Taking pin high and low

dev/null

I have a 74HC595 shift register. To prevent the latches from flickering at startup, I need a pull-up on the enable pin on the IC.
I set all latches to low, and then enable the IC.

This works, but I am wondering if the schematic is OK? (I'm newbie at eletronics)

I tried this circuit and it works. Code looks like this

  HCBits = 0                                                            ' Set all latches to low
  SHIFTOUT pHCIO, pHCCLK, MSBFIRST, [noparse][[/noparse]HCBits]      ' Send the bits.
  LOW 12                                                                ' Enable the 74HC595
  PULSOUT pHCLtc, 1                                               ' Transfer to outputs.

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dev/null

Should I put a capacitor between Vdd and 10K?

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Mike Green

You should put a capacitor (usually 0.1uF ceramic) between Vdd and Gnd on every chip, particularly a largish chip that might drive medium power loads like the 74HC595.

Your initialization "glitch" is better handled by using the /RESET line which will set all the outputs to zeros. Normally the /OE line is left with just the 10K pullup to Vdd.

If you have access to the reset pin on the Stamp, you would connect the /RESET line there so the 74HC595 would be reset whenever the Stamp is reset. The Stamp reset pin already has a pullup.

Post Edited (Mike Green) : 7/12/2009 2:57:26 PM GMT

Chris Savage

Hello, I just read a post from you regarding this IC and a pull-up on the output. Please do not double post messages. Post follow-up replies in your original thread.

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Chris Savage
Parallax Engineering

dev/null

Sure Chris. It was related. Sry. Thx Mike.

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Mike Green

Correction ... the /OE line needs to be tied to ground with a pulldown (so it's enabled). Sorry.

dev/null

I figured that Thx Mike.

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