Static CMOS Circuits
• Conventional (ratio-less) static CMOS – Covered so far • Ratio-ed logic (depletion load, pseudo nMOS) • Pass transistor logic
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Example 1 module mux(input s, d0, d1, output y);
assign y = s ? d1 : d0; endmodule
1) Sketch a design using AND, OR, and NOT gates.
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1 Example 1 module mux(input s, d0, d1, output y);
assign y = s ? d1 : d0; endmodule
1) Sketch a design using AND, OR, and NOT gates.
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Example 2
2) Sketch a design using NAND, NOR, and NOT gates. Assume ~S is available.
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2 Example 2
2) Sketch a design using NAND, NOR, and NOT gates. Assume ~S is available.
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Bubble Pushing • Start with network of AND / OR gates • Convert to NAND / NOR + inverters • Push bubbles around to simplify logic – Remember DeMorgan’s Law
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3 Example 3
3) Sketch a design using one compound gate and one NOT gate. Assume ~S is available.
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Example 3
3) Sketch a design using one compound gate and one NOT gate. Assume ~S is available.
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4 Compound Gates • Logical Effort of compound gates
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Example 4 • The multiplexer has a maximum input capacitance of 16 units on each input. It must drive a load of 160 units. Estimate the delay of the NAND and compound gate designs.
H = 160 / 16 = 10 B = 1 N = 2
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5 NAND Solution
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NAND Solution
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6 Compound Solution
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Compound Solution
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7 Example 5 • Annotate your designs with transistor sizes that achieve this delay.
Informal homework exercise (see textbook)!
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Input Order
• Our parasitic delay model was too simple – Calculate parasitic delay for Y falling • If A arrives latest? • If B arrives latest?
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8 Input Order
• Our parasitic delay model was too simple – Calculate parasitic delay for Y falling • If A arrives latest? 2 • If B arrives latest? 2.33
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Inner & Outer Inputs
• Outer input is closest to rail (B) • Inner input is closest to output (A)
• If input arrival time is known – Connect latest input to inner terminal
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9 Asymmetric Gates
• Asymmetric gates favor one input over another • Ex: suppose input A of a NAND gate is most critical – Use smaller transistor on A (less capacitance) – Boost size of noncritical input – So total resistance is same
• gA = 10/9 • gB = 2 • gtotal = gA + gB = 28/9 • Asymmetric gate approaches g = 1 on critical input • But total logical effort goes up
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Symmetric Gates
• Inputs can be made perfectly symmetric
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10 Skewed Gates • Skewed gates favor one edge over another • Ex: suppose rising output of inverter is most critical – Downsize noncritical nMOS transistor
• Calculate logical effort by comparing to unskewed inverter with same effective resistance on that edge.
– gu =
– gd =
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Skewed Gates • Skewed gates favor one edge over another • Ex: suppose rising output of inverter is most critical – Downsize noncritical nMOS transistor
• Calculate logical effort by comparing to unskewed inverter with same effective resistance on that edge.
– gu = 2.5 / 3 = 5/6
– gd = 2.5 / 1.5 = 5/3
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11 HI- and LO-Skew
• Def: Logical effort of a skewed gate for a particular transition is the ratio of the input capacitance of that gate to the input capacitance of an unskewed inverter delivering the same output current for the same transition.
• Skewed gates reduce size of noncritical transistors – HI-skew gates favor rising output (small nMOS) – LO-skew gates favor falling output (small pMOS) • Logical effort is smaller for favored direction • But larger for the other direction
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Catalog of Skewed Gates
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12 Catalog of Skewed Gates
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Catalog of Skewed Gates
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13 Asymmetric Skew
• Combine asymmetric and skewed gates – Downsize noncritical transistor on unimportant input – Reduces parasitic delay for critical input
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Best P/N Ratio
• We have selected P/N ratio for unit rise and fall resistance (μ = 2-3 for an inverter). • Alternative: choose ratio for least average delay • Ex: inverter – Delay driving identical inverter
– tpdf =
– tpdr =
– tpd =
– Differentiate tpd w.r.t. P – Least delay for P =
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14 Best P/N Ratio
• We have selected P/N ratio for unit rise and fall resistance (μ = 2-3 for an inverter). • Alternative: choose ratio for least average delay • Ex: inverter – Delay driving identical inverter
– tpdf = (P+1)
– tpdr = (P+1)(μ/P)
– tpd = (P+1)(1+μ/P)/2 = (P + 1 + μ + μ/P)/2
– Differentiate tpd w.r.t. P – Least delay for P =
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P/N Ratios
• In general, best P/N ratio is sqrt of that giving equal delay. – Only improves average delay slightly for inverters – But significantly decreases area and power
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15 Observations
• For speed: – NAND vs. NOR – Many simple stages vs. fewer high fan-in stages – Latest-arriving input • For area and power: – Many simple stages vs. fewer high fan-in stages
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Combinational vs. Sequential Logic
Ou t Lo gic In Lo gic In Ou t Circuit Circuit
State
(a)Co m b ina tion a l (b)Se qu e ntial
Outpu t= f(In) Outpu t= f(I n,Pr eviou s In)
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16 Static CMOS Circuit (Review)
At every point in time (except during the switching transients) each gate output is connected to either VDD or V ssvia a low-resistive path. The outputs of the gates assume at all times the value of the Boolean function, implemented by the circuit (ignoring, once again, the transient effects during switching periods). This is in contrast to the dynamic circuit class, which relies on temporary storage of signal values on the capacitance of high impedance circuit nodes.
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Static CMOS (Review) VD D
In 1 In 2 PUN PMO S On ly In 3
F=G
In 1 In 2 PDN NM O S On ly In 3
VSS
PUN and PDN are Du alNe tw ork s
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17 Properties of Complementary CMOS Gates (Review)
Highno isema rgins:
VOHandVOL are atVDD andGN D ,re s pec t ively .
