3
\$\begingroup\$

I would like to know if it is possible to create a programmable ROM from transistors and logic gates? Or, is it possible to create a type of persistent storage similar to Flip Flops?

\$\endgroup\$
2
  • \$\begingroup\$ Can you give some context? Why not just buy a PROM as separate component? \$\endgroup\$ – Philippe Jul 15 '13 at 9:32
  • \$\begingroup\$ I'd like to understand how it works and create it myself, avoiding the use of ICs. \$\endgroup\$ – Jacob Clark Jul 15 '13 at 9:37
5
\$\begingroup\$

If the point is to hand-build it, try a diode matrix:

http://www.cca.org/blog/20120222-Diode-Matrix.shtml

A diode matrix is an extremely low-density form of read-only memory that was used in computers in the 50s through the 70s, before EEPROMs were invented. (They are actually still used, but only inside microchips, not using discrete diodes.) Each bit in the ROM is represented by the presence or absence of one diode. The ROM is easily user-writable using a soldering iron and pair of wire cutters.

\$\endgroup\$
1
  • \$\begingroup\$ +1 I was going to suggest one until I saw the programmable part in the question and forgot the obvious method. \$\endgroup\$ – PeterJ Jul 15 '13 at 9:52
2
\$\begingroup\$

PROMs work by having fuses that are selectively blown to block or pass charge to indicate logic values. EPROMs, EEPROMs, NVRAM, and Flash all use charges trapped in semiconductor islands that selectively block or allow charges, and cannot reliably be replicated using discrete components.

\$\endgroup\$
3
  • \$\begingroup\$ So, PROMs can be replicated with discrete components and this is the type I should investigate? \$\endgroup\$ – Jacob Clark Jul 15 '13 at 9:43
  • 1
    \$\begingroup\$ You should investigate using ICs. But you could replicate a PROM with very small value fuses if you really wanted to. \$\endgroup\$ – Ignacio Vazquez-Abrams Jul 15 '13 at 9:44
  • \$\begingroup\$ Yes I understand ICs are better, however to get an understanding of how it works I'd rather create my own first! Thanks so much :) \$\endgroup\$ – Jacob Clark Jul 15 '13 at 9:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.