Radio Frequency Electronics

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Aron Dennis
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Designed the first single-wire telegraph Invented the concept relay what we now call repeaters Created

1 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

2 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

3 SPICE Results for Common Emitter 3

4 Common Base Amplifier Notice where the input signal goes 4

5 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

6 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

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

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

9 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

10 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

11 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

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

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

14 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

15 SPICE Results for Cascode 5

16 FET Cascode Common Gate Common Source 6

17 FET Cascode g m 7

18 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

19 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

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

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

22 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

23 SPICE Results for Emitter Follower 23

24 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.

25 The small-signal model 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

26 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 = = 5.66 ma V 2 r o = λi D = = 5.6K R p = r o 00K 00K=.9K C + C GD = = 20 pf ω 0 = L C + C GD = = rad s f 0 = ω 0 2π = 4.59 MHz A mid = g m R p = = 67 Q = ω 0 R p C + C GD = = 40.8 BW = f 0 Q = = 2 khz 26

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Radio Frequency Electronics

Radio Frequency Electronics Tuned Amplifiers John Battiscombe Gunn Born in 1928 in Egypt (father was a famous Egyptologist), and was Educated in England Worked at IBM s Thomas J. Watson Research Center