1/13
TSEK03 Integrated Radio Frequency Circuits 2016/Ted Johansson
Tutorial 2: LNA, solutions
Problem 1
It is preferred in current RF designs that the input of LNA be matched to 50 Ω. The
easiest way is to shunt the gate with a resistor of 50 Ω.
a) Calculate the gain, input impedance, and noise figure (NF) in absence of gate
noise. Assume that Rsh=Rs and a noisefree RL for NF derivation.
b) What are the disadvantages of shunt resistor with reference to gain and NF?
Solution:
a)
V 2 n, Rs
Rs
G
Vout
D
Rsh
V 2 n, Rsh
gm Vgs
i2 nd
RL
S
RL is noiseless
Gain:
Gain Gate = −gm RL
⎛ Rsh ⎞
A = −gm RL ⎜
⎝ Rs + Rsh ⎟⎠
Zin: When the gate capacitance is not included in the small-signal model
above, Zin is purely resistive = Rsh.
TSEK03 Integrated Radio Frequency Circuits 2016/Ted Johansson
2/13
Noise figure:
F=
Total output noise power
Output noise due to input source
V 2n,Rs = 4kTRs
V 2n,Rsh = 4kTRsh
i 2nd = 4kTγ gm
Using superposition, only one noise source is considered at a time and other sources
should be shorted (voltage noise source) / open (current noise source).
=V
V
2
no,Rs
V
2
no,Rsh
2
n,Rs
=V
⎛ Rsh ⎞
×g R ×⎜
⎝ Rs + Rsh ⎟⎠
2
n,Rsh
2
m
2
2
L
⎛ Rs ⎞
×g R ×⎜
⎝ Rs + Rsh ⎟⎠
2
m
2
2
L
V 2no,d = i 2nd × R 2L
F=
V 2no,Rs + V 2no,Rsh + V 2no,d
V 2no,Rs
= 1+
V 2no,Rsh + V 2no,d
V 2no,Rs
g 2m R 2L R 2s
4kTRsh
2
Rs + Rsh )
4kTγ gm R 2L
(
F = 1+
+
=
g 2m R 2L R 2sh
g 2m R 2L R 2sh
4kTRs
4kTRs
( Rs + Rsh )2
( Rs + Rsh )2
g 2m R 2L R 2s
Rsh
2
2
2
Rs + Rsh )
γ gm R 2L
Rsh g 2m R 2L R 2s γ gm R L ( Rs + Rsh )
(
1+
+
= 1+
+
=
g 2m R 2L R 2sh
g 2m R 2L R 2sh
Rs g 2m R 2L R 2sh
Rs g 2m R 2L R 2sh
Rs
R
( Rs + Rsh )2 s ( Rs + Rsh )2
γ ( Rs + Rsh )
R
1+ s +
R sh
g m Rs R 2sh
2
TSEK03 Integrated Radio Frequency Circuits 2016/Ted Johansson
In case of impedance match which is provided by Rs = Rsh,
γ ( Rs + Rsh )
4γ
F = 1 +1 +
=2+
2
g m Rs Rsh
gm Rs
2
b)
- Poor Noise Figure since Rsh adds extra noise.
- Input signal attenuated by the voltage divider (Rs and Rsh) => reduced gain.
- At high frequency, shunt L is needed to tune out Cgs.
3/13
4/13
TSEK03 Integrated Radio Frequency Circuits 2016/Ted Johansson
Problem 2
The inductor source degenerated amplifier shown below, presents a noiseless resistance
of 50 Ω for input power match.
a) Calculate the input impedance. How can we cancel the imaginary part of complex
input impedance so that the LNA presents 50 Ω real input resistance at input port?
b) Calculate the NF in absence of gate noise. Neglect gate-drain, gate-bulk, and gatesource capacitance. Tip: use the Q-value of the input network to determine the
gain of the circuit.
c) Cgd bridges the input and output ports. The reverse isolation of this LNA is very
poor. Why is reverse isolation important? Suggest a modification to improve the
reverse isolation.
