Question about pin fanout power.

mynet43

Part of a circuit I've designed includes 17 74HC595 8-bit shift registers.

To minimize the number of propeller control pins, I need to connect one propeller output pin to 17 input pins on the 74HC595 chips. This is true for each of three signals, SCLR, SCK and G.

Does anyone have experience with a similar situation? I.e. can one propeller pin drive 17 input pins on a 74HC chip without using a buffer chip?

Thanks for your help.

Jim

Carl Hayes

The Philips data sheet for 75HC595 specifies an input leakage current that won't exceed 6.5 microamps per device. 17 inputs, therefore, won't exceed 110 microamps -- so a Prop pin ought to be able to drive them.

However, each input has capacitance (I didn't look to see how much), and this will have the effect of slowing down the slew rate of the Prop outputs. Depending upon the risetime (and falltime) you need, that could be a problem (but I'd say it ought not to be).

Why not try it, and see?

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· -- Carl, nn5i@arrl.net

mynet43

Thanks Carl, I appreciate another set of eyes.

It looked like it should work, but 17 is a pretty big number of devices to drive with one pin.

I don't think the capacitance will be a problem, because the driver is written in spin, rather than assembly.

Better to figure it out now instead of after I get the boards back.

Thanks again,

Jim

Carl Hayes

mynet43 said...

Better to figure it out now instead of after I get the boards back.

Don't worry.· The first lot of PC boards will have the Vcc and Vss backward anyway.· Or something.· I keep a Dremel tool and very small carbide burr handy, for cutting PC board traces in the first lot.

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· -- Carl, nn5i@arrl.net

mynet43

I know. Right now it only has 143 parts on the board (really), so it should be easy, right? I usually end up with a pattern wrong or two parts that won't fit at the same time because they're too close.

I'm about to take a deep breath and do it. Thanks for the help!