EE105 - Fall 2006 Microelectronic Devices and Circuits. Some Administrative Issues

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EE105 - Fall 006 Microelectronic evices and Circuits Prof. Jan M. Rabaey (jan@eecs Lecture 8: MOS Small Signal Model Some Administrative Issues REIEW Session Next Week Tu Sept 6 6:00-7:30pm; 060 alley LSB MITERM 1 next week Th will cover everything up to and including MOSCAP Th Sept 8 6:30-8pm; Sibley Auditorium Some old midterm posted on website 1

Overview Last lecture MOS Capacitor; MOS Transistor Intro This lecture MOS Transistor (cntd Small signal model 3 MOSFET: Observed Behavior: I - I / k = 4 non-linear resistor region constant current I resistor region = 3 = For low values of drain voltage, the device is like a resistor As the voltage is increases, the resistance behaves non-linearly and the rate of increase of current slows Eventually the current stops growing and remains essentially constant (current source 4

Linear Region Current p+ NMOS > S n+ n+ p-type G Inversion layer channel 100m If the gate is biased above threshold, the surface is inverted This inverted region forms a channel that connects the drain and gate If a drain voltage is applied positive, electrons will flow from source to drain x y 5 MOSFET Linear Region The current in this channel is given by I = Wv Q y N The charge proportional to the voltage applied across the oxide over threshold Q = C ( N ox I = Wv C ( y ox If the channel is uniform density, only drift current flows v L y = μne y Ey = W I = ncox( > L μ 100m 6 3

MOSFET: ariable Resistor Notice that in the linear region, the current is proportional to the voltage I = W C ( L μ n ox Can define a voltage-dependent resistor R eq 1 L L = = = R ( I μ C ( W W n ox This is a nice variable resistor, electronically tunable! 7 Finding I = f (, Approximate inversion charge Q N (y: drain is higher than the source less charge at drain end of channel 8 4

Inversion Charge at Source/rain ( Q Q ( y = 0 + Q ( y = L / N N N Q N ( y = 0 = Cox ( Q N ( y = L = C ox ( G G = 9 Average Inversion Charge Source End rain End Cox( T + Cox( G T QN ( y Cox ( T + Cox ( T QN ( y Cox ( T CoxS QN( y = Cox( T Charge at drain end is lower since field is lower Simple approximation: In reality we should integrate the total charge minus the bulk depletion charge across the channel 10 5

rift elocity and rain Current Long-channel assumption: use mobility to find v Substituting: μn vy ( = μney ( μn( Δ/ Δ y = L I = WvQN Wμ Cox( T L W I C ( L μ ox T Inverted Parabolas 11 Square-Law Characteristics TRIOE REGION Boundary: what is I,SAT? SATURATION REGION 1 6

The Saturation Region When >, there isn t any inversion charge at the drain according to our simplistic model Why do curves flatten out? 13 Square-Law Current in Saturation Current stays at maximum (where = =,SAT W I = C ( L μ ox T W T I, sat Cox( T ( T L μ = W μcox I, sat = ( T L Measurement: I increases slightly with increasing model with linear fudge factor W μc I L ox, sat= ( T(1 +λ 14 7

Pinching the MOS Transistors epletion Region > S G p+ n+ n+ p-type NMOS Pinch-Off Point When >,sat, the channel is pinched off at drain end (hence the name pinch-off region rain mobile charge goes to zero (region is depleted, the remaining electric field is dropped across this high-field depletion region As the drain voltage is increases further, the pinch off point moves back towards source Channel Length Modulation: The effective channel length is thus reduced higher I 15 Short-Channel MOSFET Model Channel (inversion charge: neglect reduction at drain elocity saturation defines,sat =E sat L = constant rain current: -v sat / μ n I = WvQ = W ( v [ C (, SAT N sat ox E sat = 10 4 /cm, L = 0.1 μm,sat = 0.1! ], I = v WC ( (1 + λ, SAT sat ox n 16 8