No stati cpower co nsump ti on:
Ther e nev er ex ists adire c t pa th be t w e e n VD D and VSS(GN D ) inste a dy -s ta te mode.
Comp arable riseandfa lltime s : (u nder th eappr o priate sca lin gco nditions)
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Influence of Fan-In and Fan-Out on Delay
VD D
A B C D Fa n -Ou t:Nu m b e rofGa tes Co n n e c ted Every fanout (output) adds two gate capacitances (pMOS and nMOS) A B Fa n In:Qu a d r a ticTe r m du e to: C 1.Re sistan c eInc re a s ing D 2.Ca p a c itan c e Inc r ea s ing
a 2+ tp = a1 FI+ 2FI a3 FO
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18 Fast Complex Gate-Design Techniques •Tr a nsisto rSizing: As lo ng as Fa n-o utCapacita nce domin ate s
•Pro gre ssive Sizing:
Ou t
In N MN CL M1 >M2 >M3 >MN C3 In 3 M3
C In 2 M2 2
In 1 M1 C1
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Fast Complex Gate - Design Techniques •Tr a nsistorOr d ering
criticalpath criticalpath
C In L CL 3 M3 In1 M1
C In M2 2 C2 2 In2 M2
In C1 C 1 M1 In3 M3 3
(a) (b)
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19 Fast Complex Gate - Design Techniques
•Im proved Logic De sign
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Ratioed Logic
VD D VD D VD D
Resistiv e Dep le tio n PM O S Lo ad RL Lo ad VT<0 Lo ad VSS F F F In 1 In 1 In 1 In 2 PD N In 2 PD N In 2 PD N In 3 In 3 In 3
VSS VSS VSS (a )re sistiv e lo a d (b )de p letio n lo a d NM O S (c )pse u d o -N M O S
Go a l:to re d u c eth e nu m b e rofde v ic esov ercom p le me nta r y CM O S Careful design needed!
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20 Ratioed Logic
VDD •VO H=VDD
RPDN Resistive VOL = VDD RL + RPDN Lo ad RL
Desired: RL >> RPDN (to keep noise margin low) F
t = 0.69R C In1 PLH L L
In2 PDN RPDN In3 Problems: 1) Static power dissipation
2) Difficult to implement a large VSS resistor, eg 40k resistor (typical value) needs 3200 μ2 of n-diff, i.e. 1,000 transistors!
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Active Loads VDD VDD
Depletion PMOS Lo ad VT<0 Lo ad
VSS F F
In1 In1 In2 PDN In2 PDN In3 In3
VSS VSS dep letion load NM O S pseu do-N M O S • Depletion-mode transistor has negative threshold
• On if VGS = 0 • Body effect may be a problem!
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21 Pseudo-nMOS VD D
F C A B C D L
• No problems due to body effect • N-input gate requires only N+1 transistors • Each input connects to only a single transistor, presenting smaller load to preceding gate • Static power dissipation (when output is zero) • Asymmetric rise and fall times Example: Suppose minimal-sized gate consumes 1 mW of static power. 100, 000 gate-circuit: 50 W of static power (plus dynamic power)! (half the gates are in low-output state) • Effective only for small subcircuits where speed is important, eg address decoders in memories
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Pseudo-NMOS NAND Gate
VDD
GND
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22 Pass-Transistor Logic
B
Out A Sw itch Out Inputs Netwo rk B B
Is this transmission gates necessary? AND gate Need a low impedance path to ground when B = 0
•No static consumption
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Pass-Transistor Based Multiplexer
S S
F = AS + BS VDD
Out F
GND
In1 S S In2
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23 Transmission Gate XOR
B 6 transistors only! Case 1: B M 2 B = 1, M3/M4 turned off, F = AB A A Case 2: F B = 0, M3/M4 turned on, F = AB M1 M 3 / M4 B
B
F always has a path to VDD or Gnd, hence low impedance node If not, node would be dynamic, requiring refresh due to charge leakage
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Delay in Transmission Gate Networks
5 5 5 5 V 1 V V i V V n-1 V In i- 1 i+ 1 n
C C C 0 0 C 0 C 0
( a )
R R R R e q V e q e q V e q 1 V i Vi+1 n- 1 V n In
C C C C C
( b) m
Re q R e q R e q Re q R e q R e q In C C C C C C C C
( c ) Insert buffers after every m switches
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24 Delay in Transmission Gate Networks
Consider Kirchoff’s Law at node Vi Vi+1-Vi + Vi-1-Vi C dVi = dt Req Req
Therefore, dV V + V - 2V =i i+1 i-1 i dt ReqC
Propagation delay can be determined using Elmore delay analysis
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Delay Optimization
Delay can be reduced by adding buffers after m stages (tbuf = delay of a buffer)
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25 Transmission Gate Full Adder
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NMOS Only Logic: Level Restoring Transistor
VDD Le velRestore r VD D Mr B M2 X A Mn Out
M1
•Ad vanta ge :Fu llSw in g •Disa dva nta ge :Mo reCo m p le x,La rge rCa pacita nc e •Oth erap proache s:reduc edth resh old NMO S
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26 Single Transistor Pass Gate with VT=0
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Complimentary Pass Transistor Logic
A Pass-TransistorThe image cannot A be displayed. Your computer may not F B Networkhave enough B (a)
A InverseThe image cannot A be displayed. Your Pass-Transistorcomputer may not F B have enough B Network
B B B B B B
A A A B F=AB B F=A+B A F=A (b) A A A
B F= AB B F =A+B A F= A AND/NAND OR/NOR EXOR/NEXOR
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27 4 Input NAND in CPL
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