Radio Frequency Electronics

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1 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 for his whole career Arround1963, discovered a negative resistance effect named Gunn-effect. Gunn diodes allow for inexpensive microwave oscillators Worked on self-documenting computer code that relates to computer viruses Was an avid motorcyclist Somewhat eccentric family with sculptors, jazz musicians, physicists Received many awards and recognitions including IEEE Fellow, Member National Academy of Engineering, IBM Fellow Died in 2008 Image from IEEE Global History Network 1

2 Parallel RLC Circuit Consider a parallel RLC circuit driven by a current source. The source may be a FET, BJT, photodiode, Z L s = sl, Z C s = 1 sc, Z R s = R The input impedance seen by the current source is Z 11 s = V O s I i (s) = s/c s 2 + s RC + 1 LC = s/c s p 1 s + p 2 p 12 = 1 2RC ± 1 2RC 2 1 LC The poles may be real, or a complex conjugate pair. We will first consider the case where the poles are real. 2

3 Parallel RLC Circuit Real Poles Z 11 (s) = s/c s p 1 s + p 2 p 12 = 1 2RC ± 1 2RC 2 1 LC When 1 RC 2 1 LC or equivalently R ω 0 L 2 = 1 2ω 0 C then the poles lie on the negative real axis. This shows that when R is small compared to the both the reactances ( high loss), the circuit functions as a broadband network. The midband frequency range extends from p 1 to p 2. The midband is the range of frequencies where the reactance of the inductance and the capacitance is large and can be ignored compared to R. Also, Z 11m = R. At low frequencies the reactance of the inductor is small compared to C and R, and shunts the signal to ground. At high frequencies the reactance of the capacitor small compared to R and L and shunts the signal to ground. 3

4 Parallel RLC Circuit Real Poles When poles are far apart p 1 1 RC p 10 p 2 R L p 20 4

5 Parallel RLC Circuit Real Poles Z 11 s = V O s I i (s) = s/c s p 1 s + p 2 p 12 = 1 2RC ± 1 2RC 2 1 LC When real poles are far apart (p 1 >> p 2 ) then p 1 is approximated by the sum of the roots of the denominator of Z 11 (s), and p 2 is approximated by the product of the roots of the roots divided by the sum of the roots: p 1 1 RC p 10 p 2 1 LC 1 RC = R L p 20 Physically, p 10 is the frequency where the reactance of C is equal to R, and p 20 is the frequency where the reactance of C is equal to R. This leads to the simplified circuits for approximately finding the roots: Simplified circuits for finding p 1, p 2 5

6 Parallel RLC Circuit Complex Poles Z 11 s = V O s I i (s) = s/c s p 1 s + p 2 p 12 = 1 2RC ± 1 2RC 2 1 LC For the case where (1/2RC) 2 < 1 LC or R > 0 L/2, the poles are complex and take the form p 1,2 = α ± j ω 0 2 α 2 = α ± jβ Here ω 0 = 1 LC, α = 1 2RC and β = ω 0 2 α 2. The pole-zero diagram is below Physically, ω 0 is the radian frequency where the reactances of the inductor and capacitor cancels. At this frequency the circuit appears as a pure resistor R. Pole-zero diagram of Z 11 s when R > ω 0 L 2 6

7 Parallel RLC Circuit Complex Poles BW = 2α BW ω 0 = 2α ω 0 = 1 Q T BW: frequency where voltage response is 3 db down Z 11 (jω) is 2 down from it maximum value, which is R BW Q T is the so-called network quality factor ω 0 At ω 0 the phase shift is zero and the network is purely resistive 7

8 Parallel RLC Circuit Complex Poles To obtain the sinusoidal steady-state frequency response when R > ω 0 L 2, evaluate Z 11 s at s = jω and rearrange Z 11 jω = Z 11 s s=jω = s C s p 1 s + p 2 s=jω = jω C ω 0 2 ω 2 + 2jωα = R 1 + jq T ω 2 ω 0 2 ω ω 0 Here Q T = ω 0 2α = ω 0 RC = R ωl. As it turns out, Q T the network quality factor, is very widely-used in resonant circuits, and filters. A related parameter, the component que is often used to characterize inductors, and sometimes capacitors. Plot of Z 11 jω and phase of Z 11 jω as a function of frequency. Note that at resonance, the phase shift is zero and Z 11 is purely resistive. 8

