RF AmplifiersBiasing of Transistors: The Base Emitter junction should beForward biased and Base Collector junction should be reverse biased.

R2

R1

Rc

RE

RL

Rs

Vs

VCC

Cc

Cb

Equivalent Circuit:

V

Rs

Vs

RBrp

Cp

gmvbe Rc Rl

Small signal gain = gm (Rc II Rl)

Ccb CL

Extending Bandwidth in RF AmplifiersInductive load:

L

R

C

R

L

C

gmvbe

Inductive load to enhance bandwidth

Load impedance:

Z(s) = (sL +R) II 1/sC = R[sL/R+1]/[S2 LC+ sRC + 1]

If we define m=RC/[L/R], t = L/R

Z(s) = R. [ts+ 1]/[s2t2m + stm +1]

Gain with inductive load/gain wihout indutcive load= |Z(jw)|/R = [

Band width with inductive load/Bandwidth without inductive load=

Condition m=R2C/L Bandwidth boost factor Normalized Peak Freq.resMaximum bandwidth 1.41 1.85 1.19|Z|=R 2 1.8 1.03Best Magnitude Flatness 2.41 1.72 1Beat delay flatness 3.1 1.6 1 No Shunt Peaking Infinite 1 1

10V

Rc=100 ohms

BFP193

RL=50 ohms

RE=12 ohm.

Vs

Rs=50ohms

CB1, coupling capacitor,Should offerLow resistance, les parasitics.

RB1

RB2, Bias resistor

IE=10ma.12V

1V

Current through bias resistors 10 times base current. Base current is =Emitter current/beta = 0.1mA.

1k

9k

100pf

100pf

5nH

1.5pF

Lm1

Cm1

Lm2 Vout

Design Shunt InductorPeaking amplifier

Selection of Transistor

BFP 193 RF transistor, ft, unity gain frequency = 8 GHzHFE = 125 (typical). All the transistor parameters have to be entered in the model. Package equivalent circuit.

Package Equivalent Circuit:

LBO= 0.65 nH

LBI = 0.84 nH

LCI = 0.07nH

LCO = 0.42nH

Transistor Chip

LEI = 0.31nH

LEO = 0.14 nH

CCB= 19fF

CBE = 145fF

CCE=281fF

B C

E

B C

E

Design of Feedback Amplifier

Let us design the amplifier for a power gain of 10 dB.This corresponds to a voltage gain of 3.2. 10 log Pout/Pin = 10 log Vout2/vin2 = 20 log Vout/Vin = 10db. Vout/Vin = (10) 0.5 = 3.3.Av= Vout/Vin = RC/RE=- 3.3Rin =Rout =50 ohm.Rin = RF/ 1-Av = RF/1+3.3RF = 50(4.3)= 215 ohm . You can select 210 ohm or 240 ohm as the RF.Select gm. Gain = gm. Ro= 3.3 = gm.50gm = 3.3/50= 3300/50= 66 msRE=1/gm= 1/ 66ms = 50/3.3= 15 ohm. Preferable value is about 12 ohm or 10 ohms.Gm=Ic/vt , Ic= 66.25= 1.5mA. We keep Ic about 10 mA so that we get enough gain.RL=500 ohms, so that VCB=5V to reduce Base to Collector capacitance.

10V

Rc=500 ohms to get adequate reverse bias to reduce Cbc

BFP193

RL=50 ohms

RE=12 ohm.

Vs

Rs=50ohms

CB1, coupling capacitor,Should offerLow resistance, les parasitics.

RB1

RF, feedback resistor

RB2, Bias resistor

CB1, reactance 10 times less than RB2

IE=10ma

5V

.12V

.9V

Current through bias resistors 10 times base current. Base current is =Emitter current/beta = 0.1mA.

1k

.2k

3.3k

Matching Network

C

Ls

Rs

At frequency wo, The impedance of the network = jwoLs+Rs = jwoLp|| Rp= = [(woLp)2 Rp + jwoLpRp2]/Rp2+(woLp)2

LpRp

C

Rp= Rs(Q2+1), Lp=Ls(Q2+1)/Q2 = Ls if Q>>1

Cp= Cs(Q2)/(Q2+1)

L match Circuit

Rp

Ls

C

Rp= Rs(Q 2 + 1)

= Rs Q 2

= Rs(1/(woRsC)2

= (1/Rs) (Ls/C)

RsRp= Ls/C = Zo2

Rs

Downward impedance transformer

CRp

Ls

Rs

Upward impedance transformer

Tuned AmplifiersGain x bandwidth = constant

If we reduce the bandwidth, gain can be high.

G (BW) = gmR.(1/RC) = gm/C

10V

BFP193

RL=50 ohms

RE=15 ohm.

Vs

Rs=50ohms

CB1, coupling capacitor,Should offerLow resistance, les parasitics.

RB1

RB2, Bias resistor

IE=10ma.12V

1V

Current through bias resistors 10 times base current. Base current is =Emitter current/beta = 0.1mA.

1k

9k

100pf

100pf

5nH

1.5pF

Lm1

Cm1

Lm2 Vout

Strange Impedance Behaviors and Stability

Circuit Model for Base Impedance Effect:

Zb bib

Z

ib

Cbe

The impedance seen at base of the transistor,

Zb= 1/jwCbe + Z(b+1)

= 1/jwCbe + Z(-jwT / +1)w

b = ic/ib = gm vbe/ib = gm/sCbe =-j wT /wb goes to 1 at w =wT

1 = gm/wT.Cbe, wT = gm/Cbe

If Z= R, resistor Zb sees it as a capacitor

If Z is due to inductor, it appears as a resistance.

If Z is a capacitor, it appears as –ve resistance and may cause oscillations.

Impedance Looking into the Emitter Terminal:

Ze= 1/jwCbe + Z/(b+1) where Z is the impedance in the base side = 1/jwCbe + Z/ (-j wT/w +1) = 1/jwCbe + jZ(w/wT)

If Z=jwL, Ze= = 1/jwCbe - (w2/wT) L

Inductance at base appears as a negative resistance at emitter.