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I'm continuing to dissect and learn from this old (1974) calculator. One thing that surprised my was that the pins of the main IC have both negative and positive voltages 0 to ±28V. Having worked mainly with ICs like the ATMEGA line, PIC controllers and other popular chips, the voltage surprised me.

  1. Am I correct in thinking that modern IC operate at lower voltages so they can pack in more transistors without over heating?

Inside view of circuit board of Sharp ELSI 8002 Calculator Inside view of circuit board of Sharp ELSI 8002 Calculator. The mystery IC: HD3623 has no online data sheet that I could find, but MK14 at eevblog pointed me to this data sheet showing how a similar calculator works.

  1. When I once destroyed an ATMEGA128 with 12V, was the nature of the destruction heat that melted or burned the traces in the IC making it inoperable? But this old more robust IC must have thicker, more conductive traces.

  2. Is that why it is so big despite doing comparatively less than smaller modern ICs?

There is one advantage to this beefy old IC: It can drive the VFD, which seems to have a -28V grid, directly. Maybe, I should look for a programmable IC that runs at higher voltages? Or would that be frustrating?

  1. Are there IC that meet this description that can be programmed by a novice?

Maybe I should use a logic level converter of some kind instead? I want to be able to repurpose the buttons and the display within the existing case. I like the form factor and quality of the calculator, I want to make it work with a modern micro controller.

Sharp ELSI 8002 calculator. Sharp ELSI 8002 calculator. I'm fairly certain that the VFD is the Futaba 9CT06, it's not super well documented, but I can make it do what I want using bench power, so having identified the pins I just need some kind of PWM... I think.

Maybe I should table this project until I've studied more! I've found some people with similar projects, but none with the same VFD.

I've put notes by my main questions for clarity.

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    \$\begingroup\$ The first obvious advantage to using lower voltage levels is that the design is more efficient in terms of power consumption. Another reason is to do with the speed of the device. A transistor will take longer to switch a large voltage then a small one (according to the slew rate), thus high voltages limit the operational speed. \$\endgroup\$
    – Adam Z
    May 4, 2016 at 7:23
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    \$\begingroup\$ I don't believe you when you say that it works with ±28 volt. -28 volt is however common. I suspect that what you see are 28 volt tolerant open collector outputs, normally pulled "up" to ground, while the whole IC is on the -28 volt level. If you measure at random with a floating multimeter it may seem as there are both negative and positive 28 volts there. Could be wrong though. \$\endgroup\$
    – pipe
    May 4, 2016 at 7:48
  • \$\begingroup\$ @pipe you are likely correct. I have not seen +28V on any of the pins. I incorrectly assumed that since I've seen +6.5 and -28 the overall range must be ±28. What topics can I study to better grasp negative voltages? Is ±28V in some way equivalent to 0-56V? In other words, is the range indicative of anything when it comes to design? \$\endgroup\$
    – futurebird
    May 4, 2016 at 7:59
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    \$\begingroup\$ I guess the problem is that we are usually sloppy with the terminology. Voltage at one point is always in reference to another point. If you measure 28 volt between two point, and then flip the probes on your multimeter you now have -28 volt, but you haven't changed the circuit - just the point of reference. This is impossible to teach in a comment, so I suppose you just have to keep probing and searching. :) \$\endgroup\$
    – pipe
    May 4, 2016 at 8:04

3 Answers 3

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1) Partly true indeed but it is more that in modern ICs the transistors are smaller meaning the supply voltage must be lower because of the smaller structures. The limit is the field strength (Volts per distance) that can be handled so when the distance goes down the voltage must go down as well. How much power is consumed also relates to this but this can also be tuned separately by the the threshold voltage implant offering a choice between fast but high power consumption or slower but smaller power consumption.

2) If you destroy an IC it is very rarely the wires that melt, depending on what you did wrong there are many ways to damage an IC. Reversing the supply voltage for example damages the ESD protection diodes. These have nothing to do with the functionality of the IC itself but need to be there to protect it against ESD discharges.

3) what you see is the packaging, the IC itself will be only a few square millimeter in area. Fact remains that in the 1970s we could only fit a couple of thousand transistors on a few square millimeter. Nowadays we can easily fit hundreds of thousands of transistors in a few square millimeter. This is due to more modern IC manufacturing processes.

4) the -28 volt is probably needed for the VFD. You will not find modern ICs that can handle such voltages as almost no-one uses VFDs anymore as these are very power hungry.

Interfacing with this old technology is a challenge, you really have to know what you're doing. It is not for beginners.

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  • \$\begingroup\$ Cutting edge tech is even more impressive, Samsung's 10nm tech would cram several hundred million transistors into a square millimeter, but at those levels, it only takes a few volts to puncture the dielectric layers \$\endgroup\$
    – Sam
    May 4, 2016 at 8:23
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There are three reasons why ICs with such high voltage are not produced (anymore):

  • Higher voltage require larger clearance between traces. They don't necessary need bigger traces (trace width depend on the current, not voltage), but if they are too close together, the high voltage will cause electrical breakdown, causing the two traces to be shorted. So, it means the chip die must be larger to account for clearance between the interconnects. And larger chips are more expensive.

  • Higher voltage means higher power requirements (for the same resistance). Today, chip design is clearly oriented towards low power consumption, so chip designers tend to use the lowest possible voltage that still allow convenient external interfacing with the chip. Most high-end MPUs even have an internal working voltage that is lower than the one used for external interfacing (with level translators for each pin) to allow for lowest power consumption.

  • Higher voltages mean that the signal swing is larger when the logical state changes. So, if the slew rate is the same, it takes more time when changing state and thus limits operational frequency. This has to be mitigated, however: this is true when you compare a chip working at 28V to a chip working at 3.3V, but it is not true when you make the same chip working at, for example, 2.0V vs 3.3V. In this case, usually, the higher voltage can allow for faster operating frequencies (at the cost of a lot higher power consumption) because the higher voltage difference applied to fet gates makes the switching actually faster (the slew rate is a lot higher).

All in all, there is no real advantage in making the chip able to work with high voltages, except in making it easier to interconnect with fancy peripherals. So it is actually a lot cheaper to make logical chips work at low voltage and use specific circuitry to interconnect with fancy peripherals when required.

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You can actually get CD4000 logic that can go up to 15V. That being said, it replaced transistor logic that operated at 5V. CD4000 was replaced with the familiar 74HC logic that can operate at 3.3V so you can see "standard" voltage was established decades ago.

On processors and the like the move is towards smaller and smaller but quite wide variety of ICs work perfectly well with 3.3V. Lower voltages mainly give you benefit on switching losses and leakage current. Those really become an issue only with very high density logic and/or high speed interfaces.

The reason 3.3V is popular because very low voltages like 1.2V are a pain in the backside to generate and to work with. Also cutting voltage often means boosting current that may become problematic.

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