9 Parallel RLC Circuit Complex Poles Z 11 jω = jω C ω 0 2 ω 2 + 2jωα = R 1 + jq T ω 2 ω 0 2 ω ω 0 When Q T increases, the 3-dB bandwidth of the circuit decreases. In a filter application, one generally wants a narrow bandwidth to suppress noise a high-q network. On the other hand, if the network Q is too high, the bandwidth may become too narrow, and the network may attenuate some signal frequencies. Further, high Q T implies that the poles of Z 11 (s) are close to the imaginary axis circuit becomes oscillatory. 9

10 Another Interpretation of Q T Z 11 s = V O s I i (s) = s/c s p 1 s + p 2 p 12 = 1 2RC ± 1 2RC 2 1 LC When i i t = I cos ω 0 t, then v o t = IR cos ω 0 t = V 1 cos ω 0 t we have 2π (peak enery stored) energy dissipated per cycle = 2π 1 2 CV 1 2 2π ω 0 V 1 2 2R = ω 0CR = Q T In English: Q T is the ratio of the peak energy stored to the energy dissipated per cycle. High Q T poles of Z 11 are close to imaginary axis High Q T low loss in circuit High Q T circuit will ring for a long time in response to an impulse, pulse or step High Q T circuit has narrow bandwidth High Q T circuit is oscillatory 10

11 High Q T In cases where Q T is large (which means the poles of Z 11 (s) are close to the imaginary axis) one can show that a simplified expression for Z 11 (jω) is Z 11 jω = R 1 + j ω ω 0 α ω > 0 A similar expression exists for ω < 0 and Z 11 jω is symmetric around ω 0 with half-power frequencies given by ω 1,2 = ω 0 ± α. At these frequencies, Z 11 jω has decreased by a factor 1 2 from its maximum value. What value of Q T will make the simplified expression for Z 11 (jω) a good approximation? One can show that for Q T > 5 the expression is good, and for Q T > 10 the expression is outstanding. This is important, since in many practical circuits Q T > 5. 11

12 Impulse Response of RLC Circuit To obtain the response of Z 11 s to a (current) impulse, we obtain the inverse Laplace transform: Z 11 jω = R 1 + j ω ω 0 α ω > 0 For Q T > 10 the simplified expression for Z 11 (s) leads to z 11 t = ω 0 Cβ e αt cos βt + tan 1 α β u(t) z 11 t = 1 C e αt cos ω 0 t u(t) This shows that the response of a high-q parallel RLC circuit to a current impulse is an exponentially-decaying sinusoid with radian frequency ω 0 : Impulse response: low Q T Impulse response: high Q T 12

13 Single-Tuned Amplifier Tuned amplifier using a depletion-mode MOSFET Circuit for bias calculations The equivalent ac circuit The small-signal model 13

14 The small-signal model output v o s v i (s )sc GD + g m v + v o s sc + 1 = 0 r o sl R D R 3 Let R p = 1 G p where G p = 1 r o + 1 R D + 1 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 1 L C + C GD A v s A mid s ω Q s 2 + s ω Q + ω 0 2, ω 0 = 1 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 14

15 C GD A v s A mid s ω Q s 2 + s ω Q + ω 0 2, ω 0 = 1 L(C + C GD, Q = ω 0 R p C + C GD A mid = g m R P Assume λ = 0.02 V 1, I D = 3.2 ma, C GD = 20 pf g m = 2 K n I D = = 5.66 ma V 2 r o = 1 λi D = = 15.6K R p = r o 100K 100K=11.9K C + C GD = = 120 pf ω 0 = 1 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 = = 112 khz 15

16 Multiple-Tuned Circuits Tuned circuit at output Tuned circuit at input What is this? ac equivalent 16

17 Synchronous and Stagger Tuning Bandwidth of n synchronous tuned amplifiers BW n = BW n 1 17

18 Double-Tuned Amplifiers Double-tuned cascode amplifier Double-tuned C-C/CB amplifier Both these configurations provide inherent isolation between input and output 18

