Download MOS Transistors and

MOS Transistors as Switches
nMOS transistor:
Closed (conducting) when
Gate = 1 (VDD)
G
(gate)
D
(drain)
S
(source)
Open (non-conducting) when
Gate = 0 (ground, 0V)
pMOS transistor:
Closed (conducting) when
Gate = 0 (ground, 0V)
G
S
D
Open (non-conducting) when
Gate = 1 (VDD)
For nMOS switch, source is typically tied to ground and is used to pull-down signals:
Out
when Gate = 1, Out = 0, (OV)
G
when Gate = 0, Out = Z (high impedance)
S
For pMOS switch, source is typically tied to VDD, used to pull signals up:
S
when Gate = 0, Out = 1 (VDD)
G
when Gate = 1, Out = Z (high impedance)
Out
Note: The MOS transistor is a symmetric device. This means that the drain and source terminals
are interchangeable. For a conducting nMOS transistor, VDS > 0V; for the pMOS transistor, VDS
< 0V (or VSD > 0V).
The CMOS Inverter
Truth Table
VDD
I
Out
I
I
Out
0
1
1
0
Out
GND
Rin →∞
Note: Ideally there is no static power dissipation. When "I" is fully is high or fully low, no
current path between VDD and GND exists (the output is usually tied to the gate of another MOS
transistor which has a very high input impedance).
Power is dissipated as "I" transistions from 0→1 and 1→0 and a momentory current path exists
between Vdd and GND. Power is also dissipated in the charging and discharging of gate
capacitances.
Parallel Connection of Switches
Y
Y = 0, if A or B = 1
A
B
A+B
Y = 1 if A or B = 0
A
B
A+B
Y
Series Connection of Switches
Y
A
Y = 0, if A and B = 1
B
A⋅ B
A
Y = 1, if A and B = 0
A⋅ B
B
Y
NAND Gate Design
p-type transistor tree will provide "1" values of logic function
n-type transistor tree will provide "0" values of logic function
Truth Table (NAND):
B
AB
00
1
01
1
10
1
11
0
A
0
1
0
1
1
1
1
0
K-map (NAND):
NAND circuit example:
Vdd
Ptree = A + B
Ntree = A ⋅ B
Y
B
A
A
B
Y
NOR Gate Design
p-type transistor tree will provide "1" values of logic function
n-type transistor tree will provide "0" values of logic function
Truth Table:
B
AB
00
1
01
0
10
0
11
0
A
0
1
0
1
0
1
0
0
K-map:
NOR circuit example:
Vdd
A
Ptree = A ⋅ B
Ntree = A + B
B
Y
A
B
Y
What logic gate is this?
Vdd
A
Y = 1 when A ⋅ B
Y = 0 when A + B
B
Y
Answer: AND function, but poor design!
Why? nMOS switches cannot pass a logic "1" without a threshold voltage (VT) drop.
where VT = 0.7V to 1.0V (i.e.,
VDD
G
VDD
D
threshold voltage will vary)
S
VDD - VT
output voltage = 4.3V to 4.0V,
a weak "1"
The nMOS transistor will stop conducting if VGS < VT. Let VT = 0.7V,
G 5V
S →D
D→ ?
0V → 5V
0V → ?
As source goes from 0V → 5V, VGS goes from 5V → 0V.
When VS > 4.3V, then VGS < VT, so switch stops conducting.
VD left at 5V − VT = 5V − 0.7V = 4.3V or VDD − VT.
What about nMOS in series?
5V
0V → 5V
5V
0V → 4.3V
5V
0V → 4.3V
5V
0V → 4.3V
Only one threshold voltage drop across series of nMOS transistors
0V → (VDD−VT)
5V - 0.7V
4.3V
For pMOS transistor, VT is negative.
pMOS transistor will conduct if |VGS| > |VTp| (VSG > |VTp|),
or VGS < VTp
0V
G
5V
S
VTp = −0.7V
VGS = 0V − 5V = −5V
D
conducting
VGS < VTp
or
−5V < −0.7V
|VGS| > |VTp|
