Electronic Circuits Part 1: MOSFETs, BJTs Amplification Review
Explore MOSFETs and BJTs for amplification, including characteristics like current-voltage, transconductance, output resistance, and intrinsic gain. Learn about biasing in IC amplifiers and channel-length modulation effects. An example problem on finding parameters and the ideal output resistance for a current source is also provided.
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Review of MOSFETs and BJTs For amplification: MOSFET in Saturation (NMOS) 0.5 V - 0.7 V GS t v V v v V BJT in Active mode (NPN) V 0.5 - 0.6 V v ( ) BE BE ON 0.3 V v DS GS t CE 0.2 V - 0.3 V v OV 2
Current-Voltage Characteristics MOSFET in SAT BJT in Active v v BE = + V (1 ) CE V i I e v V 1 2 0 W L T = + C S 2 ( ) (1 ) DS i C v V D n ox GS t A A v BE = V I e i T = S i G B = ' V V L A A Low-Frequency Small-Signal Hybrid- Models EECE 311 - Comparison of MOSFET and BJT 3
Transconductance and Output Resistance MOSFET BJT I V = C g I m = g D T m V OV + V V V 2 V = = A CE r A O / V V + I I e V V BE T = = C S A DS r A ( ) L O ' 2 I k V W 1 2 = D OV r g m Low-Frequency Small-Signal T Models EECE 311 - Comparison of MOSFET and BJT 4
Intrinsic Gain MOSFET (gate to drain) BJT (base to collector) ( ) 0 r ( )0 r g v m gs g v v v v v v = = m v ce A = = ds A 0 0 be gs gs V V V = A g r A = A g r A 0 0 m 0 0 m 2 V T OV EECE 311 - Comparison of MOSFET and BJT 5
Example: Find , and r g A 0 o m Given: NMOS MOSFET 1 2 0.27 V OV V = V L r C I g V NPN BJT = = 0.1 mA 5 V m 4 m 0.4 m 267 A V ox C = Solution: W I C I V W L = = = = 0.1 mA 35 V 25 mV 100 I V V D C ' = = A A T 2 n V I = 350 k r A o 2 V D n ox OV C L I V g r = = 4 mA/V C g m T ' = = = 1400 V/V = 20 k A A ( ) L o 0 0 m 2 V W 1 2 n ox OV = = 0.73mA V D m OV 2 = = 14.6 V/V 0 becomes -183 V/V if A / is 50 m / 5 m W L A g r 0 0 m 6
Biasing in IC amplifiers The Basic (NMOS) MOSFET Current Source Q SAT since V V V 1 DS DS t 1 2 W L = ' 2 ( ) I k V V 1 D n GS tn 1 V V = = DD GS I I 1 D REF R W L Q SAT when V V 2 o ( ( OV ) ) L 1 2 = = ' 2 ( ) I k V V I 2 D n GS tn o 2 Matched MOSFETs have same k and Vt W I L = o 2 W I REF 1 7 EECE 311 - IC Biasing
Considering Channel-Length Modulation ( W ( )2 )1 W Vo VA2 VGS VA1 1+ Io L = IREF 1+ L Output resistance V R I V = = = 2 o r A ( ) L 2 o o ' 2 k V W 1 2 o n OV 2 What is the ideal value of output resistance for a current source? 8 EECE 311 - IC Biasing
Example Given: VDD = 3 V, IREF = 100 A, design for IO = 100 A. Q1 and Q2 are matched with L = 1 m, W = 10 m, Vt = 0.7 V, k n = 200 A/V2. What is the lowest VO for current source operation? Given V A = 20 V/ m, find the output resistance of the current source Find the change in IO when VO changes by +1V 9 EECE 311 - IC Biasing
Solution V V W L ( ) 2 = = + = = ' ' 1 20 V GS I I k V V V V L 1 2 1 1 1 D REF n GS t A A 1 1 A V ( ) 2 100 100(10) = + 0.7 1 GS V GS 20 = 1.008 V V V GS V 3 1.008 0.1 mA V = = = 19.91 K DD GS R I REF V = = = = 1.008 0.7 0.31 V V V (min) o OV GS t Output resist ance: V k 20 = = = = 208 K 2 R r A W ( ) L ( ) 10 0.31 2 o o ' 2 2 100 V 1 2 n OV Change in output current: V R I 1V 208K V R = = = = 4.8 A o o I o o o o 10 EECE 311 - IC Biasing
Simple MOSFET Current Mirror Note that this circuit and the previous one implement actually a current sink , not a current source . How do I build a current source? Question: What is the small-signal equivalent circuit? 11 EECE 311 - IC Biasing
