# Memory capacitors

Discussion in 'Electronic Basics' started by matt, Feb 20, 2006.

1. ### mattGuest

What are the general rules of thumb re memory capacitors and voltage
retention when power is switched off? Is a higher capacitance (ie 10F)
memory cap going to hold onto data for a lot longer than one rated at
1F?

Also, I assume that a 5.5 volt memory cap can't be used on a 20 year
old PCB using TTL (which is of course rated to run at a maximum of
5.25 volts).

Thanks

2. ### mattGuest

<snip>

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?

3. ### Rich WebbGuest

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.

4. ### John FieldsGuest

---
For the same current out of both capacitors, yes.
---

---
Sure it can. That 5.5V is the highest voltage to which it can be
charged, not the voltage to which it _must_ ber charged.

It'll just discharge a lot more quickly because TTL draws static
current while CMOS doesn't (well... not much, anyway)

5. ### mattGuest

That makes sense, cheers.

6. ### mattGuest

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).

7. ### John FieldsGuest

---
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.

8. ### John PopelishGuest

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.

9. ### mattGuest

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