S32K144 ADC

取消
显示结果 
显示  仅  | 搜索替代 
您的意思是: 

S32K144 ADC

3,852 次查看
raviranjankumar
Contributor III

Hi,

   I am working on S32K144 EVB. I want to configure more than 8 channel at a time. I am using PDB trigger in back to back mode. Up to 9 channel it is working fine. but after that I am not getting result. I am configuring PDB0 channel0 for first 8 channel and PDB0 channel1 for next 8 channel. To trigger  first 8 channel I am giving CH0C1.EN = FF, 

CH0C1.BB = FE (PDB0 channel0) and same for PDB0 channel1 to configure next 8 channel. Is it correct ? 

My configurations are ADC_ALTCLK1, 12-Bit Mode, Divide ratio is 4, sample time is 12.

Please help me. waiting for response....

Thanks and Regard

Ravi Ranjan Kumar.

5 回复数

3,384 次查看
frankzeng
Contributor I

Hi Ravi,

         Did you configure it successfully? And can you share your code for me? I can not configure it successfully.

0 项奖励
回复

3,384 次查看
PetrS
NXP TechSupport
NXP TechSupport

Hi Ravi,

the PDB0 ch0 to PDB0 ch1 back-to-back chaining is not possible. PDB back to back operation forming a ring is supported only on Channel 0 and so just PDB0-PDB1 ring is possible if SIM_CHIPCTL[PDB_BB_SEL]=1.

pastedImage_1.png

BR, Petr

0 项奖励
回复

3,384 次查看
raviranjankumar
Contributor III

Hi Petr,

      Thanks for reply. That means if I want to use more than 8 channel in back to back mode then I can use ADC0  eight channels and ADC1 eight channels ( I tested it is working fine) . In single ADC module I can't use more than 8 channels. Is it correct?

Thanks and Regards,

Ravi Ranjan.

0 项奖励
回复

3,383 次查看
PetrS
NXP TechSupport
NXP TechSupport

Hi Ravi,

yes, the back-to-back connection between channels within PDB module is not available. Tables 41-4/41-5 gives you a summary.

If you want to use more than 8 channels on single ADC module then you need to trigger each PDB0 channels separately with proper delay, so the chain is finished before next is started. 

BR, Petr

3,383 次查看
raviranjankumar
Contributor III

Hi Petr, 

         Thank you so much. It was really helpful for me.

Regards,

Ravi Ranjan.

0 项奖励
回复