SHIFTOUT with inverted clock ?

Bean

Is there any way to do SHIFTOUT with the clock inverted ?

I have a device that the clock pin MUST be normally held high, and clocks in data on the rising edge.

When the SHIFTOUT instruction is done the clock is left low, I must make the clock pin HIGH for the device to latch-in the new data. When I do this (make the clock pin high) the device shifts in an extra zero.

What I'm doing now is to only shift in 7 bits on the last byte of data, then set the data pin the last bit, then raise the clock line. Is there any other way ?

Bean.


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Jon Williams

No, but you could write a simple subroutine that is customized to your app:

Get_Byte:
· FOR idx =·0 TO·7
··· PULSOUT ClkPin, 5
··· temp.LOWBIT(idx) = DataPin
· NEXT

This assumes that the clock pin has been initialized high.

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Jon Williams
Applications Engineer, Parallax

Bean

Thanks Jon.

I ended up doing this:

  SHIFTOUT Dat, Clock, MSBFIRST,[noparse][[/noparse]segments(3), segments(2), segments(1), (segments(0) >> 1)\7]
  Dat = segments(0) & 1
  HIGH Clock

Bean.

▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔
"SX-Video·Module"·available from Parallax for only $28.95 http://www.parallax.com/detail.asp?product_id=30012

"SX-Video OSD module"·available from Parallax for only·$49.95 http://www.parallax.com/detail.asp?product_id=30015
Product web site: www.sxvm.com

Coming soon... Cheap 4-digit LED display with driver IC·www.hc4led.com

"Sometimes it is better to remain silent and be thought a fool, than to speak and remove all doubt."
·