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I'm trying to create a battery-backed-up non-volatile RAM using an SRAM IC (AS6C1008 from Alliance Memory, datasheet).

My idea for separating the backup battery from the main voltage source is to use two diodes like this:

Circuit

Because of the diodes, the RAM chip will only receive about 4.5 V as Vcc (depending on the diode), but that shouldn't cause any problems, since its operating range is 2.7 V to 5.5 V.

But all the logic inputs in the HIGH state would be connected to the external voltage source, 5 V, which falls into the operating range of the RAM but is higher than its Vcc.

The absolute maximum ratings in the datasheet of the RAM say nothing about this case.

Can I use it this way, or do I have to make the logic input voltages lower somehow? If I have to lower them, what would be the simplest way to do that? (The RAM has 29 logic inputs, so I would like to avoid building complex circuits for each of them). Or should I go with a different method for separating the two voltage sources?

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    \$\begingroup\$ The limititations on input voltages not being allowed above VCC have to do with the ESD protections inside the IC. Learn more about that by watching this video by Dave from the EEVBlog where he powers and IC through it's input pin! youtube.com/watch?v=2yFh7Vv0Paw \$\endgroup\$ Commented Dec 4, 2021 at 12:31
  • \$\begingroup\$ @Bimpelrekkie - Been there, seen that! Only in my case it was a broken VCC bond wire inside the chip package. \$\endgroup\$
    – SteveSh
    Commented Dec 4, 2021 at 12:44
  • \$\begingroup\$ Dallas/Maxim (as well as others) have ics specifically for this purpose. A popular choice might be the max690 series. Battery backup and watchdog. \$\endgroup\$
    – Kartman
    Commented Dec 4, 2021 at 13:01
  • \$\begingroup\$ Can all the other components in your design operate at 4.5 V? If so, just put another diode in series with their Vcc pins, so your entire circuit runs at 4.5 V. If not, put another diode in anyway, and increase your supply voltage to 5.5 V. \$\endgroup\$ Commented Dec 5, 2021 at 1:33

3 Answers 3

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The absolute maximum ratings in the datasheet of the RAM say nothing about this case.

Sometimes, it's not where you expect to see it: -

enter image description here

If Vcc is 4.5 volts, then going above 4.8 volts might be unwise.

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  • \$\begingroup\$ Thanks for pointing that out, I missed that. What could I do to decrease the voltage of the inputs? \$\endgroup\$
    – FZs
    Commented Dec 4, 2021 at 12:08
  • \$\begingroup\$ put a similar in series with the supply that powers the inputs \$\endgroup\$ Commented Dec 4, 2021 at 12:11
  • \$\begingroup\$ @FZs you might have an even bigger problem; if your battery is only (say) 3 volts, and the logic levels are still 5 volts AND the incoming 5 volt supply is missing then you are having a bigger problem. I can't resolve this. \$\endgroup\$
    – Andy aka
    Commented Dec 4, 2021 at 12:12
  • \$\begingroup\$ @Andyaka For the backup circuit to switch to the battery, the external voltage has to drop below the battery voltage. In that case, logic HIGH inputs would also be lower than the battery (they come from the same external power supply). Moreover, I will only operate the RAM when the external voltage is present, the battery is only there for data retention, so there won't be any inputs when the external voltage is gone. \$\endgroup\$
    – FZs
    Commented Dec 4, 2021 at 12:19
  • \$\begingroup\$ @FZs to analyse this correctly requires full circuit disclosure and, in all probability the raising of a new question because, although slight movements of the goal posts are tolerated on this site, nobody likes them shifting from one side of the pitch to the other! \$\endgroup\$
    – Andy aka
    Commented Dec 4, 2021 at 13:31
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With that setup the IC’s Vcc node will get pulled up when an input goes above the Vcc limit enough to forward bias the high-side protection diode. That is, the IC might get a bit of ‘phantom power’ through its pins. I say ‘might’, because the datasheet says that high-limit voltage will Vcc+0.3V, implying a Schottky bias for that protection diode. So a 5V input will pump up Vcc to 4.7V.

In reality, that bit of phantom power leakage pumping up Vcc isn’t going to cause harm, because the leakage causes Vcc to track up some, and thus the input injection current will be small - well below any possibility of latch-up or other damage.

You can minimize this by using lower-Vf Schottky diodes for the blockers, such as a BAT54C pair. These also have a 0.3V forward voltage, making Vcc also 4.7V, so the leakage from 5V I/O to Vcc should be close to nothing.

For safety, add a series resistance for the backup battery if you’re using a lithium cell.

This answer provides some more information, including a means to suppress spurious read/write during power up. See Sram battery backup

Here’s an implementation that uses a supercap: Super capacitor backup circuit

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legal input voltages typically go to VCC plus some small ammount. (there are a few exceptions, eg: LM339 can withstand +36V on its inputs at any VCC voltage)

if you can use a diode with a low voltage drop (eg schottky) it'll probably be alright. adding a resistor in series with the high input can also help.

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  • \$\begingroup\$ "adding a resistor in series with the high input can also help" - if so, wouldn't I have to have pulldown resistors as well, forming a voltage divider? \$\endgroup\$
    – FZs
    Commented Dec 4, 2021 at 12:14
  • \$\begingroup\$ @Jasen - Not sure what help those series resistors would provide in this case. The diode in series with the backup battery will prevent any current flow through the ESD diodes, except for whatever reverse bias leakage there is. \$\endgroup\$
    – SteveSh
    Commented Dec 4, 2021 at 12:45

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