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Digital Logic Synthesis for Memristors
MemristorMemristor●One type of new emerging nano-devices
●Memory-Resistor postulated by Leon Chua in 1971
●First physical implementation found by HP in 2008
Memristor
Memristor in Digital Logic
●Threshold device– Crossing threshold switches the
resistance/conductance of the memristor– Information is stored in the resistive state
●Non-volatile resistance– No refresh needed
●Can switch in nano-seconds with pico-joule energy (cite HP paper)
➔Has the potential for high density, low power logic and memory circuits
Published synthesis methods1.1. Julien Borghetti, Julien Borghetti, Gregory S. Snider, Philip J. Kuekes, J. Joshua Yang, Duncan
R. Stewart, and R. Stanley Williams. R. Stanley Williams. /‘memristive/’ switches enable /‘stateful/’ logic operations via material implication. Nature, 464:873 –876, 4 2010.
2.2. A. Chattopadhyay and Z. RakosiA. Chattopadhyay and Z. Rakosi. Combinational logic synthesis for material implication. In VLSI and System-on-Chip (VLSISoC), 2011 IEEE/IFIP 19th International Conference on, pages 200 –203, Oct. 2011. Different assumptionsDifferent assumptions
3.3. E. Lehtonen and M. LaihoE. Lehtonen and M. Laiho. Stateful implication logic with memristors. In Nanoscale Architectures, 2009. NANOARCH ‘09. IEEE/ACM International Symposium on, pages 33 –36, July 2009.
4.4. E. Lehtonen, J.H. Poikonen, and M. Laiho. E. Lehtonen, J.H. Poikonen, and M. Laiho. Two memristors suffice to compute all boolean functions. Electronics Letters, 46(3):239 –240, 4 2010.
5.5. J.H. Poikonen, E. Lehtonen, and M. Laiho. J.H. Poikonen, E. Lehtonen, and M. Laiho. On synthesis of boolean expressions for memristive devices using sequential implication logic. Computer-Aided Design of Integrated Circuits and Systems, IEEE Transactions on, 31(7):1129 –1134, July 2012.
6.6. D.B. Strukov, A. Mishchenko, and R. Brayton, D.B. Strukov, A. Mishchenko, and R. Brayton, Maximum Throughput Logic Synthesis for Stateful Logic: A Case Study, preprint. Different assumptionsDifferent assumptions
Material ImplicationMaterial Implication●Conditional logic with multiple interpretations
– If p then q– q follows p– ...
●Paradox of entailment– An argument (p->q) is false, iff the premise (p) is
true and the conclusion (q) is false
p
qp qppq = q = p p q q
IMPLY Logic
●Two memristors can perform material implication with one pulse – IMPLY
●Consider memristors as a switch with two states – Ron, Roff
●Voltage drop over P affects voltage drop over Q
●Result will be stored in Q
– Q is input and output memristor
IMPLY Logic - Notes
●Explains the conditions for Q changing its state
●Q is pre-set to “0” (low conductance / high resistance)
●Voltage level V_Rg determines voltage drop over Q
●Only if P = “0” V_Rg remains low and allows Q to change
Material Implication with Memristors
Material Implication - Notes● 5 pulses, 4 input patterns
●First two pulses to show initial input values of P and Q
●Third pulse performs operation (two pulses applied simultaneously to P and Q)
●Fourth and fifth pulses to read resulting values of P and Q
● Note different pulse amplitude during pulse three (V_set, V_cond)
Why are we interested in that?
●CMOS technology scaling is approaching limits
●Main limitation in modern CPUs is heat
●2-terminal device of 10nm size
– Allow much higher/denser device integration
●Switching between states can be done with pico Joule
Building NAND from IMPLY
●IMPLY & FALSE is a computationally complete set of operators
●2 input memristors and one work memristor can build NAND gate– Having NAND we are creating a link to known
logic synthesis algorithms
Memristors in digital logic
●IMPLY and FALSE is a complete set of operations to perform all boolean logic functions
●3 memristors can perform a NAND operation
p1p1
p2p2 p3p3
Two-input NAND needs three pulses
Examples of implication uses
((ab) 0) = (a’+b’)
((ab) (ab)) = (a’+b’) + ab
(0 ab) = (ab)
(a b) = (a’+b)
ab
a + b
a
ba + b
a0
ab ( a + b) = a * b
0
aa + b = (a * b)b
0
All these circuits assume that value of b already exists.If it does not exist, we need two inverters (from IMPLY) to create it.
ab
a + b
a
ba + b
a0
ab ( a + b) = a * b
0
a
(a * b)
b0 0
All these circuits assume that value of b already exists.If it does not exist, we need two inverters (from IMPLY) to create it.
