Memory capacitors

Oops, I meant hold onto its' rated voltage, ie if it's rated at 2.5 volts then will it continue to output 2.5 volts for longer the higher the capacitance?

Yes, all other things being equal.

Also, WRT the "5.5" issue: I'd assume that is the maximum rated voltage for that part. If it's connected to a 5 volt rail then it will only charge up to 5 V.

Thanks.

That makes sense, cheers. :)

So, on an old TTL circuit, how long would a 10F 5.5 Volt retain the voltage for? Days? Weeks? Months? Or does it depend on the size of the circuit? (It's only being used to retain the data in a 6116 RAM on a circuit which also has a dozen TTLs and some EPROMs).

--- It depends on how much current the circuit draws.

Think of the capacitor as a container full of water with a spigot on the bottom.

Voltage will be analogous to how high the water is in the container, and current will be analogous to how fast the water flows out of the spigot, so you can see that if you only open the spigot a little the water level will fall slowly, but if you open the spigot a lot, the water level will fall more quickly.

Retaining data in CMOS under static conditions might be the equivalent of about a drop of water flowing out of the spigot every hour, while for TTL it might be the equivalent of several cups of water per minute.

To determine how long you can retain data in the TTL circuitry you need to know how much current the TTL needs to operate, what the voltage across the circuitry will be when the power supply quits, what the minimum voltage is which will guarantee that the data will remain valid, and the value of the capacitor's capacitance.

Come back with that data and one (or maybe several) of us will show you how to figure out how long the cap will allow your TTL circuitry to retain data after the power supply goes away.

-- John Fields Professional Circuit Designer

More capacitance equals slower rate of change of voltage. No capacitor holds a steady voltage as current passes through it. The relationship (for an ideal capacitor, with no internal resistance) is I=C*(dv/dt), with I in amperes, C in farads, and dv/dt in volts per second.

A 5.5 volt cap can stand no more than 5.5 volts, but it can certainly stand less. The bigger question is how much current does memory retention take, That current and the value of capacitance determines the rate of change of capacitor voltage.

Thanks very much for all the replies, they've been extremely useful. :-)