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Lecture23-Amplifier Frequency Response.Pptx EE105 – Fall 2014 Microelectronic Devices and Circuits Prof. Ming C. Wu [email protected] 511 Sutardja Dai Hall (SDH) Lecture23-Amplifier Frequency Response 1 Common-Emitter Amplifier – ωH Open-Circuit Time Constant (OCTC) Method At high frequencies, impedances of coupling and bypass capacitors are small enough to be considered short circuits. Open-circuit time constants associated with impedances of device capacitances are considered instead. 1 ωH ≅ m ∑RioCi i=1 where Rio is resistance at terminals of ith capacitor C with all other v ! R $ i R = x = r #1+ g R + L & capacitors open-circuited. µ 0 π 0 m L ix " rπ 0 % For a C-E amplifier, assuming C = 0 L 1 1 ωH ≅ = Rπ 0 = rπ 0 Rπ 0Cπ + Rµ 0Cµ rπ 0CT Lecture23-Amplifier Frequency Response 2 1 Common-Emitter Amplifier High Frequency Response - Miller Effect • First, find the simplified small -signal model of the C-E amp. • Replace coupling and bypass capacitors with short circuits • Insert the high frequency small -signal model for the transistor ! # rπ 0 = rπ "rx +(RB RI )$ Lecture23-Amplifier Frequency Response 3 Common-Emitter Amplifier – ωH High Frequency Response - Miller Effect (cont.) v R r Input gain is found as A = b = in ⋅ π i v R R r r i I + in x + π R || R || (r + r ) r = 1 2 x π ⋅ π RI + R1 || R2 || (rx + rπ ) rx + rπ Terminal gain is vc Abc = = −gm (ro || RC || R3 ) ≅ −gm RL vb Using the Miller effect, we find CeqB = Cµ (1− Abc )+Cπ (1− Abe ) the equivalent capacitance at the base as: = Cµ (1−[−gm RL ])+Cπ (1− 0) = Cµ (1+ gm RL )+Cπ Chap 17-4 Lecture23-Amplifier Frequency Response 4 2 Common-Emitter Amplifier – ωH High Frequency Response - Miller Effect (cont.) ! # • The total equivalent resistance ReqB = rπ 0 = rπ "rx +(RB RI )$ at the base is • The total capacitance and CeqC = Cµ +CL resistance at the collector are ReqC = ro RC R3 = RL • Because of interaction through 1 ω = Cµ, the two RC time constants p1 ! # rπ 0 "Cπ +Cµ (1+ gm RL )$+ RL (Cµ +CL ) interact, giving rise to a dominant pole. 1 ω p1 = where rπ 0CT RL CT = Cπ +Cµ (1+ gm RL ) + (Cµ +CL ) rπ 0 ! # rπ 0 = rπ "rx +(RB RI )$ Lecture23-Amplifier Frequency Response 5 Common-Source Amplifier – ωH Open-Circuit Time Constants Analysis similar to the C-E case yields the following equations: Rth = RG RI RL = RD R3 ro RG vth = vi RI + RG RL CT = CGS +CGD (1+ gm RL ) + (CGD +CL ) Rth 1 gm ωP1 = ωP2 = RthCT CGS +CL gm ωZ = CGD Lecture23-Amplifier Frequency Response 6 3 C-S Amplifier High Frequency Response Source Degeneration Resistance First, find the simplified small -signal model of the C-S amp. Recall that we can define an effective gm to account for the unbypassed source resistance. gm g'm = 1+ gm RS Lecture23-Amplifier Frequency Response 7 C-S Amplifier High Frequency Response Source Degeneration Resistance (cont.) v R R || R Input gain is found as g G 1 2 Ai = = = vi Ri + RG Ri + R1 || R2 vd ' −gm (RD || R3 ) Terminal gain is Agd = = −gm (RiD || RD || R3 ) ≅ vg 1+ gm RS Again, we use