Solution:
a)
VS
Rs
Lg
Zin
iin
+ +
Vgs
-
io = gmVgs = gmiin ×
1
jωCgs
Substituting (2) in (1)
⎡
g L ⎤
1
Vin = iin ⎢ jω (Lg + Ls )+
+ m s⎥
jωCgs Cgs ⎥⎦
⎢⎣
Vout
gm Vgs
Vin
From model above we can write
⎛ 1 ⎞
⎟⎟ + io jω Ls
Vin = iin ( jω Lg + jω Ls ) + iin ⎜⎜
j
ω
C
⎝
gs ⎠
io
Ls
(1)
(2)
TSEK03 Integrated Radio Frequency Circuits 2016/Ted Johansson
Zin =
Vin
1
g L
= jω (Lg + Ls ) +
+ m s
iin
jωCgs Cgs
5/13
(3)
For input matching purpose, the imaginary part of (3) should be zero, which means that
Lg + Ls, should be canceled out by Cgs. Therefore, at frequency of interest, we have:
ωo (Lg + Ls ) =
And
1
1
2
⇒ ωo =
(Lg + Ls )Cgs
ωoCgs
g m Ls
= RS = 50Ω
Cgs
Notes:
- Ls is typically small and may be realized by a bond wire (higher Q than an integrated L).
- Lg can be implemented by spiral/external inductor.
b)
From part a)
Zin = jω (Lg + Ls ) +
1
g L
+ m s
jωCgs Cgs
For series RLC networks:
6/13
TSEK03 Integrated Radio Frequency Circuits 2016/Ted Johansson
We can redraw the LNA circuit in form of a series RLC network:
Rs
Lg + Ls
+
Vin
Zin
-
ωo (Lg + Ls )
Qin =
RS +
g m LS
C gs
=
⎛
⎝
g L
for match load RS = m S =>
Cgs
Qin =
g m LS
C gs
⎞
⎟C gs
⎟
⎠
1
2ωo Rs Cgs
Gain
Vgs = QinVin
gm =
I out
V gs
Gm =
I out V gs g m
=
= Qin g m
Vin
Vin
Gm = Qin g m
so,
Vout
= −Gm RL = −Qin g m RL
Vin
Noise Figure:
F=
Total noise power at output
noise power at output due to input source
For this calculation we ignore channel noise.
F=
V 2 no, Rs + V 2 no,d
V 2 no, Rs
= 1+
+
Cgs Vgs
1
ωo ⎜⎜ RS +
gm Ls
Cgs
V 2 no,d
V 2 no, Rs
7/13
TSEK03 Integrated Radio Frequency Circuits 2016/Ted Johansson
V 2 no,d = i 2 n,d R 2 L
i 2n,d = 4kTγ gm
V 2 no, RS = V 2 n, RS G 2 m R 2 L
V 2n,RS = 4kTRs Ÿ Gm = Qin g m
i 2n,d = 4kTγ gm , V 2n,Rs = 4kTRs
F =1+
in2, d RL2
Vn2,RS Qin2 g m2 RL2
=1+
γ
g m RS Qin2
Notes:
- Very good NF value.
- Narrow band matching.
- NF ↓ with Q2.
- The Q value depends on Lg + Ls. Since Ls is usually small, Q mainly depends upon Lg.
Drawbacks:
Noise of RL
VDD
VDD
VS
Rs
Vout
Lg
LD
RL generates noise so
replace RL with LD so that:
RL
1
ωo =
LD C L
Ls
Rs
VS
Lg
Ls
The CL can be considered as the input capacitance of the following mixer or filter.
c) Reverse isolation
VDD
LD
Reverse Isolation
Vout
Cgd
CL
Vout
Vb
Rs
Lg
Ls
LO
CL
(Final Design)
Reverse isolation depends upon capacitance between output and input. To make have
better isolation, the cascode device can be employed according to the figure above.
8/13
TSEK03 Integrated Radio Frequency Circuits 2016/Ted Johansson
Problem 3
In the common-source stage shown below, determine
a) Input impedance, Rin.
b) Gain.
c) Noise figure.
