EECS 40 Spring 2003 Lecture 20
S. Ross
Today we will
• Review NMOS and PMOS I-V characteristic
• Practice useful method for solving transistor circuits
• Build a familiar circuit element using a transistor
Good luck to Marquette
in the NCAA Final Four!
EECS 40 Spring 2003 Lecture 20
S. Ross
NMOS I-V CHARACTERISTIC
G
IG
ID
S
-
VDS +
D
• Since the transistor is a 3-terminal device, there is no
single I-V characteristic.
• Note that because of the gate insulator, IG = 0 A.
• We typically define the MOS I-V characteristic as
ID vs. VDS
for a fixed VGS.
• 3 modes of operation
EECS 40 Spring 2003 Lecture 20
S. Ross
NMOS I-V CHARACTERISTIC
ID triode mode
saturation mode
VGS = 3 V
VDS = VGS - VTH(N)
VGS = 2 V
VGS = 1 V
VDS
cutoff mode (VGS ≤ VTH(N))
EECS 40 Spring 2003 Lecture 20
S. Ross
NMOS I-V CHARACTERISTIC
Cutoff Mode
• Occurs when VGS ≤ VTH(N)
ID = 0
Triode Mode
• Occurs when VGS > VTH(N) and VDS < VGS - VTH(N)
W
ID 
μnCOX VGS  VTH(N)  VDS /2 VDS
L
Saturation Mode
• Occurs when VGS > VTH(N) and VDS ≥ VGS - VTH(N)
W
1
2
ID 
μnCOX VGS  VTH(N)  1 λn VDS 
L
2
EECS 40 Spring 2003 Lecture 20
S. Ross
PMOS I-V CHARACTERISTIC
G
IG
ID
S
-
VDS +
D
Symbol has “dot” at gate. NMOS does not.
ID, VGS, VDS, and VTH(P) are all negative for PMOS.
These values are positive for NMOS.
Channel formed when VGS < VTH(P). Opposite for NMOS.
Saturation occurs when VDS ≤ VGS – VTH(P). Opposite for NMOS.
EECS 40 Spring 2003 Lecture 20
S. Ross
PMOS I-V CHARACTERISTIC
VDS
cutoff mode
(VGS ≥ VTH(P))
VGS = -1 V
VGS = -2 V
VDS = VGS - VTH(P)
ID
VGS = -3 V
saturation mode
triode mode
EECS 40 Spring 2003 Lecture 20
S. Ross
PMOS I-V CHARACTERISTIC
Cutoff Mode
• Occurs when VGS ≥ VTH(P)
ID = 0
Triode Mode
• Occurs when VGS < VTH(P) and VDS > VGS - VTH(P)
W
ID   μpCOX VGS  VTH(P)  VDS /2 VDS
L
Saturation Mode
• Occurs when VGS < VTH(P) and VDS ≤ VGS - VTH(P)
W
1
2
ID   μpCOX VGS  VTH(P)  1 λp VDS 
L
2
EECS 40 Spring 2003 Lecture 20
S. Ross
SATURATION CURRENT
Since l is small or zero, current ID is almost constant in
saturation mode.
We can call this current IDSAT:


W
1
IDSAT  μnCOX VGS  VTH(N) 2
L
2


W
1
IDSAT   μpCOX VGS  VTH(P) 2
L
2
for NMOS
for PMOS
EECS 40 Spring 2003 Lecture 20
S. Ross
LINEAR AND NONLINEAR ELEMENTS
We need to find out how transistors behave as part of a circuit.
To solve a transistor circuit, obtain:
1) the nonlinear ID vs. VDS
characteristic equation for the
transistor
Linear circuit
2) The linear relationship
between ID vs. VDS as
determined by the surrounding
linear circuit
G
IG
ID
S
-
VDS +
D
Then simultaneously solve these
two equations for ID and VDS.
EECS 40 Spring 2003 Lecture 20
S. Ross
SOLVING TRANSISTOR CIRCUITS: STEPS
1) Guess the mode of operation for the transistor. (We will
learn how to make educated guesses).
2) Write the ID vs. VDS equation for this mode of operation.
3) Use KVL, KCL, etc. to come up with an equation relating ID
and VDS based on the surrounding linear circuit.
4) Solve these equations for ID and VDS.
5) Check to see if the values for ID and VDS are possible for the
mode you guessed for the transistor. If the values are
possible for the mode guessed, stop, problem solved. If the
values are impossible, go back to Step 1.
