Radio Frequency Electronics

Radio Frequency Electronics Active Components IV Samuel Morse Born in 79 in Massachusetts Fairly accomplished painter After witnessing various electrical experiments, got intrigued by electricity Designed the first single-wire telegraph Invented the concept relay what we now call repeaters Created Morse Code (digital communications?) Held several patents related to the telegraph Dies in 872 Image from Wikipedia

CE Amplifier (CS is Similar) C eq High-gain because of C E Inverting amplifier Use time constant technique: f H 2 [ r R R B S ]( C eq ) f H 2 p 2

SPICE Results for Common Emitter 3

Common Base Amplifier Notice where the input signal goes 4

Common Base Amplifier Remember, it is v BE that controls collector current in BJTs V BE I C = I S e V T v be v be v be Common Emitter Common Collector Common Base 5

Designing a Common-Base Amplifer β = 00 β = 00 I C I C R i R i Design a CE amplifier R, R 2, I C, R E, R C are determined to achieve a desired operating point A v = g m R C A v = 40I C R C =?? Ground the base Feed signal into emitter A v = Same as for CE R i = r π β = = g m =?? R i = r π = β g m = 00 40I C =?? 6

CB Amplifier This is not an inverting amplifier Thus, Miller no multiplication effect. 7

CB Amplifier (CG is Similar) These are NOT inverting amplifiers. Thus, Miller no multiplication effect. 8

A. Kruger 9 Radio Frequency Electronics The University of Iowa CB Amplifier C R R r f S E H 2 C R R f L C H ) ( 2 Equivalent input circuit Equivalent output circuit Either one could determine bandwidth (normally C μ ) Regardless, higher bandwidth than CE

Cascode Circuit Cascode amplifiers are composite amplifiers where a CE amplifier feeds a CB amplifier. One can view the CB amplifier as the load of the CE amplifier. The CB has input impedance g m. If we were to glue the CB to the CE, the CE would see a load of R C = g m. g m 0

Cascode Circuit Cascode amplifiers are composite amplifiers where a CE amplifier feeds a CB amplifier. One can view the CB amplifier as the load of the CE amplifier. The CB has input impedance g m. If we were to glue the CB to the CE, the CE would see a load of R C = g m. The gain of the CE would be g m R c = g m R C g m This means small Miller Effect The gain of the CB stage is g m R c and there is no Miller effect

BJT Cascode Amplifier Common Base 2N2222 BJTs 36 db Voltage Gain at 6 MHz 2 db Voltage Gain at 70 MHz Common Emitter 2

BJT Cascode Circuit CE is an inverting amplifier => Miller effect present CE voltage gain ~ => low Miller effect 3

BJT Cascode Circuit f H 2 R R r ( C C ) H S B M 2 ( RC RL) C2 f Either one could determine bandwidth (normally C μ ) Wide bandwidth 4

SPICE Results for Cascode 5

FET Cascode Common Gate Common Source 6

FET Cascode g m 7

Dual Gate FETs The cascode amplifier is very popular in RF and the stacked configuration has other useful applications. Consequently, semiconductor companies make dual-gate FETs where two gates squeeze the same channel Cascode amplifier with 2 separate FETs Cascode amplifier with dual gate FET 8

Dual Gate FETs D D D G G 2 G G 2 G G 2 S n-channel S p-channel S Construction Schematic Symbols 9

Dual Gate FETs Note the two gates These form back-to-back Zeners that protect the FET against damage from static electricity 20

Emitter-Follower Circuit (Source-Follower is Similar) 2

A. Kruger 22 Radio Frequency Electronics The University of Iowa ' ' ) ( 2 L m L m B S H R g C C r R g R R f ' ' ) ( L m L m B S p R g C C r R g R R Wide bandwidth ' ' ' ' L L L b R gmr sc gmr r Z

SPICE Results for Emitter Follower 23

Single-Tuned Amplifier Tuned amplifier using a depletion-mode MOSFET Circuit for bias calculations The equivalent ac circuit The small-signal model 24

The small-signal model KCL @ output v o s v i (s )sc GD + g m v + v o s + + sc + = 0 r o sl R D R 3 Let R p = G p where G p = r o + R D + R 3 (i.e., the parallel combination of the resistances au the output. Solving for the voltage transfer function v o (s) v i (s) yeilds A v s = v o s v i (s) = sc GD g m R p s 2 + s R p C + C GD s + R p C + C GD Neglecting the right-half-plane zero ( sc GD in sc GD g m ) then L C + C GD A v s A mid s ω Q s 2 + s ω Q + ω 0 2, ω 0 = L(C + C GD ), Q = ω 0R p C + C GD, A mid = g m R P Further, Q = R p ωl and BW = ω o Q 25

C GD A v s A mid s ω Q s 2 + s ω Q + ω 0 2, ω 0 = L(C + C GD, Q = ω 0 R p C + C GD A mid = g m R P Assume λ = 0.02 V, I D = 3.2 ma, C GD = 20 pf g m = 2 K n I D = 2 2.5 0 3 3.2 0 3 = 5.66 ma V 2 r o = λi D = 0.02 3.2 0 3 = 5.6K R p = r o 00K 00K=.9K C + C GD = 00 + 20 = 20 pf ω 0 = L C + C GD = 0 0 6 20 0 2 = 28.6 0 6 rad s f 0 = ω 0 2π = 4.59 MHz A mid = g m R p = 5.66 0 3.9 0 3 = 67 Q = ω 0 R p C + C GD = 28.6 0 6.9 0 3 20 0 2 = 40.8 BW = f 0 Q = 4.59 0 6 40.8 = 2 khz 26

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