5V > 0.7V
How will pMOS pass a "0"?
When |VGS| < |VTp|, stop conducting
G 0V
5V → 0V
5V → ?
S →D
D→ ?
So when |VGS| < |−0.7V|, VD will go from 5V
→ 0.7V,
a weak "0"
How are both a strong "1" and a strong "0" passed?
Transmission gate pass transistor configuration
When I = 1,
A
B
B = strong 1, if A = 1;
B = strong 0, if A = 0
I
When I = 0, non-conducting
About that AND Gate...
Vdd
No!!!
Poorly designed AND
(circuit designer fired)
A
B
Y
Instead use this,
A
B
Y
Vdd
Y
B
A
More Complex Gates
F = AB + CD
⇒
Ntree will provide 0's, Ptree will provide 1's
0's of function F is F, ⇒ F = AB + CD = AB + CD
nMOS transistors need high true inputs, so it is desirable for all input variables to be high true,
just as above.
Y
A
C
B
D
AB + CD
Likewise, a Ptree will provide 1's.
F = AB + CD,
need a form involving A, B, C, D
Apply DeMorgan's Theorem:
F = AB ⋅ CD = ( A + B) ⋅ (C + D)
A
B
C
D
Implementation ⇒
Y
Can also use K-maps:
AB
F = AB + CD
1
1
0
1
1
1
0
1
0
0
0
0
1
1
0
1
CD
For Ntree, minimize 0's; for Ptree, minimize 1's
AB
0
0
Ntree = AB + CD
CD
0
0
0
0
0
AB
1
1
1
1
1
1
1
1
1
CD
Ptree = A⋅C + A⋅D + B⋅C + B⋅D
= A (C + D) + B (C + D)
= (A + B) ⋅ (C + D)
Introduction to Static Load Inverters
1)
resistor load
R
When I = 1, inverter dissipates static power.
O
I
Switching point of inverter depends on ratio of
R to RON (on resistance of nMOS device.
VOH = 5V,
VOL close to 0V, depends on ratio R/RON
Note: output can swing from almost 0V to 5V (VDD)
2)
Again, static power dissipation occurs when
I = 1.
D
S
O
I
Load is enhancement-mode nMOS device.
Note: output swings from nearly 0V to (VDD − VTn)
Using a transistor as a load tends to require much less silicon area than a resistor.
VOH = VDD − VTn,
VOL can be close to 0V, depending on ratio of RON of two enhancement devices
Depletion-mode nMOS
nMOS device with VTn < 0V (negative threshold voltage). Device is always conducting if VGS >
0V.
3)
D
VGS = 0V always
S
O
Load device is always on, looks like a load
resistor.
I
Dissipates static power when I = 1
VOH = 5V; VOL nearly 0V, depending on ratio of RON,dep to RON,enh.
Depletion-mode devices were used before it was economical to put both p-type and n-type
devices on the same die.
4) pMOS device as static load
Here also the load device is always on
(conducting).
S
D
O
Dissipates static power when I = 1.
I
VOH = 5V; VOL nearly 0V, depending on ratio of RON,p to RON,n
Basic MOS Device Equations
Drain
Gate
Bulk (or substrate for nMOS device in n-well technology)
Source
The nMOS device is a four terminal device: Gate, Drain, Source, Bulk.
Bulk (substrate) terminal is normally ignored at schematic level, usually tied to ground for the
nMOS case. In analog applications, however, the bulk terminal may not be ignored.
Gate controls channel formation for conduction between Drain and Source. Drain at higher
potential than Source — Source usually tied to GND to act as pull-down (nMOS).
Three regions of operations — first-order (ideal) equations:
Cutoff region
ID = 0A
VGS ≤ VTn (nMOS threshold voltage)
Linear region

