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EECS 105 Fall 2003, Lecture 15 Lecture 15: Small Signal Modeling Prof. Niknejad Department of EECS University of California, Berkeley EECS 105 Fall 2003, Lecture 15 Prof. A. Niknejad Lecture Outline Review: Diffusion Revisited BJT Small-Signal Model Circuits!!! Small Signal Modeling Example: Simple MOS Amplifier Department of EECS University of California, Berkeley EECS 105 Fall 2003, Lecture 15 Prof. A. Niknejad Notation Review Large signal iC f (vBE , vCE ) I C iC f (VBE vBE ,VCE vCE ) Quiescent Point small signal (bias) DC (bias) I C ic f (VBE vbe ,VCE vce ) small signal f f (less messy!) Q (VBE ,VCE ) ic vbe vce vBE Q vCE Q transconductance Output conductance Since we’re introducing a new (confusing) subject, let’s adopt some consistent notation Please point out any mistakes (that I will surely make!) Once you get a feel for small-signal analysis, we can drop the notation and things will be clear by context (yeah right! … good excuse) Department of EECS University of California, Berkeley EECS 105 Fall 2003, Lecture 15 Prof. A. Niknejad Diffusion Revisited Why is minority current profile a linear function? Recall that the path through the Si crystal is a zig-zag series of acceleration and deceleration (due to collisions) Note that diffusion current density is controlled by width of region (base width for BJT): Density here fixed by potential (injection of carriers) Physical interpretation: How many electrons (holes) have Half go left, enough energy to cross barrier? Boltzmann distribution give half go right this number. Density fixed by metal contact Wp Decreasing width increases current! Department of EECS University of California, Berkeley EECS 105 Fall 2003, Lecture 15 Prof. A. Niknejad Diffusion Capacitance The total minority carrier charge for a one-sided junction is (area of triangle) qV 1 1 D Qn qA bh2 qA (W xdep , p )(n p 0e kT n p 0 ) 2 2 For a one-sided junction, the current is dominated by these minority carriers: qVD qADn ID (n p 0e kT n p 0 ) Wp xdep , p ID Dn Constant! Qn W x dep , p 2 p Department of EECS University of California, Berkeley EECS 105 Fall 2003, Lecture 15 Prof. A. Niknejad Diffusion Capacitance (cont) The proportionality constant has units of time Distance across Qn W p xdep , p 2 P-type base T ID Dn q W p xdep , p 2 Diffusion Coefficient T Mobility kT n Temperature The physical interpretation is that this is the transit time for the minority carriers to cross the p-type region. Since the capacitance is related to charge: Qn T I D Qn I Cd T g d T V V Department of EECS University of California, Berkeley EECS 105 Fall 2003, Lecture 15 Prof. A. Niknejad BJT Transconductance gm The transconductance is analogous to diode conductance Department of EECS University of California, Berkeley EECS 105 Fall 2003, Lecture 15 Prof. A. Niknejad Transconductance (cont) Forward-active large-signal current: iC I S e vBE / Vth (1 vCE VA ) • Differentiating and evaluating at Q = (VBE, VCE ) iC q I S e qVBE / kT (1 VCE VA ) vBE Q kT iC qI C gm vBE Q kT Department of EECS University of California, Berkeley EECS 105 Fall 2003, Lecture 15 Prof. A. Niknejad BJT Base Currents Unlike MOSFET, there is a DC current into the base terminal of a bipolar transistor: I B IC F I S F eqVBE / kT (1 VCE VA ) To find the change in base current due to change in base-emitter voltage: iB iB iB iC 1 ib vbe gm vBE Q vBE Q iC vBE Q F Q gm ib vbe F Department of EECS University of California, Berkeley EECS 105 Fall 2003, Lecture 15 Prof. A. Niknejad Small Signal Current Gain iC 0 F iB iC 0 iB ic 0ib Since currents are linearly related, the derivative is a constant (small signal = large signal) Department of EECS University of California, Berkeley EECS 105 Fall 2003, Lecture 15 Prof. A. Niknejad Input Resistance rπ iB 1 iC gm r 1 vBE Q F vBE Q F F r gm In practice, the DC current gain F and the small-signal current gain o are both highly variable (+/- 25%) Typical bias point: DC collector current = 100 A 25 mV r 100 25 k .1mA Ri MOSFET Department of EECS University of California, Berkeley EECS 105 Fall 2003, Lecture 15 Prof. A. Niknejad Output Resistance ro Why does current increase slightly with increasing vCE? Collector (n) WB Base (p) Emitter (n+) Answer: Base width modulation (similar to CLM for MOS) Model: Math is a mess, so introduce the Early voltage iC I S e vBE / Vth (1 vCE V A ) Department of EECS University of California, Berkeley EECS 105 Fall 2003, Lecture 15 Prof. A. Niknejad Graphical Interpretation of ro slope~1/ro slope Department of EECS University of California, Berkeley EECS 105 Fall 2003, Lecture 15 Prof. A. Niknejad BJT Small-Signal Model ib r vbe 1 ic g m vbe vce ro Department of EECS University of California, Berkeley EECS 105 Fall 2003, Lecture 15 Prof. A. Niknejad BJT Capacitors Emitter-base is a forward biased junction depletion capacitance: C j , BE 1.4C j , BE 0 Collector-base is a reverse biased junction depletion capacitance Due to minority