# S. RossEECS 40 Spring 2003 Lecture 21 CMOS INVERTER CMOS means Complementary MOS: NMOS and PMOS working together in a circuit D S V DD (Logic 1) D S V.

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S. RossEECS 40 Spring 2003 Lecture 21 CMOS INVERTER CMOS means Complementary MOS: NMOS and PMOS working together in a circuit D S V DD (Logic 1) D S V OUT V IN

S. RossEECS 40 Spring 2003 Lecture 21 V OUT V IN CB A E D V DD V M : Voltage when V IN = V OUT (= V M ) CMOS INVERTER RESPONSE

S. RossEECS 40 Spring 2003 Lecture 21 triode mode saturation mode V DS IDID V GS = 3 V V GS = 1 V V DS = V GS - V TH(N) LAST TIME: SINGLE TRANSISTOR CIRCUIT Linear I D vs V DS given by surrounding circuit X X

S. RossEECS 40 Spring 2003 Lecture 21 ANALYSIS OF INVERTER CIRCUT Obtain: 1)the two nonlinear I D vs. V DS equations for the transistors: I D(N) vs. V DS(N) and I D(P) vs. V DS(P) 2)A linear relationship between I D(N) and I D(P) (e.g., via KCL) 3)An independent linear relationship between V DS(N) and V DS(P) (e.g. via KVL) Using the above, write: I D(P) vs. V DS(P) in terms of I D(N) vs. V DS(N) (or vice-versa) Solve the two transistor equations simultaneously.

S. RossEECS 40 Spring 2003 Lecture 21 ANALYSIS OF INVERTER CIRCUIT: UNLOADED D S V DD (Logic 1) D S V OUT V IN 1)Transistor equations: I D(N) = f N (V DS(N) ) I D(P) = f P (V DS(P) ) 2)I D(P) +I D(N) = 0 3)V DS(N) -V DS(P) = V DD Rewrite 1) as I D(N) = -f P (V DS(N) -V DD ) Find simultaneous solution to: I D(N) = f N (V DS(N) ) I D(N) = -f P (V DS(N) -V DD )

S. RossEECS 40 Spring 2003 Lecture 21 D S V DD (Logic 1) D S V OUT V IN ANALYSIS OF INVERTER CIRCUIT: UNLOADED Also note: V GS(N) = V IN V GS(P) = V IN - V DD V OUT = V DS(N) + V GS(N) - + V GS(P) - + V DS(N) _

S. RossEECS 40 Spring 2003 Lecture 21 V DS IDID V GS(P) < V TH(N) - V DD V GS(N) < V TH(N) CMOS INVERTER: REGION A V DS(P) = V GS(P) - V TH(P) NMOS cutoff mode PMOS triode mode V DD No current flow in Region A!

S. RossEECS 40 Spring 2003 Lecture 21 V DS IDID V GS(P) = V TH(N) +  - V DD V GS(N) = V TH(N) +  V DS(N) = V GS(N) - V TH(N) CMOS INVERTER: REGION B V DS(P) = V GS(P) - V TH(P) NMOS saturation mode PMOS triode mode V DD

S. RossEECS 40 Spring 2003 Lecture 21 V DS IDID V DS(N) = V GS(N) - V TH(N) CMOS INVERTER: REGION C V DS(P) = V GS(P) - V TH(P) NMOS saturation mode PMOS saturation mode V DD

S. RossEECS 40 Spring 2003 Lecture 21 V DS IDID V GS(P) = V TH(P) -  V GS(N) = V DD + V TH(P) -  V DS(N) = V GS(N) - V TH(N) CMOS INVERTER: REGION D V DS(P) = V GS(P) - V TH(P) NMOS triode mode PMOS saturation mode V DD

S. RossEECS 40 Spring 2003 Lecture 21 V DS IDID V GS(P) > V TH(P) V GS(N) > V TH(P) + V DD CMOS INVERTER: REGION E V DS(N) = V GS(N) - V TH(N) NMOS triode mode PMOS cutoff mode V DD No current flow in Region E!

S. RossEECS 40 Spring 2003 Lecture 21 IDID V IN CB A E D V DD CMOS INVERTER RESPONSE: CURRENT FLOW

S. RossEECS 40 Spring 2003 Lecture 21 No I D current flow in Regions A and E if nothing attached to output; current flows only during logic transition If resistor or diode attached to output, current will flow through PMOS when input is low (output is high) If another inverter (or other CMOS logic) attached to output, transistor gate terminals of attached stage do not permit current: current flows only during logic transition D S V DD D S V OUT1 V IN D S V DD D S V OUT2

S. RossEECS 40 Spring 2003 Lecture 21 EXAMPLE: RESISTIVE LOAD D S V DD = 5 V D S V OUT 1)Transistor equations: I D(N) = 0 A (NMOS cutoff) V IN = 0 V Find the power absorbed by the resistor and the inverter. Power absorbed by inverter: P = I D(P) V DS(P) + I D(N) V DS(N) Let W/L  C OX = 1 mA, V TH(N) = -V TH(P) = 1 V, = 0. 1 k 

S. RossEECS 40 Spring 2003 Lecture 21 EXAMPLE: RESISTIVE LOAD D S V DD = 5 V D S V OUT 2)I D(N) and I D(P) relationship: I D(P) +I D(N) = -V OUT / 1 k  3)V DS(N) and V DS(P) relationship: V DS(N) -V DS(P) = V DD V IN = 0 V 1 k  4) Substitute into PMOS transistor equation:

S. RossEECS 40 Spring 2003 Lecture 21 EXAMPLE: RESISTIVE LOAD D S V DD = 5 V D S V OUT V IN = 0 V 1 k  5)Solutions: V DS(P) = {-8.87 V, -1.13 V} V DS(P) = -1.13 V agrees with triode mode I D(P) = -3.24 mA Power absorbed by inverter: I D(P) V DS(P) + I D(N) V DS(N) = 3.66 mW Power absorbed by resistor: R I 2 = (1 k  )(-3.24 mA) 2 = 10.5 mW

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