Week 11

Report
ECE 551
Digital System Design &
Synthesis
Lecture 11
Verilog Design for Synthesis
Topics
Optimization from the Design Level



Interaction of Description and Synthesis
Critical Path Optimization
High-Level Architectures for Datapaths
2
Overview

In the previous lecture, we looked at ways the
synthesis tool can automatically optimize our logic

In this lecture, we will look at the ways the
designer who is writing the HDL code can optimize
and manage trade-offs.
3
Overview

How you implement something in Verilog can have
a profound effect on what is actually synthesized
(and the effort required to do it!)
 Functionally identical ≠ identical hardware

To be effective, you need to
 Know what it is that you are trying to describe (i.e. not
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

viewing Verilog as an abstract language)
Know how the desired hardware should be organized
Know how the synthesis tools will be likely to implement
a given description
Describe the hardware in a way that causes the synthesis
tools to do what you want
4
Knowing what you want to describe
Case Study: Multiplier
5
4-Input Multiplier

What does the below code describe?
module mult(output reg [31:0] out,
input [31:0] a, b, c, d);
[email protected](*) begin
out = ((a * b) * c) * d;
end
endmodule
6
Multiplier Implementation


Area:
Delay:

How can we improve the delay and/or area?
47381
8.37
7
Multiplier Redux


What are we describing?
How will it compare in speed and area?
module multtree(output reg [31:0] out,
input [31:0] a, b, c, d);
[email protected](*) begin
out = (a * b) * (c * d);
end
endmodule
8
Tree Multiplier


Area:
Delay:
47590 vs. 47381
5.75 vs. 8.37
9
Multiplier – once again...

How can we reduce the area?
module multtree(output reg [31:0] out,
input [31:0] a, b, c, d);
[email protected](*) begin
out = (a * b) * (c * d);
end
endmodule
10
Shared Multiplier [1]
module multshare(output reg [31:0] out,
input [31:0] in, input clk, rst);
reg [31:0] multval;
reg [1:0] cycle;
always @(posedge clk) begin
if (rst) cycle <= 0;
else cycle <= cycle + 1;
out <= multval;
end
always @(*) begin
if (cycle == 2'b0) multval = in;
else multval = in * out;
end
endmodule
11
Shared Multiplier [2]


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Area:
15990 vs. 47590
Critical Path Delay: 3.14
Latency: 3.14 * 4
= 12.56 vs. 5.75
12
Shared Multiplier (cont)

Given that only one multiplier will be allowed for
the implementation, could we have done better
on the latency than the previous example did?
At what cost?
module multtree(output reg [31:0] out,
input [31:0] a, b, c, d);
[email protected](*) begin
out = (a * b) * (c * d);
end
endmodule
13
Knowing what you want to describe
Lesson: You need to think about what
sort of hardware you want to design
from the very beginning of the process.
Synthesis tools will only do so much with
the descriptions you give them.
14
Knowing what you are describing
Case Study: Mixed Flip-Flops
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Mixing Flip-Flop Styles (1)


Say we don’t need to reset q2
What will this synthesize to?
module badFFstyle (output reg q2, input d, clk, rst_n);
reg q1;
always @(posedge clk)
if (!rst_n) q1 <= 1'b0;
else begin
q1 <= d;
q2 <= q1;
end
endmodule
16
Flip-Flop Synthesis (1)


Area = 59.0
Slack = 0.53 (clock = 1ns, input delay 0.2)

Q2 now has to implement a load enable that is
connected to the reset
17
Mixing Flip-Flop Styles (2)
module goodFFstyle (output reg q2, input d, clk, rst_n);
reg q1;
always @(posedge clk)
if (!rst_n) q1 <= 1'b0;
else q1 <= d;
always @(posedge clk)
q2 <= q1;
endmodule
18
Flip-Flop Synthesis (2)


Area = 50.2 (85% of original area!)
Slack = 0.53 (unchanged)

Without the load enable function, flip flop Q2 is
smaller.
Use reset and enable only when you need them!

