CMOS Static NAND Gate
■
Second switching condition: VA = VDD and VB switches from 0 to VDD
VDD
VDD
M3
ID
M4
VGS1 = VDD
VM
VM
M2
+
VGS2 = VM − VDS1
ID1 = ID2
M1 VDS1
_
VDS
(a)
(b)
At VB = VM, the current through M1 and M2 is higher than when VA = VB since
the gate voltage on M1 is now VDD and its VDS1 must be smaller --> VGS2 is
larger. Effective kn is increased.
At VB = VM, only M4 is conducting current --> only half the current as for
previous switching condition. Effective kp is that of device M4
■
Result: VM is 0.3-0.5 V lower than for VA = VB switching condition, for typical
dimensions
EE 105 Spring 1997
Lecture 17
NAND Gate Transfer Functions
■
SPICE is useful to solve for the transfer functions under the various switching
conditions (see Ex. 5.7). Note that the backgate effect means that the curves
when VA switches and when VB switches are not identical.
VOUT
(V)
Transfer Functions
5.0
4.0
B SWITCHES
A SWITCHES
A & B SWITCH
3.0
2.0
1.0
0.0
0.0
■
1.0
2.0
3.0
4.0
5.0
VIN (V)
Results: setting kn = 2 kp results in VM approximately VDD/2.
EE 105 Spring 1997
Lecture 17
CMOS NAND Gate Transient Analysis
■
Worst-case situation for low-to-high transition: only one of the p-channel
transistors is switching (say M4):
kp
2
Ð I Dp = Ð I D4 = ----- ( V DD + V Tp )
2
■
■
For high-to-low transition, consider M1 and M2 in series with effective length at
2Ln (worst-case since current is lowest with VA = VB)
kn
2
2
I Dn = I D = I D2 = µ n C ox [ W n ⁄ 2 ( 2L n ) ] ( V DD Ð V Tn ) = ----- ( V DD Ð V Tn )
1
4
For equal propagation delays, we require IDn = -IDp
kn
kp
----- = ----- --> kn = 2kp
4
2
The factor of 2 mobility difference between the p and n channels indicates that
(W/L)n = (W/L)p (2 input NAND gate)
■
For an M-input NAND gate, we find that
(W/L)n = (M/2) (W/L)p
Note: NOR gates suffer from a factor of 2M between the n- and p-channel ratios
which makes them unattractive for large fan-in gates
EE 105 Spring 1997
Lecture 17
CMOS Dynamic Logic
■
Static NOR gate
VDD
A
M4
B
M3
Q
A
M1
B
M2
CL
Idea: n-channel and p-channel devices separately perform the same logic function.
replace p-channels with a resistor -->
Q = A+B
replace n-channels with a resistor -->
Q = AB
... two functions are identical by DeMorganÕs Theorem
■
Let n-channels perform the logic and get rid of the pull-up devices
(or vice versa)
EE 105 Spring 1997
Lecture 17
n-Channel CMOS Dynamic Logic
clock signal φ(t) charges up load capacitance through MP (P = precharge) when
it transitions from high to low; ME (E = evaluate) is cutoff and prevents any
discharge path of CL through logic function transistors.
■
clock signal goes high --> MP is cutoff, ME conducts --> CL discharges if one of
the logic transistors has a high input.
■
VDD
φ
φ
A
B
MP
C
M
CL
+
VOUT
VDD
−
evaluate
evaluate
precharge
0
Logic
Function
φ
ME
(a)
■
t
(b)
Payoffs:
1. large fan-in NOR gates without huge p-channel load devices (also, avoids
backgate effect on loads)
2. tends to be fast due to smaller load capacitances
■
Drawback:
1. clock is essential to refresh logic level stored on CL, which complicates the
design
EE 105 Spring 1997
Lecture 17
n-Channel Dynamic Logic Propagation Delays
■
Consider ÒtPLHÓ to be the time required to pre-charge the output node
Precharge Circuit (tPLH)
φ=0
VDD
MP
VOUT (t = 0) = 0 V
CL
A
B
φ=0
■
M
ME (cutoff)
Charging current
kp
2
Ð I Dp = ----- ( V DD + V Tp )
2
EE 105 Spring 1997
Lecture 17
n-Channel Dynamic Logic Propagation Delays
■
Consider ÒtPHLÓ to be the worst-case time to evaluate the logical function after
clock goes high.
Evaluate Circuit (tPHL)
(only one input high)
φ=5V
VDD
MP (cutoff)
VOUT (t = 0) = 0 V
CL
5V
0V
φ=5V
■
0V
ME
Discharging current: assume (W/L)E = (W/L)A = ... (W/L)M and note that the
transistors are in series -->effective value is kn / 2
kn
2
I D = µ n C ox  ----- ( V DD Ð V Tn )
 4
n
EE 105 Spring 1997
Lecture 17
Boolean Functions in Dynamic Logic
■
Examples:
VDD
φ
VDD
φ
MP
ME
Q
A
A
B
C
D
B
C
Q
φ
φ
ME
(a)
■
MP
(b)
(a) n-channel dynamic logic
Q = ( A + B )C
■
(b) p-channel dynamic logic
The output is Òpre-dischargedÓ to zero by MP and is only charged if there is a
path through the logic transistors when the clock goes low and ME conducts.
Q = AB + C + D
EE 105 Spring 1997
Lecture 17
CMOS Transmission Gates
■
Need: ÒgateÓ signals by having a series switch that can be shorted or opencircuited.
C
C
IN
OUT
OUT
IN
C
C
(a)
■
(b)
Why n-channel and p-channel in parallel? Only one device (say, n-channel):
canÕt pass an input voltage > VDD - VTn, since device will enter the cutoff region
.
G
G
5V
S
D
5V
B
0V
(a) charge-up
5 V − VTn
0V
D
S
0V
0 V − VTp
B
5V
(b) discharge
EE 105 Spring 1997
Lecture 17
Pass Transistor Logic
■
Advantages: reduced transistor count and higher speed compared with static
CMOS
■
Disadvantage: reduced noise margins
B
B
A
B
B
B
OUT
OUT
A
B
(a)
(b)
Switching
Network
(c)
EE 105 Spring 1997
Lecture 17