Assume that the channel-length modulation is NOT neglected. Also assume matching at
the input.
VDD
Vb
M2
RF
RS
M1
+
Vin -
Vout
Rin
Solution:
Small signal model:
Ix
RF
+
+
gm1Vx
Vx
-
ro1||ro2 Vout
-
a)
V x − I x RF
+ g m1Vx = I x
ro1 || ro 2
=>
Rin =
b)
Vout
1 + g m1 (ro1 || ro 2 )
R
=1−
⋅ RF = 1 − F
Vx
RF + (ro1 || ro 2 )
RS
AV =
Vout 1 ⎛ RF
= ⎜1 −
Vin 2 ⎜⎝ RS
⎞
⎟⎟
⎠
Vx
R + (ro1 || ro 2 )
= F
= RS
I x 1 + g m1 (ro1 || ro 2 )
9/13
TSEK03 Integrated Radio Frequency Circuits 2016/Ted Johansson
c)
i) Noise of RS:
2
V n,out
RS
= 4kTRS ( AV )
2
⎛1
R ⎞
= 4kTRS ⎜ (1 − F )⎟
R ⎠
⎝2
2
S
ii) Noise of RF:
Vn
RF
+
RS
gm1Vx
ro1||ro2 Vout
-
V 2 n ,out
RF
⎛
⎜
1 − g m1RF
= 4kTRF ⎜
⎜
RF + RS
⎜ 1 − g m1 RS − r || r
o1
o2
⎝
⎞
⎟
⎟
⎟
⎟
⎠
2
iii) Noise of M1 and M2 :
VX
RF
RS
KCL at Vout:
I out = g m1VX +
Vout
Vout
+
ro
RS + RF
VX =
RS
Vout
RS + RF
=>
Rout =
Vout
RF + RS
=
I out 1+ g R + RF + RS
m1 S
ro1 || ro2
Iout
gm1VX
ro
Vout
Rout
10/13
TSEK03 Integrated Radio Frequency Circuits 2016/Ted Johansson
V 2 n,out
M1 & M 2
2
= 4kTγ ( g m1 + g m 2 ) Rout
=>
F=
Total noise power at output
V 2 n ,out total V 2 n ,out
= 2
=
noise power at output due to input source V n,out RS
2
F = 1+
V 2n,out
RF
+V 2n,out
2
V n,out
RS
M1 &M 2
RS
+ V 2 n ,out
V
2
RF
+ V 2 n,out
n ,out RS
⎛
⎞
⎜
⎟
1− gm1RF
2
⎟ + 4kTγ Rout
4kTRF ⎜
(gm1 + gm2 )
R
+
R
⎜ 1− g R − F
S ⎟
⎜
⎟
m1 S
ro1 || ro2 ⎠
⎝
= 1+
2
⎛1
RF ⎞
4kTRS ⎜ (1− )⎟
RS ⎠
⎝2
M1 &M 2
Problem 3, another solution, by PhD student
Advanced RF ICs Design Course
Ch.5 LNAs
Solutions to problems 5.6, 5.8, 5.10, 5.12
Dai Zhang
Problem 5.6
For the CS stage of Fig 5.13(a), determine the closed-loop gain and noise figure if channellength modulation is not neglected. Assume matching at the input.
VDD
Vb
M2
RF
Vout
RS
M1
VX
Vin
Solutions:
1.To calculate the closed-loop gain
First, calculate Rin based on the small-signal model:
I X = g m1VX +
VX − I X RF
ro
1+
!
Rin =
RF
ro
VX
=
≈
1
IX
+ g m1
ro
1+
RF
ro
g m1
Assume input matching, we thus have
1+
RS =
RF
ro
g m1
!
1+
RF
= RS g m1
ro
1
Secondly, calculate the voltage gain from Vx to Vout based on the above small-signal model.
KCL at Vout:
VX − Vout
V
= g m1VX + out
RF
ro
!
