Lecture 010 4430 Review I (12/29/01) Page 0101 LTUR 010 4430 RVIW I (RAIN: HLM hap. 1) Objective The objective of this presentation is: 1.) Identify the prerequisite material as taught in 4430 2.) Insure that the students of 6412 are adequately prepared Outline Models for Integratedircuit Active evices ipolar, MO, and imo I Technology ingletransistor and MultipleTransistor Amplifiers Transistor urrent ources and Active Loads
Lecture 010 4430 Review I (12/29/01) Page 0102 MOL FOR INTRATIRUIT ATIV VI PN Junctions tep Junction arrier potentialψ o = kt q ln N A N n i 2 = V t ln N A N n i 2 = U T ln N A N n i 2 epletion region widths W 1 = W 2 = 2ε si (ψ o v )N qn A (N A N ) 2ε si (ψ o v )N A qn (N A N ) W 1 N epletion capacitance j j = A ε si qn A N 2(N A N ) 1 ψ o v = j0 1 v ψ o Fig. 01001 j0 0 ψ 0 v
Lecture 010 4430 Review I (12/29/01) Page 0103 PNJunctions raded Junction raded junction: N Above expressions become: epletion region widths W 1 = 2ε si (ψ o v )N qn (N A N ) m 1 W 2 = 2ε si (ψ o v )N A W qn (N A N ) m N m epletion capacitance j = A ε si qn A N 1 2(N A N ) m j0 ψ o v m = 1 v where 0.33 m 0.5. ψ o m 0 x N A j j0 3 2.5 2 1.5 Fig. 01002 1 m=0.333 0.5 m=0.5 0 10 8 6 4 2 0 2 v φ 0 Fig. 01003
Lecture 010 4430 Review I (12/29/01) Page 0104 Large ignal Model for the JT in the Forward Active Region Largesignal model for a npn transistor: v i β F i i = Is v exp β F Vt Largesignal model for a pnp transistor: Assumes v is a constant and i is determined externally V (on) β F i Fig.01004 v arly Voltage: i β F i i = Is exp Vt β F v Assumes v is a constant and i is determined externally i V (on) Modified large signal model becomes i = I 1 v V A β F i exp v V t Fig.01005
Lecture 010 4430 Review I (12/29/01) Page 0105 The bersmoll quations The reciprocity condition allows us to write, α F I F = α R I R = I ubstituting into a previous form of the bersmoll equations gives, i = I exp v V t 1 I α R exp v V t 1 and i = I α F exp v V t 1 I exp v V t 1 These equations are valid for all four regions of operation of the JT. Also: ependence of β F as a function of collector current The temperature coefficient of β F is, T F = 1 β F βf Τ 7000ppm/
Lecture 010 4430 Review I (12/29/01) Page 0106 imple mall ignal JT Model Implementing the above relationships, i c = g m v i g o v ce, and v i = r π i b, into a schematic model gives, i b i c v i rπ g m v i ro v ce Fig. 01006 Note that the small signal model is the same for either a npn or a pnp JT. xample: Find the small signal input resistance, R in, the output resistance, R out, and the voltage gain of the common emitter JT if the JT is unloaded (R L = ), v out /v in, the dc collector current is 1mA, the arly voltage is 100V, and β ο = 100 at room temperature. g m = I Vt = 1mA 26mV = 1 26 mhos or iemans R R out = r o = V A I = 100V 1mA = 100kΩ v out v in in = r π = β o g m = 100 26 = 2.6kΩ = g m r o = 26m 100kΩ = 2600V/V
Lecture 010 4430 Review I (12/29/01) Page 0107 omplete mall ignal JT Model r µ r b ' µ r c π r π v 1 g m v 1 r o cs r ex The capacitance, π, consists of the sum of je and b. π = je b Fig. 01007
