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In the following schematic you can see that Q1 and Q4 both are connected to a Arduino's Digital Input Pin. So if Q1 is high and Q4 is low, would there be current flowing from Q1 through Q4 (and through the chip) to Ground - or is there already some protection inside?

schematic without diode

Are the output pins on the 74HC595 shift register protected from connected voltages in the realms of their own VCC? Or do you need to put a diode in?

schematic including diode

I feel like this is a very basic question, which makes it very weird that I didn't find an answer to this, already... :-( Am I looking into the wrong resources? I had a look into the provided datasheet, but couldn't find any information. Didn't get any results from searching for existing posts on Stack Exchange or other sites.

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    The technical term for the top illustration is "shorting outputs". Since the 595 doesn't have open collector outputs, the chip won't stand a short circuit between its output pins. Use diodes instead, for instance, as st2000 suggests.
    – user16306
    Jun 28, 2017 at 13:02

1 Answer 1

Reset to default
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In this data sheet it says:

When the output-enable (OE) input is high,the outputs are in the high-impedance state.

However on page 3 you will notice that the *OE pin controls all outputs at the same time. So if you connect 2 output pins and one is high and the other is low and you hold *OE low ... you will likely burn out the driver for one of the 2 output pins if not the chip its self.

If you add diodes to BOTH outputs you are essentially creating a wired or gate. (You should add a resistor. See the examples in the provided link.) This is fine as long as you understand how a wired or gate works.

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  • Thank you :-) However, this causes a follow-up question: under 10.2.2 in this datasheet it says "Outputs should not be pulled above VCC." I understood this as: outputs (in LOW) can be pulled up (for example by another output in HIGH) as long as it's not above VCC. If that's not what this sentence from the datasheet means, what does it mean then?
    – Liam
    Jun 28, 2017 at 17:05
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    it means that when the output is in a high impedance state (OE pulled high) if you take the pin to a higher voltage than vcc you will damage the IC. if the OE pin is enabled and you try and drive it to any state other than the voltage its alreay at, you are creating a short and would also damage the IC.
    – James Kent
    Jun 29, 2017 at 15:05
  • @JamesKent Ah, of course... I forgot about OE, again. (It's always pulled LOW in my circuit.) So that makes sense now. Excellent. Thank you very much. :-)
    – Liam
    Jun 29, 2017 at 15:27

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