Document 13436441
advertisement
Lecture 22 Frequency Response of Amplifiers (II) VOLTAGE AMPLIFIERS Outline 1. 2. 3. Full Analysis Miller Approximation Open Circuit Time Constant Reading Assignment: Howe and Sodini, Chapter 10, Sections 10.1-10.4 6.012 Spring 2009 Lecture 22 1 Common Emitter Amplifier V+ iSUP iOUT RS + + V − s RL vOUT − + V V − BIAS V− • Operating Point Analysis – vs=0, RS = 0, ro → ∞, roc → ∞, RL → ∞ – Find VBIAS such that IC=ISUP with the BJT in the forward active region 6.012 Spring 2009 Lecture 22 2 Frequency Response Analysis of the Common Emitter Amplifier Cµ RS + + Vs + − rπ Vπ ro ⎢⎢roc gmVπ Cπ − RL Vout − (a) π π π • Frequency Response – Set VBIAS = 0. π – Substitute BJT small signal model (with capacitors) including RS, RL, ro, roc – Perform impedance analysis 6.012 Spring 2009 Lecture 22 3 1. Full Analysis of CE Voltage Amplifier Cµ RS + + Vs + − Vπ rπ ro ⎢⎢roc gmVπ Cπ RL Vout − − (a) Replace voltage π π source and resistance with current source and resistance using Norton Equivalent π π π π π Cµ + + π Is R'in Vπ gmVπ Cπ R'out − − R'in = RS ⎢⎢rπ Node 1: Vout R'out = ro ⎢⎢roc ⎢⎢RL (b) Vπ + jωC π Vπ + jωCµ (Vπ − Vout ) Is = R ′in Node 2: Vout g m Vπ + = jωCµ (Vπ − Vout ) R′out 6.012 Spring 2009 Lecture 22 4 Full Frequency Response Analysis (contd.) • Re­arrange 2 and obtain an expression for Vπ • Substituting it into 1 and with some manipulation, we can obtain an expression for Vout / Is: ( ) − Rin ′ R′out gm − jωCµ Vout = Is 1 + jω R′outCµ + Rin ′ Cµ + Rin ′ Cπ + g m R′out Rin ′ Cµ − ω 2 R′out Rin ′ Cµ Cπ ( ) Changing input current source back to a voltage source: rπ Cµ ′ −gm R R′out 1 − jω R + r gm Vout S π = Vs 1 + jω (Rout ′ C µ + Rin′ C µ (1 + gm Rout ′ ) + Rin′ C π )− ω 2 Rout ′ Rin′ C µ C π where R ′in = R S || rπ and R ′out = ro || roc || R L We can ignore zero at gm/Cµ because it is higher than ωT. The gain can be expressed as: Vout Avo Avo = = Vs (1 + jωτ1 )(1 + jωτ2 ) 1 − jω(τ1 + τ2 ) − ω2 τ1τ2 where Avo is the gain at low frequency and τ1 and τ2 are the two time constants associated with the capacitors 6.012 Spring 2009 Lecture 22 5 Denominator of the System Transfer Function Vout V s dB − 20 decade Avo − 40 1/τ1 1/τ2 dB decade ω� log scale τ1 + τ2 = Rout ′ C µ + Rin ′ Cµ (1+ gm Rout ′ ) + Rin ′ C ππ τ1 • τ2 = R R′out ′ R R′in ′ Cµ C π We could solve for τ1 and τ2 but is algebraically complex. •However, if we assume that τ1 >> τ2 ,⇒ τ1 + τ2 ≈ τ1. •This is a conservative estimate since the true τ1 is actually smaller and hence the true bandwidth is actually larger than: [ ] τ1 ≈ Rin ′ Cπ + C µ (1 + g m Rout ′ ) + Rout ′ Cµ Then: ω 3dB = 6.012 Spring 2009 1 τ1 = 1 Rin′ Cπ + Cµ (1 + gm Rout ′ ) + Rout ′ Cµ Lecture 22 6 2. The Miller