1. RF Amplifiers Biasing of Transistors: The Base Emitter junction should be Forward biased and Base Collector junction should be reverse biased. R2 R1 Rc RE RL Rs Vs VCC Cc Cb

2. Equivalent Circuit: V Rs Vs RB rr CC gmvbe Rc Rl Small signal gain = gm (Rc II Rl) Ccb CL

3. Extending Bandwidth in RF Amplifiers Inductive load: L R C R L C gmvbe

4. Inductive load to enhance bandwidth Load impedance: Z(s) = (sL +R) II 1/sC = R[sL/R+1]/[S 2 LC+ sRC + 1] If we define m=RC/[L/R],  = 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.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

5. 10V Rc=100 ohms BFP193 RL=50 ohms RE=12 ohm. Vs Rs= 50ohms CB1, coupling capacitor, Should offer Low 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 5nH 1.5pF Lm1 Cm1 Lm2 Vout Design Shunt Inductor Peaking amplifier

6. Selection of Transistor 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: 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 BC E B C E

7. 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=- 3.3 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.

8. 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 offer Low 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

9. Matching Network C Ls Rs At frequency  o, The impedance of the network = j  Ls+Rs = j  Lp|| Rp = = [(  Lp) 2 Rp + j  oLpRp 2 ]/Rp 2 +(  oLp) 2 Lp Rp C Rp= Rs(Q 2 +1), Lp=Ls(Q 2 +1)/Q 2 = Ls if Q>>1 Cp= Cs(Q 2 )/(Q 2 +1)

10. L match Circuit Rp Ls C Rp= Rs(Q 2 + 1) = Rs Q 2 = Rs(1/(  oRsC) 2 = (1/Rs) (Ls/C) RsRp= Ls/C = Zo 2 Rs Downward impedance transformer C Rp Ls Rs Upward impedance transformer

11. Tuned Amplifiers Gain x bandwidth = constant If we reduce the bandwidth, gain can be high. G (BW) = gmR.(1/RC) = gm/C

12. 10V BFP193 RL=50 ohms RE=15 ohm. Vs Rs= 50ohms CB1, coupling capacitor, Should offer Low 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 5nH 1.5pF Lm1 Cm1 Lm2 Vout

13. Strange Impedance Behaviors and Stability Circuit Model for Base Impedance Effect: Zb  ib Z ib Cbe The impedance seen at base of the transistor, Zb= 1/j  Cbe + Z(  +1) = 1/j  Cbe + Z(  j  T /   = ic/ib = gm vbe/ib = gm/sCbe =-j  T /   goes to 1 at  =  T 1 = gm/  T.Cbe,  T = 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.

14. Impedance Looking into the Emitter Terminal: Ze= 1/j  Cbe + Z/(  +1) where Z is the impedance in the base side = 1/j  Cbe + Z/ (-j  T/  +1) = 1/j  Cbe + jZ(  /  T) If Z=j  L, Ze= = 1/j  Cbe - (  2 /  T) L Inductance at base appears as a negative resistance at emitter.