问题
I'd like to read differential voltage values from an MCP3304 (5v VDD, 3.3v Vref, main channel = 7, diff channel = 6) connected to an RPi 2 b+ as close as possible to the MCP3304's max sample rate of 100ksps. Preferably, I'd get > 1 sample per 100µs (> 10 ksps).
A kind user recently suggested I try porting my code to C for some speed gains. I'm VERY new to C, so thought I'd give Cython a shot, but can't seem to figure out how to tap into the C-based speed gains.
My guess is that I need to write a .pyx file that includes a more-raw means of accessing the ADC's bits/bytes via SPI than the python package I'm currently using (the python-wrapped gpiozero package). 1) Does this seem correct, and, if so, might someone 2) please help me understand how to manipulate bits/bytes appropriately for the MCP3304 in a way that will produce speed gains from Cython? I've seen C tutorials for the MCP3008, but am having trouble adapting this code to fit the timing laid out in the MCP3304's spec sheet; though I might be able to adapt a Cython-specific MCP3008 (or other ADC) tutorial to fit the MCP3304.
Here's a little .pyx loop I wrote to test how fast I'm reading voltage values. (Timing how long it takes to read 25,000 samples). It's ~ 9% faster than running it straight in Python.
# Load packages
import time
from gpiozero import MCP3304
# create a class to ping PD every ms for 1 minute
def pd_ping():
cdef char *FILENAME = "testfile.txt"
cdef double v
# cdef int adc_channel_pd = 7
cdef size_t i
# send message to user re: expected timing
print "Runing for ", 25000 , "iterations. Fingers crossed!"
print time.clock()
s = []
for i in range(25000):
v = MCP3304(channel = 7, diff = True).value * 3.3
# concatenate
s.append( str(v) )
print "donezo" , time.clock()
# write to file
out = '\n'.join(s)
f = open(FILENAME, "w")
f.write(out)
回答1:
There is probably no need to create an MCP3304 object for each iteration. Also conversion from float to string can be delayed.
s = []
mcp = MCP3304(channel = 7, diff = True)
for i in range(25000):
v = mcp.value * 3.3
s.append(v)
out = '\n'.join('{:.2f}'.format(v) for v in s) + '\n'
If that multiplication by 3.3
is not strictly necessary at that point, it could be done later.
来源:https://stackoverflow.com/questions/38231123/trying-to-read-an-adc-with-cython-on-an-rpi-2-b-via-spi-mcp3304-mcp3204-or-m