charge injection into base, we have to account for the diffusion capacitance as well Cb F g m Department of EECS University of California, Berkeley EECS 105 Fall 2003, Lecture 15 Prof. A. Niknejad BJT Cross Section Core Transistor External Parasitic Core transistor is the vertical region under the emitter contact Everything else is “parasitic” or unwanted Lateral BJT structure is also possible Department of EECS University of California, Berkeley EECS 105 Fall 2003, Lecture 15 Prof. A. Niknejad Core BJT Model Reverse biased junction Base Collector g m v Fictional Resistance (no noise) Reverse biased junction & Diffusion Capacitance Emitter Given an ideal BJT structure, we can model most of the action with the above circuit For low frequencies, we can forget the capacitors Capacitors are non-linear! MOS gate & overlap caps are linear Department of EECS University of California, Berkeley EECS 105 Fall 2003, Lecture 15 Prof. A. Niknejad Complete Small-Signal Model “core” BJT Reverse biased junctions Real Resistance (has noise) External Parasitics Department of EECS University of California, Berkeley EECS 105 Fall 2003, Lecture 15 Prof. A. Niknejad Circuits! When the inventors of the bipolar transistor first got a working device, the first thing they did was to build an audio amplifier to prove that the transistor was actually working! Department of EECS University of California, Berkeley EECS 105 Fall 2003, Lecture 15 Prof. A. Niknejad Modern ICs Source: Intel Corporation Used without permission Source: Texas Instruments Used without permission First IC (TI, Jack Kilby, 1958): A couple of transistors Modern IC: Intel Pentium 4 (55 million transistors, 3 GHz) Department of EECS University of California, Berkeley EECS 105 Fall 2003, Lecture 15 Prof. A. Niknejad A Simple Circuit: An MOS Amplifier Input signal VDD RD Supply “Rail” vo I DS vs vGS VGS vs Output signal VGS Department of EECS University of California, Berkeley EECS 105 Fall 2003, Lecture 15 Prof. A. Niknejad Selecting the Output Bias Point The bias voltage VGS is selected so that the output is mid-rail (between VDD and ground) For gain, the transistor is biased in saturation Constraint on the DC drain current: VDD Vo VDD VDS IR RD RD All the resistor current flows into transistor: I R I DS , sat Must ensure that this gives a self-consistent solution (transistor is biased in saturation) VDS VGS VT Department of EECS University of California, Berkeley EECS 105 Fall 2003, Lecture 15 Prof. A. Niknejad Finding the Input Bias Voltage Ignoring the output impedance W 1 I DS , sat nCox (VGS VTn )2 L 2 Typical numbers: W = 40 m, L = 2 m, RD = 25k, nCox = 100 A/V2, VTn = 1 V, VDD = 5 V VDD W 1 I RD I DS , sat nCox (VGS VTn ) 2 2 RD L 2 5V μA 1 100μA 20 100 2 (VGS 1) 2 50k V 2 .1 (VGS 1) 2 VGS 1.32 VGS VT .32 VDS 2.5 Department of EECS University of California, Berkeley EECS 105 Fall 2003, Lecture 15 Prof. A. Niknejad Applying the Small-Signal Voltage Approach 1. Just use vGS in the equation for the total drain current iD and find vo vGS VGS vs vs vs cos t ˆ W1 vO VDD RDiDS VDD RD nCox (VGS vs VT )2 L 2 Note: Neglecting charge storage effects. Ignoring device output impedance. Department of EECS University of California, Berkeley EECS 105 Fall 2003, Lecture 15 Prof. A. Niknejad Solving for the Output Voltage vO W1 vO VDD RDiDS VDD RD nCox (VGS vs VT )2 L 2 2 W 1 2 vs vO VDD RDiDS VDD RD nCox (VGS VT ) 1 L 2 VGS VT I DS 2 vs vO VDD RD I DS 1 VGS VT VDD 2 Department of EECS University of California, Berkeley EECS 105 Fall 2003, Lecture 15 Prof. A. Niknejad Small-Signal Case Linearize the output voltage for the s.s. case Expand (1 + x)2 = 1 + 2x + x2 … last term can be dropped when x << 1 2 vs 2v s vs 2 1 + ------------------------- = 1 + ------------------------- + ------------------------- - - - V G S – V Tn V GS – V Tn V – V GS Tn Neglect 2vs vO VDD RD I DS 1 VGS VT Department of EECS University of California, Berkeley EECS 105 Fall 2003, Lecture 15 Prof. A. Niknejad Linearized Output Voltage For this case, the total output voltage is: VDD 2vs vO VDD 1 2 VGS VT VDD vsVDD vO 2 VGS VT “DC” Small-signal output The small-signal output voltage: vsVDD vo Av vs VGS VT Voltage gain Department of EECS University of California, Berkeley EECS 105 Fall 2003, Lecture 15 Prof. A. Niknejad Plot of Output Waveform (Gain!) Numbers: VDD / (VGS – VT) = 5/ 0.32 = 16 output input mV Department of EECS University of California, Berkeley EECS 105 Fall 2003, Lecture 15 Prof. A. Niknejad There is a Better Way! What’s missing: didn’t include device output impedance or charge storage effects (must solve non-linear differential equations…) Approach 2. Do problem in two steps. DC voltages and currents (ignore small signals sources): set bias point of the MOSFET ... we had to do this to pick VGS already Substitute the small-signal model of the MOSFET and the small-signal models of the other circuit elements … This constitutes small-signal analysis Department of EECS University of California, Berkeley