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EE40 Lec 20 MOS Circuits Reading: Chap. 12 of Hambley

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Presentation on theme: "EE40 Lec 20 MOS Circuits Reading: Chap. 12 of Hambley"— Presentation transcript:

1 EE40 Lec 20 MOS Circuits Reading: Chap. 12 of Hambley
Supplement reading on MOS Circuits

2 Small-signal equivalent circuits Examples:
OUTLINE Bias circuits Small-signal equivalent circuits Examples: Common source amplifier Source follower Common gate amplifier Digital Gates CMOS EE40, Fall Prof. White

3 Bias Circuits Use load line to find Quiescent operating point.
Remember no current flow through the gate. Fixed-plus Self-Bias CKT VDD RD VG+vin VDD RD R1 R2 RS

4 Steps for MOSFET Circuit Analysis
1) Look at DC case to find Q point Use load line technique All capacitors are open circuit, Inductors are short circuit Determine Q-point, get gm and rd for small signal AC model 2) AC Small signal analysis DC source is ac ground (because there is no AC signal variation). All capacitors are approximated as short circuit (unless otherwise specified).

5 Example: Common Source Amplifier
VDD RD C R1 + - vo C RL VG + - vin + - v(t) R2 RS C

6 Step 1: find Q point VDD RD C R1 + C vo VG RL VDS - + + vin R2 v(t) RS
Not connected for DC component RD C R1 + - vo C VG RL VDS + - vin + - v(t) R2 RS C Not connected for DC component

7 Load line to determine Q Point by graphical method
Loadline to determine VGSQ Loadline to determine VDSQ VGSQ From load lines, we get ID  and hence gm and rd

8 Load line to determine Q Point by analytical method
Solve VGSQ assume saturation region first IDQ is known, then solve VDSQ Check VDSQ value is consistent with saturation region ( i.e. VDS> VGSQ-Vt) From load lines, we get ID  and hence gm and rd

9 Determination of gm and rd graphically
Example: Q point is known to be VGS=2.5V, VDS=6V

10 Determination of gm and rd by Analytical Models
In Saturation Region In Triode Region

11 Small Signal Model Inverting For output impedance Rout:
Turn off all independent sources. Take away load impedance RL

12 Example: Source Follower
VDD R1 C VG C + - vin RL + - vo + - v(t) R2 RS

13 Step 1: find Q point VDD R1 C VG C + - vin RL + - vo + - v(t) R2 RS

14 Small Signal Model Rout is small For output impedance Rout:
Non-inverting, Voltage Gain <1 Rin high Current gain can be high For output impedance Rout: Turn off all independent sources. Take away RL Add Vx and find ix Rout is small

15 Example: Common Gate Amplifier
VDD RD C + - vo VG RL C + - vin + - v(t) RS -VSS

16 Step 1: find Q point VDD RD C + - vo VG RL C + - vin + - v(t) RS -VSS

17 Load line The only difference in all three circuits are the intercepts at the axes. Again from load lines, we get ID  and hence gm and rd

18 Small Signal Model Non-inverting For output impedance Rout:
Turn off all independent sources. Take away RL Add Vx and find ix

19 Logic Gates : Pull-Up and Pull-Down
PMOS or Resistor NMOS or Resistor

20 Inverter = NOT Gate Vin Vout Ideal Transfer Characteristics Vout Vin

21 NMOS Inverter: Resistor Pull-Up
Circuit: Voltage-Transfer Characteristic vOUT VDD F A iD vIN = VDD increasing vGS = vIN > VT vIN VT VDD VDD/RD VDD A F 1 vDS vGS = vin  VT

22 Output is low only if both inputs are high
NMOS NAND Gate Output is low only if both inputs are high VDD RD F A Truth Table A B F 1 B

23 Output is low if either input is high
NMOS NOR Gate Output is low if either input is high VDD RD F A B Truth Table A B F 1

24 Disadvantages of NMOS Logic Gates
Large values of RD are required in order to achieve a low value of VLOW keep power consumption low Large resistors are needed, but these take up a lot of space.

25 CMOS Inverter: Intuitive Perspective
CIRCUIT SWITCH MODELS VDD VIN VOUT S D G VDD VDD Rp VOUT VOUT VOL = 0 V VOH = VDD Rn Low static power consumption, since one MOSFET is always off in steady state VIN = VDD VIN = 0 V

26 The CMOS Inverter: Current Flow
N: sat P: sat VOUT i N: off P: lin C V DD VDD S G N: sat P: lin D I VIN VOUT A B D E D G N: lin P: sat S N: lin P: off VIN VDD

27 Power Dissipation: Direct-Path Current
VDD V VDD-VT DD vIN: S G VT D i vIN vOUT Ipeak D G i: S tsc time Energy consumed per switching period:

28 CMOS NAND Gate VDD A B F 1 A B Notice that the pull-up network is related to the pull-down network by DeMorgan’s Theorem! F A NMOS, Pull-down PMOS, Pull-up B

29 CMOS NOR Gate VDD A B F 1 A Notice that the pull-up network is related to the pull-down network by DeMorgan’s Theorem! B F NMOS, Pull-down PMOS, Pull-up B A

30 Multiple Input NOR Gate

31 Features of CMOS Digital Circuits
The output is always connected to VDD or GND in steady state Full logic swing; large noise margins Logic levels are not dependent upon the relative sizes of the devices (“ratioless”) There is no direct path between VDD and GND in steady state no static power dissipation


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