AX =
Vout
VX
1
− g m1
1 − g m1RF
1
R
RF
=
≈
=
− F
1
1
RS g m1
RS g m1 RS
+
ro RF
Hence, the voltage gain from Vin to Vout is:
AV =
1
1
1
R
= (
− F)
2 AX 2 RS g m1 RS
2. To calculate the noise figure
The noise contributions are from M1 and M2, Rs, and RF.
a. calculate the output noise power of M1 and M2
First, calculate the equivalent output resistance Rout based on the small-signal model, where the
input source is shorted.
KCL at Vout:
I out = g m1VX +
Vout
Vout
+
ro
RS + RF
RS
VX =
Vout
RS + RF
!
Since in practice, RF>>RS, and RS g m1 = 1 +
Rout =
Rout =
Vout
1
=
I out 1 + RS g m1 + 1
RS + RF ro
(1)
RF
, hence (1) can be further approximated to:
ro
1
RF ro
=
R
2( RF + ro )
2+ F
1
ro
+
RF
ro
Hence, the output noise power of M1 and M2 is:
Vn2,out
M 1, M 2
2
= 4kTγ ( g m1 + g m 2 ) Rout
2
b. calculate the output noise power of RF
The voltage gain from the thermal noise voltage of RF to output is AX. So the output noise power
of RF is:
Vn2,out
RF
= 4kTRF AX2
The noise of Rs is multiplied by the gain when referred to the output, and the result is divided by
the gain when referred to the input. We thus have:
2
2
2
1 Vn, out
4kTRF AX2 4kTγ ( g m1 + g m 2 ) Rout
RF γ ( g m1 + g m 2 ) Rout
NF =
= 1+
+
= 1+
+
4kTRS AV2
4kTRS AV2
4kTRS AV2
4 RS
RS AV2
Since we have calculated Rout =
RF ro
1
1
1
R
= (
− F ) , so NF can be
and AV =
2( RF + ro )
2 AX 2 RS g m1 RS
further expressed as:
2
2
1 Vn , out
4kTRF AX2 4kTγ ( g m1 + g m 2 ) Rout
RF
γ ( g m1 + g m 2 ) RF2 ro2 RS
NF =
= 1+
+
= 1+
+
4kTRS AV2
4kTRS AV2
4kTRS AV2
4 RS ( R + r ) 2 ( 1 − R ) 2
F
o
F
g m1
3
11/13
TSEK03 Integrated Radio Frequency Circuits 2016/Ted Johansson
Homework
Determine the noise figure of the common-gate (CG) circuits shown below. Neglect
channel-length modulation and the body effect. Assume that R1 is the loss of inductance
L1 .
VDD
L1
VDD
L1
Vout
Vout
Vb1
RS
C1
Vb1
RS
M1
Vb2
Vin
C1
M1
Vin
RB
M2
a)
b)
Answer:
a) CG biased with current source M2:
Calculate Rin and Av.
VDD
L1
R1
Vout
Vb1
RS
Vin
M1
VX
Rin
Rin =
1
and AX = gm1R1
g m1
Assume input matching, RS =
R1
1
and AV =
g m1
2RS
12/13
Now we calculate the noise figure. The noise contributions are from M1, M2, R1, and RS.
TSEK03 Integrated Radio Frequency Circuits 2016/Ted Johansson
a. The output noise power of M1:
(similar to Problem 5.8 in the book or refer to the book on page 277)
2
n , out M 1
V
R1
R12
4kTγ
2
=
(
) = kTγ
g m1 R + 1
RS
S
g m1
b. Calculate the output noise power from R1
Vn2,out
R1
= 4kTR1
c. Calculate the output noise power from M2
AV =
2
Vn,out
gm2 R1
2
M2
= kT γ gm2 R12
TSEK03 Integrated Radio Frequency Circuits 2016/Ted Johansson
13/13
b) CG biased with resistor RB:
Conclusion: If RB is large, then there is less noise from a circuit using a resistor for
biasing compared to a current source realized using a transistor.