Now we assume Now we assume that all inputs that all inputs
must be created must be created with Stateful with Stateful
IMPLY technology IMPLY technology from scratch.from scratch.
NOT & NOT & OROR
NOT & ORNOT & ORNOT OR with two inputs
A
0
B
x
A
Bx
A
0x
A x
B
00
NOT
A
0x
A x
B
C
A
0
NOT is a one WM NOT is a one WM gate gate
AA
B
2-input OR is a 2-input OR is a two WM gatetwo WM gate
AA0
B
X=A+BB0
0
A+BA+B
BB
00
BB
2-input OR2-input OR
AA
BB
00
AA
00
00
NAND & AND
NAND NAND & AND& AND
NAND & AND
NAND
A
0
B
x
A
Bx
NAND & ANDNAND & AND
B
C
A
0
00
2-input AND is a 2-input AND is a two ancilla gatetwo ancilla gate
2-input NAND is 2-input NAND is a one ancilla a one ancilla
gategate
0 NAND(a,b)
AND(a,b)
00
NAND & ANDNAND & AND
0
b
c
d
Working (memorizing) memristor (ba) =b’+a
(c(ba) c) = (c’+(b’+a)=(bc)’+a
(bcd)’+0
0
y
z
v
(yzv)’ + 0
0
Realization of a Sum Realization of a Sum of positive Productsof positive Products
bcd+yzv
Imply serves as inverter
NAND(b,c,d)
NAND(y,z,v)
1
2
1
2
Two Two Working Working MemristorsMemristors
SOPSOP
Inhibit gate Inhibit gate A * B’ = (A’ + B)’ 2 gates
B
C
A
0
00
2-input INHIBIT 2-input INHIBIT is a two WM gateis a two WM gate
A’ A’ Two working bits
B’ B’
A’ + B
00
A * B’ = (A’ + B)’
NOR NOR GatesGates
NORNORB
C
A
A0 00
0
NOR is a two WM bit NOR is a two WM bit gategate
A A+B
B00
(A+B)’
EXOR EXOR GatesGates
EXOR = EXOR = 8 literals in NAND 8 literals in NAND = 8 IMPLY= 8 IMPLY
B
A
00
EXOR is a EXOR is a three WM gatethree WM gate
00
A’B + A B’
A’A’
A’A’
B’B’
BB
AA
AAA + B’A + B’
00 A’BA’B
00B’B’
BB B + A’B + A’00
SYNTHESIS WITH SYNTHESIS WITH EXORS WITH NO EXORS WITH NO
LIMIT ON LIMIT ON NUMBER OF NUMBER OF
ANCILLA BITSANCILLA BITS
B
A
00
00
A’B + A B’
A’A’
AAA + B’A + B’
00 A’BA’B
00 B’B’
BB B + A’B + A’00
C
00
00
A’B + A B’
A’A’
AAA + B’A + B’
00 A’BA’B
00 B’B’
BB B + A’B + A’00
First Working Memristor
Second Working Memristor
Third Working Memristor
Fourth Working Memristor
This circuit has 4 This circuit has 4 working working memristors and 16 memristors and 16 IMPLY gatesIMPLY gates
16 IMPLY gates, 4 WM16 IMPLY gates, 4 WM
B
A
00
00
A’B + A B’
A’A’
AAA + B’A + B’
00 A’BA’B
00 B’B’
BB B + A’B + A’00
MUXMUX
MUXMUXB
0
MUX is MUX is a three a three ancilla ancilla gategate
00
A
C
(AB)’(AB)’
AB + A’CAB + A’C
AA
A’A’
BB
CC
(A’C)’(A’C)’00
00A’A’
AA
(A+C’)’(A+C’)’
7 WM expected
Circuits from reversible gates versus circuits from memristor material
implications
SimilaritiesNo fanout
In-gate memory exists
DifferencesNo inverter
Different gates
Examples of Examples of typical multi-typical multi-input gatesinput gates
B
0
C
A
A A + B A + B + C = (ABC)
Realization of positive product Realization of positive product
(negated) which is (negated) which is multi-input multi-input NANDNAND
0
A A + B A + B + C
Realization of positive product Realization of positive product
(negated) which is (negated) which is multi-input multi-input OROR
AA
BB
CC
Area and DelayArea and Delay
●Area– In CMOS: number of gates– Memristive logic: number of input + work memristors
●Delay– In CMOS: number of logic levels– Memristive logic: number of gates + number of
FALSE operations
Our Understanding of Lehtonen’s Our Understanding of Lehtonen’s Algorithm: Algorithm: Synthesis with K-mapsSynthesis with K-maps
