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
Equivalent Circuit: V Rs Vs RB rr CC gmvbe Rc Rl Small signal gain = gm (Rc II Rl) Ccb CL
Extending Bandwidth in RF Amplifiers Inductive 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]/[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 |Z|=R Best Magnitude Flatness Beat delay flatness No Shunt Peaking Infinite 1 1
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
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
Design of Feedback Amplifier Let us design the amplifier for a power gain of 10 dB. This corresponds to a voltage gain of 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.
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
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)
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
Tuned Amplifiers Gain 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 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
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.
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.