It means, use memory of 0 AEMIF range beginning from 0x800000 (started from beginning of this range) and with size 0x4000000
On Nov 20, 2013, at 1:03 PM, ivan.khoronzhuk <ivan.khoronzhuk@xxxxxx> wrote:
On 11/20/2013 08:21 PM, Jean-Christophe PLAGNIOL-VILLARD wrote:+ the chip select signal.
+ Minimum value is 1 (0 treated as 1).
+
+- ti,cs-wsetup: write setup width, ns
+ Time between the beginning of a memory cycle
+ and the activation of write strobe.
+ Minimum value is 1 (0 treated as 1).
+
+- ti,cs-wstrobe: write strobe width, ns
+ Time between the activation and deactivation of
+ the write strobe.
+ Minimum value is 1 (0 treated as 1).
+
+- ti,cs-whold: write hold width, ns
+ Time between the deactivation of the write
+ strobe and the end of the cycle (which may be
+ either an address change or the deactivation of
+ the chip select signal.
+ Minimum value is 1 (0 treated as 1).
+
+If any of the above parameters are absent, current parameter value will be taken
+from the corresponding HW reg.
+
+The name for cs node must be in format csN, where N is the cs number.
this is wired we should use reg instead to represent the cs as done for SPI
or a an other property
Best Regards,
J.
Ok, I will add new property cs-chipselect like following :
ti,cs-chipselect: number of chipselect. Indicates on the
aemif driver which chipselect is used
for accessing the memory.
For compatibles "ti,davinci-aemif" and
"ti,keystone-aemif" it can be in range [0-3].
For compatible "ti,omap-L138-aemif" range is [2-5].
Is it OK?
Why do you need this? As it was mentioned just use reg:
So you’d have something like:
memory-controller@21000A00 {
…
nand:cs2@2 {
reg = <2 0 0>;
ranges;
...
}:
};
However, I’m confused by the example in which you have:
+ nand@0,0x8000000 {
+ compatible = "ti,davinci-nand";
+ reg = <0 0x8000000 0x4000000
+ 1 0x0000000 0x0000100>;It means, use memory of 1 AEMIF range + 0x0000000 and with size 0x0000100
+Any of them is not cs number. 0 - is just a memory range number.
+ .. see davinci-nand.txt
+ };
What chipselects is this on 0 & 1?
- k
--
Regards,
Ivan Khoronzhuk
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