Linear I- Characteristics: short-channel MOSFET 17 I versus 6 x 10-4 5 4 Resistive =.5 Saturation =.0-4.5 x 10 1.5 =.5 =.0 I (A 3 = - T = 1.5 I (A 1 = 1.5 1 = 1.0 0.5 = 1.0 0 0 0.5 1 1.5.5 ( Long Channel 0 0 0.5 1 1.5.5 ( Short Channel 18 9

A Simple Circuit: An MOS Amplifier v in Input signal R Supply Rail v O = O + v o v = + v s v s i = I + i ds Output signal v o 19 Large Signal Analysis OUT = = I R Load Line 0 10

Sweep Input oltage v IN + MOSFET is saturated high slope v O = v MOSFET is triode low slope 0 (negative supply, SS = 0 v IN = v 1 Finding the Input Bias oltage Hand calculation: use I,SAT I W μ C = L n ox, SAT ( Typical numbers: W = 40 μm, L = μm, R = 5 kω μ n C ox = 100 μa/, = 1, = 5 I 1 = = = I, SAT = 10 100 ( 1 R Want ~.5: I,SAT = /R = 0.1mA, = 1.3 11

Applying the Small-Signal oltage Approach 1. Just use v IN in the equation for the total drain current i and find v OUT v = v = + v v s IN ( t = cos( ωt sm s Result: v OUT = R i = R W μnc L ox ( + v s 3 Solving for the Output oltage v OUT v OUT = R W μ ( 1 ncox vs + L I v OUT = R I 1+ vs 4 1

Output oltage v O = v t 0 v IN = v t 5 Small-Signal Case Linearize the output voltage for the small-signal case: Expand (1 + x = 1 + x + x last term can be dropped when x << 1 1+ 1+ vs v s = 1+ v s + vs 6 13

Linearized Output oltage For this case, the total output voltage is v O = = ( R I = O R + v o I vs 1+ RI The output voltage is the sum of the C voltage O and the smallsignal voltage v o. RI vo = v RI A = s = A v s v s 7 Plot of Output Waveform (Gain! Numbers: / ( T = 5/ 0.3 = 16 output input m 8 14

There is a Better Way! What s missing: didn t include device output impedance or charge storage effects (must solve nonlinear differential equations Approach. o problem in two steps. C voltages and currents (ignore small signals sources: set bias point of the MOSFET... we had to do this to pick already Substitute the small-signal model of the MOSFET and the smallsignal models of the other circuit elements This constitutes small-signal analysis 9 An MOS Amplifier Input signal R Supply Rail v o v s I v = + vs Output signal 30 15

Small-Signal Analysis Step 1. Find C Bias ignore small-signal source I,Q,BIAS 31 Which Operating Region? TRIOE = = 3 3 SAT OFF 3 16

I / k Changing One ariable at a Time Square Law Saturation Region = 3 = 1 T Linear Triode Region Slope of Tangent: Incremental current increase Assumption: >,SAT = (square law 33 The Transconductance g m efined as the change in drain current due to a change in the gate-source voltage, with everything else constant W μc I L ox, sat = ( T (1 +λ Δi i W g = = = μc ( (1 + λ m ox T Δv v L,, 0 W gm = μcox ( T L Gate Bias W I W g = μc = μc I L W μc L ox L m ox ox g m I = ( T rain Current Bias rain Current Bias and Gate Bias 34 17

Output Resistance r o efined as the inverse of the change in drain current due to a change in the drain-source voltage, with everything else constant Non-Zero Slope δ I δ 35 Evaluating r o W μc i L ox = ( T (1 +λ o = v, r i r0 = W μc L ox 1 1 ( λ 1 r0 λi T 36 18

Total Small Signal Current i ( t = I + i ds i i i = v + v ds gs ds vgs vds 1 i = g v + v r ds m gs ds o Transconductance Conductance 37 Putting Together a Circuit Model 1 i = g v + v r ds m gs ds o 38 19