Lecture 010 4430 Review I (12/29/01) Page 0108 xample 1 erive the complete small signal equivalent circuit for a JT at I = 1mA, V = 3V, and V = 5V. The device parameters are je0 = 10fF, n e = 0.5, ψ 0e = 0.9V, µ0 = 10fF, n c = 0.3, ψ 0c = 0.5V, cs0 = 20fF, n s = 0.3, ψ 0s = 0.65V, β o = 100, τ F = 10ps, V A = 20V, r b = 300Ω, r c = 50Ω, r ex = 5Ω, and r µ = 10β o r o. olution ecause je is difficult to determine and usually an insignificant part of π, let us approximate it as 2je0. je = 20fF µ = µ0 1 V = ne ψ 0c 10fF 1 3 0.5 0.3 = 5.6fF and cs = cs0 1 V = ns ψ 0s g m = I Vt = 1mA 26mV = 38mA/V b = τ F g m = (10ps)(38mA/V) = 0.38pF π = b je = 0.38pF 0.02pF = 0.4pF 20fF 5 = 10.5fF 0.3 1 0.65 r π = β o g m = 100 26Ω = 2.6kΩ, r o = V A I = 20V 1mA = 20kΩ and r µ = 10β ο r o = 20MΩ
Lecture 010 4430 Review I (12/29/01) Page 0109 Transition Frequency, f T f T is the frequency where the magnitude of the shortcircuit, commonemitter current =1. ircuit and model: i o rb µ r c i i i i r π v 1 π g m v 1 r o cs i o Fig.01008 Assume that r c 0. As a result, r o and cs have no effect. r π I o (jω) V 1 1r π ( π µ )s I i and I o g m V 1 I i (jω) = g m r π β o ( π µ )s = ( π µ )s 1g m r π g m 1 β o g m Now, β(jω) = I o(jω) β o I i (jω) = ( π µ )jω 1 β o g m At high frequencies, g m β(jω) jω ( π µ ) When β(jω) =1 then ω g m T = π µ or f T = 1 g m 2π π µ
Lecture 010 4430 Review I (12/29/01) Page 01010 JFT Large ignal Model Large signal model: v I (1 v V p ) 2 Fig. 01009 Incorporating the channel modulation effect: i = I 1 v 2 V (1λv p ), v v V p igns for the JFT variables: Type of JFT V p I v pchannel Positive Negative Normally positive nchannel Negative Positive Normally negative
Lecture 010 4430 Review I (12/29/01) Page 01011 Frequency Independent JFT mall ignal Model chematic: i d v gs g m v gs ro v ds Parameters: g m = di dv Q = 2I V p where 1 V V p = g m0 Fig.01010 1 V g m0 = 2I V p r o = di dv Q = λi 1 V 2 1 V p λi Typical values of I and V p for a pchannel JFT are 1mA and 2V, respectively. With λ = 0.02V1 and I = 1mA we get g m = 1mA/V or 1m and r o = 50kΩ. V p
Lecture 010 4430 Review I (12/29/01) Page 01012 Frequency ependent JFT mall ignal Model omplete small signal model: gss v gs gs gd g m v gs ro i d v ds Fig.01011 All capacitors are reverse biased depletion capacitors given as, gs0 gs = 1 V (capacitance from source to top and bottom gates) 1/3 ψ o gd0 gd = 1 V (capacitance from drain to top and bottom gates) 1/3 ψ o gss0 gss = 1 V (capacitance from the gate (pbase) to substrate) 1/2 ψ o f T = 1 2π g m gs gd gss = 30MHz if g m = 1mA/V and gs gd gss = 5pF
Lecture 010 4430 Review I (12/29/01) Page 01013 imple Large ignal MOFT Model Nchannel reference convention: Nonsaturationi = Wµ o ox L (v V T )v v 2 2 (1 λv ), 0 < v < v V T aturationi = Wµ o ox (v V T )v (sat) v (sat) 2 2 (1 λv ) L = Wµ o ox 2L (v V T ) 2 (1 λv ), 0 < v V T < v where: µ o = zero field mobility (cm 2 /volt sec) ox = gate oxide capacitance per unit area (F/cm 2 ) λ = channellength modulation parameter (volts 1 ) V T = V T0 γ 2 φ f v 2 φ f V T0 = zero bias threshold voltage γ = bulk threshold parameter (volts 0.5 ) 2 φ f = strong inversion surface potential (volts) For pchannel MOFTs, use nchannel equations with pchannel parameters and invert current. v i v v Fig. 01012
