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SN74LV139A: Open-Drain Operation

Kyle Pawlowski
Prodigy 20 points
Part Number: SN74LV139A
Other Parts Discussed in Thread: SN74LVC07A

I am using this decoder chip in my design. This design requires the decoder to be powered down while the outputs are connected to CS pins of battery powered chips that stay powered on. These pins are externally pulled-up with 4.7k resistors to 3.3V power. I was having some trouble getting the chip to work the way I wanted to. The pull-ups were backfeeding through to the Vcc pin and preventing the outputs to go into high-Z mode when the chip lost power. I ended up connecting the Vcc pin to a really strong pull-down to get it to go into high-Z mode.

Looking for an alternative way to connect the chip, I found that the chip works like an open-drain device when I leave the Vcc pin floating. It seems to work exactly how I'd like it to. I'm nervous though because this doesn't seem like defined behavior that is spelled out in the datasheet. Are there any big issues I could run into using the decoder in this way?

  • 0
    Guru 316440 points

    Outputs that drive high are connected to VCC. To get Ioff, you'd have to actively force VCC to 0 V (and this requires discharging any decoupling capacitors).

    WIth VCC disconnected, it still gets powered through any high outputs. It looks like open drain because the outputs that drive high cannot be stronger than the pull-up resistors. There is no guarantee that the chip can power up this way, and any other devices connected to the same VCC will be powered, too.

    If possible, move the SN74LV139A into the part of the circuit that stays powered. Alternatively, add an open-drain buffer (e.g., SN74LVC07A).

  • 0 in reply to Clemens Ladisch
    Prodigy 20 points

    Ok, that gives me a hint. I think the value pull-down I need to force Vcc to 0V depends on the values of the pull-ups on the outputs. I had a really strong pull-down before because I was trying to beat the pull-ups. 

    If I do want to power it through the outputs, is there a way to guarentee that the chip will power this way? What will prevent it from working? It is the option that requires fewer components. 

  • 0 in reply to Kyle Pawlowski
    Prodigy 20 points

    Status update: it looks like the value of the pull-ups on my outputs have little to do with the value of the pull-down that is required on Vcc to send it into high-Z mode. Is there a spec that I can point to that should determine the value of that pull-down resistor for the Vcc line? That would make me feel better than just experimenting to find a value that works. 

  • 0 in reply to Kyle Pawlowski
    Guru 316440 points

    The power consumption of the decoder itself is negligible. You need to overpower all pull-up resistors, and be able to discharge any decoupling capacitors.

    I cannot recommend this. Add an open-drain buffer.

  • 0 in reply to Clemens Ladisch
    Prodigy 20 points

    How do people normally drive the Vcc pin to 0V to make use of Ioff? Really, the reason I picked out this chip was to have a part in-between the battery-powered and non-batter-powered parts of my circuit that has defined behavior when not powered. If I need to add another part to get it to work, I'll probably just try to find a different part that I can use in a similar way.