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DMT 231 / 3 ELECTRONICS II Lecture II Power Amplifiers (Class AB & Class C) POWER AMPLIFIER Crossover distortion can be virtually eliminated by applying small quiescent bias on each transistor (See Figure) If Qn and Qp are matched, each emitter-base junction is biased with VBB/2 when vI is zero. Hence vO is also zero. The quiescent collector currents are; iCn iCp I S eVBB / 2VT – Class-AB Operation POWER AMPLIFIER – Class-AB Operation As vI increases, the voltage at the base of Qn increases and vO increases. Qn operates as an emitter follower supplying current to RL. The output voltage is given by; VBB vO vI vBEn 2 The collector current of Qn is; iCn iL iCp (Neglecting the base currents) POWER AMPLIFIER Since iCn must to supply the load current, vBEn increases which causes vBEp to decrease because VBB is constant. The decrease in vBEp results in a decrease in iCp. – Class-AB Operation POWER AMPLIFIER When vI goes negative, the base voltage of Qp decreases followed by a decrease in vO. Qp operates as emitter follower, sinking the load current. As iCp increases vEBp increases causing a decrease in vBEn and iCn. – Class-AB Operation POWER AMPLIFIER – Class-AB Operation Transfer characteristics (vO versus vI) POWER AMPLIFIER – Class-AB Operation POWER AMPLIFIER – Class-AB Operation – Class-AB Operation POWER AMPLIFIER iCn iCp relationship vBEn vEBp vBB Using the relationship I C I S e can be written as; VBE / VT , the above expression iCn iCp I CQ VT ln VT ln 2VT ln IS IS IS Hence; iCniCp I 2 CQ The product of iCn and iCp is constant, therefore if iCn increases iCp decreases but does not to zero POWER AMPLIFIER – Class-AB Operation Example III Mn and Mp are matched transistors with the following parameters; VT 1 V; K 0.20 A/V 2 If VDD = 10 V and RL = 8 , find the bias voltage VBB/2 for IDQ = 0.05 A. Find also VGSn, VSGp andvI if vO = 5 V. POWER AMPLIFIER – Class-AB Operation Example III – Solution iD K vGS VT 2 Since the MOSFETs are matched, at quiescent point; iD I DQ and vGS VGSQ VSGQ Hence; I DQ VBB 2 VBB K VT 2 2 POWER AMPLIFIER – Class-AB Operation Example III – Solution (cont’d) Substituting values; VBB 0.05 0.2 1 2 which yields; 2 VBB 1 .5 V 2 From the expression iDn K vGSn VT we have; 2 iDn vGSn VT K POWER AMPLIFIER – Class-AB Operation Example III – Solution (cont’d) When vO 5 V; iDn vO iL RL 5 0.25 A; 20 and iDn 0.25 vGSn VT 1 2.12 V K 0.2 POWER AMPLIFIER – Class-AB Operation Example III – Solution (cont’d) Since; VBB vGSn vSGp then; vSGp VBB vGSn 3 2.12 0.88 V And; VBB 3 vI vO vGSn 5 2.12 5.62 V 2 2 POWER AMP Class-AB Qn & Qp are assumed matched transistors Small biasing voltage to eliminate dead band POWER AMP Class-AB Various techniques are used in obtaining the bias voltage VBB in class AB power amplifier circuit. POWER AMP Class-AB with Input Buffer Transistors R1, R2 and the emitter-followers Q1 and Q2 establish the required quiescent bias. R3 and R4 (usually of low values) are incorporated to provide thermal stability. The output voltage is approximately equal to the input voltage (emitter-follower) POWER AMP Class-AB with Input Buffer Transistors When the input voltage vI increases, the base voltage of Q3 increases and the output voltage vO increases. The emitter current of Q3 increase to supply the load current iO. The base current of Q3 increases. The increase in base voltage of Q3 reduces the voltage across, and the current through R1. This means iB1 and iE1 also decrease. POWER AMP Class-AB with Input Buffer Transistors Also when the input voltage vI increases, the voltage across R2 increases and iE2 and iE2 increase. The input current iI accounts for the reduction in iB1 and the increase in iB2 i.e. iI iB 2 iB1 (Kirchhoff’s Current Law) POWER AMP Class-AB with Input Buffer Transistors Neglecting vR 3 , vR 4 , iB 3 and iB 4 we have; vI VBE V iB 2 1 n R2 and; V vI VEB iB1 1 p R1 POWER AMP Class-AB with Input Buffer Transistors If; V V , VBE VEB , R1 R2 R and; n p then; vI VBE V V vI VEB iI 1 R 1 R 2vI 1 R POWER AMP Class-AB with Input Buffer Transistors Since the voltage gain is approximately unity, the output current is; vO vI iO RL RL The current gain is; iO 1 R Ai iI 2 RL which is quite substantial. A large current gain is desirable since the output stage must meet the power requrements. POWER AMP Class-AB with Input Buffer Transistors EXAMPLE IV (a) Determine the quiescent bias currents in all transistors; (b) Calculate all the currents labeled in the figure and the current gain when vI = 10 V. POWER AMP Class-AB with Input Buffer Transistors EXAMPLE IV – Solution (a) For vI = 0 (quiescent currents); iR1 iR 2 iE1 iE 2 15 0.6 7.2 mA 2 Assuming all transistors are matched, the bias currents in Q3 and Q4 are also approximately 7.2 mA since the base-emitter voltages of Q1 and Q3 are equal and those of Q2 and Q4 are equal. POWER AMP Class-AB with Input Buffer Transistors EXAMPLE IV – Solution (cont’d) (b) For vI = 10; Because the voltage gain is approx. unity; vO vI iO RL RL 10 100 mA 100 iE 3 iO 100 mA iB 3 iE 3 100 1.64 mA 1 61 POWER AMP Class-AB with Input Buffer Transistors EXAMPLE IV – Solution (cont’d) V VBE vI iR1 R1 15 0.6 10 2.2 mA 2 iE1 iR1 iB3 2.2 1.64 0.56 mA iE1 0.56 iB1 9.18 μA 1 61 POWER AMP Class-AB with Input Buffer Transistors EXAMPLE IV – Solution (cont’d) Since Q4 tends to turn off when vI increases, iB4 is negligible. Therefore; iE 2 iR 2 vI vEB V R2 10 0.6 15 2 12.2 mA iB 2 iE 2 12.2 mA 200 μA 1 61 POWER AMP Class-AB with Input Buffer Transistors EXAMPLE IV – Solution (cont’d) The input current; iI iB 2 iB1 200 9.18 191 μA The current gain; iO 100 Ai 524 iI 0.191 POWER AMP Class-AB with Input Buffer Transistors EXAMPLE IV – Solution (cont’d) If the previous expression i.e. Ai we have; Ai 1 R 2 RL is used, 1 602 610 2 0.1 The higher gain is due the fact that the base currents of Q3 and Q4 are neglected in deriving the expression. POWER AMPLIFIER – Class-C Operation Transistor conducts for less than half a cycle of input signal • • • Tuned circuit is required. Used for RF amplifier. Efficiency > 78.5% B – E junction is reverse-biased to obtain Q-point beyond cut-off.