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Download Class B, PA - Huji cse Moodle 2014/15
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Classification of PAs: linear vs. switching a. class A, b. class B, c. class AB, d. class C 1 RF PA types • Linear PAs – Classes A, B, AB, C(?) • Switching mode PAs– Classes D, E, F2, F3, S Class A Power Amplifiers Highest Lin. But ≤ 1/2 vIN θ VDD RFC vD θ vIN θ VTH vOUT θ iOUT θ iD θ vD θ iD θ vD θ VDD vOUT θ θ iOUT θ vOUT θ 0 θ 0 iOUT θ Vom η Drain 2 iD θ θ VDD RFC θ θ 2 2 P 2 R Vom 1 RFout 2 2 PDC 2 VDD 2VDD R Class B, PA Operates ideally with zero IQ. DC power is small ≤ 78.53% single ended and push-pull versions 3 4 5 CLASS-AB & B • low IQ – no IQ for vin=0 •Crossover Distortion • Non-linear effects 6 CLASS C Vdd vO iD Vin RLoad Vt t iD t It is biased so that the transistor conduction angle is significantly less than 180°. Tuned circuit is mandatory, its Q determines the BW. It is useful for providing a high-power CW or FM signal. 7 CLASS C class C amplifier is NL- it does not directly replicate the input signal. It requires one transistor- topologically similar to the class A except for the dc bias levels. A tuned output (filter) is mandatory. Its Q determines the bandwidth of the amplifier. 8 General structure 9 CLASS C Due to the RFC (large inductance RF coil), only DC current is drawn from the power supply. When the transistor is on, the bias supply current flows through the transistor and the output voltage is approximately 90% of VCC. When Q is off, the current from the power supply flows into the capacitor. A major problem is the breakdown voltage and the leakage current reducing the efficiency. 10 PA’s Summary 11 PA example In this example the power handling of a MOSFET with, Imax=1A and Vmax=10V is examined. Let’s assume Ron= 2 for all values of VGS and ro=∞ so that the transistor’s I-V characteristics can be simplified as shown. ID I.Find PL, max in class A operation. Imax II.Find RL required to achieve PL, max. III. Determine the drain efficiency for PL, max and find its maximum value, and find the corresponding RL. IV.Determine the attainable efficiency in case of class A operation at 1/2 PL, max. V.Repeat Section I for a fixed VDD=3.4 V. 2 VDS PA example VDS VDD I D IQ RL ROL PL , RF 13 2VDS VDD I max I max VDD Vmin 4 PA example I. The maximum RF power that can be delivered to the load PL ,max 1 1 1 V ptp I ptp 2 2 2 1 1 Vmax Vmin I max 10 2 1 1W 8 8 II. The optimal value of the load resistance: V 10 2 RL 8 I 1 PA example III. The drain efficiency for PL,max of part I. pL,out 1W 1 A V 33.3% I Q VQ 0.5 6 3 The maximum value of drain efficiency is obtained for Vmin=0 1 Vmax Vmin 1 2 Vmax Vmin Vmin 0 2 15 PA example IV. The drain efficiency for PL=0.5 PL,max A new value of RL should be taken to satisfy the following equation the same Vmax 1 2 W 1 V ptp I ptp 8 1 Vmax Ron I max I max 8 I max 0.44 A, RL 20.8, 42% 16 PA example V. In this case, Imax=1A should be maintained. So, Vmin=2V, and Vmax= 2VDD-Vmin=23.4 – 2 =4.8 V 1 1 1 PL ,max V ptp I ptp 2 2 2 (4.8 2) 0.35W 8 17 Matching Networks (MNs) Objective: to optimize coupling of RF power to load. RF power supply Zg Zin MN Zout Zant For optimum power transfer, we require: Zg = Zin* 18 and Zout = Zant* Antenna DC Resistive Circuits the concept of available power RS RL ES When RL = RS, the power delivered to the load is a maximum: 2 PL ,max ES 4 RS PL,max is called the available power of the generator. 19 AC Circuits ZS VOC Z S RS jX S , For Z L Z S* , 20 ZL Z L RL jX L PL PL ,max 2 VOC 8RS AC Circuits Notes: 21 Since reactances are frequency dependent, perfect impedance match is obtained only at a single frequency. At all other frequencies, the matching becomes progressively worse, and eventually non-existent. For broad-band applications, the match bandwidth must be increased. The L networks These are two-element MN consisting of one capacitor and one inductor arranged in an L-shape orientation. There are 4 possible arrangements 22 Example A 100 generator is connected to a resistive load of RL = 1000. Find the ratio of the actual load power and the available power. 100 vs 1000 Solution Let's assume vs = A cos t Available power is obtained for a load of RL = 100. 23 For RL = 100, the input (at the load terminals) current and voltage are: A cos t iL 2 RL 1 v L A cos t , 2 Thus: A 2 cos 2 t PL (t ) i L (t )v L (t ) 4 RL PL t 24 max A2 A2 8RL 800 Actual power: 10 A 1000 cos t vL A cos t 1000 100 11 A A cos t cos t iL 1000 100 1100 A2 cos 2 t PL (t ) iL (t )vL (t ) 1210 A2 PL t Actual 2420 PActual 800 0.33 4.8dB 2420 Pmax 25 In this example, without impedance matching, about 4.8 dB of the available source power would be lost. The matching network cancels this loss. The parallel combination of the load resistor and the MN capacitor is calculated by Z 26 jX C RL jX C RL For XC=333 Z 100 j300 ohm This is a series combination of a 100-ohm resistor and a j300-ohm capacitor. A 1000 -j333 Please note that Rp > Rs 100 -j300 B To complete the Z-match, all we must do is to add an equal and opposite (i.e. +j300 ohm) reactance in series with the source resistor. 27 L - Network Design Procedure QS Q P 28 QS XS RS QP RP XP RP 1 RS QS = Q of the series leg. QP = Q of the shunt leg. RP = The shut resistance. RS = The series resistance. XP = The shunt reactance. XS = The series reaction. Example Design an L-shape matching network for ZS=100 ohm and ZL=1000 ohm at 100 MHz. Assume that a dc voltage must also be transferred from source to the load. 29 Solution The need for dc path dictates an inductor in the series leg. 1000 1 3 100 X S QS RS 3 100 300 ohm (inductive) QS QP RP 1000 XP 333 ohm (capacitive) QP 3 Component values for f 100 MHz : X 300 L S 477 nH 6 2 100 10 1 C 4.8 pF X P 30 Example Design a circuit to match a source of ZS=100 +j126 ohm to a load consisting a parallel combination of a 1000-ohm resistor and 2 pF capacitor, at f=100MHz. 31 Solution First, we'll totally ignore the reactance and simply match the real part of ZS to the real part of ZL. Design for a parallel capacitor and a series inductor. 32