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Tutorial-1 Low Noise Amplifier (LNA) Design Complied by Rashad M. Ramzan Objective: Low noise amplifiers are one of the basic building blocks of any communication system. The purpose of the LNA is to amplify the received signal to acceptable levels with minimum self-generated additional noise. Gain, NF, non-linearity and impedance matching are four most important parameters in LNA design. The objective of this tutorial is to outline the basic tradeoffs between different amplifying topologies w.r.t gain, NF and impedance matching. After this comparison it is concluded that inductor degenerated common source topology gives the best performance to meet the gain, NF, and impedance matching goals with minimum power consumption in case of narrow band designs.

Goals: After this tutorial, students should be able to •

Calculate the gain, input impedance and NF of common gate, common source, and shunt feedback amplifiers.

•

Understand the basic equations and tradeoff between different LNA topologies.

• Perform the calculation for inductor degenerated common source topology and understand the tradeoff between the gain, NF, and impedance matching. A supplement tutorial LNA lab is also part of this course which guides through different analyses to design a practical LNA.

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Spring 2010: Radio Frequency Integrated Circuits (TSEK03)

Problem-1.1(Tutorial) NMOS transistor is racing horse in LNA design arena due to its higher mobility compared to PMOS transistors. Calculate the IP3 of NMOS CS amplifier shown below. Assume that NMOS transistor is in saturation.

a) Consider simplified square law model. (HW)

ID =

Kn (VGS − VT ) 2 2

b) Consider the short channel effects as:

K n  (VGS − VT ) 2    2 1 + θ (VGS − VT )  θ = Velocity Saturation, Mobility Degradation ID =

VGS − VT = 0.2V

and θ = 0.1V −1

Observe that this transistor is not a very “short channel” device as θ >1

Av ,tot =

− g m RL ≅ − g m RL RS 1 + g m RS + 1 + RL + RF RF

iin Vin

Av ,tot ≅ − g m RL Also

Z in =

RF + RL 1 + g m RL

By ignoring Cgs, we have considered real part only.

Electrical Engineering Department (ISY), Linköping University

Rs Vgs

RF

Vout gmVgs

RL

Spring 2010: Radio Frequency Integrated Circuits (TSEK03)

For source resistance

V 2 nRS ,out = A 2 v ,tot V 2 nRS

---------------(1)

For feedback resistance

RF V

Rs

i

2

V 2 RF ,out

RF

gmVgs

Vgs

RL

Vgs = −iRS = iRF − VRF + VRF ,out VRF ,out = RL (i − g mVgs ) VRF ,out = VRF

V

2

n , RF ,out

1 1+

RS + R F RL (1 + g m Rs )

= VRF

R  = V n , RF  L (1 + g m RS )  RF 

RL (1 + g m RS ) RF

2

2

---------------------(2)

Similarly

Rs

i

VnD ,out RL

Vgs

gmVgs

+ I nD + g mVgs +

VnD ,out

Vgs = RS

VnD ,out =

V 2 nD ,out

RF

RS + R F

I 2 n,D

RL

=0

VnD ,out RS + R F

I n,D g R 1 1 + + m S R L RS + R F RS + R F

≈ I nD RL

So,

V 2 nD ,out = I 2 nD RL

2

------------------------------------------(3)

Combaining (1) (2) & (3) Electrical Engineering Department (ISY), Linköping University

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Spring 2010: Radio Frequency Integrated Circuits (TSEK03)

F = 1+

R  V 2 n , RF  L (1 + g m R S )  RF 

2

A 2 v ,tot V 2 n , RS

+

I 2 nD R 2 L A 2 v ,tot V 2 n , RS

Av ,tot = − g m RL , V 2 n , RS = 4kTRS ∆f , V 2 M ,RF = 4kTRF

 1 1 + g m RS 

R F = 1+ S RF

&

I 2 nD = 4kTγg m

2

 γ  + g m RS 

b). NF

↓

gmRS ↑ & RF ↑ usually

RS = 50Ω

- Better performance than CS amplifier - RF induces noise - At higher f

↑

a shunt inductor needed to tune out Cgs

- Broadband Amp @ Lower frequency - To make NF

↓

RF > RS and gmRS >> 1

Problem-1.4 (HW) Common gate amplifier also offers 50 Ω input impedance match and solves the input matching problem.

