UNIVERSITY OF CALIFORNIA AT BERKELEY
College of Engineering
Department of Electrical Engineering and Computer Sciences
Discussion Notes: Large-signal PMOS behavior
EE 105
Prof. Wu
1
Spring 2008
Large signal PMOS behavior (long-channel)
Figure 1 shows a PMOS transistor with the source, gate, and drain labeled. Note that ID is defined to be
flowing from the source to the drain, the opposite as the definition for an NMOS. As with an NMOS, there
are three modes of operation: cutoff, triode, and saturation. I will describe multiple ways of thinking of the
modes of operation of a PMOS. I’ll only focus on long-channel devices in these notes, but everything can
easily be generalized to short-channel devices.
ID
S
G
VG
+
−
+ V
− S
D
Figure 1: PMOSFET
Note first that for a PMOS, the following equations generally hold for the devices and modes of operation
we care about:
VT H < 0
VGS < 0
VDS < 0
VDSat = VGS − VT H < 0
This is in contrast to the behavior of an NMOS, which typically has all of these voltages as positive values.
Also note the following definitions (you should know these by know, but just in case):
VGS ≡ VG − VS
VDS ≡ VD − VS
VSG ≡ VS − VG
VSD ≡ VS − VD
All of the following ways of formulating the large-signal equations for PMOSFETs are completely equivalent. Use whichever you feel most comfortable with.
1.1
Method A
For this method, I will describe the PMOS behavior in terms of VGS , VDS , and VT H .
1
Cutoff
VGS > VT H
ID = 0
Triode
VGS < VT H
VDS > VGS − VT H = VDSat
W
VDS
ID = µp Cox
(VGS − VT H ) −
VDS
L
2
Note that since VDS > VGS − VT H (and both values are negative), the term (VGS − VT H ) −
negative. However, since VDS is also negative, the current ID is positive.
VDS
2
is
Saturation
VGS < VT H
VDS < VGS − VT H = VDSat
1
W
2
ID = µp Cox
(VGS − VT H ) [1 − λ (VDS − VDSat )]
2
L
Note that VGS − VT H is negative, but since we square it, the term ends up being positive. Also note that
since VDS < VDSat (and both values are negative), the term multiplied by λ is negative, so the minus sign
makes the last term positive. Thus, ID is positive.
1.2
Method B
This method involves flipping the subscripts of the voltages VGS and VDS and flipping the signs of VT H and
VDSat . That means we’ll use the positive voltages VSG and VSD and the negative voltages VT H and VDSat
in the large signal equations.
Cutoff
VSG < −VT H
ID = 0
Triode
VSG > −VT H
VSD < VSG + VT H
VSD
W
(VSG + VT H ) +
VSD
ID = µp Cox
L
2
Saturation
VSG > −VT H
VSD > VSG + VT H
1
W
ID = µp Cox
(VSG + VT H )2 [1 + λ (VSD + VDSat )]
2
L
1.3
Method C
Here, we use all positive voltages, some with flipped subscripts, some with absolute value signs: VSG , VSD ,
|VT H |, and |VDSat |.
2
Cutoff
VSG < |VT H |
ID = 0
Triode
VSG > |VT H |
VSD < VSG − |VT H |
VSD
W
(VSG − |VT H |) +
VSD
ID = µp Cox
L
2
Saturation
VSG > |VT H |
VSD > VSG − |VT H |
1
W
2
ID = µp Cox
(VSG − |VT H |) [1 + λ (VSD − |VDSat |)]
2
L
1.4
Method D
Finally, we can use all positive voltages but modified entirely with absolute value signs: |VGS |, |VDS |, |VT H |,
and |VDSat |. Note that these formulas are identical to the NMOS formulas aside from the absolute value
signs and the usage of hole mobility (µp ) rather than electron mobility (µn ).
Cutoff
|VGS | < |VT H |
ID = 0
Triode
|VGS | > |VT H |
|VDS | < |VGS | − |VT H |
W
|VDS |
ID = µp Cox
(|VGS | − |VT H |) +
|VDS |
L
2
Saturation
|VGS | > |VT H |
|VDS | > |VGS | − |VT H |
1
W
2
ID = µp Cox
(|VGS | − |VT H |) [1 + λ (|VDS | − |VDSat |)]
2
L
3