Synthesis with K-maps
21 IMPLY gates, 2 WM21 IMPLY gates, 2 WM
Variants of synthesis algorithms
1. Find all groups in every level (Lehtonen).
1. Find only groups corresponding to essential primes and secondary essential primes.
1. Find minimum SOP cover of F and SOP cover of F’ and use groups that correspond to primes from these covers.
1. Use cost heuristics related to best prime selection.
All primesSecondary essential primes in red
Secondary essential primes in red
Kernels of Primary and Secondary essential primes
0 0 0
00
All essential primes of F’ in red
0
All kernels of essential primes of F and F’
01 1
We take kernel of the first level (in red)
We do not take another kernel of the first level because it was not a kernel of an essential implicant
00
X
X
We invert the function
0
X
X
1
11
1
1 1
11
0 0
0
0
0
We select the group being kernel of essential prime of F’
0
X
X
1
11
1
1 1
11
0 0
0
0
0
Do not select this group
0
X
X
1
11
1
1 1
11
0 0
0
0
0
We select the group being kernel of essential prime of F’
0
X
X
1
11
1
1 1
11
0 0
0
0
0
Replace with don’t cares
0
X
X
1
1X
1
1 1
X1
0 0
0
0
0
We invert the function
0X
X
X
X
0X
X
X
X
We select the group being kernel of essential prime of F’
0
X
X
1
1
1
1 1
11
0 0
0
0
We invert the function
X
X
X
X
0X
X
X
X
We select the group being kernel of essential prime of F’
X
X
0
X
X
1
1
1
1 1
11
0 0
0
0
X
X
X
X
0
X
X
1
1
1
X X
1X
0 0
0
0
X
X
X
X
0X
X
X
X
X
X
We invert the function
X X
X
0X
X
X
X
X
X
X X
X
We select the group being kernel of essential prime of F’
0
X
X
1
1
1
X X
1X
0 X
X
0
X
X
X
X
We invert the function
0X
X
X
X
X
X
X X
X
We select the group being kernel of essential prime of F’
X
X
0
X
X
1
1
1
X X
1X
0 X
X
0
X
X
X
X
0
X
X
1
1
1
X X
1X
0 X
X
0
X
X
X
X
Groups selected
Number of levels does not count, number of pulses counts to cost
3 pulses each 4 pulses
2 pulses2 pulses 1 pulse
Our method replaces primes from minimal cover with bigger positive groups
1
0
Groups selected
0
1. We have more groups than in SOP2. But groups have less literals3. We have more inverters for layers4. We have no inverters for primes5. This tradeoff causes big differences between costs of SOP
and our method for various functions6. Interesting research topic
ABC and Automated Logic SynthesisABC and Automated Logic Synthesis
1. Can we use existing tools to perform synthesis?
1. How do we integrate memristor logic to these tools?
1. Are the results valid with respect to memristor logic specifics (area, delay)?
ABC
●From Alan Mishchenko, UC Berkeley
●System for synthesis and verification of binary sequential logic circuits
●AIG based
Technology Mapping
IPMPLY in ABC
●Genlib file
●Technology Mapping
ABC Output● Delay is the number of gates + number of
memristor initializations
● Area is the number of input + work memristors
– Cannot be controlled from ABC
● One issue that is special in memristor logic: Fanout
p
qp qppq = q = p p q q
NotationNotationp q
p q
Parallel Fanout
Green line – previous node will not be overwritten
Red line – previous node will be overwritten
Computing n9 first, overwrites n8
Computing n10 first, overwrites n7
One node (n7 or n8) has to be copied/recomputed
Solving Parallel Fanout
What is cheaper?
Copy or recompute?
Which node is cheaper to recompute?
Copying might require one additional work memristor and two pulses
Recomputing n8 requires one pulse
Recomputing n7 requires 4 pulses➔ Recompute n8
Post Processing ABC resultsPost Processing ABC results
●Fanout requires post processing
●Two strategies:– Store each value with fanout in an
additional memristor (2 inverter)– Recompute the whole sub-circuit
that caused fanout
●Area/Delay trade-off
●Fanout is increasing delay and likely the area as well
Post Processing
ABC results
BenchmarksBenchmarks
For more results, comparison with other SOP and ESOP based methods see poster by Anika Raghuvanshi
Benchmarks - NotesBenchmarks - Notes● Pulse Count (Anika)
●Solution for minimum number of work memristors (minimum area)
●Follows similar approach as presented with the k-maps
●Pulse count = delay● Gate count
●Number of IMPLY gates as computed by ABC●Can contain fan-out that has to be post-processed●Is not area optimized (more than 2 work
memristors)● Pulse Count (ABC)
●Post processed to avoid harmful fan-out●Still not area optimized
ConclusionsConclusions1. Very little published on synthesis with IMPLY gates
1. Very little published on synthesis with memristors.
1. Although logic synthesis for memristors may seem similar to standard SOP or multi-level combinational synthesis, it is different because of assumption of minimal level number of Working Memristors?