c) Calculate the gain, input impedance and NF in absence of gate noise. Neglect gate drain and gate to bulk and gate to source capacitance. a) What are the disadvantage of common gate amplifier with reference to gain and NF?

Problem-1.5 (Tutorial) The disadvantages of the amplifiers discussed in Problem-2, 3 & 4 can be circumvented by using the source degenerated LNA shown below.

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Spring 2010: Radio Frequency Integrated Circuits (TSEK03)

a) Calculate the input impedance. This inductor source degenerated amplifier presents a noiseless resistance for 50Ω for input power match. How we can cancel the imaginary part of complex input impedance so that the LNA presents 50Ω real input resistance at input port. b) Calculate the NF in absence on gate noise. Neglect gate drain and gate to bulk and gate to source capacitance. c) Cgd bridges the input and output ports. The reverse isolation of this LNA is very poor. Why reverse isolation is important? Suggest the modification to improve reverse isolation.

Solution: a). VDD VS

Rs

Lg

iin

io

Vout

gmVgs Zin Vgs

Rs

VS

Vin Ls

Lg

RL Vout Ls

(Biasing not shown) From model above we can write

 1  Vin = iin ( jωLg + jωLs ) + iin   + io jωLs ---------------(1)  jωc  1 io = g mVgs = g m iin × --------------------------------------(2) jωC gs Substituting (2) in (1)

 g L  1 Vin = iin  jω (Lg + Ls ) + + m s jωC gs C gs  

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Spring 2010: Radio Frequency Integrated Circuits (TSEK03)

Z in =

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Vin g L 1 = jω (Lg + Ls ) + + m s iin jωC gs C gs

Z in = jω (Lg + Ls ) +

g L 1 + m s jωC gs C gs

For matching Lg + Ls are canceled out by Cgs. So at frequency of interest

ωo (Lg + Ls ) = And

1 1 2 ⇒ ωo = ωo C gs (Lg + Ls )C gs

RS = 50Ω =

gm Ls C gs

Notes: a). Ls is typically small and may be realized by the bond wire for source. b). Lg can be implemented by spiral/external inductor.

b). Reference: For series RLC Circuit R L

From part a)

Z in = jω (Lg + Ls ) +

g L 1 + m s jωC gs C gs

We can draw this circuit as

Rs Vin Here

Q in =

gm

ω o (L g + L s ) RS +

Qin =

g m LS C gs

gm C gs

=

Ls C gs

ωo  RS + 

1 L ωo L 1 = = R C R ωo RC and VC = QSVin Qs =

Cgs Vgs

ω o (L g + L s ) R S + ω T LS

frequency of current gain equal 1

1 

C VC

Lg + Ls Zin

Q ωT ≅

Vin

g m LS C gs

 C gs  

for match load

RS =

g m LS C gs

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Spring 2010: Radio Frequency Integrated Circuits (TSEK03)

1 2ω o Rs C gs

Qin =

Gain

Vgs = QinVin

Vout Rs

I g m = out V gs

Vin

Lg Vgs Zin

Ls

RL

I out V gs g m = = Qin g m V in V in

Gm =

G m = Qin g m

so,

V out = −G m R L where Gm = Qin g m 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 nRS ,OUT + V 2 nD ,OUT V 2 nRS ,OUT