19 Tapped Inductor/Auto Transformer Tapped inductor Representation by an ideal transformer 19

20 Autotransformer Autotransformer at output What R D, C 2 sees 20

21 Autotransformer What R 1, C 1 sees 21

22 RLC Circuit Impulse Response Revisited Impulse response: high Q T z 11 t = 1 C e αt cos ω 0 t u(t) How do we use this information to infer the response to other common signals? For example, a narrow pulse, a step function, a series of narrow pulses, etc.? From linear systems theory, we know that once we know the impulse response, the response to a signal x(t) from the convolution of the impulse response and x(t): y t = x t z 11 t = x τ z 11 t τ dτ For the high-q RLC circuit and x(t) the unit step, this becomes y t = x t z 11 t = z 11 t τ dτ 0 22 = 1 C e α t τ cos ω 0 t τ dτ 0

23 RLC Circuit Impulse Response Revisited Step response for high-q RLC circuit y t = 1 C e α t τ cos ω 0 t τ dτ 0 The integral has this form: e ax cos bx dx = eax a cos bx + a sin bx a 2 + b 2 One can perform this integration to fins the response to a unit step. One can use add/subtract unit- and delayed unit steps to determine the response to a pulse or pulses. However, this becomes rather tedious. Alternatively, one can use Laplace transforms to ease the effort. However, this too can become tedious. Simpler, and more instructive, is to refer to the circuit and imagine what happens when the current source is, say a narrow pulse, rather than an impulse. 23

24 RLC Circuit Impulse Response Revisited With a narrow pulse, the RLC circuit will Ring at ω 0. Decay according to e αt In fact, the general form of the response will be similar, except that the amplitude will be different and there will be a slight delay before the onset of the ringing. The narrower the pulse are the, the better it will approximate the impulse response: z 11 t = 1 C e αt cos ω 0 t u(t) Consider what happens when a narrow current pulse with amplitude I m and duration D is applied to the RLC circuit at t = 0 +, the inductor is an open circuit and the capacitor is a short circuit, and the output voltage is zero. Then the capacitor starts to charge and its voltage rises linearly, and after a time D it has a voltage v c = v o t = I m D C across it. Then the current is turn off, and the circuit starts to ring. Thus, we expect a linear increase in voltage for the duration of pulse and then a cosine that decays according to e αt Inductor is open at t =

25 RLC Circuit Impulse Response Revisited Inductor is open at t = 0 + z 11 t = 1 C e αt cos ω 0 t u(t) v c = v o = DI m C I m D t 25

26 RLC Circuit Impulse Response Revisited z 11 t = 1 C e αt cos ω 0 t u(t) v c = v o = DI m C I m D t 26

27 Response of RLC circuit to a Pulse Train What is f 0? What is Q T? What is the response to a train of pulses? Intuitively, after each pulse, the circuit rings at the natural frequency and the amplitude decays as e αt Response of the RLC circuit with f 0 = khz to a 28.6 khz pulse train where pulses are 2 μs wide. Note that after every pulse the circuit rings at f 0 and dies away until the next pulse. 27

28 Response of RLC circuit to a Pulse Train Driven at 91 khz If the pulse frequency is at or close to resonance, the amplitude will grown and then stabilize where the energy added balances the energy lost. Thus high-q circuits will higher amplitude. Driven at 100 khz Response of a parallel RLC circuit with f 0 = khz to a 100 khz pulse train where pulses are 2 μs wide. Note that the resulting waveform is close to sinusoidal. 28

29 I C /I D Versus Time Characteristics Class-A (always on) Class-B (on during one half cycle) Class-AB (hybrid) Class-C (on less than one half cycle) 29

30 Class-A CE Amplifier Q-point was chosen for maximum output swing Consumes power even when there is no input signal Don t design this into MP3 player/cell phone etc. (max) signal load power supply power 25% 30

31 BJT Saturation Voltage In both circuit, as I B increases, I C increases and V CE decreases. However, it reaches a point where V CE will not decrease, regardless how much I B is. This is the saturation voltage V CE(sat) For low power BJTs, V CE(sat) is often 200 mv~ 600 mv. However, saturation voltage can be higher. For example, for the PN2222 it is ~ 1 V For power transistors this can be significant. For example, for the 2N3055 BJT, V CE(sat) can be as high as A. Thus, P D = 4.4 W in saturation. 31

32 I C /I D Versus Time Characteristics Class-A (always on) Class-B (on during one half cycle) Class-AB (hybrid) Class-C (on less than one half cycle) 32

