Omitting a 74HC595 in a DOUTx4 chain

[WickedBlade]

Hello,

in my project I need more than 4x8 digital outputs, so I hooked up 2 DOUTx4 modules (which I will call DOUT1 and DOUT2 for brevity). As it happens, I don't need all 8 chips on them, only 7 out of 8. Since I have already soldered the output pins that I need, I find myself with an unused chip on DOUT1. I was wondering if I could omit plugging the chip in the socket, maybe with some simple additional wiring on the DOUT1 PCB.

I would then have to make slight adjustments to my program, but that's not a problem.

Any hint as to what to do? I don't suppose I can simply omit the chip?

[Wilba]

You can take out a chip and insert a wire in the socket (or preferably solder a wire on the bottom of the PCB), thus continuing the chain.

You need to connect pin 9 and pin 14 of the IC socket without a 74HC595. Refer to the schematic: http://ucapps.de/mbhp/mbhp_doutx4.pdf

This means that the IC sockets following it get shifted as well... i.e. if you took out the 4th IC of DOUT1, the 1st IC of DOUT2 becomes the 4th IC of DOUT1 (from the code's point of view).

[WickedBlade]

Thank you for your answer. That's what I kinda figured out myself but I wasn't too sure. I'll try that, thanks again!