EECS 40 Spring 2003 Lecture 21
S. Ross
CMOS INVERTER
CMOS means Complementary MOS:
NMOS and PMOS working together in a circuit
VDD (Logic
S1)
D
VIN
D
S
VOUT
EECS 40 Spring 2003 Lecture 21
S. Ross
CMOS INVERTER RESPONSE
VOUT
VDD
VM: Voltage when
VIN = VOUT (= VM)
A
B
C
D
E
VIN
VDD
EECS 40 Spring 2003 Lecture 21
S. Ross
LAST TIME: SINGLE TRANSISTOR CIRCUIT
ID triode mode
saturation mode
VGS = 3 V
X
VDS = VGS - VTH(N)
Linear ID vs VDS given
by surrounding circuit
X
VGS = 1 V
VDS
EECS 40 Spring 2003 Lecture 21
S. Ross
ANALYSIS OF INVERTER CIRCUT
Obtain:
1) the two nonlinear ID vs. VDS equations for the transistors:
ID(N) vs. VDS(N) and ID(P) vs. VDS(P)
2) A linear relationship between ID(N) and ID(P) (e.g., via KCL)
3) An independent linear relationship between VDS(N) and
VDS(P) (e.g. via KVL)
Using the above, write:
ID(P) vs. VDS(P) in terms of ID(N) vs. VDS(N) (or vice-versa)
Solve the two transistor equations simultaneously.
EECS 40 Spring 2003 Lecture 21
S. Ross
ANALYSIS OF INVERTER CIRCUIT: UNLOADED
VDD (Logic
S1)
D
VIN
VOUT
D
S
1) Transistor equations:
ID(N) = fN(VDS(N))
ID(P) = fP(VDS(P))
2) ID(P)+ID(N) = 0
3) VDS(N)-VDS(P) = VDD
Rewrite 1) as
ID(N) = -fP(VDS(N)-VDD)
Find simultaneous solution to:
ID(N) = fN(VDS(N))
ID(N) = -fP(VDS(N)-VDD)
EECS 40 Spring 2003 Lecture 21
S. Ross
ANALYSIS OF INVERTER CIRCUIT: UNLOADED
VDD (Logic
S1)
Also note:
VGS(N) = VIN
VOUT
D
VIN
D
S
+
VDS(N)
_
VGS(P) = VIN - VDD
VOUT = VDS(N)
EECS 40 Spring 2003 Lecture 21
S. Ross
CMOS INVERTER: REGION A
ID
VDS(P) = VGS(P) - VTH(P)
VGS(N) < VTH(N)
VGS(P) < VTH(N) - VDD
No current flow
in Region A!
NMOS cutoff mode
PMOS triode mode
VDS
VDD
EECS 40 Spring 2003 Lecture 21
S. Ross
CMOS INVERTER: REGION B
ID
VGS(N) = VTH(N) + e
VDS(P) = VGS(P) - VTH(P)
VGS(P) = VTH(N) + e - VDD
NMOS saturation mode
PMOS triode mode
VDS(N) = VGS(N) - VTH(N)
VDS
VDD
EECS 40 Spring 2003 Lecture 21
S. Ross
CMOS INVERTER: REGION C
ID
NMOS saturation mode
PMOS saturation mode
VDS(P) = VGS(P) - VTH(P)
VDS(N) = VGS(N) - VTH(N)
VDS
VDD
EECS 40 Spring 2003 Lecture 21
S. Ross
CMOS INVERTER: REGION D
ID
VGS(N) = VDD + VTH(P) - e
VGS(P) = VTH(P) - e
VDS(N) = VGS(N) - VTH(N)
NMOS triode mode
PMOS saturation mode
VDS(P) = VGS(P) - VTH(P)
VDS
VDD
EECS 40 Spring 2003 Lecture 21
S. Ross
CMOS INVERTER: REGION E
ID
VGS(N) > VTH(P) + VDD
VGS(P) > VTH(P)
VDS(N) = VGS(N) - VTH(N)
No current flow
in Region E!
NMOS triode mode
PMOS cutoff mode
VDS
VDD
EECS 40 Spring 2003 Lecture 21
S. Ross
CMOS INVERTER RESPONSE: CURRENT FLOW
ID
A
B
C
D
E
VIN
VDD
EECS 40 Spring 2003 Lecture 21
S. Ross
• No ID 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
VDD
VDD
S
D
VIN
S
VOUT1
D
D
D
S
S
VOUT2
EECS 40 Spring 2003 Lecture 21
S. Ross
EXAMPLE: RESISTIVE LOAD
VDD = 5 V
S
VOUT
D
VIN = 0 V
Find the power absorbed by
the resistor and the inverter.
Power absorbed by inverter:
P = ID(P)VDS(P) + ID(N)VDS(N)
D
1 kW
S
Let W/L m COX = 1 mA,
VTH(N) = -VTH(P) = 1 V,
l = 0.
1) Transistor equations:
ID(N) = 0 A (NMOS cutoff)
W
ID(P)   μpCOX VGS(P)  VTH(P)  VDS(P) /2 VDS(P)
L



EECS 40 Spring 2003 Lecture 21
S. Ross
EXAMPLE: RESISTIVE LOAD
VDD = 5 V
S
VOUT
D
2) ID(N) and ID(P) relationship:
ID(P)+ID(N) = -VOUT / 1 kW
3) VDS(N) and VDS(P) relationship:
D
VIN = 0 V
1 kW
VDS(N)-VDS(P) = VDD
S
4) Substitute into PMOS transistor equation:

5  VDS(P)
1 kW



 1 mA - 5V  ( 1V )  VDS(P) /2 VDS(P)
EECS 40 Spring 2003 Lecture 21
S. Ross
EXAMPLE: RESISTIVE LOAD
VDD = 5 V
5)Solutions:
S
VDS(P) = {-8.87 V, -1.13 V}
VOUT
D
D
VIN = 0 V
1 kW
VDS(P) = -1.13 V agrees with
triode mode
ID(P) = -3.24 mA
S
Power absorbed by inverter: ID(P)VDS(P) + ID(N)VDS(N) = 3.66 mW
Power absorbed by resistor: R I2 = (1 kW)(-3.24 mA)2 = 10.5 mW