Approximation Effect of Cµ on the Input Impedance: It Cµ + Vt + − gmVt R'out = ro ⎢⎢roc ⎢⎢RL Vout − The input impedance Zi is determined by applying a test voltage Vt to the input and measuring It: Vout = −gm Vt Rout ′ + I t Rout ′ The Miller Approximation assumes that current through Cµ is small compared to the transconductance generator I t << g m Vt Vout ≈ −gm Vt Rout ′ We can relate Vt and Vout by It Vt − Vout = jωC µ 6.012 Spring 2009 Lecture 22 7 The Miller Approximation (contd.) After some Algebra: Vt 1 1 = Z eff = = It jω Cµ (1+ gm R′out ) jω C 1 − A µ vC ( µ ) The effect of Cµ at input is that Cµ is “Miller multiplied” by (1­AvCµ) Generalized “Miller Effect” Z + Vi − � Avo + Vi + Vo Avo Zeff − − + Vo − �Z�eff = Z /(1 − A� vo) • An impedance connected across an amplifier with voltage gain Avo can be replaced by an an impedance to ground … divided by (1­Avo) • Avo is large and negative for common­emitter and common­source amplifiers • Capacitance at input is magnified. Zeff = 6.012 Spring 2009 Z (1− Avo ) Lecture 22 8 Frequency Response of the CE Voltage Amplifier Using Miller Approximation RS + + Vs + − rπ Vπ CM C π gmVπ R'out Vout − − CM = Cµ(1 + gm R'out) • The Miller capacitance is lumped together with Cπ, which results in a single pole low pass filter at the input rπ Vout = − g m Vs rπ + RS Rout ′ 1 1 + jω (C π + C M )(RS || rπ ) • At low frequency (DC) the small signal voltage gain is rπ Vout Rout ′ = −gm Vs rπ + RS • The frequency at which the magnitude of the voltage gain is reduced by 1/√2 is 1 1 1 ω3dB = = ′ )Cµ (Rs || rπ )(Cπ + CM ) (Rs || rπ ) Cπ + (1 + gm Rout 6.012 Spring 2009 Lecture 22 9 3. Open Circuit Time Constant Analysis Assumptions: • No zeros • One “dominant” pole (1/τ1 << 1/τ2 , 1/τ3 … 1/τn ) • N capacitors Vout Avo = Vs (1 + jωτ1 )(1 + jωτ2 )(1 + jωτn ) The example shows a voltage gain; however, it could be Iout/Vs or Vout/Is. Multiplying out the denominator: Vout Avo = Vs 1 + b1( jω) + b2 ( jω)2 + ... + bn ( jω)n where b1 = τ1 + τ2 + τ3 +….+ τn It can be shown that the coefficient b1 can be found exactly [see Gray & Meyer, 3rd Edition, pp. 502­506] N N b1 = ∑ RTi Ci = ∑ τ Cio i=1 i • τCio is the open­circuit time constant for capacitor Ci • Ci is the ith capacitor and RTi is the Thevenin resistance across the ith capacitor terminals (with all capacitors open­circuited) 6.012 Spring 2009 Lecture 22 10 Open Circuit Time Constant Analysis Estimating the Dominant Pole The dominant pole of the system can be estimated by: b1 = τ1 + τ2 + τ3 +….