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ESE319 Introduction to Microelectronics 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 2008 Kenneth R. Laker updated 29Nov10 1 ESE319 Introduction to Microelectronics 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 iC waveform due to input vI waveform 2008 Kenneth R. Laker updated 29Nov10 2 ESE319 Introduction to Microelectronics Amplifier Classifications VCC iC RC RB vI VB Class A amplifier – amplifier BJT conducts for entire vI cycle. For all vI: v I V B ≥ 0.7 V where V B max v I 0.7 V Ic = amplitude of current due to vi. RE IC = dc current Transistor cut off (iC = 0) if: v I V B 0.7 V 2008 Kenneth R. Laker updated 29Nov10 NOTE: when vI = 0, iC = IC 3 ESE319 Introduction to Microelectronics Amplifier Classifications - cont. Class B – Amplifier BJT conducts positive-half of vI cycle. Amp BJT conducts for all vI s.t.: V B =0 V ⇒ v I ≥0.7 V Transistor cut off (iC = 0) if: v I V B 0.7 V 2008 Kenneth R. Laker updated 29Nov10 NOTE: 1. when vI < 0.7V, iC = 0 2. a 2nd class B BJT is needed to conduct for the negative vI cycle. 4 ESE319 Introduction to Microelectronics Amplifier Classifications - cont. Class AB – Amplifier BJT conducts for positive vI swing + part of negative vI swing s.t.: v I V B ≥ 0.7 V where 0V B max v I 0.7 V Conducts for: v I ≥0.7−V B Cut-off for rest of negative vI swing: Transistor cut-off (iC = 0) if: v I V B 0.7 V 2008 Kenneth R. Laker updated 29Nov10 NOTE: 1. when vI = 0, iC = IC 2. a 2nd class AB BJT is needed to conduct for interval slightly larger than the negative vI cycle. 5 ESE319 Introduction to Microelectronics 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. 2008 Kenneth R. Laker updated 29Nov10 6 ESE319 Introduction to Microelectronics Current Biased Class A Emitter Follower +VCC Q3 diode connected transistor I -VCC -VCC current mirror 2008 Kenneth R. Laker updated 29Nov10 A current mirror establishes the bias current. To operate reliably, 1. Q1 and Q2 must be forward active. 2. Current Mirror Q3 and Q2 need to be matched as well as possible and be at the same ambient temperature. 7 ESE319 Introduction to Microelectronics Class A Amplifier Analysis Consider the case when vI ≥ vBE1 = 0.7 V (pos. swing of vI): if vCE1 < VCE1-sat => Q1 sat. + vCE1 - i L =i E1 −I vO =v I −0.7V =i L R L vCE1=V CC −v O For Q1 ≠ sat: v CE1V 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 vCE2 =v OV CC -VCC -VCC v I ≥v BE1=0.7V vO =v I −v BE1=v I −0.7 V vCE1=V CC −v O 2008 Kenneth R. Laker updated 29Nov10 Max values: Q1 ≠ sat. v O vO−max =V CC −V CE1−sat v I v I −max=V CC −V CE1− sat 0.7V 8 ESE319 Introduction to Microelectronics Class A Amplifier Analysis - cont. Consider the case where vI < 0.7 V (neg. swing of vI): vCE1=V CC −v O i L =i E1 −I if iE1 = 0 => Q1 off i E1=i L I 0 v CE2V CE2−sat vO v CE2=v O V CC V CE2−sat i L = −I RL ⇒ v O −V CC V CE2− sat v O −I R L v I −V CC V CE2− sat 0.7 v I −I R L 0.7 V Min values: Q1 and Q2 forward-active vO =v I −0.7V =i L R L vCE2 =v OV CC + vCE2 -VCC -VCC if vCE2 < VCE2-sat => Q2 sat. v I 0.7 V v O v O−min =max {−IR L ,−V CC V CE2−sat } vO =v I −0.7 & vO =i L R L v I v I −min=max {−IR L0.7 V ,−V CC V CE2−sat 0.7 V } vCE2 =v OV CC =v I −0.7V CC 2008 Kenneth R. Laker updated 29Nov10 NOTE: max means least negative 9 ESE319 Introduction to Microelectronics Class A Amplifier VTC – Plot vO Q1 saturated VCC - VCE1-sat max vO swing iff RL ≥ or Iff −IR L−V CC −V CE2−sat V CC −V CE2−sat I V CC −V CE2−sat I≥ RL Q1 cutoff Q2 saturated Bias current I & RL set limits on negative vO = vO-min swing vO =v I −V BE1 slope = 1 VBE1 -IRL vI V CC −V CE2− sat R L I -VCC + VCE2-sat v O v O−min=−IR L ⇒ V CC −V CE2−sat R L I If −IR L−V CC −V CE2−sat v O v O−min=−V CC V CE2−sat ⇒ V CC −V CE2−sat R L I v O =v I −0.7 where max {−IR L ,−V CC −V CE2−sat }v O V CC −V CE1−sat 2008 Kenneth R. Laker updated 29Nov10 10 ESE319 Introduction to Microelectronics Class A Stage VTC Simulation I =120 mA IR L =V CC =12 V ⇒ R L =100 11.8 V Q1 cutoff ideal current source i.e. no Q2 Q1 saturated - 12 V VEE = - VCC Q1 forward active => no clipping v O v O−max =V CC −V CE1−sat =11.8V Q1 not Sat v O v O−min =−IR L=−12 V Q1 not Cut-off 2008 Kenneth R. Laker updated 29Nov10 11 ESE319 Introduction to Microelectronics Class A Stage VTC Simulation - cont. I =120 mA R L =75 IR L =9V V CC −V CE1−sat Q1 saturated 11.8 V Q1 cutoff -9V 2008 Kenneth R. Laker updated 29Nov10 12 ESE319 Introduction to Microelectronics Quick Review Class A Amp VTC vCE1=V CC −v O max vo swing i.f.f. or I≥ V CC −V CE2−sat RL v O =v I −0.7 where max {−IR L ,−V CC −V CE2−sat }v O V CC −V CE1−sat 2008 Kenneth R. Laker updated 29Nov10 13 ESE319 Introduction to Microelectronics +VCC = +15 V Example I R L =1 k -VCC = -15 V 2008 Kenneth R. Laker updated 29Nov10 Let VCE1-sat = VCE2-sat = 0.2 V, VBE1 = VBE2 = 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 vO swing. 2. Determine the resulting output voltage swing and the maximum and minimum Q1 emitter currents. 14 ESE319 Introduction to Microelectronics Example cont. I SOLUTION: R L =1 k -VCC 1. For maximum output voltage swing: IR L =V CC −V CE2−sat where RL = 1 kΩ V CC −V CEsat 15 V −0.2V I= = =14.8 mA RL 1k R= 2008 Kenneth R. Laker updated 29Nov10 V CC −V BE 15 V −0.7V = =0.97 k I 14.8 mA 15 ESE319 Introduction to Microelectronics Example - cont. From Part 1: I =14.8 mA SOLUTION: 2. Output voltage swing: V o− peak = I R L =14.8 V ⇒−14. 8V v O 14.8 V -VCC = -15 V −14.8 V v O 14. 8V ⇒− I i L I 0−V BE2−V CC I= R Max and min Q1 emitter currents: 2008 Kenneth R. Laker updated 29Nov10 i E1= I i L ⇒ 0 mAi E1 2I=29.6 mA 16 ESE319 Introduction to Microelectronics Instantaneous and Average Power The source of power to the amplifier load, RL, comes from the supplies,VCC and – VCC. The supplies deliver power, and the load and the transistors absorb it. Instantaneous power absorbed by resistor RL: 2 2 pa t =v ab t ia b t =iab t R L =va b t / R L i ab RL Instantaneous power delivered by battery VCC: pd t=v ab t i ba t=V CC I Average power: i ab =I ab− peak sin t for period T T 2 2 I V 1 ab− peak ab− peak P ab av = ∫ i 2ab t R L dt =I 2ab−rms R L= R L= T 0 2 2 RL P D av =V CC I 2008 Kenneth R. Laker updated 29Nov10 i ba = I V CC 17 ESE319 Introduction to Microelectronics 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 1 PVCC = ∫ V CC i C1 dt where iC1 =I Ic sin t T 0 T Ic 1 P VCC =V CC I ∫ 1 sin t dt=V CC I T 0 I Total delivered power: iC -VCC -VCC P D =P D av =P−VCC PVCC =2 V CC I II. Average power to the load: 2 2 2 V V / 2 V o−rms o− peak o− peak v O =V o− peak sin t P L av = = = RL RL 2 RL 2008 Kenneth R. Laker updated 29Nov10 18 ESE319 Introduction to Microelectronics Class A Power Conversion Efficiency Using the power delivered to transistors from the batteries PDav and the power delivered to the load PLav: P D av =2V CC I and V 2o− peak P L av = 2 RL Note: 1. Average currents and PD av from the power supply do not change with the signal level Vo-peak. 2. PL av increases with the square of the signal level Vo-peak. power conversion efficiency 2 2 P L av V o− /2 R V 1 o− peak 1 V o− peak V o− peak peak L = = = = P D av 2V CC I 4 V CC I R L 4 I R L V CC 2008 Kenneth R. Laker updated 29Nov10 19 ESE319 Introduction to Microelectronics Power Conversion Efficiency 2 V 1 o− peak 1 V o− peak V o− peak = = 4 I R L V CC 4 I R L V CC Since V o− peak V CC and V o− peak I R L : Maximum power conversion efficiency is realized when V o− peak =V CC =I R L Hence: ignoring the VCE1-sat and VCE2-sat 1 V CC V CC 1 max = = or 25 % 4 V CC V CC 4 2008 Kenneth R. Laker updated 29Nov10 20 ESE319 Introduction to Microelectronics 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 R L 98 Ω P L av =692.36 mW P D av=1.33W 1.44 W =2.77 W P L av 0.69 W = = =0.249≈0.25 P D av 2.77 W 2008 Kenneth R. Laker updated 29Nov10 R L =98 21 ESE319 Introduction to Microelectronics Class A Power Simulation - cont. V o− peak =V i− peak −0.7 V o− peak =6−0.7=5.3 V V CC −V CE −sat = I R L 98 Ω P L av =182.28 mW P D av=1.33W 1.44 W =2.77 W P L av 0.18 W = = =0.0650.25 P D av 2.77 W 2008 Kenneth R. Laker updated 29Nov10 R L =98 22 ESE319 Introduction to Microelectronics 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. 2008 Kenneth R. Laker updated 29Nov10 23