rf circuits design 6ue.pwr.wroc.pl/.../lecture/rf_circuits_design_6.pdfamplifier block diagram rf...

RF circuits designGrzegorz BeziukGrzegorz Beziuk

RF Amplifier design

References[1] Tietze U., Schenk C., Electronic circuits : handbook for design and applications, Springer

2008

[2] Pozar D. M., Microwave engineering –3rd edition, 2005, John Wiley and Sons, Inc.

[3] Bowick C., RF circuit design, 1997, Newnes, Elsevier Science

[4] Wadell B. C., Transmission line design handbook, 1991, Artech House, Inc.

[5] Agilent Technology, A comparison of various bipolar transistor biasing circuits, Agilent

Technology Application Note 1293

[6] Avago Technologies, Microwave transistor bias consideration, Avago Technologies

Application Note 944 – 1.

Amplifier block diagram

RF

transistor

[S]

Output

matching

circuit

Input

matching

circuit

BIAS

circuit

ZG

ZL

Amplifier parameters

- Gain – maximum or specified

- Bandwidth – wide bandwidth or single

frequency

- Noise Figure

- Source impedance

- Load Impedance

- Output power for a power amplifier

Transistor selection

- Gain – maximum

- Cut off frequency (fT)

- Noise characteristic

- Maximum power for a power amplifier

Datasheet and/or *.s2p (measured for

selected operating point) file delivered by

transistor manufacturer

Transistor selection - datasheet


Transistor selection –

datasheet/*.s2p file

Transistor S matrix can be taken from a datasheet

or downloaded from transistor manufacturer

website as a *s2p file.

The *.s2p file can be created also with the

transistor S parameters taken from datasheet.

Transistor selection

– *.s2p file

! Filename: BFR92AC.S2P Version: 2.0

! Philips part #: BFR92A Date: May 1990

! Bias condition: Vce=5V, Ic=10mA

!

# MHz S MA R 50

! Freq S11 S21 S12 S22 !GUM [dB]

40 .669 -16.6 23.024 164.2 .008 79.9 .970 -8.2 ! 42.1

100 .593 -39.2 19.631 145.4 .020 72.1 .873 -17.5 ! 34.0

200 .449 -67.6 14.753 125.9 .033 68.7 .716 -24.7 ! 27.5

300 .356 -86.7 11.181 113.9 .043 66.2 .620 -26.8 ! 23.7

400 .313 -99.2 8.899 106.3 .052 68.0 .562 -27.2 ! 21.1

500 .275 -113.9 7.512 99.9 .060 67.4 .531 -26.9 ! 19.3

600 .243 -120.1 6.336 95.7 .070 69.7 .512 -26.9 ! 17.6

700 .219 -127.0 5.513 92.2 .078 71.3 .499 -26.9 ! 16.3

800 .213 -139.5 4.868 88.8 .089 72.4 .494 -27.0 ! 15.2

900 .194 -144.7 4.391 85.8 .098 72.5 .488 -27.2 ! 14.2

1000 .182 -156.2 3.953 82.4 .106 73.1 .479 -27.4 ! 13.2

1200 .190 -163.4 3.339 77.5 .125 73.5 .472 -28.3 ! 11.7

1400 .193 -175.5 2.945 72.4 .144 73.3 .465 -30.1 ! 10.6

1600 .185 -179.0 2.626 69.3 .163 73.6 .466 -31.4 ! 9.6

1800 .163 163.1 2.316 65.0 .181 73.3 .465 -32.8 ! 8.5

2000 .179 156.0 2.174 61.7 .200 72.6 .462 -34.0 ! 7.9

2200 .210 142.6 1.977 57.6 .219 72.5 .443 -35.4 ! 7.1

2400 .216 144.8 1.857 54.6 .240 71.1 .429 -38.9 ! 6.5

2600 .230 143.8 1.762 52.1 .258 70.5 .424 -42.4 ! 6.0

2800 .213 136.1 1.657 49.3 .279 70.0 .427 -44.9 ! 5.5

3000 .240 127.1 1.571 45.5 .299 68.6 .419 -46.0 ! 5.0

Ansoft Designer

Transistor

selection –

*.s2p file

( )

++

++=

=

jsimsrejsimsre

jsimsrejsimsre

ss

ssS

)()()()(

)()()(

22222121

12121111

2221

1211

Transistor selection – stability check

Test for unconditional

stability:

(eq. 1 and eq. 2)

<⋅−⋅=∆

>∆+−−

=

1

12

1

21122211

2112

22

22

2

11

ssss

ss

ssK

Transistor selection – gain

definitionsPower Gain = G = PL/Pin – the ratio of the power dissipated in the

load ZL to the power delivered to the amplifier input

Available Power Gain = GA = Pavn/Pavs – the ratio of the power

available from the amplifier to the power available from the

source. This assumes conjugate matching of both the source and

the load. Depends on ZS and ZL

Transducer Power Gain = GT = PL/Pavs – the ratio of the power

delivered to the load to the power available from the source.