V 
ID = ß  (VGS − VTn )VDS − DS 

2 

2
0 < VDS < VGS − VTn
(
Note: ID is linear with respect to (VGS − VTn) only when VDS
Saturation region
ID =
ß
(VGS − VTn )2 0 < VGS − VTn < VDS
2
2
)
2 is small.
Device parameters:
ß=
transistor gain factor, dependent on process parameters and
device geometry (Kn)
process dependent, constant
ß=
µε W
tox L
under control of the designer
As W/L increases, effective RON of device decreases
µ = surface mobility of the carriers in the channel
ε = permittivity of the gate insulator
tox = thickness of the gate insulator
See Figure 2.5, 2.8 concerning µ, ε, and tox
SPICE represents ß by a factor given by
K' = µCox = µ t
ε
= KP
ox
So,
ID =
K' W
(VGS − VTn )2 ;
2 L
saturation region
VI characteristic
ID
VDS
VGS
boundary between
linear & saturation
regions (dashed line)
|VGS - V T| = |V DS |
ID
V GS4
V
LINEAR
GS
V GS5
SATURATION
V GS3
V GS2
V GS1
CUTOFF
V DS
Things to note:
In the "linear" region, ID becomes less and less linear with VGS as VDS becomes large.
(
This is because the VDS
2
)
2 term in the linear region grows large.
Higher VGS values increase channel conductance allowing for higher values of ID for a
given VDS.
*MOSFET Characteristics
Vds 1 0 DC 10
Vgs 2 0 DC -.723
Vdummy 3 0 DC 0
M1 1 2 3 3 Mfet
.MODEL Mfet NMOS(KP=3686U VTO=2.30 LAMBDA=0.137)
.DC Vds 0 10 .2 Vgs 2.5 5 .5
.probe
.end
What do W and L physically look like?
nMOSFET layout:
Source
n+ diffusion
Gate (polysilicon)
Drain
n+ diffusion
W
L
In digital logic, typically will draw all transistors with the minimum gate length and vary the
width.
Larger W ⇒ larger transconductance (more current flow for given gate voltage), higher gate
capacitance
During fabrication process, the actual width and length of the channel can be reduced by
diffusion from the bulk, source, and drain into the device channel.
SPICE has some MOSFET model parameters to account for this effect, LD and WD, where the
actual the actual length and width is calculated as
Leffective = Ldrawn - 2 × LD
Weffective = Wdrawn - 2 × WD
If LD, WD parameters not specified in the model, then SPICE assumes they are 0.
Ideal Inverter
Vout
VDD
switching
point
VDD
2
Vin
Actual Inverter Characteristics, some definitions
Vout (V)
VOH
VOL
Vin (V)
VIL
VIH
Vth
•
VIL represents the maximum logic 0 (LOW) input voltage that will guarantee a logic 1
(HIGH) at the output
•
VIH represents the minimum logic 1 (HIGH) input voltage that will guarantee a logic 0
(LOW) at the output
Noise Margin
Illustration of Noise Margin:
Vin
Vout
VDD
VDD
Output logic 1
VOH
Input logic 1
NMH
VIH
VIL
NML
Input logic 0
VOL
0V
Output logic 0
0V
Calculate noise margin using
NML = VIL - VOL
How do we determine VIL, VOL, VOH, and VIH?
We must exam the inverter's transfer characteristic.
NMH = VOH - VIH
CMOS Inverter Regions of Operation
A
B
D
E
1.5 10 -4
5
4
1.2 10 -4
3
9 10 -5
C
6 10 -5
2
I DD (A)
V out (V)
Vout
IDD
1
3 10 -5
0
0 10 0
0
1
2
3
4
5
V in (V)
Region A:
0 ≤ Vin < VTn
⇒
pMOS nonsaturated; nMOS cutoff
• nMOS is cutoff because Vin < VTn
Why is the pMOS device in the linear region?
Linear region ≡
VSDp < VSGp - |VTp|
(5 − 5)V < (5 − 0)V − |−0.7|V
[for VDD = 5V and VTp = −0.7V]
0V < 4.3V
Note that the pMOS device can be in linear region even if IDp ≅ 0A!
Region B:
VTn ≤ Vin < Vth
⇒
Why is nMOS saturated?
