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.
Vdd
RL
Vout
Vin M1
Solution:
There are two thermal noise sources as shown below.
RL Vin
Vout
M1
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:
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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.
VDD
I1
M2 Vout
RS
+ M1
Vin
–
Solution:
Gain:
Vout
gm2Vout gm1Vx
Vx
RS
+
Vin
–
There is no “ac” path from M1 => Vx = 0
Vout 1
Av = =
Vin RS g m
Noise of M2:
Vout
In,M2 gm2Vout gm1Vx
Vx
RS
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I n,M 2 = −gm2Vn,M 2 out
2
2 I n,M 2 4kTgm2γ 4kTγ
Vn,M 2 out
= 2
= 2
=
gm2 gm2 gm2
Noise of M1:
Vout
gm2Vout gm1Vx In,M1
Vx
RS
There is no “ac” path from M1 => g m1Vx + I n , M 1 = 0 (1)
Also,
g m 2Vout RS = Vx (2)
From (1) and (2):
I n,M1 1
Vout = ⋅
gm1 gm2 RS
⇒
2
I n,M1 1 4kTγ g 4kTγ
V2 = 2
⋅ 2 2 = 2 2 m12 = 2
n,M1 out gm1 gm2 RS gm1gm2 RS gm1gm2 RS2
2
2 I n,M 4kTgm2γ 4kTγ
Vn,M 2 out
= 22= 2
=
gm2 gm2 gm2
γ γ
+ 2
g gm1gm2 RS2 1
F = 1+ m2 ⋅
1 RS
2
gm2 RS2
γ
= 1+ γ RS gm2 +
gm1RS
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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 .
Vdd Vdd
R1 R2
RS Vout
VS M1 M2
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:
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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)
Vn2 1
F = 1+ 2
⋅
Av 4kTRs
2
Vn,in
= 1+ (1)
4kTRs
NF = 10 log F = 3dB ⇒ F = 10 0.3 ≈ 2 (2)
From (1) and (2), we have:
2
Vn,in 2
F = 1+ =2 ⇒ Vn,in = 4kTRs ⋅
4kTRs
Since Vn2,in = Vn2,R , 50% of the total output noise is due to the input source, Rs.
s
b)
NF = 10 log F = 1dB ⇒ F = 10 0.1 ≈ 1.26 (3)
From (1) and (3), we have:
2
Vn,in 2
F = 1+ = 1.26 ⇒ Vn,in = 0.26 × 4kTRs ⋅
4kTRs
Since Vn2,in =0.26 Vn2,R , the total output noise due to the input source, Rs is
s
1
×100 = 79.4%
1.26
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Homework
Determine the noise figure of the stages below, with respect to a source impedance of
Rs. Neglect channel-length modulation and body effect.
VDD
I1
Vout
M2 Vb
Vin M1
Answer:
Vout 1
Av = =
Vin g R − ( g R + 1)( gm2
m2 s m2 s )
g m 2 − g m1
2 4kTγ
Vn,M 2 =
gm2
2
⎡ ⎛ 1 ⎞⎤
⎢ ⎜ gm2 + ⎟⎥
2 Rs
Vn,M1 = ⎢4kTγ gm1 ⎜ ⎟⎥
⎢ ⎜ g (g + 2g + 1 ) ⎟⎥
⎢⎣ ⎜ m2 m1 m2 ⎟
⎝ Rs ⎠⎥⎦
2 2
Vn,M1 +Vn,M 2 1
F = 1+ 2
⋅
Av 4kTRs
