Radio Frequency Integrated Circuit Design - Webs

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Voltage-Controlled Oscillators This can be solved for v�′ noting that g m ≈ 1/re . v�′ = i i j�C 2 Another equation can be written for vce. vce = i i + g mv�′ j�C 1 Substituting (8.41) into (8.42) gives vce =� 1 j�C1��ii + g m ii j�C2� 263 (8.41) (8.42) (8.43) Now using (8.41) and (8.43) and solving for Z i = vi /ii with some manipulation, Z i = vi ii = v�′ +vce = ii 1 j�C1 + 1 j�C2 − g m � 2 C 1C2 (8.44) this is just a negative resistor in series with the two capacitors. Thus, a necessary condition for oscillation in this oscillator is rs < g m � 2 C 1C2 (8.45) where rs is the equivalent series resistance on the resonator. It will be shown in Example 8.4 that the series negative resistance is maximized for a given fixed total series capacitance when C 1 = C 2. An identical expression to (8.45) can be derived for the Colpitts common-collector circuit. 8.8.2 Negative Resistance for Series and Parallel Circuits Equation (8.44) shows the analysis results for the oscillator circuit shown in Figure 8.18 when analyzed as an equivalent series circuit of C 1, C 2, and R neg. Since the resonance is actually a parallel one, the series components need to be converted back to parallel ones. However, if the equivalent Q of the RC circuit is high, the parallel capacitor C p will be approximately equal to the series capacitor C s , and the above analysis is valid. Even for low Q, these simple equations are useful for quick calculations.
264 Radio Frequency Integrated Circuit Design Figure 8.18 (a) Colpitts oscillator circuit; and (b) equivalent series model. Example 8.3 Negative Resistance for Series and Parallel Circuits Assume that, as before, L = 10 nH, R p = 300�, C 1 = 2.5 pF, C 2 = 10 pF, and the transistor is operating at 1 mA, or re = 25� and g m = 0.04. Using negative resistance, determine the oscillator resonant frequency and apparent frequency shift. Solution As before, C T is 2 pF and � = 1 √ LC T 1 = 7.07 Grad/sec √10 nH × 2pF= or frequency f o = 1.1254 GHz. The series negative resistance is equal to g m rs =− � 2 C 1C2 Then Q can be calculated from Q =− 1 = �rs C T −0.04 (7.07107 GHz) 2 � 2.5 pF � 10 pF =−32.0� = −1 7.07107 GHz � 32 � 2pF =−2.2097 rpar = rs (1 + Q 2 ) =−32(1 + 2.2097 2 ) =−188� We note that the parallel negative resistance is smaller in magnitude than the original parallel resistance, indicating that the oscillator should start up successfully. The above is sufficient for a hand calculation; however, to complete the example, the equivalent parallel capacitance can be determined to be
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Radio Frequency Integrated Circuit
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Radio Frequency Integrated Circuit
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Contents Foreword xv Acknowledgment
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Contents 3.8 Base Shot Noise Discus
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Contents 5.25 Packaging 135 5.25.1
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Contents 7.12 General Design Commen
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Contents 9.5 Some Simple Image Reje
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Foreword I enjoyed reading this boo
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Foreword estimates of parasitic cap
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xx Radio Frequency Integrated Circu
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2 Radio Frequency Integrated Circui
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2 Issues in RFIC Design, Noise, Lin
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Issues in RFIC Design, Noise, Linea
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The input noise is Issues in RFIC D
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2.5 Filtering Issues Issues in RFIC
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3 A Brief Review of Technology 3.1
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A Brief Review of Technology IC = I
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3.4 Small-Signal Model A Brief Revi
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3.6 High-Frequency Effects A Brief
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A Brief Review of Technology If thi
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A Brief Review of Technology Soluti
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A Brief Review of Technology 3.8 Ba
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A Brief Review of Technology • Pi
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A Brief Review of Technology 3.16,
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A Brief Review of Technology The tr
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4 Impedance Matching 4.1 Introducti
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Impedance Matching Z 2 = Z in + j
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Figure 4.5 A Smith chart. Impedance
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4.3 Impedance Matching Impedance Ma
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Impedance Matching Figure 4.9 Which
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Impedance Matching Figure 4.11 Lowp
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Impedance Matching section, convers
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Impedance Matching Much as in the p
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Impedance Matching Note that dots i
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Impedance Matching Thus, in the fir
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Impedance Matching The exact resist
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Impedance Matching 4.10 Quality Fac
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Impedance Matching Example 4.7 Matc
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Impedance Matching line to change w
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Impedance Matching S22 = b 2 a2 | a
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Impedance Matching without the need
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96 Radio Frequency Integrated Circu
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98 Radio Frequency Integrated Circu
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100 Radio Frequency Integrated Circ
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Voltage-Controlled Oscillators Figu
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Voltage-Controlled Oscillators from
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Voltage-Controlled Oscillators [10]
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10 Power Amplifiers 10.1 Introducti
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Power Amplifiers where G is the pow
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Power Amplifiers Figure 10.4 Optima
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Power Amplifiers Figure 10.8 Power
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Power Amplifiers Figure 10.9 Curren
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Power Amplifiers The efficiency is
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Power Amplifiers sinusoid), this pe
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Power Amplifiers Note that this agr
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Power Amplifiers stabilization. The
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Power Amplifiers Solution The first
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Power Amplifiers Figure 10.22 Class
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Figure 10.24 Class E waveforms. Pow
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Power Amplifiers B = 0.1836 0.81Q R
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Power Amplifiers Figure 10.25 Initi
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Power Amplifiers added to the funda
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Power Amplifiers 10.8.1 Variation o
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Example 10.4 Class F Power Amplifie
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Figure 10.36 Class H amplifier. Pow
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Power Amplifiers ered quite good. O
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Power Amplifiers Figure 10.39 Time-
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Figure 10.42 High-Q matching circui
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Figure 10.44 Multiple transistors.
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Power Amplifiers Figure 10.48 Combi
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Power Amplifiers have very narrow b
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Power Amplifiers Figure 10.53 Feedf
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Power Amplifiers in class A (input
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About the Authors John Rogers recei
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Index 1-dB compression point, 30-32
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Index Damped resonator, 247-48 Emit
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Index Local oscillator, 5 Mixing co
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Index Quadrature phase shift keying
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