MK64FX interface with TFT(LCD) and NOR flash(SST39VF802C)

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MK64FX interface with TFT(LCD) and NOR flash(SST39VF802C)

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NEIL4
Contributor II

I have done interfacing TFT(LCD) with flexbus. Used Chip Select-CS0.

Now Im interfacing NOR Flash(SST39VF802C) in  16bit multiplexed mode via Latch, Have some couple of questions.

1.Chip Select guide, I need FB_RW# signal on both the interface.

2. FB_AD[0] connection with latch and the flash.

3. How to execute code from the NOR flash.

4. How set CSAR: 

0x6000_0000–0x7FFF_FFFF      FlexBus (External Memory - Write-back)

0x8000_0000–0x9FFF_FFFF      FlexBus (External Memory - Write-through)

 

Thank You

 

 

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jingpan
NXP TechSupport
NXP TechSupport

Hi @NEIL4 ,

1. yes, FB_RW can shared by all devices on FlexBus.
2. FB_AD[0] connect to Flash_D[0] only. Needn't connect to address line because SST39VF801 is word address only.
3. The attachment is a small test project which copy a small piece of code to MRAM and execute in MRAM. I think it can run in NOR flash also (but of course NOR flash erase/program is different).
4. CSAR[BA] is base address. BA is set to 0x6000 if base address is 0x60000000, and BA is set to 0x8000 if base address is set to 0x80000000. CSMR[BAM] is the memory size, it means how many 64k space is occupied by this device. For example, if it is 1M byte, then BAM=16.

 

Regards,

Jing

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1,144 次查看
jingpan
NXP TechSupport
NXP TechSupport

Hi @NEIL4 ,

1. yes, FB_RW can shared by all devices on FlexBus.
2. FB_AD[0] connect to Flash_D[0] only. Needn't connect to address line because SST39VF801 is word address only.
3. The attachment is a small test project which copy a small piece of code to MRAM and execute in MRAM. I think it can run in NOR flash also (but of course NOR flash erase/program is different).
4. CSAR[BA] is base address. BA is set to 0x6000 if base address is 0x60000000, and BA is set to 0x8000 if base address is set to 0x80000000. CSMR[BAM] is the memory size, it means how many 64k space is occupied by this device. For example, if it is 1M byte, then BAM=16.

 

Regards,

Jing

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