1
2
gm2
2
RS
2
=
4kTγ
gm1gm2
2
RS
2
Vn,M 2 out
2
=
In,M 2
2
gm2
2
=
4kTgm2γ
gm2
2
=
4kTγ
gm2
F =1+
γ
gm2
+
γ
gm1gm2
2
RS
2
1
gm2
2
RS
2
⋅
1
RS
= 1+γRSgm2 +
γ
gm1RS
TSEK03 Integrated Radio Frequency Circuits 2017/Ted Johansson 4/6
Problem 3
A two-stage amplifier is shown below. Determine the noise factor of this amplifier.
Consider only the thermal noise sources and ignore the gate noise of the transistors.
Assume that R1 and R2 are noiseless and ignore all the parasitics. Furthermore,
assume that .
VS
Vdd
R1
M1
Vdd
R2
M2
Vout
RS
Solution:
The gain of the first stage is A1 = gm1R1 and the gain of the second stage is A2 =
gm2R2. There are two noise sources contributed by the transistors. Thus the total noise
at the output is:
The total noise at the output due to the source is:
Based on these expressions the noise figure is:
TSEK03 Integrated Radio Frequency Circuits 2017/Ted Johansson 5/6
Problem 4
A circuit exhibits a noise figure of 3 dB.
a) What percentage of the output noise power is due to the source resistance, Rs?
b) Repeat the problem for NF = 1 dB.
Solution:
a)
F =1+
Vn
2
Av
2
⋅
1
4kTRs
= 1+
Vn,in
2
4kTRs
(1)
NF =10logF = 3dB ⇒ F =100.3
≈ 2 (2)
From (1) and (2), we have:
F = 1+
Vn,in
2
4kTRs
= 2 ⇒ Vn,in
2
= 4kTRs ⋅
Since 2
,in
n
V = 2
, s
R
n
V , 50% of the total output noise is due to the input source, Rs.
b)
)
3
(
26
.
1
10
1
log
10 1
.
0
≈
=
⇒
=
= F
dB
F
NF
From (1) and (3), we have:
F = 1+
Vn,in
2
4kTRs
=1.26 ⇒ Vn,in
2
= 0.26× 4kTRs ⋅
Since 2
,in
n
V =0.26 2
, s
R
n
V , the total output noise due to the input source, Rs is
%
4
.
79
100
26
.
1
1
=
×
TSEK03 Integrated Radio Frequency Circuits 2017/Ted Johansson 6/6
Homework
Determine the noise figure of the stages below, with respect to a source impedance of
Rs. Neglect channel-length modulation and body effect.
Vin
Vout
VDD
M1
M2
I1
Vb
Answer:
)
)(
1
(
1
1
2
2
2
2
m
m
m
s
m
s
m
in
out
v
g
g
g
R
g
R
g
V
V
A
−
+
−
=
=
Vn,M 2
2
=
4kTγ
gm2
Vn,M1
2
= 4kTγgm1
gm2 +
1
Rs
gm2 (gm1 + 2gm2 +
1
Rs
)
⎛
⎝
⎜
⎜
⎜
⎜
⎞
⎠
⎟
⎟
⎟
⎟
⎡
⎣
⎢
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
⎥
2
F =1+
Vn,M1
2
+Vn,M 2
2
Av
2
⋅
1
4kTRs

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10.pdf

  • 1. TSEK03 Integrated Radio Frequency Circuits 2017/Ted Johansson 1/6 Tutorial 1: Noise, solutions Problem 1 In the amplifier schematic shown below, determine the total input-referred noise voltage. Consider only the thermal noise sources and ignore the gate noise of the transistors. Neglect channel-length modulation and body effect. Vin Vdd RL M1 Vout Solution: There are two thermal noise sources as shown below. RL M1 Vout Vin Since: , then because of uncorrelation we can add the noise currents to get: To get the input-referred noise voltage we should divide the total output noise by the gain square as:
  • 2. TSEK03 Integrated Radio Frequency Circuits 2017/Ted Johansson 2/6 Problem 2 (5.4 Course book) Determine the noise figure of the stages below, with respect to a source impedance of Rs. Neglect channel-length modulation and body effect. Vin Vout VDD M1 M2 I1 + – RS Solution: Gain: gm2Vout Vin + – RS gm1Vx Vout Vx There is no “ac” path from M1 => Vx = 0 m S in out v g R V V A 1 = = Noise of M2: gm2Vout RS gm1Vx Vout Vx In,M2
  • 3. TSEK03 Integrated Radio Frequency Circuits 2017/Ted Johansson 3/6 In,M 2 = −gm2Vn,M 2 out Vn,M 2 out 2 = In,M 2 2 gm2 2 = 4kTgm2γ gm2 2 = 4kTγ gm2 Noise of M1: gm2Vout RS gm1Vx Vout Vx In,M1 There is no “ac” path from M1 => 0 1 , 1 = + M n x m I V g (1) Also, x S out m V R V g = 2 (2) From (1) and (2): Vout = In,M1 gm1 ⋅ 1 gm2RS ⇒ V n,M1 out 2 = In,M1 2 gm1 2 ⋅ 1 gm2 2 RS 2 = 4kTγgm1 gm1 2 gm2 2 RS 2 = 4kTγ gm1gm2 2 RS 2 Vn,M 2 out 2 = In,M 2 2 gm2 2 = 4kTgm2γ gm2 2 = 4kTγ gm2 F =1+ γ gm2 + γ gm1gm2 2 RS 2 1 gm2 2 RS 2 ⋅ 1 RS = 1+γRSgm2 + γ gm1RS
  • 4. TSEK03 Integrated Radio Frequency Circuits 2017/Ted Johansson 4/6 Problem 3 A two-stage amplifier is shown below. Determine the noise factor of this amplifier. Consider only the thermal noise sources and ignore the gate noise of the transistors. Assume that R1 and R2 are noiseless and ignore all the parasitics. Furthermore, assume that . VS Vdd R1 M1 Vdd R2 M2 Vout RS Solution: The gain of the first stage is A1 = gm1R1 and the gain of the second stage is A2 = gm2R2. There are two noise sources contributed by the transistors. Thus the total noise at the output is: The total noise at the output due to the source is: Based on these expressions the noise figure is:
  • 5. TSEK03 Integrated Radio Frequency Circuits 2017/Ted Johansson 5/6 Problem 4 A circuit exhibits a noise figure of 3 dB. a) What percentage of the output noise power is due to the source resistance, Rs? b) Repeat the problem for NF = 1 dB. Solution: a) F =1+ Vn 2 Av 2 ⋅ 1 4kTRs = 1+ Vn,in 2 4kTRs (1) NF =10logF = 3dB ⇒ F =100.3 ≈ 2 (2) From (1) and (2), we have: F = 1+ Vn,in 2 4kTRs = 2 ⇒ Vn,in 2 = 4kTRs ⋅ Since 2 ,in n V = 2 , s R n V , 50% of the total output noise is due to the input source, Rs. b) ) 3 ( 26 . 1 10 1 log 10 1 . 0 ≈ = ⇒ = = F dB F NF From (1) and (3), we have: F = 1+ Vn,in 2 4kTRs =1.26 ⇒ Vn,in 2 = 0.26× 4kTRs ⋅ Since 2 ,in n V =0.26 2 , s R n V , the total output noise due to the input source, Rs is % 4 . 79 100 26 . 1 1 = ×
  • 6. TSEK03 Integrated Radio Frequency Circuits 2017/Ted Johansson 6/6 Homework Determine the noise figure of the stages below, with respect to a source impedance of Rs. Neglect channel-length modulation and body effect. Vin Vout VDD M1 M2 I1 Vb Answer: ) )( 1 ( 1 1 2 2 2 2 m m m s m s m in out v g g g R g R g V V A − + − = = Vn,M 2 2 = 4kTγ gm2 Vn,M1 2 = 4kTγgm1 gm2 + 1 Rs gm2 (gm1 + 2gm2 + 1 Rs ) ⎛ ⎝ ⎜ ⎜ ⎜ ⎜ ⎞ ⎠ ⎟ ⎟ ⎟ ⎟ ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ ⎥ 2 F =1+ Vn,M1 2 +Vn,M 2 2 Av 2 ⋅ 1 4kTRs