gulu2065
Member
I use the SPI work on slave and read data from another chip. Here is my code.
HardwareSPI spi(1);
...
void setup() {
....
spi.beginSlave(MSBFIRST, 0);
}
void loop() {
byte response = spi.read();
// Print out the response received.
SerialUSB.println(response, BIN);
}
It read data at the speed of 1.6Mhz.
Thanks
mbolivar
LeafLabs
If you have an unread byte, and the SPI master sends another one, then you get an "overrun condition". The second byte sent is lost.
You can check if this has happened by checking the OVR flag in the status register (SR) for the SPI peripheral you're interested in, using this function:
bool hasOverrunOccurred() {
return SPI1_BASE->SR & SPI_SR_OVR;
}hasOverrunOccurred() will return true if you've missed a byte, and false otherwise.
If that happens, you need to clear the OVR flag in the status register. If you don't, then you won't be able to detect later overruns anymore -- hasOverrunOccurred() will just return true forever.
You clear the OVR flag by reading the first unread byte, then reading the status register again. Here's some (untested) code that should do the trick:
// Read the unread byte
byte unread = spi.read();
// Read the status register
uint32 sr = SPI1_BASE->SR;
mbolivar
LeafLabs
No, unfortunately, the USB serial port is not very fast. The USB peripheral is quite fast, but we don't have the necessary library support for fast transfers over USB yet.
And I'm sorry, but I don't know enough about your application to know when you have to tear down your SPI port. Is it very important never to miss a byte? If so, I guess you'd need to stop as soon as the first overrun condition happens. If it's not that important, then maybe you can recover from the error somehow.
gbulmer
Moderator
gulu2065 - I had understood that your use case would allow the program to capture the data, store it, then upload it.
Is that not the case?
If you can store it, try something like:
byte buff[16422];
int i=0;
void loop() {
buff[i++] = spi.read();
if (i>= sizeof(buff)) {
// Print out the response received.
for (int j=0; j<i; j++) {
SerialUSB.println(response, BIN);
}
i=0;
}
}(usual WARNINGS: I have not compiled or tested this code)
gulu2065
Member
Hi glulmer,
It seems the RAM is not that big. When I compile this, it show this error. I guess the buffer is too big, right?
elf section .bss' will not fit in regionram'
c:/users/guo/desktop/maple/hardware/tools/arm/bin/../lib/gcc/arm-none-eabi/4.4.1/../../../../arm-none-eabi/bin/ld.exe: region `ram' overflowed by 1096 bytes
Can I store the data in the flash, I see it is much bgiger than the RAM.
gulu2065
Member
gbulmer,
Thank, that code works!
Now there is no doubt that it is overflow. I am trying to use DMA so solve this problem. But I can't figure out how to do it.
Can you have a look at this ?
thanks
http://forums.leaflabs.com/topic.php?id=1072
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