Download Pass-Transistor Logic

Pass-Transistor
Logic
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Pass-Transistor Logic
Inputs
B
Switch
Out
A
Out
Network
B
B
• N transistors
• No static consumption
Primary inputs drive the gate terminals + source-drain terminals. In contrast to static CMOS –
primary inputs drive gate terminals.
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Example: AND Gate
B
When B is “1”, top device turns on and copies the input
A to output F. When B is low, bottom device turns on
and passes a “0”.
The presence of the switch driven by B is essential to
ensure that the gate is static – a low-impedance path
must exist to supply rails.
Adv.: Fewer devices to implement some functions.
Example: AND2 requires 4 devices (including inverter to
invert B) vs. 6 for complementary CMOS (lower total
capacitance).
A
B
F = AB
0
NMOS is effective at passing a 0, but poor at pulling a node to Vdd. When the pass transistor a
node high, the output only charges up to Vdd-Vtn. This becomes worse due to the body effect.
The node will be charged up to Vdd – Vtn (Vs)
Vs = Vdd − (Vtn0 + γ ( 2Φ f + Vs − 2Φ f ))
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NMOS-Only Logic
3.0
In
s
0.5µ m/0.25µ m
1.5µ m/0.25 µm
Out
0.5µm/0.25 µm
Voltage [V]
VDD
In
2.0
Out
s
1.0
0.0 0
0.5
1
1.5
2
Time [ns]
Vs is initially 0. Vs will initially charge up quickly, but the tail end of the transient is slow. The
current drive of the transistor (gate-to-source voltage) is reduce significantly as Vs
approaches Vdd-Vtn (the current available to charge up node “s” is reduced drastically.
For cascading, the output of a pass transistor (#1) should not drive the gate of another MOS
device (#2). This will produce an output = Vdd-Vtn1-Vtn2
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Energy Consumption
In
VDD
s
0.5µ m/0.25µ m
1.5µ m/0.25 µm
Out
0.5µm/0.25 µm
Pass transistors require lower switching energy to charge up a node, due to the reduces
voltage swing. The output node charges from 0 -> Vdd-Vtn, and the energy drawn from the
power supply for charging the output of a pass transistor is given by CL.Vdd(Vdd-Vtn )
While lower switching power is consumed, it may consume static power when output is high –
the reduced voltage level may be insufficient to turn off the PMOS transistor of the subsequent
CMOS inverter.
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Complementary Pass Transistor Logic
A
A
B
B
Pass-Transistor
Network
F
To accept and produce true
and complementary inputs
and outputs.
(a)
A
A
B
B
B
Inverse
Pass-Transistor
Network
B
B
A
F
B
B
A
A
B
F=AB
A
B
F=A+B
F = AB
AND/NAND
A
F=A ⊕Β
(b)
A
A
B
B
F =A+B
B
OR/NOR
A
F = A ⊕Β
EXOR/NEXOR
• Since circuit is differential, complimentary inputs and outputs are available. Although generating differential
signals require extra circuitry, complex gates such as XORs, MUXs and adders can be realized efficiently.
• CPL is a static gate, because outputs are connected to Vdd or GND through a low-resistance path (high noise
resilience).
• Design is modular – all gates use same topology; only inputs are permuted. This facilitates the design of a
library of gates.
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Main Problems of NMOS-only Switch
C = 2.5V
C = 2.5 V
M2
A = 2.5 V
A = 2.5 V
B
CL
B
Mn
M1
VB does not pull up to 2.5V, but 2.5V - VTN
Threshold voltage loss causes
static power consumption + slower transition
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NMOS Only Logic:
Level Restoring Transistor
VDD
VDD
Level Restorer
Mr
B
A
Mn
M2
X
Out
M1
• Advantage: Full Swing. Eliminates static power in inverter + static power through level restorer and pass
transistor, since restorer is only active when A is high.
• Restorer adds capacitance, takes away pull down current at X – contention between Mn and Mr (slower
switching). Hence Mr must be sized small. Mn and Mr must be sized such that the voltage at node X drops
below the threshold of the inverter VM, which is a function in the sizes of M1 and M2.
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Solution 2: Single Transistor Pass Gate with
VT~0
V
Use very low threshold values for NMOS
pass transistors, and standard highthreshold devices for inverters.
DD
0V
Note: Body effect will still cause an
increase in the threshold voltage.
2.5V
VDD
0V
VDD
Out
2.5V
WATCH OUT FOR LEAKAGE CURRENTS
(DC Sneak path)
While these leakage paths are not critical
when the device is switching constantly,
they do pose a large energy overhead
when the circuit is in the ideal state.
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Solution 3: Transmission Gate
C
NMOS passes a strong “0”
PMOS passes a strong “1”
C
A
Transmission gates enable rail-to-rail swing
A
B
C
C
These gates are particularly efficient in
implementing MUXs
S
A
B
C = 2.5 V
A =2.5 V
VDD
B
CL
M2
C= 0 V
F
S
M1
B
S
F=(AS+ BS)
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6 devices vs. 8 for complementary CMOS
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Another Example: Transmission Gate XOR
6 devices (including inverter for B) vs. 12 for
complementary CMOS
B
For B=1, M3 & M4 are off, M1 & M2 are on. F = AB
For B=0, M1 & M2 are off. M3 & M4 are on. F = AB
Regardless of the value of A & B, node F is
connected to Vdd or GND (static gate)
B
M2
A
A
F
M1
B
M3/M4
B
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Resistance of Transmission Gate
30
2.5 V
Resistance, ohms
Rn
20
Rp
2.5 V
Rn
Vout
Rp
10
0
0.0
0V
R n || Rp
1.0
Vout, V
2.0
Rn {(Vdd-Vout)/In} & Rp {(Vdd-Vout)/Ip} are in parallel. The currents through devices are
dependent on value of Vout and hence the operating mode of the transistors. During the lowto-high transition, the pass transistors traverse through a number of operation modes.
Since Vd and Vg = Vdd, the NMOS is either in saturation or off. The PMOS changes from
saturation to linear during the transient.
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30
2.5 V
Resistance, oh ms
Rn
20
q
q
2.5 V
Vout
Rp
10
0
0.0
q
Rp
Rn
0 V
Rn || R p
1.0
Vout , V
2.0
Vout < |Vtp|: NMOS and PMOS saturated
|Vtp| < Vout < Vdd – Vtn: NMOS saturated, PMOS linear
Vdd -Vtn < Vout: NMOS cutoff, PMOS linear
Req is relatively constant. Thus when analyzing transmission-gate networks, the
simplifying assumption that the switch has a constant resistive value is
acceptable.
Rn =
Vdd − Vout
=
In
Similarly,
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Vdd − Vout

