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7721 Datasheet(PDF) 381 Page - Renesas Technology Corp

Part # 7721
Description  16-BIT SINGLE-CHIP MICROCOMPUTER
PDF  577 Pages
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Manufacturer  RENESAS [Renesas Technology Corp]
Direct Link  http://www.renesas.com
Logo RENESAS - Renesas Technology Corp

7721 Datasheet(HTML) 381 Page - Renesas Technology Corp

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DMA CONTROLLER
13-102
13.9 DMA transfer time
7721 Group User’s Manual
(2)
Repeat transfer mode
In the repeat transfer mode of burst transfer (edge sense), the method of terminating DMA transfer
___
is only the forced termination by the TC input. Therefore, the time from the CPU’s relinquishing the
___
right to use bus until regaining the right depends on the timing of the TC input.
Transfer

Right to use
bus
CPU
DMAC
CPU
Ž
TC input
1 block
Transfer
Ž
Transfer

Ž
Œ
Transition


Transition
Fig. 13.9.5 Repeat transfer mode (burst transfer mode and edge sense selected)
Œ Transition of the right to use bus from CPU to DMAC: 1 cycle
 DMA transfer per 1 block:
• In 2-bus cycle transfer···(Read cycle + Write cycleg1)
× the number of transfersg2
g1: Add a value which satisfies the read/write conditions. Refer to “Table 13.4.1.”
g2: When the transfer unit is 16 bits, the number of transfers = the number of transfer
bytes/2
When the transfer unit is 8 bits, the number of transfers = the number of transfer bytes
• In 1-bus cycle transfer···Refer to “Table 13.4.5.”
Ž Terminate processing: 3 cycles
___
 DMA transfer of the block at the TC input: above 
The number of transfers is assumed to be up to the DMA transfer of 1-unit transfer which was in
___
progress at the TC input.
 Transition of the right to use bus from DMAC to CPU: 1 cycle
[Example]
External data bus width = 16 bits, 2-bus cycle transfer, transfer unit =16 bits, the number of the
transfer bytes = 10 bytes, and under the following conditions:
• Transfer source: address direction = forward, start address of data = even, with Wait
• Transfer destination: address direction = backward, start address of data = even, without Wait
___
• TC is input when the “m”th byte (m = even) of the “n”th block is in transfer.
Œ + (n – 1) ( + Ž) +  +  = 1 + (n – 1){5(3 + 4) + 3} + m/2 + 1 = 38n + m/2 – 36 cycles



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