the Miller effect CeqG = CGD (1− Agd )+CGS (1− Ags ) to find the equivalent " % capacitance at the gate as: (−gm RL ) gm RS = CGD $1− '+CGS (1− ) # 1+ gm RS & 1+ gm RS " % gm (RD || R3 ) CGS = CGD $1+ '+ # 1+ gm RS & 1+ gm RS Lecture23-Amplifier Frequency Response 8 4 C-S Amplifier High Frequency Response Source Degeneration Resistance (cont.) The total equivalent resistance ReqG = RG RI = Rth at the gate is The total capacitance and C = C +C resistance at the drain are eqD GD L R = R R R ≅ R R = R Because of interaction through eqD iD D 3 D 3 L CGD, the two RC time constants interact, giving rise to the dominant pole: 1 ω p1 = CGS gm RL RL Rth[ +CGD (1+ )+ (CGD +CL )] 1+ gm RS 1+ gm RS Rth And from previous analysis: ' ' gm gm +gm +gm ω p2 = = ωz = = (CGS +CL ) (1+ gm RS )(CGS +CL ) CGD (1+ gm RS )(CGD ) Lecture23-Amplifier Frequency Response 9 C-E Amplifier High Frequency Response Emitter Degeneration Resistance Analysis similar to the C-S case yields the following equations: ! # Rπ 0 = Rth "rπ +(βo +1) RE $ Rth = RI RB = RI R1 R2 RL = RiC RC R3 ≅ RC R3 1 ωP1 = Rπ 0CT 1 ωP1 = Cπ gm RL RL Rπ 0[ +Cµ (1+ )+ (Cµ +CL )] 1+ gm RE 1+ gm RE Rπ 0 ' gm gm ωP2 = = (Cπ +CL ) (1+ gm RE )(Cπ +CL ) ' +gm +gm ωz = = Cµ (1+ gm RE )(Cµ ) Lecture23-Amplifier Frequency Response 10 5 Gain-Bandwidth Trade-offs Using Source/Emitter Degeneration Resistors Adding source resistance to the C-S (or C-E) amp causes vd gm (RD || R3 ) gain to decrease and dominant Agd = = − pole frequency to increase. vg 1+ gm RS 1 ω p1 = CGS gm RL RL Rth[ +CGD (1+ )+ (CGD +CL )] 1+ gm RS 1+ gm RS Rth However, decreasing the gain gm also decreased the frequency ω p2 = of the second pole. (1+ gm RS )(CGS +CL ) Increasing the gain of the C-E/C-S stage causes pole -splitting, or increase of the difference in frequency between the first and second poles. Lecture23-Amplifier Frequency Response 11 High Frequency Poles Common-Base Amplifier 1 gm 1 Ai ≅ = (1 gm ) + RI 1+ gm RI Aec = gm (RiC RL ) ≅ gm RL " $ RiC = ro #1+ gm (rπ RE RI )% Since Cµ does not couple input and output, input and output poles can be found directly. C C eqE = π CeqC = Cµ +CL 1 R = R R R = R R ReqC = RiC || RL ≅ RL eqE iE E I g E I m 1 1 −1 ω = ≅ (" 1 % + g p2 m (RiC || RL )(C +CL ) RL (C +CL ) ωP1 ≅ *$ RE RI 'Cπ - ≅ > ωT µ µ )# gm & , Cπ Lecture23-Amplifier Frequency Response 12 6 High Frequency Poles Common-Gate Amplifier Similar to the C-B, since CGD does not couple the input and output, input and output poles can be found directly. C C eqS = GS CeqD = CGD +CL 1 R = r !1+ g R || R # ReqS = || R4 || RI iD o " m ( 4 I )$ gm ReqD = RiD || RL ≅ RL 1 gm ω p1 = ≅ 1 1 ! 