Comments on: Mystery EEPROM circuit

By: spiffed

spiffed — Wed, 10 Feb 2010 12:28:46 +0000

While it doesn’t provide any insight, this is a defacto standard circuit for toner cartridge page counters.

By: x893

x893 — Sun, 07 Feb 2010 20:02:14 +0000

I was shocked by the question about R1 and R4

By: Brad Gilbert

Brad Gilbert — Sun, 07 Feb 2010 16:10:23 +0000

D2 and R4 are to make sure that the input voltage on Pin1 remains below some given threshold. R4 is mainly to reduce the current that D2 would otherwise have to maintain.

D1 allows any positive voltage on Pin1 to go to Vcc. C1 helps to hold Vcc to a usable voltage. R1 is probably there to help bring the system to a “safe” state, after it is done being used.

R2 and R3 are of course there to reduce the current to SCL and SDA.

There would be a problem if Pin1 was connected directly to a voltage source that was higher than D2 would normally allow. So I would assume that Pin1 might be connected through a resistor on the other side of the connector.

By: Travis

Travis — Sun, 07 Feb 2010 14:17:35 +0000

The diode D1 is to prevent a power supply being hooked up wrong, the D2 zener diode will create a voltage drop across it of a certain amount depending on what zener diode it is, but say it’s a 3.6 V, so then 3.6 volts goes to D1, which charges the R1 and C1 circuit which powers the eeprom. As long as constant data is flowing to this device it will work. The zener diode D2, is probably dropping a 5V supply down to a lower voltage because the eeprom can run on a lower voltage. R4 helps to create the 3.6V, because a zener diode can go into avalanche mode and destroy itself, the R4 will help control the current I believe, but I’m not positive on that part.

This is a neat little trick to not have to power a sensor, I might keep this in my back pocket for something useful some day.

Travis