Lecture 010 4430 Review I (12/29/01) Page 01014 MOFT mallignal Model omplete schematic model: where g m di dv Q = β(v V T ) = 2βI and g mbs = ι v Q = implified schematic model: v gs v bs g m v gs i v v v Q = i v T v T v Q = g mbs v bs g ds di dv Q = r ds i d v ds Fig. 01013 λi 1 λv λi g m γ 2 2 φ F V = ηg m v gs g m v gs r ds i d v ds Fig. 4.22 xtremely important assumption: g m 10g mbs 100g ds
Lecture 010 4430 Review I (12/29/01) Page 01015 MOFT epletion apacitors and Model: J A = 1 v MJ JW P P 1 v MJW, v F P P and J A = 1 F 1MJ 1 (1MJ)F MJ V P JW P 1 F 1MJW 1 (1MJW)F MJW V P, v> F P where A = area of the source P = perimeter of the source JW = zero bias, bulk source sidewall capacitance MJW = bulksource sidewall grading coefficient For the bulkdrain depletion capacitance replace "" by "" in the above equations. ource io 2 Polysilicon gate A ulk rain bottom = A rain sidewall = AF F H AH v F P H rain F F P P Fig. 01014 v F P v Fig 01015
Lecture 010 4430 Review I (12/29/01) Page 01016 MOFT Intrinsic apacitors, and utoff Region: utoff = 2 2 5 = ox (W eff )(L eff ) 2O(L eff ) = 1 ox (L)W eff ) = O(W eff ) = 3 ox (L)W eff ) = O(W eff ) aturation Region: = 2 5 = O(L eff ) = 1 (2/3) 2 = ox (L0.67L eff )(W eff ) = O(W eff ) 0.67 ox (W eff )(L eff ) = 3 ox (L)W eff ) = O(W eff ) Active Region: = 2 5 = 2O(L eff ) Active = 1 0.5 2 = ox (L0.5L eff )(W eff ) = (O 0.5 ox L eff )W eff = 3 0.5 2 = ox (L0.5L eff )(W eff ) = (O 0.5 ox L eff )W eff V = 0 p p substrate aturated V = 0 p p substrate V = 0 p p substrate V = 0 V < V T V > 0 Polysilicon n n V = 0 V >V T V >V V T n n Inverted Region ;;; Polysilicon V = 0 Polysilicon V >V T V <V V T n n Inverted Region Fig 01016
Lecture 010 4430 Review I (12/29/01) Page 01017 mallignal Frequency ependent Model The depletion capacitors are found by evaluating the large signal capacitors at the operating point. The charge storage capacitors are v gs gs constant for a specific region of operation. gb ainbandwidth of the MOFT: Assume V = 0 and the MOFT is in saturation, f T = 1 g m 2π gs gd 1 g m 2π gs Recalling that gs 2 3 W oxwl and g m = µ o ox L (V V T ) gives f T = 3 µ o 4π L2 (V V T ) gd g m v gs v bs r ds v ds g mbs v bs bs i d bd Fig 01017
Lecture 010 4430 Review I (12/29/01) Page 01018 ubthreshold MOFT Model Weak inversion operation occurs when the applied gate voltage is below V T and pertains to when the surface of the substrate beneath the gate is weakly inverted. V n nchannel n iffusion urrent psubstrate/well Fig. 01018 Regions of operation according to the surface potential, φ. φ < φ F : ubstrate not inverted φ F < φ < 2φ F : hannel is weakly inverted (diffusion current) 2φ F < φ : trong inversion (drift current) rift current versus diffusion current in a MOFT: log i iffusion urrent 106 rift urrent 1012 0 V T V Fig. 01019
Lecture 010 4430 Review I (12/29/01) Page 01019 Largeignal Model for ubthreshold Model: i = K x W L ev /nvt(1 e v /V t )(1 λv ) where K x is dependent on process parameters and the bulksource voltage n 1.5 3 and V t = kt q If v > 0, then 1µA i V =V T i = K x W L ev /nvt (1 λv ) V <V T mallsignal model: gm = i v Q = qi nkt 0 0 1V v Fig 01020 gds = i v Q I V A
Lecture 010 4430 Review I (12/29/01) Page 01020 Models Largesignal mallsignal omponents pn Junction JT MOFT trong inversion Weak inversion JFT apacitors epletion Parallel plate UMMARY