Is VDSn > VGSn - VTn?
pMOS nonsaturated, nMOS saturated
Because (VDSn = Vout) > Vth and (VGSn = Vin) < Vth ,
then
VDSn > VGSn - VTn
Vout > Vin - VTn
[B-1]
Why is pMOS in linear region?
It started out in linear and will remain in linear as long as
VSDp < VSGp - |VTp|
(VDD - Vout) < (VDD - Vin) - |VTp|
Vin < Vout - |VTp|
[B-2]
Vout in the above expression (Eqn. [B-2]) is decreasing towards Vth and Vin is increasing
towards Vth. When Eqn. [B-2] no longer holds, then the pMOS device will become
saturated.
For the pMOS device, then
regions A ⇒ B ⇒ C correspond to
linear ⇒ linear ⇒ saturated, respectively.
How can you predict the output voltage for region B?
The nMOS is saturated, so IDn =
ßn
ß
(Vin − VTn ) 2 = n (VGSn − VTn ) 2
2
2
The pMOS is linear, so
IDp =
IDp =
ßp
2
ßp
2
(2(V
− | VTp |)VSDp − (VSDp ) 2 )
(2(V
− Vin − | VTp |)(VDD − Vout ) − (VDD − Vout ) 2 )
SGp
DD
Can solve for Vout since
IDn = IDp
VDD
S
IDp
D
D
IDn
S
GND
Vout
Equivalent circuit for region B ⇒
IDn
Region C:
Vin = Vth
⇒
pMOS saturated, nMOS saturated
In order for nMOS to be saturated, need
VDSn > VGSn − VTn
Vout > Vin − VTn
In order for pMOS to be saturated, need
VSDp > VSGp − |VTp|
VDD − Vout > VDD − Vin − |VTp|
Vout < Vin + |VTp|
So Vout in region C,
Vin − VTn < Vout < Vin + |VTp|
The CMOS inverter has very high gain in region C so small changes in Vin produce large
changes in Vout. No closed form equation for Vout. Somewhere in this region, Vout = Vin,
which is the switching point for this gate.
Equivalent circuit for region C:
VDD
IDp
Vout
IDn
What is Vin in region C?
In region C, both devices in saturation so
IDp =
ßp
2
IDn =
(VDD − Vin − | VTp |) 2
ßn
(Vin − VTn ) 2
2
So, using IDn = IDp, Vin can be solved for (more on this later....)
Region D:
Vth < Vin ≤ VDD − |VTp|
Hence,
⇒
IDp =
IDn =
ßp
2
pMOS saturated, nMOS linear
(VDD − Vin − | VTp |) 2
(
ßn
2
2(Vin − VTn )Vout − Vout
2
)
Again, since IDp = IDn, we can solve for Vout:
Vout2 − 2(Vin − VTn)Vout +
using
x=
ßp
ßn
(VDD − Vin − | VTp |) 2 = 0
− b ± b 2 − 4ac
2a
and, recognizing from above,
a = 1, b = −2(Vin − VTn), c =
we get
ßp
ßn
(VDD − Vin − | VTp |) 2
Vout = (Vin − VTn) −
(Vin − VTn ) 2 −
ßp
ßn
(Vin − VDD − | VTp |) 2 .
IDp
Vout
Equivalent circuit for region D ⇒
Region E:
Vin > VDD − |VTp|
⇒
pMOS is cutoff, nMOS is linear mode
Since VSGp = VDD − Vin (< |VTp|),
∴ Vout ≅ 0V
due to nMOS acting as pull-down while pMOS in cutoff.
CMOS Inverter Transfer Characteristic
A
B
D
E
1.5 10 -4
5
4
1.2 10 -4
3
9 10 -5
C
6 10 -5
2
I DD (A)
V out (V)
Vout
IDD
1
3 10 -5
0
0 10 0
0
2
3
4
5
V in (V)
Analysis:
VOH:
1
Vin < VTn, the nMOS transistor is in cutoff while the pMOS transistor is turned-on
(inversion layer established). The result is
VOH ≅ VDD.
VOL:
(VDD − Vin) < |VTp|, the pMOS is in cutoff while the nMOS is on and providing a
conduction channel to ground. Hence,
VOL ≅ 0V.
VIL:
Input low voltage, here the nMOS transistor is saturated and the pMOS is nonsaturated.
Equating the currents provides
ß
ßn
(VIL − VTn ) 2 = p (2(VDD − VIL − | VTp |)(VDD − Vout ) − (VDD − Vout ) 2 ) .
2
2
VIL:
(continued) Since two unknowns exist, Vin = VIL and Vout, a second equation is needed.
Use the unity-gain condition to obtain this second equation,
dVout (∂I Dn /∂Vin ) − (∂I Dp /∂Vin )
=
= −1,
dVin
(∂I Dp /∂Vout )
provides