1 $ C ω = ≅ || R || R C GS p2 # 4 I & GS (RiD || RL )(CGD +CL ) RL (CGD +CL ) " gm % Lecture23-Amplifier Frequency Response 13 High Frequency Poles Common-Collector Amplifier ve gm RL Abe = = vb 1+ gm RL " % v R gm RL b in C = C (1− A )+C (1− A ) = C (1− 0)+C 1− Ai = = eqB µ bc π be µ π $ ' v R R # 1+ gm RL & i i + in € Cπ CeqB = Cµ + 1+ gm RL CeqE = Cπ +CL € ReqB = (Rth + rx ) || RiB = [(RI || RB )+ rx ]||[rπ + (βo +1)RL ] = (Rth + rx ) ||[rπ + (β +1)RL ] " % 1 (Rth + rx ) ReqE = RiE || RL ≅ $ + ' || RL # gm βo +1 & Lecture23-Amplifier Frequency Response 14 7 High Frequency Poles Common-Collector Amplifier (cont.) The low impedance at the output makes the input and output time constants relatively The feed-forward high-frequency well decoupled, leading to two poles. path through Cπ leads to a zero 1 in the C-C response. Both the ω p1 = zero and the second pole are ! C $ π quite high frequency and are (Rth + rx ) || [rπ + (βo +1)RL ]#Cµ + & " 1+ gm RL % often neglected, although their 1 effect can be significant with ω p2 ≅ large load capacitances. (! $ + 1 Rth + rx *# + & || RL -(Cπ +CL ) g 1 )" m β + % , gm ωz ≅ Cπ Lecture23-Amplifier Frequency Response 15 High Frequency Poles Common-Drain Amplifier Similar the the C-C 1 g m amplifier, the C-D high ω p1 = ωz ≅ CGS C R (C + ) GS frequency response is th GD 1+ g R dominated by the first pole m L 1 1 due to the low impedance at ω = ≅ the output of the C-C p2 " % (RiS || RL )(CGS +CL ) 1 amplifier. $ || RL '(CGS +CL ) # gm & Lecture23-Amplifier Frequency Response 16 8 Frequency Response Cascode Amplifier There are two important poles: the input pole for the C-E and the output pole for the C-B stage. The intermediate node pole can usually be neglected because of the low impedance at the input of the C-B stage. RL1 is small, so the second term in the first pole can be neglected. Also note the RL1 is equal to 1/gm2. 1 1 ω = = pB1 g R R rπ 0CT m1 L1 L1 rπ 01([Cµ1(1+ )+Cπ1]+ [Cµ1 +CL1]) 1+ gm1RE1 rπ 0 1 1 ω = ≅ pB1 g 1/ g m1 m2 rπ 01(2Cµ1 +Cπ1) rπ 01([Cµ1(1+ )+Cπ1]+ [Cµ1 +Cπ 2 ]) gm2 rπ 01 1 ω pC2 ≅ RL (Cµ 2 +CL ) Lecture23-Amplifier Frequency Response 17 Frequency Response of Multistage Amplifier • Problem: Use open-circuit and short-circuit time constant methods to estimate upper and lower cutoff frequencies and bandwidth. • Approach: Coupling and bypass capacitors determine the low -frequency response; device capacitances affect the high -frequency response • At high frequencies, ac model for the multi-stage amplifier is as shown. Lecture23-Amplifier Frequency Response 18 9 Frequency Response Multistage Amplifier Parameters (example) Parameters and operation point information for the example multistage amplifier. Lecture23-Amplifier Frequency Response 19 Frequency Response Multistage Amplifier: High-Frequency Poles High-frequency pole at the gate of M1: Using our equation for the C-S input pole: 1 1 f = P1 ! $ 2π RL1 Rth1 #CGD1 (1+ gm RL1) +CGS1 + (CGD1 +CL1)& " Rth1 % RL1 = RI12 (rx2 + rπ 2 ) ro1 = 598Ω (250Ω+ 2.39kΩ) 12.2kΩ = 469 Ω CL1 = Cπ 2 +Cµ 2 (1+ gm2 RL2 ) R R R r R r r 1 R R r 3.33 k L2 = I 23 in3 o2 = I 23 ( x3 + π 3 +(βo3 + )( E3 L )) o2 = Ω ! $ CL1 = Cπ 2 +Cµ 2 (1+ gm2 RL2 ) = 39pF +1pF"1+ 67.8mS(3.33kΩ)% = 266 pF 1 1 f = = 689 kHz P1 2π ! 469Ω $ 9.9kΩ 1pF"!1+ 0.01S(3.33kΩ)%$+ 5pF + (1pF + 266pF) "# 9.9kΩ %& Lecture23-Amplifier Frequency Response 20 10 Frequency Response Multistage Amplifier: High-Freq.
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