Datasheet ee-47 Application Notes (Analog Devices)

Engineer To Engineer Note EE-47

Notes on using Analog Devices’ DSP, audio, & video components from the Computer Products Division

Phone: (800) ANALOG-D or [phone], FAX: [phone], EMAIL: [e-mail]

ADSP-2106x Link Ports - Maximum
Throughput

Last Modified: 04/23/01

This EE Note will present some benchmarks for
Link Port throughput on the ADSP-2106x
floating point family DSPs. General benchmark
calculations will be shown for 6 dedicated link
ports. Also to be discussed is a design that used
3 link ports for transmitting and receiving while
other IOP activity is happening at the same time.

Three calculations are shown below for
maximum sustaining performance through all 6
link ports if no other IOP activity is occurring
on the DSP:

#1
At LCLK = 40MHz, 4 bits/transfer with all 6
link ports enabled, 32 or 48 bit words:
Æ 40MHz * 3bytes/cycle = 120Mbytes/sec
sustained.

#2

At LCLK = 80MHz (LCLKX2x = 1), the link
port can transmit/receive 2x more data for each
core clock cycle. With all 6 link ports enabled,
48 bit words:
Æ 40MHz * 6bytes/cycle = 240Mbytes/sec
sustained.

#3
At LCLK = 80MHz, with all 6 link ports
enabled, 32 bit words, 4 link ports are always
operating at one time for sustained throughput, 2
are always waiting for IOD (I/O Data) service:

Æ 40 MHz * 4bytes/cycle = 160Mbytes/sec
sustained.

Question:

Why can only 4 link ports be sustained if I
transmit 32-bit word at LCLK = 80MHz?

Answer:

4 transfers are needed to transmit a full 32-bit
word at 8bits/cycle. Table1 shows the
progression of #bytes left to be sent at each
clock cycle for each link port. At clock cycle 1,
LP0 starts the transfer and has 4 bytes to send.
At cycle 2, LP1 initiates its first byte and LP0
has 3 bytes remaining. By cycle 4, LP0 has
transmitted out the entire 32-bit word already
but LP4 has not started transmitting yet. There
are no more data for LP0 to sustain a continuous
transfer. The word size is the limiting factor in
the number of link ports that can be
continuously sustained. In 32-bit data word
transfer at LCLK = 80MHz, only 4 link ports
can be sustained.

Cycle 1 2 3 4 5 6

LP0 4 3 2 1
LP1 4 3 2 1
LP2 4 3 2 1
LP3 4 3 2
LP4 4
LP5
Table 1

In the case where the word size is 48 bits, 6
transfers are needed to transmit the entire word.
Table 2 shows the #bytes left to be transmitted
at each clock cycle for all link ports. At cycle 4,
LP0 has still 3 bytes to send and LP4 is
initiating its first byte. At cycle 6, LP0 finishes
sending its 48-bit word while LP5 has already
started sending its first byte. Therefore all 6 link
ports can be sustained (case #2) in transmission
of 48-bit words at LCLK = 80MHz.

a

Cycle

LP0 6 5 4 3 2 1
LP1
LP2
LP3
LP4
LP5
Table 2

1 2 3 4 5 6

6 5
6

6 5 4 3 2

6 5 4 3

6 5 4

-----------------------------------------------------------

As another example, let’s look at a six SHARCs
system that is to be designed.

Five of the SHARCs will be number crunchers,
connected together with the External Port for 32
bits transfers. The last SHARC will act as a
dedicated I/O processor, where 3 of its link ports
are connected up as inputs, while the other 3 are
outputs. The I/O SHARC will then be connected
up with the other SHARCs on the external bus.
It then will use a broadcast write function to
write to the other 5 processors on the board. The
I/O SHARC will not have any other tasks.

Question:

What will be the practical maximum throughput
through the link ports if data comes in 1K
blocks at any of the link ports in a random
fashion?

Answer:

The key thing is that the DMA IOP bus can
sustain a one word per cycle transfer with the
RAM (independent of any core accesses). If
chaining is used, after each DMA completes 8
words (for links ports) must be transferred over
the IOP bus to reinitialize the DMA channel.
These 8 words will share the IOP bus bandwidth
with the Link DMAs and must be subtracted out
of the Max Link bandwidth.

Assume:

• 25ns, 40MHz CLKIN rate

• 32-bit data words sent over Links

• 2x Link Clock mode, gives 40Mbytes/s

per link or 10Mwords/s per link

• Max IOP bus (DMA) bandwidth with
on-chip RAM is 1 word/cycle

• DMA chaining is used to auto initialize
new DMA block transfers

• Each Link port DMA chain takes 8
cycles of IOP bus overhead to restart a
Link port

• 1024 words blocks are DMAed over the

links

Æ Max IOP bus bandwidth = 40Mwords/s
Æ #Link ports that can be sustained = (40Mw/s)

/ (10Mw/s) = 4 Link ports

Therefore, the IOP bus is the bottleneck (for 32­bit words link transfers)

Æ #IOP bus transfers = (1024 cycles/block) + (8
cycles chain overhead/block) = 1032
cycles/block

Æ #Words transferred = (1024 cycles/block)
Æ Overhead for chaining = 1024/1032

Æ Overall Link port transfer rate is (1024/1032)

* 40Mwords/s = 39.68992248062 Mwords/s

If DMA chaining is not used and the core needs
to reinitialize the Link port transfers then there
will be some amount of overhead for interrupt
routines, setup of new DMA pointers, and
enabling of the transfer.

Similar calculations can be applied to the
ADSP-2116x to find the maximum throughput.

Note: There is an anomaly with the revision 2.0
ADSP-21062 link ports. Basically link Buffer 3
can only be used to transmit and not receive.
Remember that there are 6 link buffers and 6
link ports. Any buffer can be logically assigned
to any link port so link buffer 3 can be assigned
to a link port that is only transmitting. Not all

EE-47 Page 2

Notes on using Analog Devices’ DSP, audio, & video components from the Computer Products Division

Phone: (800) ANALOG-D or [phone], FAX: [phone], EMAIL: [e-mail]

rev 2.0 parts will exhibit this problem but due to
the characteristics of the problem we cannot test
and reliably screen out problem parts. Please
refer to the latest ADSP-21062 anomaly list that
describes this problem.

EE-47 Page 3

Notes on using Analog Devices’ DSP, audio, & video components from the Computer Products Division

Phone: (800) ANALOG-D or [phone], FAX: [phone], EMAIL: [e-mail]