+ τn N b1 = ∑ RTi Ci i=1 ≈ τ1 = 1 ω1 RTiCi is the open­circuit time constant for capacitor Ci Power of the Technique: • Estimates the contribution of each capacitor to the dominant pole frequency separately • Enables the designer to understand what part of a complicated circuit is responsible for limiting the bandwidth of amplifier • The approximate magnitude of the Bode Plot is 6.012 Spring 2009 Lecture 22 11 Common Emitter Amplifier Analysis Using OCT Cµ RS + + Vs + − Vπ rπ gmVπ Cπ ro ⎢⎢roc RL − Vout − (a) From the Full Analysis rπ Cµ 1− π jω ′ π −g mRπ out R + r gm S π Vout = Vs 1+ jω Rout ′ Cµ + Rin ′ Cµ (1+ gm Rout ′ )+ Rin ′ Cπ − ω2 Rout ′ Rin ′ CµCπ ( ) π where R ′in = RS || rπ and R ′out = ro || roc || RL b1 = Rout ′ Cµ + Rin ′ C µ (1 + g m Rout ′ ) + Rin ′ Cπ ω3dB ≈ 6.012 Spring 2009 1 1 = b1 Rout ′ C µ + Rin ′ Cµ (1+ +g m Rout ′ ) + Rin ′ Cπ Lecture 22 12 Common Emitter Amplifier Analysis Using OCT—Procedure 1. 2. 3. Eliminate all independent sources [e.g. Vs → 0] Open­circuit all capacitors Find the Thevenin resistance by applying it and measuring vt. Time Constant for Cπ Result obtained by inspection RT π = RS || rπ τ C = RT πC π πo 6.012 Spring 2009 Lecture 22 13 Common Emitter Amplifier Analysis Using OCT—Time Constant for Cµ Using the same procedure Let R ′in = RS || rπ and R ′out = ro || roc || RL vπ −it = Rin ′ vt + v π it = + g m vπ Rout ′ Eliminate vπ: vt = RTµ = Rout ′ + Rin ′ (1 + g m Rout ′ ) it τ Cµo = RT µ Cµ = [Rout ′ + Rin ′ (1 + g m Rout ′ )]Cµ 6.012 Spring 2009 Lecture 22 14 Common Emitter Amplifier Analysis Using OCT—Dominant Pole Summing individual time constants b1 = RT πC π + RTµ Cµ b1 = Rout ′ Cµ + Rin ′ C µ (1 + g m Rout ′ ) + Rin ′ Cπ Assume τ1 >> τ2 b1 = τ1 + τ2 ≈ τ1 b1 = Rout ′ Cµ + Rin ′ C µ (1 + g m Rout ′ ) + Rin ′ Cπ ω3dB ≈ 1 1 = b1 Rout ′ C µ + Rin ′ Cµ (1+ gm Rout ′ ) + Rin ′ Cπ This result is very similar to the Miller Effect calculation Additional term R’outCµ taken into account 6.012 Spring 2009 Lecture 22 15 Compare the Three Methods of Analyzing the Frequency Response of CE Amplifier Full Analysis— ω3dB ≈ 1 τ1 = 1 Rout ′ Cµ + Rin ′ Cµ (1 + gm Rout ′ ) + Rin ′ Cπ Miller Approximation— 1 1 ω3dB = ′ Cπ + (1 + gm Rout ′ )Cµ Rin Open Circuit Time Constant— ω3dB ≈ 6.012 Spring 2009 1 1 = b1 Rout ′ C µ + Rin ′ Cµ (1+ gm Rout ′ ) + Rin ′ Cπ Lecture 22 16 What did we learn today? Summary of Key Concepts • Full Analysis – Assumes that τ1 + τ2 ≈ τ1 ω3dB ≈ 1 τ1 = 1 R′outCµ + Rin ′ Cµ (1 + gm R′out ) + Rin ′ Cπ • Miller Approximation – Does not take into account R’out 1 1 ω3dB = ′ Cπ + (1 + gm Rout ′ )Cµ Rin • Open Circuit Time Constant (OCT) – Assumes a dominant pole as full analysis ω3dB ≈ 6.012 Spring 2009 1 1 = b1 Rout ′ C µ + Rin ′ Cµ (1+ gm Rout ′ ) + Rin ′ Cπ Lecture 22 17 MIT OpenCourseWare http://ocw.mit.edu 6.012 Microelectronic Devices and Circuits Spring 2009 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.