Depends on ZS and ZL

Transistor selection – gain definitions

RF

transistor

[S]

G0

Output

matching

circuit

GL

Input

matching

circuit

GS

ZG

ZL

ΓS

Γin

Γout

ΓL

2

11

2

1

1

S

S

S

sG

Γ−

Γ−=

2

210 sG =

2

22

2

1

1

L

L

L

sG

Γ−

Γ−=

(eq. 4)(eq. 3)

(eq. 5)

0

0

ZZ

ZZ

S

SS

+

−=Γ

0

0

ZZ

ZZ

L

LL

+

−=Γ

(eq. 6) (eq. 7)

unilateral case (s12 = 0):

LSTU GGGG 0max = ][][][][ 0max dBGdBGdBGdBG LSTU ++=or (eq. 9)(eq. 8)

-----------------------------------------------------------------------------------------------------------------

Transistor input and output reflection

coefficients

RF

transistor

[S]

G0

Output

matching

circuit

GL

Input

matching

circuit

GS

ZG

ZL

ΓS

Γin

Γout

ΓL

L

L

in

inin

s

sss

ZZ

ZZ

Γ−

Γ+=

+

−=Γ

22

211211

0

0

1S

S

out

outout

s

sss

ZZ

ZZ

Γ−

Γ+=

+

−=Γ

11

211222

0

0

1

Remark: every reflection coefficient definition is related to Z0. Thus, Z0

( = usually 50 Ω) is treated as reference impedance.

(eq. 10) (eq. 11)

Maximum power transfer -

impedance matching

Lg

Lgout

RR

RVU

+=

gL

g

outRR

VI

+=

( )22

gL

Lg

outoutoutRR

RVIUP

+==

( ) ( )( )

0max =→L

outout

Rd

PdP

( )( ) ( ) ( )3

2

2

2 2

gL

gL

gL

g

L

out

RR

VR

RR

V

Rd

Pd

+−

+=

( )( ) gL

L

out RRRd

Pd=⇒= 0

Maximum power transfer -

conjugate matching idea

⇒

⇓gL XX −=

(for the specific frequency)

⇐

Transistor input and output

conjugate matching

Maximum power transfer from input to transistor and from

transistor to output occur for:

*

Sin Γ=Γ *

Lout Γ=Γand

The input and output must be matched simultaneously:

Γ−

Γ+=Γ

Γ−

Γ+=Γ

S

SL

L

LS

s

sss

s

sss

11

211222

*

22

211211

*

1

1

(eq.12) (eq.13)


conjugate matching

From the system of equations we obtain quadratic equation

for ΓS (or ΓL):

( ) ( ) ( ) 01 22

**

11

2

22

2

11

22*

2211 =∆−+Γ−+−∆+Γ∆− ssssss SS

The solution is:

1

2

1

2

11

2

4

C

CBBS

−±=Γ

2

2

2

2

22

2

4

C

CBBL

−±=Γ

22

22

2

111 1 ∆−−+= ssB

22

11

2

222 1 ∆−−+= ssB

*

22111 ssC ∆−=

*

11222 ssC ∆−=

The signs in eq. 14 and 15 are opposite to signs of B1 and B1, respectively.

(eq.14)

(eq.15)


conjugate matchingThen, from (eq.3 and eq.4) we can calculate the

impedances Zin and Zout that satisfy conjugate matching at

transistor input and output:

*

0

*

1

1

Γ−

+Γ==

S

SSin ZZZ

*

0

*

1

1

Γ−

+Γ==

L

LLout ZZZ(eq.16) (eq.17)

Now, we have to choice the matching circuit implementation

(type): lumped elements, distributed elements, single or

double stubs tuning technique, tappered line a. s. o..