=1+

V 2 nD ,OUT V 2 nRS ,OUT

V 2 nD ,OUT = i 2 n ,D R 2 L

i 2 n , D = 4kTγg m ∆f

V 2 nRS ,OUT = V 2 n ,RS G 2 m R 2 L

V 2 n , RS = 4kTRS ∆f

i n2, D R L2

F = 1+ V

F = 1+

2 n , RS

2 in

2 m

Q g R

2 L

&

Gm = Qin g m

i 2 n , D = 4kTγg m , V 2 n , RS = 4kTRS

γ g m R S Qin2

Notes: - Very good NF value - Narrow band matching - NF

↓ with Q 2

- The Q is dependent upon Lg + Ls, Ls usually small so Q depends mainly upon Lg Electrical Engineering Department (ISY), Linköping University

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Spring 2010: Radio Frequency Integrated Circuits (TSEK03)

C). Drawbacks i). VDD

VS

Rs

VDD RL generates noise so replace RL with LD so that’s

RL Vout

Lg

ωo =

1 LD C L

LD VS

Rs

CL

Lg

Ls

Ls

The CL can be considered the input capacitance of the following mixer or filter.

VDD

ii).

LD Reverse Isolation

Cgd

CL

Vb Rs

Vout

Lg Ls

Lo

(Final Design)

Reverse isolation depends upon capacitance between output and input. To make it less the cascode architecture can be used.

Problem-1.6 (HW) Fill-in the Table below, use the data from Problem-1.2, 1.4, 1.3 and 1.5

Type of LNA Shunt Resistor

Zin Rsh

Noise Factor

2+

4γ g m RS

Common Gate Shunt Feedback Source Degenerated

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Gain

− g m RL 2

NF (dB)

Spring 2010: Radio Frequency Integrated Circuits (TSEK03)

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a) Calculate the NF for all above amplifiers. Assume γ=2, gm = 20mS, Rs = 50Ω, RF = 500Ω, and Qin = 2. b) Which is the best topology for Narrow Band LNA design at high frequency?

Problem-1.7 (Tutorial) Real Design: We will design the inductor-source-degenerated LNA shown in Fig below to meet the specification outlined for IEEE802.11b standard. The first cut approximate values are calculated as a starting point for simulation. LNA Specification: NF < 2.5 dB, Gain > 15dB, IIP3 > -5dBm, Centre Frequency = 2.4 GHz Load Capacitance = 1pF Technology Parameters for 0.35um CMOS:

Leff = 0.35µ m, µ n Cox = 170 µ A V 2 , Cox = 4.6 mF m2 , µ p Cox = 58 µ A V 2 , γ = 2

δ = 4, C = 0.395, α = 0.85

Solution:

Technology 0.35µm CMOS:

µ o Cox = 170 µA V 2 , µ p Cox = 58 µA V 2 ,    2 Cox = 4.6 mF m , γ = 2, Leff = 0.35µm  δ = 4, C = 0.395, α = 0.85

Design Parameters NF < 2.5 dB, Gain > 15dB, IIP3 > -5dBm, f0 = 2.4 GHz

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Spring 2010: Radio Frequency Integrated Circuits (TSEK03)

VDD RREF

LD

M3 RBIAS

M2 Vout

Lg M1

Vin

RS

CB

CL = 10pF

Ls

Component Description Ls – Matches input impedance Lg – Sets the Resonant Frequency fO = 2.4 GHz M3 – Biasing transistor which forms current mirror with M1 Ld – Tuned output increases the gain and also work as band pass filter with CL M2 – Isolates tuned input from output to increase reverse isolation, also reduces the effect of Miller capacitance Cgd CB – BC blocking capacitor chosen to have negligible reactance at fO = 2.4 GHz RBIAS – Large enough so that its equivalent current noise is small enough to be ignored. (Don’t consider it as voltage noise source. Why??)