2. We created methods to synthesize circuits with minimum number of working memristors
3. We created methods to synthesize circuits with small but not minimal (3, 4) working memristors.
1. Research question: “how important is this assumption for future memristor technologies?”
Future worksFuture works1. Synthesize for given fixed number of working
memristors
2. Compare various synthesis methods:
1. SOP
2. ESOP
3. TANT
4. NAND Tree
5. Bi-decomposition
6. Ashenhurst-Curtis decomposition
3. Analyze tradeoffs between various methods for various types of functions (symmetric, unate, linear, self-dual, etc).
Future worksFuture works1. Synthesis of pipelined, systolic circuits
2. Synthesis of Finite State Machines and sequential circuits built from blocks.
3. Fuzzy and multiple-valued circuits.
4. Exact synthesis
The important characteristic of a memristor is shown in the graph in Figure 2(b), where the steep curve shows the low resistance, as shown by line AB (the ‘on’ state of the memristor) and the flatter curve shows the high resistance (the ‘off’ state of the memristor) as shown by line interval CD.
Memristor’s state described by interval AB can also be called as ‘closed’ or in binary state definitions as ‘1’ or ‘true’.
Similarly, the state described by line interval CD can also be called ‘open’ or in binary state definitions as ‘0’ or ‘false’.
When voltage is increased beyond certain point, shown as Vopen, the state of the memristor changes from closed to open (transition point B to C in the diagram).
Now as the voltage is decreased and goes through the zero point, the resistance stays the same until the negative voltage exceeds Vclose.
At this point the state changes from open to closed (shown by transition from point D to A).
If the voltage remains between VClose and VOpen, then there is no change in the state of the memristor.
The change from state open to closed and closed to open, allows memristor to act as a binary switch.
And the fact that the state remains the same when the voltage is between Vopen and Vclose provides the important ‘memory’ property.
Even when the voltage is removed, the state will remain the same, and is remembered.
Observe that while a transistor is a three-terminal device, a memristor is only a two-terminal device which simplifies the layout.
Figure 4(a) shows the circuit when the state of memristor P is ‘0’ (open).
P has a high resistance, and can be thought of as disconnected, which implies that the voltage across grounding resistor is zero.
This means that the voltage across the memristor Q is equivalent to Vset.
As shown in Figure 2(b), Vset is greater that Vclose.
The high voltage causes the state of Q to become ‘1’ regardless of Q’s original state (‘0’ or ‘1’).
Figure 4(b) shows the circuit when the state of memristor P is ‘1’.
Figure 4: Workings of IMPLY gate using two Memristors. (a) Output when P=0, (b) Output when P=1
Now P has a low resistance, and can be thought of as a wire, which implies that the voltage across the grounding resistor is now the same as Vcond, the voltage applied at P input.
This means that the voltage across Q is equivalent to Vset-Vcond.
Refering to Figure 2(b) again, the magnitude of Vset-Vcond is less than Vclose, and is not enough to switch the state of Q irrespective of its previous state.
This means that if Q’s state was ‘0’, it will remain ‘0’. If the state was ‘1’, it will remain ‘1’.
Figure 4(b) shows the circuit when the state of memristor P is ‘1’.
Logic Synthesis for Memristors
●Find all product implicants with positive literals only
●Replace with don't care
●Invert
●Repeat until all '1' covered
examples
((abc)’+ (ab))’ + (bcd) =
((a’ + b’ + c’) + ab)’ + bcd =
(abc) * (ab)’ + bcd =
(abc) * (a’ + b’) + bcd =
(b’c) + bcd
c, Experimental
direct-current current–voltage switching characteristics (four-probe
method).
Traces b–f are offset.
Trace a shows a closed-to-open transition,
trace b shows stability and trace c shows an open-to-closed transition.
Traces d–f repeat this cycle.
d, Switch toggling by pulsed voltages (2 ms long; VSET525V and VCLEAR519 V). Non-destructive reads at 20.2V test the switch state.
• Figure 2 Illustration of the IMP operation for the four input values of p and q.
• a, IMP is performed by • two simultaneous voltage pulses, VCOND
and VSET, • applied to switches P and Q,
respectively, • to execute conditional toggling on
switch Q depending on the state of switch P.
• b, The truth table for
the operation q’ q’ p p IMP IMP qq. .
p q
Figure 2c, The blue and red curves show the voltagesapplied and the absolute value of the currents read at junctions P and Qbefore and after the IMP voltage pulses.
The measured low- and high-currentvalues reproduce the IMP truth table.