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I'm building a small battery powered device which contains an ATTiny85 micro controller, a dual common anode 7 segment display and 2x595 shift registers to drive the 7 segment display.

The device will spend most of its time sleeping. While in sleep mode, the 7 segment display will be off. The 595s have a worst case quiescent current of 160uA so with both of them on its a full 320uA of wasted current.

What would be the most efficient method of turning off the 595s when they are not used?

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    \$\begingroup\$ I read this as "2595 shift registers" and thought it was a bit much. But anyways, obviously you want to remove the power from the 595s, in whatever way you are going to do it, keep in mind that some ICs can be powered by non-vcc pins too (A while ago there was an eevblog video about it) \$\endgroup\$
    – PlasmaHH
    Jul 15, 2016 at 10:25
  • \$\begingroup\$ Have you considered using a larger MCU that can drive the display directly? \$\endgroup\$ Jul 15, 2016 at 10:37
  • \$\begingroup\$ @IgnacioVazquez-Abrams I only have attiny85s on hand at the moment. I could get a larger MCU but I would have to wait a few days for shipping. \$\endgroup\$
    – bitshift
    Jul 15, 2016 at 10:52
  • \$\begingroup\$ FYI: Have you considered low-power micro processors with built in LCD drivers that can stay on even when most of the micro processor is in sleep mode? That is, the LCD will always be displaying. Even when the processor is not awake to change it. \$\endgroup\$
    – st2000
    Jul 15, 2016 at 12:54

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What would be the most efficient method of turning off the 595s when they are not used?

You will have to disable their outputs, this is done by making the not(OE) (output enable) pin HIGH.

Disconnecting the supply lines of the chip will work ONLY if you can guarantee that none of the other pins of the 595 will be high. Because if only one pin is high, it will supply power to the chip. See this EEVBlog video for an explanation on that.

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  • \$\begingroup\$ It really depends on the particular IC. A lot will not be powered through inputs. For obvious example consider a simple CMOS gate. \$\endgroup\$
    – jbord39
    Jul 15, 2016 at 14:50
  • \$\begingroup\$ @jbord39 It really depends on the particular IC No it does not. It's not the CMOS gate itself, it's the ESD protection diodes that cause this effect. All modern ICs have such ESD protection on their inputs, it is a must or they can never meet ESD requirements. Your remark shows that you know little about this subject or you would not have made that statement. Not all datasheets mention the ESD protection explicitly but that does not mean it's not there. \$\endgroup\$ Jul 15, 2016 at 16:04
  • \$\begingroup\$ Fair enough for the CMOS example, I'm still not convinced all logic families would suffer from this though. \$\endgroup\$
    – jbord39
    Jul 15, 2016 at 16:34
  • \$\begingroup\$ Thanks for clarification. I do have some experience doing design and layout in 22nm, 14nm, and 10nm technology; and your statements are overly broad. For example if the n-substrate used for the PMOS are electrically isolated between the ESD diode and circuit power; and only electrically connected during high power. There are a huge number of different process technologies that implement their ESD in different ways. And I do not mean CMOS in particular. Although it does seem like most of the IC's would have this behaviour based on your info. \$\endgroup\$
    – jbord39
    Jul 15, 2016 at 16:57
  • \$\begingroup\$ I'm still not convinced all logic families would suffer from this though. Show me an example where there are no ESD diodes between an input and VCC. For 99% of ICs, especially simple logic, ESD protection will be with diodes to VCC and VSS as it is proven technology. For the latest 10 nm processor sure it can be different but that often also makes them more sensitive to ESD damage. \$\endgroup\$ Jul 15, 2016 at 21:00
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Try using a lower power device like the MC74VHC595 - it specifies a maximum supply current of only 4 uA at 25 degC. At up to 125 degC its supply current is less than 40 uA. This applies to the ON semi and Fairchild parts.

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You could use a small P-Channel Enhancement Mode MOSFET in the supply line of the shift registers to switch the supply. A BSS84 will work fine with 3.3V and 5V logic levels.

If you do this, make sure that you don't drive any of the shift-register inputs while the shift-register is turned off. Otherwise you may power the chip via it's inputs.

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  • \$\begingroup\$ Even worse, the chip can be powered from its outputs as well. \$\endgroup\$ Jul 15, 2016 at 10:52
  • \$\begingroup\$ On the output he could use some diodes.Like Schottky diode. \$\endgroup\$ Jul 15, 2016 at 10:55
  • \$\begingroup\$ The 7 segment display I was planning on using is a common anode one. To prevent an output from powering the 595s, could I somehow use an n channel mosfet to switch ground? \$\endgroup\$
    – bitshift
    Jul 15, 2016 at 10:55
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    \$\begingroup\$ You can switch Ground to the 595's, just make sure that all their inputs are high when doing it. \$\endgroup\$ Jul 15, 2016 at 11:59
  • \$\begingroup\$ This is like a power gating headswitch in conventional VLSI design. \$\endgroup\$
    – jbord39
    Jul 15, 2016 at 14:53

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