Output Stages Power amplifier classification Class A amplifier circuits Class A Power conversion efficiency Class B amplifier circuits Class B Power conversion efficiency Class AB amplifier circuits Class AB Power conversion efficiency 1
Output Stage Functions Provide amplifier with low output resistance Handle large signals with low THD Deliver power to the load efficiently Output stages are classified according to the i C waveform due to input v I waveform 2
Amplifier Classifications V CC Class A amplifier amplifier BJT conducts for entire v I cycle. For all v I : i C R C v I V B 0.7V where V B max v I 0.7V R B I c = amplitude of current due to v i. v I V B R E I C = dc current Transistor cut off (i C = 0) if: v I V B 0.7V NOTE: when v I = 0, i C = I C 3
Amplifier Classifications - cont. Class B Amplifier BJT conducts positive-half of v I cycle. Amp BJT conducts for all v I s.t.: V B =0V v I 0.7V Transistor cut off (i C = 0) if: v I V B 0.7V NOTE: 1. when v I < 0.7V, i C = 0 2. a 2 nd class B BJT is needed to conduct for the negative v I cycle. 4
Amplifier Classifications - cont. Class AB Amplifier BJT conducts for positive v I swing + part of negative v I swing s.t.: v I V B 0.7V where 0 V B max v I 0.7V Conducts for: v I 0.7 V B Cut-off for rest of negative v I swing: Transistor cut-off (i C = 0) if: v I V B 0.7V NOTE: 1. when v I = 0, i C = I C 2. a 2 nd class AB BJT is needed to conduct for interval slightly larger than the negative v I cycle. 5
Class A Power Amplifier Design Used as op amp output stage and some audio output power amps. Basic considerations for low (audio) frequency operation. 1. Power usually delivered to a low impedance load. 2. Signal usually has little, preferably no dc content. 3. May have low frequency content, as low as 20 Hz. Emitter follower circuit has best power transfer efficiency, since its output impedance is low. As a bonus, its input impedance is relatively high. Principal advantage lower distortion than Class B & AB. Principal disadvantage lower power efficiency than Class B & AB. 6
Current Biased Class A Emitter Follower Q 3 diode connected transistor +V CC A current mirror establishes the bias current. I current mirror To operate reliably, 1. Q 1 and Q 2 must be forward active. 2. Current Mirror Q 3 and Q 2 need to be matched as well as possible and be at the same ambient temperature. 7
Class A Amplifier Analysis Consider the case when v I v BE1 = 0.7 V (pos. swing of v I ): if v CE1 < V CE1-sat => Q 1 sat. + - v CE1 i L =i E1 I v O =v I 0.7V =i L v CE1 =V CC v O For Q1 sat: v CE1 V CE1 sat v CE1 =V CC v O V CE1 sat v O V CC V CE1 sat v O V CC V CE1 sat v O max =V CC V CE1 sat v CE2 =v O V CC v I v BE1 =0.7V v O =v I v BE1 =v I 0.7 V v CE1 =V CC v O Max values: Q 1 sat. v O v O max =V CC V CE1 sat v I v I max =V CC V CE1 sat 0.7V 8
v CE1 =V CC v O ESE319 Introduction to Microelectronics Class A Amplifier Analysis - cont. Consider the case where v I < 0.7 V (neg. swing of v I ): + - v CE2 i L =i E1 I v O =v I 0.7V =i L if v CE2 < V CE2-sat => Q 2 sat. if i E1 = 0 => Q 1 off v CE2 =v O V CC v CE2 V CE2 sat v CE2 =v O V CC V CE2 sat v O V CC V CE2 sat v I V CC V CE2 sat 0.7 Min values: Q 1 and Q 2 forward-active i E1 =i L I 0 i L = v O I v O I v I I 0.7V v I 0.7 V v O v O min =max { I, V CC V CE2 sat } v O =v I 0.7 & v O =i L v I v I min =max { I 0.7 V, V CC V CE2 sat 0.7 V } v CE2 =v O V CC =v I 0.7 V CC NOTE: max means least negative 9
V CC - V CE1-sat max v O swing iff V CC V CE2 sat I or I V CC V CE2 sat Q 1 cutoff Q 2 saturated ESE319 Introduction to Microelectronics Class A Amplifier VTC Plot v O V BE1 -I + V CE2-sat Q 1 saturated v O =v I V BE1 slope = 1 V CC V CE2 sat I v I Bias current I & set limits on negative v O = v O-min swing Iff I V CC V CE2 sat If v O v O min = I V CC V CE2 sat I I V CC V CE2 sat v O v O min = V CC V CE2 sat V CC V CE2 sat I v O =v I 0.7 where max { I, V CC V CE2 sat } v O V CC V CE1 sat 10
