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Speeding up thermalization -- practical?

On Apr 15, 5:31 pm, whit3rd wrote in

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> > > Can similar data recovery be performed on volatile RAM chips even > > after the power is offed. > > Similar, no. Recovery, yes. The volatility has a time decay constant > of a second or so, and it takes a long, temperature-dependent, delay > after power-off to thermalize the information to nonexistence.

Let's say that after the power supply is cut-off from the volatile RAM chip, the RAM chip is heated to the hottest it can get without suffering any physical damage. Will this speed up the rate at which data is lost? What is the maximum physically-safe temperature to erase all info? What would be the best way of heating the chips?

Reply to
GreenXenon
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ant

I get your concept, but I'm pretty certain you are using the word "thermalization" incorrectly. The term has nothing to do with the termperature cycling of memory chips. It more correctly describes the process undergone by high-energy (fast) neutrons as they lose energy by collision.

As to your question: I suspect there is no official answer to be found in the manufacturer's datasheets. Most likely, it will just specify absolute maximum limits and not guarantee any performance parameters beyond those ratings. Additionally, it will probably recommend that those limits not be approached in your design.

Reply to
mpm

stant

I can think of 3 approaches to this problem of data retention in VRAM chips:

  1. Low power induction heating
  2. Low power microwaves
  3. Low power lasers

I say low power because the heat generated needs to be small enough not to damage the VRAM chip.

Reply to
GreenXenon

Nominal MIL-spec temp range is -55C to +125C so one can safely use

+125C for this operation. That said, most semiconductors easily tolerate 150C. Remember, ions, static charges, and chemical reactions speed up by a factor of two for every 10C increase in temperature.
Reply to
Robert Baer

Ions speed up? Static charges? Explain?

John

Reply to
John Larkin

Oh yes! Put a charge in a CMOS device or even take those trapped under / inside and measure the capacitance as a function of voltage. Bake at 250 with one polarity present then re-measure after cooling (voltage present while cooling always), bake with opposite polarity present then re-measure after cooling (voltage present while cooling always), bake with a short and re-measure. With the data, one can calculate amount of positive charges and amount of negative charges trapped. Over sufficient time and temp, a goodly amount of those charges do dissipate.

Reply to
Robert Baer

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