VIL 1 +
ß n 
= 2Vout +
ß p 
 ßn 
  VTn − VDD − |VTp|.
ß 
 p
Now the two equations needed to solve for VIL and Vout exist.
VIH:
Input high voltage, here the nMOS is nonsaturated and the pMOS is saturated.
Equating the drain currents yields
)
(
ßp
ßn
2
2(VIH − VTn )Vout − Vout =
(VDD − VIH − | VTp |) 2 ,
2
2
the first of two equations needed to solve two unknowns, Vin = VIH and Vout. Use the
unity-gain condition to get the second,
dVout (∂I Dp /∂Vin ) − (∂I Dn /∂Vin )
=
= −1.
dVin
(∂I Dn /∂Vout )
This provides

VIH 1 +

ßp 
ß
 = 2Vout + VTn + p (VDD − | VTp |) ,
ßn 
ßn
the second equation needed to solve for the two unknowns.
Vth:
At the CMOS inverter's switching point, or inverter threshold, Vth = Vin = Vout and both
the pMOS and nMOS transistors are saturated. Again, equating the drain currents,
ß
ßn
(Vth − VTn ) 2 = p (VDD − Vth − | VTp |) 2
2
2
is obtained which can be easily solved to provide Vth,
VTn +
Vth =
Note: switching point of gate (Vth) is
ßp
ßn
(VDD − | VTp |)

ßp 
1 +


ß n 

VDD
2
-if-
ßn
= 1 and VTn = −VTp.
ßp
So, switching point of inverter is function of the ratio of the nMOS/pMOS gains and the
threshold voltages of the nMOS, pMOS transistors.
βn/βp Ratio
The ßn (gain of nMOS) / ßp (gain of pMOS) ratio determines the switching point of the CMOS
inverter.
Vout (V)
Equal pull-up/pull-down
"strength"
5
Strong
pull-up
4
3
ßn = 10
ßp
ßn = 1
ßp
ßn = 0.1
ßp
2
1
0
0
Strong
pull-down
1
2
3
4
5
Vin (V)
VDD
2
Switching point = VDD/2
if ßn/ßp = 1 and VTn = |VTp|
Recall that
µε W
ß=t L .
ox
If we assume that the nMOS and pMOS transistors have equal W/L ratios, then
µnε Wn
tox Ln
µn electron mobility
ßn
=
=
ßp
µp = hole mobility .
µpε Wp
tox Lp
In silicon, the ratio µn/µp is usually between 2 to 3.
This means, that if
Ln = Lp,
then
Wp must be 2 to 3 times Wn
in order for
ßn = ßp .
if
Vout
Wp Wn
ß
=
because n > 1
Lp
Ln
ßp
5
4
3
2
1
Vin
0
0
1
2
VDD
2
3
4
5
Calculate the switching point of a static load inverter as function of ßn/ßp:
In region C, already know nMOS device is saturated from previous analysis.
VDD
Vout
Vin
For pMOS to be saturated need:
VSDp > VSGp − |VTp|
VDD − Vout > VDD − 0V − |VTp|
Vout < |VTp|
Not true!!!
(If Vout in region C is about
VDD
and
2
|VTp|
(typically this is true))
∴ pMOS must be in linear region
VDD
2
>
Then
ß
ß
I Dn = n (VGSn − VTn ) 2 = n (Vin − VTn ) 2
2
2
and
I Dp =
I Dp =
(2(V
2
ßp
SGp − | VTp
|)VSDp − VSDp
2
)
(
2(VDD − | VTp |)(VDD − Vout ) − (VDD − Vout ) 2 )
2
ßp
Equate IDn = IDp and solve for Vout.
ß
Vout = | VTp | + (VDD − | VTp |) 2 − n (Vin − VTn ) 2
ßp
Can also solve for ßn/ßp,
2
2
ß n (VDD − | VTp |) − (Vout − | VTp |)
=
ßp
(Vin − VTn ) 2
Consider again
2
2
ß n (VDD − | VTp |) − (Vout − | VTp |)
=
ßp
(Vin − VTn ) 2
for the pseudo-nMOS inverter.
Let |VTp| = VTn = 0.2VDD and Vin = Vout =
VDD
. Then, for VDD = 5V,
2
ßn
≅ 6.1 !!!
ßp
Note that this is very different result from the CMOS inverter case!
If VDD = 3.3V, but the value of VTn = |VTp| is unchanged (i.e., 1V in the above example),
then
ßn
≅ 11.5
ßp
for a switching point equal to
VDD
.
2
The ßn/ßp ratio depends on the absolute value of VDD! This means that the operation of
the pseudo-nMOS inverter will NOT scale with VDD (for a given CMOS technology).
For the CMOS inverter, the ßn/ßp ratio for a switching point of VDD/2 is independent of
VDD so its operation will scale with supply voltage. This is a another big advantage of
CMOS technology.
Not unusual for static CMOS circuits to operate over a very large range of power supply
voltages, i.e., 2.0V to 6.0V is common.