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Mixing Flip-Flop Styles

Would an asynchronous reset have fixed it?
module badFFstyle2 (output reg q2, input d, clk, rst_n);
reg q1;
always @(posedge clk, negedge rst_n)
if (!rst_n) q1 <= 1'b0;
else begin
q1 <= d;
q2 <= q1;
end
endmodule
20
Flip-Flop Synthesis (3)

Using asynchronous reset instead
 Bad: Area = 58.0, slack = 0.57
 Good: Area = 49.1, slack = 0.57
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Knowing what you are describing
Lesson: If you don’t know the rules of the
language, it’s easy to describe
something different than what you
intended.
Following coding style guidelines makes
this easier.
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Knowing the interpretation
Case Study: Conditional Multiplier
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Conditional Multiplier [1]
module multcond1(output reg [31:0] out,
input [31:0] a, b, c, d, input sel);
always @(*) begin
if (sel) out = a * b;
else out = c * d;
end
endmodule
What would you expect this to generate?
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Conditional Multiplier [2]


Area:
Delay:
15565
3.14
Two 32-bit muxes and one multiplier!
25
Selected Conditional Multiplier [1]
module multcond2(output reg [31:0] out,
input [31:0] a, b, c, d, input sel);
wire [31:0] m1, m2;
assign m1 = a * b;
assign m2 = c * d;
always @(*) begin
if (sel) out = m1;
else out = m2;
end
endmodule
What do you expect here compared to the previous one?
26
Selected Cond. Mult. [2]


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Area:
30764 vs. 15565
Delay:
3.02 vs. 3.14
Why is the area larger and delay
lower?
2 multipliers and a 64-bit mux!
So why did that happen?
27
Resource Sharing Rules

Can happen automatically if variable is assigned
by multiple expressions (if/else) with the same
operation and bit widths
 NO combinational feedback can be caused
 Inputs may be reordered to reduce mux area

The Verilog HDL Compiler operates according to
the following rules for automatic sharing
 No sharing in conditional operators


x = s ? (a+b) : (a+c); //will use two adders
If/else will permit sharing
Manual control is also available – see reading.
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Conditional Multipler – One More Time


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If you know ahead of time that you want two
muxes and one multiplier, describe that directly!
Don’t rely on the synthesis tool to improve
inefficient HDL; describe what you want first.
Caveat: You have to know what you want.
module multcond2(output reg [31:0] out,
input [31:0] a, b, c, d, input sel);
wire [31:0] op1, op2;
assign op1 = sel ? a : c;
assign op2 = sel ? b : d;
always @(*) begin
out = op1 * op2;
endmodule
29
Knowing the interpretation
Lesson: Different ways of describing the
same behavior in Verilog may lead to
different results.
Understanding how the synthesis tool
interprets different Verilog constructs is
a valuable skill to becoming an expert
designer.
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Knowing the Synthesis Tool
Case Study: Decoder Synthesis
31
Decoder Synthesis

Parameterized decoders are commonly written in
one of two ways in Behavioral Verilog
 Use the select input as an index to assert only the


desired output after negating all outputs
Test the select input in a loop for all decoder outputs,
and only asserted the matching output
Will this choice affect
 Circuit delay?
 Circuit area?
 Compiler time?
 Surprisingly, the answer is: Yes, quite a lot, even though
we are trying to describe the exact same hardware!
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Decoder Using Indexing
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Decoder Using Loop
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Decoder Verilog: Timing Comparison
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Decoder Verilog: Area Comparison
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Decoder Verilog: Compile Time Comparison
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Knowing the Synthesis Tool
Lesson: Never forget that in the end, you
are at the mercy of the synthesis tool.
Even when something is part of the
Verilog Standard, you can’t always be
sure it will be supported (or supported
well) by every tool.
This knowledge comes with time.
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Putting it all Together
 If we
 Know what hardware we want
 Know how to describe what we want
 Can interpret the results we get from the
synthesis tool
 Now we can begin making low-level
optimizations
39
Late-Arriving Signals
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
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After synthesis, we can identify the critical path(s)
that are controlling the overall circuit speed, and
which signals are responsible for those path(s).
Assume that one signal to a block of logic is known
to arrive after the others. To deal with this:
Circuit reorganization
 Rewrite the code to restructure the circuit in a way that
minimizes the delay with respect to the late arriving
signal