Class A Stage VTC Simulation I=120 ma I =V CC =12 V =100 11.8 V Q1 saturated Q1 cutoff - 12 V ideal current source i.e. no Q 2 V EE = - V CC Q1 forward active => no clipping v O v O max =V CC V CE1 sat =11.8V v O v O min = I = 12 V Q1 not Sat Q1 not Cut-off 11
Class A Stage VTC Simulation - cont. I =120 ma =75 I =9V V CC V CE1 sat 11.8 V Q1 saturated Q1 cutoff - 9 V 12
Quick Review Class A Amp VTC v CE1 =V CC v O max v o swing i.f.f. or I V CC V CE2 sat v O =v I 0.7 where max { I, V CC V CE2 sat } v O V CC V CE1 sat 13
+V CC = +15 V Example I = -15 V =1 k Let V CE1-sat = V CE2-sat = 0.2 V, V BE1 = V BE2 = 0.7 V and 1 = 2 =large. 1. Determine the value for resistor R that will set the bias current I sufficiently large to allow the largest possible output voltage v O swing. 2. Determine the resulting output voltage swing and the maximum and minimum Q 1 emitter currents. 14
Example cont. I =1 k SOLUTION: 1. For maximum output voltage swing: I =V CC V CE2 sat I = V CC V CEsat R= V CC V BE I 15 V 0.2V = 1 k where = 1 kω =14.8 ma 15V 0.7V = 14.8 ma =0.97 k 15
Example - cont. From Part 1: I =14.8 ma SOLUTION: 2. Output voltage swing: V o peak =I =14.8V 14. 8V v O 14.8V = -15 V I= 0 V BE2 V CC R 14.8V v O 14. 8V I i L I Max and min Q 1 emitter currents: i E1 =I i L 0 ma i E1 2I=29.6 ma 16
Instantaneous and Average Power The source of power to the amplifier load,, comes from the supplies,v CC and V CC. The supplies deliver power, and the load and the transistors absorb it. Instantaneous power absorbed by resistor : p a t =v ab t i ab t =i 2 ab t =v 2 ab t / Instantaneous power delivered by battery V CC : p d t =v ab t i ba t =V CC I i ba =I i ab Average power: P ab av = 1 T i 2 2 ab t dt=i ab rms = I ab peak 0 2 P D av =V CC I T i ab =I ab peak sin t for period T 2 2 = V ab peak 2 V CC 17
Emitter Follower Power Relationships I. Average power delivered by the batteries: For the current mirror transistor side: P VCC =V CC I For the amplifier transistor side: T P VCC = 1 V T CC i C1 dt where i C1 =I I c sin t 0 P VCC =V CC I 1 T I 1 c T 0 I sin t dt=v CC I Total delivered power: P D =P D av =P VCC P VCC =2V CC I i C II. Average power to the load: v O =V o peak sin t P L av = V 2 o rms = V 2 o peak / 2 = V 2 o peak 2 18
Class A Power Conversion Efficiency Using the power delivered to transistors from the batteries P Dav and the power delivered to the load P Lav : P L av = V 2 o peak P D av =2V CC I and 2 Note: 1. Average currents and P D av from the power supply do not change with the signal level V o-peak. 2. P L av increases with the square of the signal level V o-peak. power conversion efficiency = P L av = V 2 o peak /2 = 1 P D av 2V CC I 4 2 V o peak V CC I = 1 4 V o peak I V o peak V CC 19
Power Conversion Efficiency = 1 4 2 V o peak = 1 I V CC 4 V o peak I V o peak V CC Since V o peak V CC and V o peak I : Maximum power conversion efficiency is realized when V o peak =V CC =I ignoring the V CE1-sat and V CE2-sat Hence: max = 1 4 V CC V CC V CC V CC = 1 4 or 25 % 20
Class A Power Efficiency Simulation V o peak =V i peak 0.7 V o peak =12 0.7=11.3V V CC V CE sat =I 98 Ω P L av =692.36 mw P D av =1.33W 1.44 W =2.77 W = P L av P D av = 0.69W 2.77W =0.249 0.25 =98 21
Class A Power Simulation - cont. V o peak =V i peak 0.7 V o peak =6 0.7=5.3V V CC V CE sat =I 98 Ω P L av =182.28 mw P D av =1.33W 1.44 W =2.77W = P L av P D av = 0.18 W 2.77 W =0.065 0.25 =98 22
Conclusions 1. The class A amplifier provides the most nearly linear amplification of its input, but this comes at a price: The best power conversion efficiency that can be obtained is 25%. 2. That is 75% of the power supplied by the sources is dissipated in the transistors. This is a waste of power, and it leads to a potentially serious heating problems with the transistors. All of this constant battery power is dissipated in the transistors even when no signal is applied zero percent efficiency! Next we will consider a much more efficient amplifier configuration the class B amplifier. 23