# How to generate +5 V and -9 V voltages to read an Intel 1702A EPROM via Arduino?

I have some simple experience with TTL logic and Arduino, but I'm only a hobbyist and am bit stymied by my current project, which is attempting to read the data from an old Intel 1702A EPROM. The actual reading part should be simple: set the address lines and look at the data lines.

But this EPROM wants both +5 V (Vcc) and -9 V (Vdd, Vgg) supply voltages in its read mode. (Here's the data sheet.) I have a DC supply that can give me +9 V, which can power the Arduino and as a nice side effect creates a +5 V supply at the Arduino that I can then use for the EPROM's Vcc, but I'm stuck on how to find a minus 9 V supply, such that the Arduino logic and the 1702A are all relative to each other correctly.

I know this is not an advanced question, but I'd appreciate some guidance. I know I can't simply take the ground connection on the power supply (which is -9 V relative to its positive terminal) because it's not correct relative to the +5 V Vcc on the chip.

Can someone point me to some simple setup (voltage divider? zener diodes? or?) that I can use to end up with the -9 V? If my DC input needs to be something other than +9 V to make the difference work, then let me know, and I'll need to power the Arduino somehow too.

Thank you!

• Search term for you: voltage inverter. Aug 12, 2021 at 3:24
• use a 9 V battery Aug 12, 2021 at 3:58

An inverting charge pump could be appropriate here, although you'd need to provide a fairly large current (as high as 60 mA depending on the device variant) for the device. One example with enough current capacity in a hobbyist-friendly package is the LT1054, which gives an example circuit for generating your negative rail1 on page 10 of its datasheet (image reproduced below):

With everything sharing a common ground on the negative terminal of the power supply, you'll have a +9V rail (power supply positive terminal) which can feed VIN on the Arduino and VIN on the charge pump circuit, a +5V rail on the 5V pin of the Arduino (it's the output of a voltage regulator), and a -9 V rail (the -Vout of the charge pump circuit).

1 It's a much more versatile chip than just a means of generating an equal-voltage negative rail - the datasheet shows tons of other example uses.

• Perfect! Thanks for this suggestion and context. Looking forward to getting one of these and figuring it out. Aug 12, 2021 at 4:34
• A small observation for future travelers-- I went to order one of these and discovered that I need the LT1054 (without the "L" suffix) to get to 9V-- the low voltage variant tops out at 7V. Aug 12, 2021 at 15:18
• @BenZotto my bad, thanks for catching that typo! Aug 12, 2021 at 15:19

A quicker and easier approach than trying to figure out how to use a voltage converter was suggested by jsotola. Get a 9-volt battery, connect the + to ground and the - to -9. Hardly elegant, but I assume once you've read the data from the 1702 (and where in God's name did you find an EPROM that old?) you'll never look at the chip again. Although you'll need to check the actual voltage level out of the battery. The tolerance on the -9 supply for the 1702 is 5%, or +/- 0.45 volts.

Of course, if you're looking for a good excuse to increase your knowledge level, by all means do it the hard way.

• Thanks for this clarification. That was among my original thoughts, but all of the other lines (addr in data out) would be TTL at +5, tied to the Arduino's lines. I'm very fuzzy on the basic principles around essentially using two different power supplies like this with the same EPROM... don't they all need to be relative to some fixed ground level? And if so, how would I ensure I'd achieve that with a 9V battery? Aug 12, 2021 at 16:04
• As answered, the battery's + connects to the circuit's ground. Aug 12, 2021 at 16:05
• @BenZotto - "don't they all need to be relative to some fixed ground level?" Yes, but the 1702 is weird (by modern standards) in that there is no ground connection to the chip. +5 and -9 must be referenced to the same ground which defines the voltage of the other signals, but that ground is not connected to the EEPROM. And you ensure this by connecting one lead or the other for each power supply to the circuit (Arduino) ground. Note that this requires that the 9 volt supply be isolated (that is galvanically isolated), and a battery works perfectly in this case. Just make sure the voltage is OK Aug 12, 2021 at 23:16
• @BenZotto - 2nd part first - right. You got it. First part. Determining whether a power supply is isolated is not necessarily obvious. In general, you should not assume that cheap supplies, especially wall-warts are isolated. Generally, if they are positive supplies the negative will be tied to power line ground. So if you have two of them plugged into the wall, their negative lines will be tied together. If you then connect one positive to one negative you'll short it out. The way to test is to plug both in, then take a meter and check for no voltage between -'s, and infinite resistance. Aug 13, 2021 at 13:17
• @BenZotto - But also, as I've stated before, check the battery voltage before you connect. In your case you need a voltage between 8.55 and 9.45 volts. Also check after you connect the chip, just in case the load changes the voltage. Aug 13, 2021 at 13:19