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## PowerPoint Slideshow about ' RF Amplifiers' - mingan

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Biasing of Transistors: The Base Emitter junction should be

Forward biased and Base Collector junction should be reverse biased.

VCC

Rc

R2

Cc

Cb

RL

Rs

R1

RE

Vs

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 wihoutindutcive load

= |Z(jw)|/R = [

Band width with inductive load/Bandwidth without inductive load=

Condition m=R2C/L Bandwidth boost factor Normalized Peak Freq.res

Maximum bandwidth 1.41 1.85 1.19

|Z|=R 2 1.8 1.03

Best Magnitude Flatness 2.41 1.72 1

Beat delay flatness 3.1 1.6 1

No Shunt Peaking Infinite 1 1

Design Shunt Inductor

Peaking amplifier

5nH

CB1, coupling capacitor,

Should offer

Low resistance, les parasitics.

9k

Rc=100 ohms

Lm2

100pf

Vout

RB2, Bias resistor

BFP193

1V

Lm1

100pf

.12V

IE=10ma

RL=50 ohms

Rs=

50ohms

RB1

1k

RE=12 ohm.

1.5pF

Vs

Cm1

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

BFP 193 RF transistor, ft, unity gain frequency = 8 GHz

HFE = 125 (typical). All the transistor parameters have to be entered in the model.

Package equivalent circuit.

Package Equivalent Circuit:

CCB= 19fF

C

B

LBO= 0.65 nH

LBI =

0.84 nH

B

Transistor Chip

LCI =

0.07nH

C

LCO =

0.42nH

E

LEI =

0.31nH

CCE=281fF

CBE = 145fF

LEO =

0.14 nH

E

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=- 3.3

Rin =Rout =50 ohm.

Rin = RF/ 1-Av = RF/1+3.3

RF = 50(4.3)= 215 ohm . You can select 210 ohm or 240 ohm as the RF.

Select gm.

Gain = gm. Ro= 3.3 = gm.50

gm = 3.3/50= 3300/50= 66 ms

RE=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.

CB1, reactance 10 times less than RB2

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

RF, feedback resistor

CB1, coupling capacitor,

Should offer

Low resistance, les parasitics.

3.3k

.2k

RB2, Bias resistor

5V

BFP193

.9V

.12V

IE=10ma

RL=50 ohms

Rs=

50ohms

RB1

1k

RE=12 ohm.

Vs

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

At frequency wo,

The impedance of the network = jwoLs+Rs =jwoLp|| Rp

= = [(woLp)2Rp + jwoLpRp2]/Rp2+(woLp)2

Ls

C

Rs

C

Lp

Rp

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

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

Ls

Rp= Rs(Q 2 + 1)

= RsQ 2

= Rs(1/(woRsC)2

= (1/Rs) (Ls/C)

RsRp= Ls/C = Zo2

Rs

Rp

C

Downward impedance transformer

Ls

Rp

Rs

C

Upward impedance transformer

Gain x bandwidth = constant

If we reduce the bandwidth, gain can be high.

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

5nH

CB1, coupling capacitor,

Should offer

Low resistance, les parasitics.

9k

Lm2

100pf

Vout

RB2, Bias resistor

BFP193

1V

Lm1

100pf

.12V

IE=10ma

RL=50 ohms

Rs=

50ohms

RB1

1k

RE=15 ohm.

1.5pF

Vs

Cm1

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

Strange Impedance Behaviors and Stability

Circuit Model for Base Impedance Effect:

ib

- The impedance seen at base of the transistor,
- Zb= 1/jwCbe + Z(b+1)
- = 1/jwCbe + Z(-jwT/w +1)
- b = ic/ib = gm vbe/ib
- = gm/sCbe =-j wT/w
- goes to 1 at w =wT
- 1 = gm/wT.Cbe, wT = gm/Cbe

Cbe

Zb

bib

Z

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.

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