Logic duplication
 This is the classic speed-area trade-off. By duplicating
logic, we can move signal dependencies ahead in the
logic chain.
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Original Code
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Original Synthesis
What can we do if A is the late-arriving signal?
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Reorganized: Operator In if
Changed the operation from
(A + B) < 24 to A < (24 – B)
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Reorganized: New Hardware
What’s
going on
here?
44
Duplication Example: Original Design
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Original Hardware
ADDRESS
PTR
ADDR
OFFSET
What if control is the late arriving signal?
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Data Duplication : New HDL Code
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Duplication: New Hardware
ADDR1
COUNT1
OFFSET1
ADDR2
COUNT1
COUNT2
OFFSET2
48
Exercise

Assume we are implementing the below code, and
cin is the late arriving signal. How can we
optimize the resulting hardware for speed? At
what cost?
reg [30:0] a, b;
reg [31:0] y;
reg cin;
[email protected](*)
y = a + b + cin;
49
Exercise

Rewrite the code below to
 1. Minimize area
 2. Best performance if sel is late-arriving
reg [3:0] x [3:0];
reg [1:0] sel;
reg [3:0] y, sum;
[email protected](*)
y = sum + x[sel];
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Exercise

Revise to maximize performance wrt late
reg [3:0] state;
reg late, y, x1, x2, x3;
[email protected](*)
case(state)
SOME_STATE:
if (late) y = x1;
else y = x2;
default:
if (late) y = x1;
else y = x3;
endcase
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First, consider how it will synthesize
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Optimized Example
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If you have a small number of case items, the
case select signal will be shorter path, but may be
a long path with a lot of case items.
For non-parallel case statements, the body of first
case item may have a much shorter path than
that of the default case.
If it is a parallel case statement, the case select
signal will be a short path.
Strategy: If possible, move the late signal to the
case select or limit it to the first case item.
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Dealing with late signals in Case
reg [3:0] state;
reg late, y, x1, x2, x3;
[email protected](*)
case(late)
1’b0: if(state == SOME_STATE)
y = x2;
else
y = x3;
1’b1: y = x1;
endcase
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High-Level Datapath Strategies
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Low-level optimizations can be very valuable, but
from a design perspective, the most important
decisions are made at a high level.

Next we will look at three different ways of
architecting a datapath and evaluate their tradeoffs
 Single-cycle
 Multi-cycle
 Pipelined
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Single-cycle Multiplier
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Complete a single computation in one cycle.
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Multi-cycle Multiplier
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Spread one operation over multiple cycles.
One active computation.
Share parts of the datapath to reduce area
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Pipelined Multiplier
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Spread one operation over multiple cycles.
Multiple active computations.
Need extra pipeline registers.
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Evaluating Tradeoffs
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Why might we choose one of these over the other?
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Area – self-explanatory
Throughput – What is the rate of results?
 Product of Frequency and Results/cycle
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Latency – How long does it take to produce one
result?
 Product of Frequency and Cycles/computation
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Single-cycle Multiplier
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Assume the following delays:
32-bit Mult: 6 ns, 64-bit mult 10 ns, Reg Setup: 2 ns
Compute the Throughput and Latency
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Multi-cycle Multiplier
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Assume Control Logic not on critical path
128-bit mux: 3 ns, hybrid multiplier: 7 ns
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Pipelined Multiplier
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Summary
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
High-Level Strategies for tradeoffs between Area,
Latency, and Throughput
Single cycle
 Good: latency – (one long cycle)
 Mixed: throughput - (one output per cycle, but low freq)
 Bad: area

Multi-cycle
 Good: area – (share hardware)
 Bad: throughput, latency – (<1 output per cycle)

Pipelined
 Good: throughput – (one output per cycle, high freq)
 Bad: latency, area – (multiple cycles, extra registers)
63
Conclusions

The designer is responsible for some optimizations
that cannot be achieved by the synthesis tool.

It takes a lot of knowledge to be an expert
designer
 Hardware Design
 HDL
 Synthesis Tool

One of the largest roles of the designer is to
understand tradeoffs and make appropriate
decisions
64

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