The Basic NPN BJT Current Source Neglecting the base current and as long as Vo > 0.3 V V I = V CC BE REF R V BE = = V I I e I T 1 1 C S REF V BE = I I = V I I e I T 2 2 C S o I I A A A A = = = 2 2 let C S 2 2 m E E 1 1 1 1 C S E E = I I m o REF = = Q 's emmiter area Q 's emmiter area A A 1 1 E 2 2 E 12 EECE 311 - IC Biasing
Considering the base current with m = 1 (equal BJT areas) I = + 2 C I I REF C = I I o C 1 I = o 2 I + 1 REF Error term is 2/ Considering the base current and different BJT areas I m m I + = o + 1 1 REF How do we include the Early Effect? 13 EECE 311 - IC Biasing
Simple BJT Current Mirror Again, this is a current sink , not current source . How do I build a current source? Hint: PNP What is the small-signal equivalent of this circuit? 14 EECE 311 - IC Biasing
Common Source Amplifier with Resistive Load (Review from 310) Voltage gain, input resistance, output resistance
Common Source Amplifier with Active Load Replace resistor with current source Small-Signal Analysis By inspection, R R r = = i o o ( ) = v m gs g v r o o v v = = = = o v v A g r A 0 gs i v m o i 16
Small-Signal Analysis of CMOS Common-Source Amplifier Small-signal equivalent of Q3-Q2 is simply ro2 vSG2 is pure DC = = 0 0 v g v 2 2 2 gs m gs = = R R i // r r 1 2 o o o v v ( ) = = / / o A g r r 1 1 2 v m o o i 18
CMOS Common-Source Amplifier Voltage at gate of Q2 is a pure DC level (VBIAS) In small-signal analysis, Q2 appears as simply ro2 The resistance seen looking into MOSFET drain, with gate and source at DC levels is equal to ro 19
CMOS CS Output Resistance The small-signal resistance seen looking into MOSFET drain, with gate and source at DC levels, is equal to ro ro2 Ro = ro1 // ro2 ro1 20
Example: CMOS Common-Source Amp Given 3V DD V k = = = = ' 2 200 A/V n ' p 2 65 A/V k 10 for all MOSFETs W L V = = 0.6V V tn tp = 20V V An = 10V V Ap = 100 A I REF Find: Output resistance of amplifier Small-signal voltage gain vo/vi Limits of vo to keep Q1 and Q2 saturated 21
= = = = Solution: 100 A I I I 1 2 D D REF = = 200 (10) ' 2 2 100 A = I k V V W L 1 2 1 2 1 1 1 D n OV OV 0.316V V 1 OV 100 I V = = = 0.633 mA/V 1 g D 1 m 0.5 0.316V 20 200K 100 1 2 1 OV V I = = = An r 1 o 1 D V 10 Ap = = = 100K r 2 o 100 I 2 D Output resistance: // Voltage gain: A g = We will neglect channel- length modulation in DC analyses. = = 66.67K R r r 1 2 o o o ( ) = // 42.2 V/V r r 1 1 2 v m o o Limits of output voltage swing: ( ) 2 ( ) 2 = = = 65 (10) ' p 100 A 0.6 I k V V V W L 1 2 1 2 REF SG tp SG = = = 1.155V 1.155 0.6 0.555V V V 2 SG OV enters SAT when: V V = Q v Q v 2 3 0.555 = = 2.445 V 2 O DD OV enters SAT when: 0.316 V OV V = = 1 22 1 O
CMOS Common-Source Cascade Assume identical stages What is the overall low-frequency gain if the per-stage gain is 42? What is the input resistance? What is the output resistance?
Common Emitter Amplifier with Active Load Small-Signal Analysis By inspection, R r R r = = i o o ( ) = v g v r o m o v v = = = o v v A g r i v m o i 24
BJT Version Common-Emitter Amplifier VCC VCC VBIAS Q3 Q2 Q2 vo vo IREF Rsig vi vi Q1 Q1 vsig 25
BJT CE Output Resistance The small-signal resistance seen looking into BJT collector, with base and emitter at DC levels, is equal to ro VCC Ro = ro1 // ro2 VBIAS Q2 ro2 Av = -gm(ro1 // ro2) vo ro1 vi Q1 What is the input resistance? resistance looking into base, with emitter at DC level, is equal to r . 26