Matching with shunt single stub

technique

General circuit

YL

d

lopen or

shorted

stub

Y0

Y0

Y0 = 1/Z

0 ZL = R

L + jX

L

d

lopen

stub

Z0

Z0

Z0

d

lshorted

stub

Z0

Z0

Z0

ZL = R

L + jX

L

Zg = R

g = Z

0

Zg = R

g = Z

0

⇒

Matching circuit fabricated with

microstrip technique


technique

( )

( )( )

<+

≥

=

0,2

1

0,2

1

tfortarctg

tfortarctgd

ππ

π

λ

( )[ ]

=−

≠−

+−±

=

0

0

0

0

0

22

0

,2

,/

ZRforZ

X

ZRforZR

ZXRZRX

t

LL

L

L

LLLL

A distance from load to the stub:

(eq.18)

(eq.19)


techniqueStub length:

Open: ( )BZarctglopen

02

1

πλ−=

=

BZarctg

lshorted

0

1

2

1

πλShorted:

( )( )( )[ ]20

2

0

00

2

tZXRZ

tZXtXZtRB

LL

LLL

++

+−−=

(eq.20)

(eq.21)

The other microstrip matching circuit types are described in details in [2]. The

references [1,3] describe the matching techniques, as well: [1] briefly, [3] for RF

frequency region f <1 GHz (circuits with lumped elements).

RF Transistor Bias CircuitsTransistor RF properties (S parameters vs. frequency) depends on its

operating point.

The operating point is set up by appropriate transistor bias circuit delivered IB,

IC and UCE to RF transistor.

For low and middle frequency region of RF frequency band (f > 2GHz) the

biasing circuits usually are similar to bias circuits use at low frequencies

(passive, resistive), described in details on Electronic Circuits course and in

[5,6]. One difference is that the bias circuit is connected to the transistor

terminals through RF chokes.

For middle and high frequency band (f > 1GHz) active bias circuits (current

sources or mirrors) are applied to improve a temperature stability of the

transistor parameters. They were considered on Electronic Circuit course, as

well.

RF Transistor Bias CircuitsThe separation between DC bias amplifier circuit and RF amplifier part.

BIAS

CIRCUIT

RF CHOKE RF CHOKE

BYPEPASS

CAPACITOR

BYPEPASS

CAPACITOR

INPUT

MATCHING

CIRCUIT

OUTPUT

MATCHING

CIRCUIT

DC

BLOCK

DC

BLOCK

RF

OUTPUT

RF

INPUT

RF Transistor Bias CircuitsThe separation between DC bias amplifier circuit and RF amplifier part in

microstrip technique.

Radial Stub dimentions

can be calculated with

[4] or using web

calculator :

http://www.flambda.com/

php/stub/stub.phpRF CHOKE

BYPEPASS

CAPACITOR

RADIAL

STUB

λ/4

to bias

circuit

or

λ/4

λ/4

to bias

circuit

As narrow as possible

(high Z) microstrip

transmission lines

Uto transistor to transistor

Example of RF amplifier design

Task: Design RF amplifier for maximum gain at 1.5 GHz

using single, open stub, shunt matching section; Zg = ZL =

Z0 = 50Ω. The amplifier will be fabricated on FR4

substrate:εr = 4.5, H = 1.5mm, copper thickness 35 µm.

Supply voltage UCC = 9V.

1. Transistor selection (on the basis of datasheet) : BFR92A

(βDC = 90, GUM = 14 dB (1 GHz), fT = 5GHz)

2. With Ansoft Designer you can determine S parameters at

1.5GHz, for selected operating point.

Example of RF amplifier design2a. Create schematic as below (use N-port element instead of

transistor).

2b. Choose transistor operating point UCE = 5V, IC = 10 mA. Download to N-

port element the BFR92AC.s2p file (S parameters measured for selected

operating point).

2c. Simulate circuit for a frequencies 0.1 – 3GHz, with step 0.01GHz.

2d. Read out magnitude and angle values of every sXX parameter at 1.5

GHz.


Mag(S)


Angle(S)


°−∠=°∠=

°∠=°−∠==

77.3047.08.7078.2

46.7315.031.17719.0

2221

1211

ss

ssS

S matrix:

3. With eq. 1 and eq. 2 calculate transistor stability factor

=⋅−⋅=∆

=∆+−−

=

328.0

02.12

1

21122211

2112

22

22

2

11

ssss

ss

ssK

Because K>1 and |∆|<1 at f = 1.5 GHz transistor is unconditional stable.

3a. Check transistor stability frequency range with Ansoft.


Example of RF amplifier designTransistor is stable for frequencies from 0.95 GHz to 2.48 GHz. Stability coefficient

K at 1.5 GHz is equal to 1.01; the difference between calculated and simulated K

values is caused by a precision of S matrix reading out from the sxx traces.

