Lecture 040 CE and CS Output Stages (1/11/04) Page ECE Analog Integrated Circuits and Systems II P.E. Allen

Transcription

1 Lecture 040 CE and CS Output Stages (1/11/04) Page 0401 LECTURE 040 COMMON SOURCE AND EMITTER OUTPUT STAGES (READING: GHLM 8498, AH 181) Objective The objective of this presentation is: Show how to design stages that 1.) Provide sufficient output power in the form of voltage or current..) Avoid signal distortion..) Be efficient 4.) Provide protection from abnormal conditions (short circuit, over temperature, etc.) Outline Common source stage Common emitter stage Summary Lecture 040 CE and CS Output Stages (1/11/04) Page 040 COMMON SOURCE OUTPUT STAGE Current source load inverter V GG IQ i D1 M i OUT M1C L I Q I Q i D dominates as the load line I Q Fig Cutoff Triode A Class A circuit has current flow in the MOSFETs during the entire period of a sinusoidal signal. Characteristics of Class A amplifiers: Unsymmetrical sinking and sourcing Linear Poor efficiency (peak) (peak) P RL v Efficiency = P = Supply ( )I = Q ( ) (V OUT (peak) DD ) = Maximum efficiency occurs when (peak) = = which gives 5%.

2 Lecture 040 CE and CS Output Stages (1/11/04) Page 040 Specifying the Performance of a Common Source Amplifier Output resistance: 1 1 r out = g ds1 g ds = (λ1λ)i D Current: Maximum sinking current is, IŌUT = K 1W 1 L 1 ( V T1 ) I Q Maximum sourcing current is, I OUT = K W L ( V GG V T ) I Q Requirements: Want r out << I OUT > C L SR I OUT > (peak) The maximum current will be determined by both the current required to provide the necessary slew rate (C L ) and the current required to provide a voltage across the load resistor ( ). Lecture 040 CE and CS Output Stages (1/11/04) Page 0404 SmallSignal Performance of the Class A Amplifier Although we have considered the smallsignal performance of the Class A amplifier as the current source load inverter, let us include the influence of the load. The modified smallsignal model: C 1 v in g m1 v in rds1 r ds The smallsignal voltage gain is: v out g m1 v in = g ds1 g ds G L The smallsignal frequency response includes: A zero at z = g m1 C gd1 and a pole at (g ds1 g ds G L ) p = C gd1 C gd C bd1 C bd C L C vout Fig. 0400

3 Lecture 040 CE and CS Output Stages (1/11/04) Page 0405 Example 5.51 Design of a Simple ClassA Output Stage Use the values of K N =110µA/V, K P =50µA/V, V TN =0.7V and V TP =0.7V and design the W/L ratios of M1 and M so that a voltage swing of ± volts and a slew rate of 1 volt/µs is achieved if = 0 kω and C L = 1000 pf. Assume that = = volts and V GG = 0 volts. Let the channel lengths be µm and assume that C gd1 = 100fF. Solution Let us first consider the effects of and C L. i OUT (peak) = ±V 0kΩ = ±100µA and C L SR = = 1000µA Since the slew rate current >> the current for, we can safely assume that all of the current supplied by the inverter is available to charge C L. Using a value of ±1 ma, W 1 (I OUT I Q ) L 1 = K N ( V TN ) = 4000 µm 110 (5.) µm and W I OUT L = K P ( V GG V TP ) = µm 50 (.) µm The smallsignal performance of this amplifier is, A v = 8.1 V/V (includes = 0kΩ) Lecture 040 CE and CS Output Stages (1/11/04) Page 0406 Broadband Harmonic Distortion The linearity of an amplifier can be characterized by its influence on a pure sinusoidal input signal. Assume the input is, V in (ω) = V p sin(ωt) The output of an amplifier with distortion will be V out (ω) = a 1 V p sin (ωt) a V p sin (ωt)... an V p sin(nωt) Harmonic distortion (HD) for the ith harmonic can be defined as the ratio of the magnitude of the ith harmonic to the magnitude of the fundamental. For example, secondharmonic distortion would be given as HD a = a 1 Total harmonic distortion (THD) is defined as the square root of the ratio of the sum of all of the second and higher harmonics to the magnitude of the first or fundamental harmonic. Thus, THD can be expressed as THD = [a a... a n]1/ a 1 The distortion of the class A amplifier is good for small signals and becomes poor at maximum output swings because of the nonlinearity of the voltage transfer curve for largesignal swing.

4 Lecture 040 CE and CS Output Stages (1/11/04) Page 0407 COMMON EMITTER OUTPUT STAGE Common Emitter Class A Output Stage I R1 Q1 V CC Q Large signal characteristic: I Q i OUT ic1 Q1 V EE i OUT = I Q i C1, = i OUT, v IN = I s1 exp V I t Q R Q saturates L1 Q1 cutoff 0 <1 V EE V CE1 (sat) V CC V CE (sat) V BE1 v IN and i C1 = I s1 exp I Q Fig Lecture 040 CE and CS Output Stages (1/11/04) Page 0408 Harmonic Distortion in the Common Emitter Output Stage Assume the input signal is = V BE1 v in Substituting this in the expression on the last slide gives, v IN V BE1 v in v in = I s1 exp I Q = R L I s1 exp exp I Q = I Q exp 1 Using the expansion of exp(x) 1 x x/ x/6 gives v in = I Q 1 v in 1 v in 6 = a1 v in a v in a v in where a 1 = I Q, a = I Q and a = I Q 6 Suppose v in (t) = V p sinωt, then (t) = a 1 V p sinωt a V p sin ωt a V p sin ωt = a 1 V p sinωt a V p (1cosωt) a V p 4 (sinωt sinωt) HD = a V p 1 a 1 V = a V p p a = V p 1 4 V and HD t = a V p 4 For V p = 0.5, HD = 1.5% and HD 1% 1 a 1 V = a V p p 4a 1 = 1 V p 4

5 Lecture 040 CE and CS Output Stages (1/11/04) Page 0409 Small Signal Performance of the Common Emitter Output Stage Let r o1 r o = r o, then R in i in iout R out B C v in r π g m v in ro1 r o v out E E Fig R in = r π1 = β ο g m1, R out = r o r o R L, v out v in = g m1r o r o R L g m1 and i out i in = β o r o r o If V out (peak) = 0.6V, = 1kΩ and I Q = 1.86mA, then A v g m1 = I C Vt = = 70.6V/V Vp = A v = = 8.5mV (peak) HD = = 0.08 and HD = = Where does the distortion come from? The ac gain at the negative peak output voltage is = 94.6V/V The ac gain at the positive peak output voltage is = 48.5V/V Note the emitter follower is much more linear because of the inherent negative feedback. Lecture 040 CE and CS Output Stages (1/11/04) Page SUMMARY Requirements of Output Stages The objectives are to provide output power in form of voltage and/or current. In addition, the output amplifier should be linear and be efficient. Low output resistance is required to provide power efficiently to a small load resistance. High source/sink currents are required to provide sufficient output voltage rate due to large load capacitances. Types of output stages considered: Common emitter and common source Maximum efficiency is 5% Secondharmonic distortion can be significant