EECS 40 Spring 2003 Lecture 20
S. Ross
CHECKING THE ANSWERS
NMOS
• VDS must be positive
• ID must be positive
• VDS < VGS – VT(N) in triode
VDS ≥ VGS – VT(N) in saturation
• VGS > VT(N) in triode or saturation
VGS ≤ VT(N) in cutoff
PMOS
• VDS must be negative
• ID must be negative
• VDS > VGS – VT(P) in triode
VDS ≤ VGS – VT(P) in saturation
• VGS < VT(P) in triode or saturation
VGS ≥ VT(P) in cutoff
EECS 40 Spring 2003 Lecture 20
S. Ross
EXAMPLE
1) Guess the mode:
1.5 kW
D
4V
ID
+
_
3V
G
+
_
Since VGS > VTH(N), not in cutoff
mode. Guess saturation mode.
+
2) Write transistor ID vs. VDS:
VDS
_
ID  IDSAT
ID  250  106 A / V 2 3 V  1 V 2
S
VTH(N) = 1 V,
½ W/L mnCOX = 250 m A/V2,
l = 0 V-1.
ID  1 mA
3) Write ID vs. VDS equation
using KVL:
- VDS - 1.5 kW ID  4V  0
EECS 40 Spring 2003 Lecture 20
S. Ross
EXAMPLE
4) Solve:
1.5 kW
D
4V
ID
+
_
3V
G
+
_
+
VDS
_
S
VTH(N) = 1 V,
½ W/L mnCOX = 250 m A/V2,
l = 0 V-1.
ID = 1mA
VDS = 2.5 V
5) Check:
ID and VDS are correct sign,
and VDS ≥ VGS-VT(N) as
required in saturation mode.
EECS 40 Spring 2003 Lecture 20
S. Ross
WHAT IF WE GUESSED THE MODE WRONG?
1) Guess the mode:
Since VGS > VTH(N), not in cutoff
mode. Guess triode mode.
1.5 kW
D
4V
ID
+
_
3V
G
+
_
2) Write transistor ID vs. VDS:
+
VDS
_
S
VTH(N) = 1 V,
½ W/L mnCOX = 250 m A/V2,
l = 0 V-1.
ID = 2·250·10-6(3 – 1 – VDS/2)VDS
3) Write ID vs. VDS equation
using KVL:
- VDS - 1.5 kW ID  4V  0
EECS 40 Spring 2003 Lecture 20
S. Ross
WHAT IF WE GUESSED THE MODE WRONG?
4) Solve for VDS with quadratic
equation:
1.5 kW
D
4V
ID
+
_
3V
G
+
_
+
VDS
_
S
VTH(N) = 1 V,
½ W/L mnCOX = 250 m A/V2,
l = 0 V-1.
VDS = {4 V, 2.67 V}
5) Check:
Neither value valid in triode mode!
VDS > VGS – VT(N) not allowed in
triode mode.
EECS 40 Spring 2003 Lecture 20
S. Ross
GUESSING RIGHT
How do you guess the right mode?
Often, the key is the value of VGS.
(We can often find VGS directly without solving the whole circuit.)
ID
ID
VGS ≤ VT(N)
VGS = VT(N) + e
probably saturation
definitely cutoff
VDS
VGS - VT(N) = e
VDS
EECS 40 Spring 2003 Lecture 20
S. Ross
GUESSING RIGHT
When VGS >> VTH(N), it’s harder to guess the mode.
ID
triode mode
saturation mode
VGS - VTH(N)
If ID is small, probably triode mode
VDS
EECS 40 Spring 2003 Lecture 20
S. Ross
A CLOSER LOOK
In this circuit, the transistor
delivered a constant current
IDSAT to the 1.5 kW resistor.
1.5 kW
D
4V
ID
+
_
3V
G
+
_
This circuit acts like a
constant current source, as
long as the transistor
remains in saturation mode.
IDSAT does not depend on the
attached resistance if
saturation is maintained.
+
VDS
_
S
IDSAT
1.5 kW
EECS 40 Spring 2003 Lecture 20
S. Ross
A CLOSER LOOK
The circuit will go out of
saturation mode if
• VGS < VT(N)
or
• VDS < VGS – VT(N)
IDSAT does depend on VGS;
one can adjust the current
supplied by adjusting VGS.
RL
D
VDD
ID
+
_
VGS
G
+
_
This can happen if VGS is too
large or too small, or if the
load resistance is too large.
+
VDS
_
S
IDSAT
RL
EECS 40 Spring 2003 Lecture 20
S. Ross
ANOTHER EXAMPLE
1) Guess the mode:
1.5 kW
2 kW
4V
ID
+
_
D
G
+
6 kW
VDS
_
What is VGS?
No current goes into/out gate.
VGS = 3 V by voltage division.
Guess saturation (randomly).
2) Write transistor ID vs. VDS:
ID  IDSAT
ID  250  106 A / V 2 3 V  1 V 2
ID  1 mA
S
VTH(N) = 1 V,
½ W/L mnCOX = 250 m A/V2,
l = 0 V-1.
3) Write ID vs. VDS equation
using KVL:
- VDS - 1.5 kW ID  4V  0
Effectively the same circuit as previous example: only 1 source.