Design Procedure Size of M1: From the noisy two-port theory (see the course book or lecture notes) the optimal input matching and minimum noise figure is given by:

Gopt = αω C gs Fmin = 1 +

(

)

δ 1 2 1− C = 5γ 50Ω

2 ω 5 ωT

(

γC 1 − C

2

----------------------(A)

) = 1 + 2.3 ωω

-------(B)

T

From (A)

C gs ≅ 2.7 pF ⇒ WM 1 ≈ 3C gs / 2COX Leff ≈ 2.5mm ( not feasible – huge size, huge power ! ) Conclusion: We will not go for the global minimum noise figure. Instead, we will look into the constraint power design approach. Solution: LNA NF will be optimized for given power which is higher than the global minimum NF. Electrical Engineering Department (ISY), Linköping University

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Spring 2010: Radio Frequency Integrated Circuits (TSEK03)

In this case the optimum transistor width is given by:

Wopt =

1 3ωo Leff Cox RS

while the minimum power-constraint NF :

Fmin, p = 1 + 2.4

γ ω ω ⇒ Fmin, p = 1 + 5.6 α ωT ωT

---------------(C)

(B) is the global minimum noise figure. (C) is the minimum NF for a given power consumption. In practice the difference is usually 0.5dB to 1dB (no big deal for Lower Power)

Step - 1:

I1 = I 2 = 5mA (Limited Power consumption) Step - 2:

WM 1 =

1 3ω 0 L eff C ox R S

1 3 × 0.35µ × 4.6m × 50 × ωo

WM 1 =

 RS = 50Ω, Cox = 4.6 mF m 2 ,    µ n Cox = 170 µA V , Leff = 0.35µm,  ω = 2πf , f = 2.4GHz  o o  o 

WM 1 = 3.9 × 10 −4 WM 1 = 3.9 × 10 −4 = 390µm Step - 3:

2 C gs1 = WM 1 Leff Cox 3 C gs1 =

2 × 390µ × 0.35µ × 4.6m = 0.41 pF 3

W  g m1 = 2 µ n C ox   I DM 1  L M1

or

g m1 =

2 I DM 1 V GS − VT

 390  g m1 = 2 × 170µ ×   × 5m = 43 mA V  0.35 

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(for short channel model)

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Spring 2010: Radio Frequency Integrated Circuits (TSEK03)

ωT ≈

g m1 43 mA V = = 104G rad Sec C gs1 0.41 pF

Assuming Now

γ =2

Fmin = 1 + 5.6

Fmin = 1 + 5.6

ωo ωT

2π 2.4G ≈ 2.55dB 104G

NF ≈ 2.55dB This NF is very close to the specified value. If we increase ID then ωT should increase slightly as well and hence, a lower NF value can be achieved at expense of more power. Step - 4: Source and gate inductance such that they cancel Cgs and set

50Ω input impedance

ωo = 2πf o = 2π 2.4 = 15G rad Sec From previous problem R S = RTransformed =

LS =

RS

ωT

=

g m LS ≅ ω T LS C gs

50 ≅ 0.5nH 100G

LS = 0.5nH can be implemented using the bond wire. 1 L + L = g s Now ω 02 C gs1

Lg + Ls =

1 = 10.81nH (15G ) × 0.41 pF 2

Lg ≈ 10nH Step - 5:

Ld =

1 ωo 2C L

Ld =

1 ≅ 4.4nH (15G ) × 1 pF

Q C L = 1 pF

2

Ld = 4.4nH Electrical Engineering Department (ISY), Linköping University

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Spring 2010: Radio Frequency Integrated Circuits (TSEK03)

Step - 6: Size of M3 is chosen to minimize power consumption

W M 3 = 70µm, R REF = 2kΩ ⇒ I 3 = 0.6mA R BIAS = 2kΩ

(Large enough so that it’s equivalent current noise can be neglected)

C B = 10 pF ( X C ≈ 6.6Ω

so good value @ 2.4G

XB =

Step - 7: Size M2 = M3 So that they can have shared Drain Area.. (Note: You will simulate same LNA circuit in LAB # 2)

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1 = 6.6Ω ) 2πf o C B