33 Inductively Coupled Class-A Amplifier Short at DC (establish Q-point). Open at AC DC and AC load lines (max) 50% This is double that of a Class A with collector resistor 33

34 Transformer-Coupled Common Emitter A transformer gives us flexibility to transform the load resistance to a value that is easier to work with. a is the turns ratio What do these dots mean? R L = a 2 R L (max) 50% 34

35 Transformer-Coupled Emitter Follower R L = a 2 R L (max) 50% 35

36 Class-A Characteristics Power hungry efficiency is 25%, or 50% but then inductors/transformers are needed (costly, heavy) Simple Inherently low distortion 36

37 I C /I D Versus Time Characteristics Sect 8.3 Class-A (always on) Class-B (on during one half cycle) Class-AB (hybrid) Class-C (on less than one half cycle) 37

38 Class-B Complementary Push-Pull Output Stage Note dual power supplies With v i = 0, both transistors off => no current consumption Note dual power supplies 38

39 Class-B Complementary Push-Pull Output Stage On Note dual power supplies Off Note dual power supplies 39

40 Class-B Complementary Push-Pull Output Stage Note dual power supplies Off On Note dual power supplies 40

41 Voltage Transfer Characteristics: Complementary Push-Pull Output Stage Transistors off until v i reaches V BE(ON) Crossover distortion 41

42 Effective Load Line: Ideal Class-B Output Stage 42

43 Average Power Dissipation vs. Peak Output Voltage: Class-B Output Stage P Qn 2 VCC (max) 2 R (max) 78.5% L Remember, for Class A, η = 25% (fixed) 43

44 Bipolar Class-AB Output Stage Supply bias voltages so that both transistors are turned on, with small quiescent current 44

45 Characteristics of Class-AB Output Stage Input Signal Transistor Currents Load Current 45

46 Class-AB Output Stage: Q-Point Established by Diodes How? Place diodes in same thermal environment as transistors 46

47 Class-AB Output Stage: Darlington Pair Mimics a PNP Darlington Why not use PNP Darlington (no mimic)? 47

48 MOSFET Class-AB Output Stage Supply bias voltages so that both FETs are turned on, with small quiescent current MOSFETs V TO s vary much more than BJT s V BE 48

49 Matched and Complementary Transistors Push-pull stages work best if the complementary output transistors have matching characteristics. That is, β n = β p, V n(sat) = V p(sat), etc. or V TP = V TN, K n = K p, etc. This hard to achieve in discrete parts, but there are several so-called complemenary-pairs on the market. It is easier to achieve close matching in ICs 49

50 Class C Amplifiers and Multipliers The Fourier series representation of a periodic signal x(t) is Where Note that this is the average ( dc) value These are zero if the signal is odd These are zero if the signal is even 50

51 Class C Amplifiers and Multipliers Consider a pulse train with period T and pulse duration k. The duty cycle is d = k T. Time-domain Frequency Domain Harmonics Fundamental b n = 0 because the function is even 51

52 Class C Amplifiers and Multipliers Consider a pulse train with period T and pulse duration k. The duty cycle is d = k T. Time-domain Frequency Domain Bandpass filter selects fundamental Outputs is a sine at f 0 Input is pulse train at f 0 Amplitude of sine changes as input pulses amplitude changes 52

53 Class C Amplifiers and Multipliers 5 V Input 50 V Output Drain current Conduction angle 180 FET is biased so that it conducts for only a fraction of the input sine. If conduction angle 180, then we have class C operation, and η = 90% possible. Question: what are maximum efficiencies for class A and class B operation? 53

54 Class C Amplifiers and Multipliers 54

55 Class C Amplifier and Multipliers What is going on here? Match output to load for for maximum power transfer Match source to BJT for maximum power transfer 15 W, 100 MHz Class C Amplifier 55

56 Class C Amplifier and Multipliers Consider a pulse train with period T and pulse duration k. The duty cycle is d = k T. Time-domain Frequency Domain Bandpass filter selects harmonic Outputs is a sine at 2f 0 Input is pulse train at f 0 Amplitude of sine changes as input pulses amplitude changes 56

57 Multipliers For efficient multiplication, we want narrow pulses 57

58 58

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