Author Topic: in PID Help to calculate the RAM Information  (Read 7459 times)

sdecorme

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in PID Help to calculate the RAM Information
« on: February 18, 2013, 09:40:22 am »
Hi,
Actually I'm trying to make a card compatible with HID and using the PID . I've succeded to make my device recognized correclty but now I'm loosing my mind to understand the RAM managing.
I've understand that the size of the RAM is proportionnal of the effect. But I don't understand how to calculate the RAM pool , there is some example in the pid documentation but it is really hard to understand .
My device have a single axis effect with all the 12 effects activated.
Could you help me to determine how to calculate and manage this POOL ?
Thanks

Jan Axelson

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Re: in PID Help to calculate the RAM Information
« Reply #1 on: February 18, 2013, 12:46:16 pm »

sdecorme

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Re: in PID Help to calculate the RAM Information
« Reply #2 on: February 18, 2013, 02:55:24 pm »
Thanks,
Hi Jan thanks for your book , it help me a lot ...
Of course I've read this post like all the microchip forum post about the FFB  ;D.
My descriptor work fine and my device is recognized as a FFB with all the FFB applications.
I can receive and send all the frame but now I have to finish my stack and I need to understand how the RAM pool work.
On the microchip post they don't speak about this ?
Thanks


Jan Axelson

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Re: in PID Help to calculate the RAM Information
« Reply #3 on: February 18, 2013, 05:50:27 pm »
The Microchip post talks about how developers have abandoned using HID PID due to lack of support under Windows.

I don't know of any reference or examples other than the spec.

If you have a specific question, maybe someone can help.

I'm glad to hear you found USB Complete useful!

sdecorme

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Re: in PID Help to calculate the RAM Information
« Reply #4 on: February 19, 2013, 01:57:16 am »
But the HID and PID is still under Windows 8 and I think it is a good introduction to make my electronic working, before going to driver develloppement.
Of course your book is usefull , it is a very introduction in the USB wonderful and undocumented world  ??? >:(

Thanks