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I'm using an Alliance parallel SRAM IC, and under the Electrical Characteristics table, it says VCC can be min 3.0 V, max 5.5 V. In the Absolute Maximum Characteristics table, it says terminal voltage must be no more than 7 V with respect to VSS (GND in my case).

What happens when operating in the no-man's-land between 5.5 V and 7 V, say, around 6 V supply to VCC?

datasheet

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3 Answers 3

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No one can tell for sure in general, except the manufacturer.

Providing those two sets of values is industry standard. The rationale behind is that beyond absolute maximum ratings the manufacturer makes absolutely no guarantee about the device: it can break, it can malfunction, who knows! Its behavior also depends on how much you go beyond those max ratings.

Below those values, but outside the normal operating conditions ("no-man's land") the manufacturer also makes no promise, except one: you can stay in "no-man's land" for as long as you want and when you return into normal operating range, the device will behave as specified.

This means that the manufacturer guarantees that the device won't be damaged or degraded. However, be careful with that promise: that promise is made assuming the device is connected to almost nothing and the external world is "benevolent".

If you drive the device in no-man's land and it starts acting weird, in a real scenario other devices connected to the misbehaving chip could in turn damage the device or other part of the circuit indirectly.

For example, imagine a power supervisor chip driven outside its operating range. It could go a little crazy and mess up the power sequencing of a board, activating power lines in the wrong sequence. This could lead to PN-junctions turning on that never meant to be turned on, causing short circuits in the system.

As another example, consider an opamp with supply lines slightly above the normal operating conditions (but not above the max). Maybe the offset voltage rises dramatically and this causes a major error in the output signal. Perhaps this opamp is part of a PID controller and because of that unexpected rise of offset voltage a motor goes out of control, making a robotic arm acting erratically.

In both cases the fact that the device is not damaged when it is driven back into normal operating conditions is irrelevant, since the system has already been damaged by its crazy behavior.

Bottom line: never design a circuit where in normal operation a device could be driven outside normal operating conditions.

Pro tip

Sometimes manufacturers are very conservative about what is specified in the publicly available datasheet. After all, testing a device for every possible condition is expensive, so corner cases are often neglected. By testing a device yourself you could determine that a device could be able to work reliably outside what is specified in the datasheet.

In this case you could contact the manufacturer and ask for official confirmation of your findings and even ask for a better characterized device (at a cost, of course).

However, even if you do in-house testing, never rely only on those. The results may be applicable only to the specific batch of devices you own, and a small process variation in manufacturing could make all your test results wrong for another batch. Always ask the manufacturer!

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  • \$\begingroup\$ Thanks for the detailed answer! The device seems to operate normally when I do this, but I don't plan on actually using it as such. I'm mysteriously getting ~5.8V on the VCC pin and it got me wondering about what the difference between the ratings are. Replacing the 1008 with a similar, but smaller 62256 causes the supply to go down to ~5V as I expect it to (diode-or of 5V supply + backup battery at 3V) so something weird is going on. \$\endgroup\$
    – Nick U.
    Apr 17 at 15:41
  • \$\begingroup\$ @NickU. How did you measure that 5.8V? Did you use a DMM or an oscilloscope? If you measured it with a DMM the excess voltage may be due to some oscillations or ripple on Vcc. Maybe the bypass caps are not enough for that device or the power supply is not filtered enough. \$\endgroup\$ Apr 17 at 15:47
  • \$\begingroup\$ I used both a meter and an o'scope. Not a lot of ripple on the supply. When /CE is LOW, the supply drops to ~5V as expected, but when /CE is HIGH, the VCC charges up to ~5.8V again. I've got a 22uF aluminum and 100nF ceramic on the supply. Replacing the 1008 with a 62256, and VCC behaves all the time. So there's some difference (besides pin count) between the chips. \$\endgroup\$
    – Nick U.
    Apr 17 at 16:01
  • \$\begingroup\$ To be clear, I can make the 32-pin 1008 act like a 28-pin 62256 by tying A15/A16 to GND, and tying CE2 to VCC. So when I have that configuration, the VCC rises to 5.8V when /CE is not active. Swapping the chip out for a 28-pin 62256 without changing any wiring (so CE2 on the 1008 is now VCC on the 62256) and VCC stays 5V or below. \$\endgroup\$
    – Nick U.
    Apr 17 at 16:17
  • \$\begingroup\$ @NickU. That's weird. Are those caps placed near the chip? Does the power rail float to 5.8V even with no load connected? If this is the case, maybe you should redesign the power supply so that it gives out 5.0V (or a tad higher, say 5.1V) with no load and check the size of your PCB traces, because 0.8V voltage drop when a device becomes active is quite a lot. Maybe you have too much resistive drop on your PCB traces. Or maybe it is also inductive drop: if the drop reduces when you reduce the clock frequency that could be a symptom of too little bypass caps (or them being placed too far). \$\endgroup\$ Apr 17 at 16:56
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Generally, the specs and performance is only guaranteed in the conditions listed in the characteristics table.

Above that (and below Absolute Maximum Ratings), performance and many aspects of functionality are not guaranteed. In fact you should not apply those voltages for any sustained amount of time as the device may degrade and become unreliable. In practice, it likely works for most IC samples, but would not be recommended for production designs, or designs that critically rely on the performance.

Above Absolute Maximum Ratings, the device may have been damaged or degraded and no specs are guaranteed, even back in the characteristics conditions.

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The data sheet explicitly says that data retention is from 1.5 volts to 5.5 volts: -

enter image description here

So, either do what the DS says and reduce your supply or speak with the manufacturer.

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  • \$\begingroup\$ Good call, I totally missed that part. Btw I'm not intentionally supplying it with 6V, I have a diode-ored supply to the chip, one 5V and one 3V backup battery. For some reason, during operation, when /CE is high, this voltage will charge up to ~5.8V. Pulling /CE low causes it to drop to 5V as expected. I only seem to have this issue with the 1008, if I replace with a 62256 it doesn't happen. I don't intend to actually use the 1008 out of spec it just got me thinking about what the difference between the ratings are. (If you've got any ideas why I'm getting above 5V on this, I'm all ears haha) \$\endgroup\$
    – Nick U.
    Apr 17 at 15:20
  • \$\begingroup\$ @NickU. That's likely because the 1008 has a lower stand-by current, low enough to "exceed" the minimum load required for your power supply to maintain regulation. \$\endgroup\$
    – pipe
    Apr 18 at 22:24
  • \$\begingroup\$ There's a more detailed explanation of the issue in the comments on the answer above, it's not the 5V supply causing the problem, I've scoped it and the 5V is very stiff. \$\endgroup\$
    – Nick U.
    Apr 18 at 23:10

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