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| 型號(hào)： | 935263151557 |
| 廠商： | NXP SEMICONDUCTORS |
| 元件分類： | 消費(fèi)家電 |
| 英文描述： | SPECIALTY CONSUMER CIRCUIT, PQFP240 |
| 封裝： | 32 X 32 MM, 3.40 MM HEIGHT, MSQFP-240 |
| 文件頁(yè)數(shù)： | 176/518頁(yè) |
| 文件大?。?/td> | 7111K |
| 代理商： | 935263151557 |
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Philips Semiconductors
Arbiter
File: arb.fm5, modified 7/23/99
PRELIMINARY INFORMATION
19-7
For example, with the same settings as in the example of
Section 19.5.1, then
E2 is (3 + 2) / 2 = 2.5
EVO is (3 + 7) / 3 * 2.5 = 8.33,
which gives
BVO = (80 - 1.216) * 64 / [ 20*8.33 + 16*2 / (5/4) ]
resulting in 26.23 million bytes per second which corre-
sponds to 25.01 MB/s.
Note: In order to compute the latency Bx when a unit is
not enabled its weight has to be considered as 0 in the
E{2,3,4,5,6} equations and in E{AI,AO,VLD} for AI, AO or
VLD.
The maximum amount of requests Ax for device x al-
lowed during Mcycles period is:
Ax = floor(Bx / S)
Where floor(X) is the greatest integral value less than or
equal to X.
Note: This number does not take into account the worst
case pattern for request acknowledgment. Thus if the pe-
riod is too small Ax is not accurate.
19.6
EXTENDED BEHAVIOR ANALYSIS
The following sections try to give a more accurate behav-
ior of the TM1100 arbitration system.
19.6.1
Extended Bandwidth Analysis
The minimum bandwidth allocation derived from the ar-
biter settings is accurate if one of the two following con-
ditions are true:
The units emit requests all the time (i.e. do back-to-
back requests).
After a request has been acknowledged, the unit
emits a new request before the new arbitration point.
The arbitration is decided around every 16 cycles.
This time depends on the direction of the transac-
tions (read/write).
In TM1100 the only unit able to almost sustain back-to-
back requests is the data cache. The other units will post
a request and wait for the data before the next request is
posted. This behavior makes the bandwidth computa-
tion:
almost accurate if the unit is down in the arbiter hier-
archy (true if the units placed above are enabled).
rather inaccurate if large weights are used for a unit.
Since no back-to-back requests are implemented the
worst case is that a unit can only get one request out of
three if all the others are asking. This limits the use of
large weights for other units than the data cache.
However some units might be able to catch one request
out of two. This depends on the way requests interleave,
since the arbitration point is dependent on the type of the
request (read or write) as well as on the CPU ratio.
This makes it almost impossible to describe the behavior
precisely.
The exact bandwidth necessary for units like VO, VI, AO
or AI (see dedicated sections in each corresponding
chapter) are well known. If the arbiter settings allocate
more bandwidth for these units than they can use, the
extra bandwidth can be used by the units that are located
below these units (VO, VI) or at the same level (AO and
AI) in the arbiter hierarchy.
As an example, with the default settings, VO gets 25% of
the available bandwidth and the CPU gets 50%. If the
SDRAM clock speed is 100 MHz, then 100 MB/s are al-
located to VO. If VO runs at 27 MHz (NTSC or PAL
mode), then VO will not use all this allocated bandwidth.
Thus any of the units that are below VO can potentially
use the remaining allocated bandwidth.
In other words - even if only 10% are allocated to one unit
like the CPU, PCI or the ICP, it may use more.
19.6.2
Extended Latency Analysis
Some units (VO and VI) have a latency/bandwidth re-
quirement and their behavior needs to be simulated in or-
der to find out the correct settings. For example the re-
quirement for VO (in image mode 4:2:2 or 4:2:0 without
up scaling, overlay disabled) is:
During 128 VO clock cycles, VO block needs to
have 2 requests acked ([2 Ys, one U and one V]/2).
The default value 0 for ARB_BW_CTL leads to a bus al-
location of 50% for CPU, 25% for VO and 25% for L3
blocks.
The worst case arbitration for VO is then: CPU L3 CPU
VO, CPU L3 CPU VO to which the refresh (K), internal
delays (T) and E for the first CPU request need to be
added.
The first VO request will require 129 SDRAM cycles
(DVO = 5 or from the worst case pattern 19 + 10 + 20 + 4
* 20).
The arbitration pattern shows that the following request
will require (in the worst case) an extra 4 * 20 SDRAM cy-
cles. Thus VO clock speed cannot be greater than
61.24% (128 / [129 + 80]) of the SDRAM clock speed.
By changing the settings to 33% for the CPU, 33% for VO
and 33% for L3 blocks (i.e. CPUweight = 1, L2weight = 2,
VOweight = 1, L3weight = 1), the new SDRAM/VO clock
percentage becomes 75.74% (128 / [109 + 60]) corre-
sponding to a worst case arbitration pattern of CPU L3
VO, CPU L3 VO.
Before changing the settings the minimum SDRAM
speed to run VO at 74.25 MHz (High Definition speed)
was 122 MHz. After the new allocation 100 MHz is fine.
Note that here DVO remains equal to 5.
E
5
VI
weight
L5
weight
+
L5
weight
------------------------------------------------
E
4
×
=
E
6
PCI
weight
L6
weight
+
L6
weight
----------------------------------------------------
E
5
×
=
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