4. For maximum gain you should design matching sections for a conjugate match.

Using eq. 14 and eq. 15 determine source and load reflection coefficients:

°−∠=−±

=Γ 32.1787831.02

4

1

2

1

2

11

C

CBBS

°∠=−±

=Γ 9.318532.02

4

2

2

2

2

22

C

CBBL


5. Maximum (unilateral) amplifier gain can be calculated with eq.: 5,6,7,8, and 9.

5859.21

12

11

2

=Γ−

Γ−=

S

S

S

sG

7284.72

210 == sG

7579.01

12

22

2

=Γ−

Γ−

L

L

L

sG

1454.150max == LSTU GGGG

][8.11][][][][ 0 dBdBGdBGdBGdBG LSTUma =++=

)log(10][ GdBG =Remember:

remark: The gain of amplifier will be less than calculated GTUmax because

transistor is not unilateral (s12 ≠ 0).

Example of RF amplifier design6. Now you can calculate input and output transistor impedances (using eq.

16 and eq. 17):

jZZZS

SSin 72.00826.6

1

1*

0

* +=

Γ−

+Γ==

jZZZL

LLout 4.16168.48

1

1*

0

* −=

Γ−

+Γ==

7. Next, calculate the parameters of the input and the output matching

circuits using eq. 18, 19 and 20.

Example of RF amplifier design7a. Input matching circuit calculation:

( )[ ]3652.0

/

0

0

22

0

1 −=−

+−+=

ZR

ZXRZRXt

L

LLLL

in

( )[ ]3324.0

/

0

0

22

0

2 =−

+−+=

ZR

ZXRZRXt

L

LLLL

in

( ) °=== 39.180511.02

12 λπλ

tarctgd in

( )( ) °==+= 94.159443.02

11 λππλ

tarctgd in

Stub distance from the amplifier input:

Stub length:

( )( )( )[ ] 0504.0

2

10

2

0

1010

2

1 =++

+−−=

inLL

inLinLLin

tZXRZ

tZXtXZtRB

( )( )( )[ ] 0504.0

2

20

2

0

2020

2

2 −=++

+−−=

inLL

inLinLLin

tZXRZ

tZXtXZtRB

( )

°=+−=

=−=−=

67.1115.01898.0

1898.02

110

1

λλ

λπλ

in

inopenBZarctg

l

( ) °==−= 32.681898.02

120

2 λπλ

in

inopenBZarctg

l

Example of RF amplifier designnote1: if some od calculated dimension of d or l is negative just add to it

0.5λ.

note2: you got two solutions od the matching circuit dimensions, both of

them are valid.

7b. Output matching circuit calculation:

Let’s choose the solutions: d2in = 18.39o, lopen2in = 68.32

o, d2out = 89.76o, lopen2out = 73

o.

6209.11 =outt °== 32.58162.01 λλoutd

0654.01 =outB°=+−=

=−=

99.1065.02028.0

2028.01

λλ

λλ

outopenl

31.2432 =outt °== 76.892493.02 λλoutd

0654.02 −=outB °== 732028.02 λ

λ

outopenl


Amplifier circuit without DC biasing circuit. The Transmission Lines (TL) impedance

is equal to 50Ω. TL width (2.77 mm) was calculated with Ansoft TL calculator.



Example of RF amplifier design8. Bias circuit.

Bias circuit was calculated for:

UCE = 5V, IC = 10mA, UCC = 9V,

UBE = 0.65V, βDC = 0.65V. RB3

= 1k was assumed. After the

resistors calculations the circuit

Pspice simulation was carried

out. Pspice BFR92A model

requires modification of BF

parameter from 120 to 102 in

order to simulate with betaDC =

90. Finally, UCE = 4.95V and IC =

10.1mA operating point was

achieved.

Example of RF amplifier design9. RF chokes.

Dimentions of Radial Stub,

with assumption that RI =

0.5W, were calculated with

calculator on:

http://www.flambda.com/php/st

ub/stub.php.

The length of l\4 line was

calculated with Ansoft TL

Calculator. The line width was

assumed 0.4mm.

P = λ/4=29.2mm

U

W = 0.4mm

RI = 0.15mm

RL = 15.53mm

A = 90o


Final

amplifier

circuit

rf circuits design 6ue.pwr.wroc.pl/.../lecture/rf_circuits_design_6.pdfamplifier block diagram rf...

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