(Vdd − Vout ) 2 
k n (Vdd − Vtn )(Vdd − Vout ) −

2


V −V
1
Rp = dd out ≈
Ip
k p (Vdd − Vtp )
≈
1
k n (Vdd − Vtn )
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Delay in Transmission Gate Networks
2.5
2.5
V1
In
Vi
Vi-1
C
0
2.5
Vi+1
C
0
2.5
0
Vn-1
C
C
Vn
0
C
(a)
Req
In
Req
V1
Req
Vi
C
C
Vi+1
Vn-1
C
C
Req
Vn
C
(b)
n
t p = 0.69∑ CReq k = 0.69CReq
k =0
n( n + 1)
2
tp is proportional to n 2 and increases rapidly with the number of switches in the
chain.
Solution: Insert buffers.
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Delay Optimization
2.5
2.5
V1
In
2.5
Vi
Vi-1
C
0
2.5
C
0
Vn-1
Vi+1
C
0
Vn
C
C
0
(a)
Req
Req
V1
In
Req
Vi
C
Vn-1
Vi+1
C
C
Req
Vn
C
C
(b)
m
Req
Req
Req
Req
Req
Req
In
C
CC
C
C
CC
C
(c)
m( m + 1)   n 
n
t p = 0.69 CReq
+  − 1tbuf
Linear dependence on n instead of n2

2
m
 m 
t buf
∂t p
To find mopt, then
= 0 yielding mopt = 1.7
CReq
∂m
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