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64-bit DMA transaction for PCIe interface

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Old 04-12-2007, 05:03 AM   #1
Edward Hsu
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64-bit DMA transaction for PCIe interface

Dear All,

I am porting AVStream driver to a TV capture addon card with PCIe
interface. This PCIe card supports 64-bit DMA transaction with ScatterGather
feature and it is a DMA bus master. I used IoGetDmaAdapter function to get
DMA_ADAPTER structure. However, the WDK document about "DEVICE_DESCRIPTION"
said that
************************************************** *********
If the ScatterGather member is set to TRUE and the InterfaceType member is
set to PCIBus, the Dma32BitAddresses member is ignored and the device is
assumed to support 32-bit DMA addresses.
************************************************** *********
Does it mean that I cannot use the 64-bit DMA transaction of this PCIe card?
How can I enable 64-bit DMA transaction?

Thanks,
Edward
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Old 04-12-2007, 05:07 AM   #2
Alexander Grigoriev
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Re: 64-bit DMA transaction for PCIe interface

The device is assumed to support _at least_ 32 bit address. If you set
Dma64BitAddresses, this means your board will get 64 bit physical addresses
in a scatter-gather list..

"Edward Hsu" <EdwardHsu@discussions.microsoft.com> wrote in message
news:632DDFF6-90D5-40BE-BC4C-981FF4764000@microsoft.com...
> Dear All,
>
> I am porting AVStream driver to a TV capture addon card with PCIe
> interface. This PCIe card supports 64-bit DMA transaction with
> ScatterGather
> feature and it is a DMA bus master. I used IoGetDmaAdapter function to get
> DMA_ADAPTER structure. However, the WDK document about
> "DEVICE_DESCRIPTION"
> said that
> ************************************************** *********
> If the ScatterGather member is set to TRUE and the InterfaceType member is
> set to PCIBus, the Dma32BitAddresses member is ignored and the device is
> assumed to support 32-bit DMA addresses.
> ************************************************** *********
> Does it mean that I cannot use the 64-bit DMA transaction of this PCIe
> card?
> How can I enable 64-bit DMA transaction?
>
> Thanks,
> Edward



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