PIC power backup with "super capacitor"

[flotulopex]

Thanks Sayzer,

Am I right to calculate the capacitor's capacity like this (sorry but I don' t know the english technical names):

Assuming a very bad case where my PIC consumes 1mA and I would like my PIC to stay powered 2 hours, it would give this:
Q(cap load) = I(amps) x t(seconds)
this would give: 1mA x 7200seconds = 7,2C(Coulomb - capacity)

Then, to find out the cap's capacity, I would calculate the time like this:
C(Capacity) = Q(cap load) / U(Volt)
this would give: 7,2C / 5V = 1,44F

This is theory but please let me know if I'm wrong and if yes, where.

What about the low power programming?

[sayzer]

flotulopex,

I have no idea about that formula and the calculation.

I am sure someone will jump in and help you out.

But, why don't you try it in a real circuit?

So simple to test in a real application. With your own eyes, you can see how long it will last.

------------------------

[keithdoxey]

Working from memory here because it is a long time since I went to college!!!

Isnt the charge/discharge curve of a capacitor non linear an also as the cap discharges the voltage drops so you will hit brownout long before the cap is discharged. Might be better just with a lithium or rechargeable battery as the backup supply.

[dhouston]

Super caps are frequently used with RTC chips so you can find a lot of online data about how to select the right size. Maxim has an online calculator that you can probably use.

• http://www.maxim-ic.com/tools/calcul...lc_id/supercap

[paul borgmeier]

In simplified terms, the governing equation for a “perfect” capacitor is:
I = C dV/dt
which states that the current is equal to the capacitance times the change in voltage with respect to time. In layman's terms, you can get a first approximation like this:
I = C (Vinitial – Vfinal)/time
Example, If:
C = .47F
Vfinal = 5.0 Volts
Vfinal = 3.3 Volts
time = 7200 seconds
the capacitor would be able to supply
I = .47 (5.0-3.3)/7200 = .000111 amps = .11mA during that time frame (the voltage would change from 5.0 V to 3.3 V after 7200 seconds while supplying the shown current)

However, the equation works both ways – if your circuit draws more current, for example .001 amps, you can determine how long you can keep your capacitor voltage at 3.3 or above like this
I = 0.001 = (5.0-3.3)/time
solving for time gives time = 1700 seconds (about 28 minutes).

I picked a lower limit of 3.3 V – you need to determine this for your setup. Also, with the nanoWatt controllers, you should be able to get your current consumption much lower, like several uA or less, which means you could have a backup for days.

Last, no system is perfect – there are leaks and losses and other things going on that you need to be aware of – design your power backup to last longer than you really need it – and test it!

[Luciano]

Hi,

See this link:
http://www.qsl.net/zl1bpu/MICRO/CLOCK/index.htm

From the above link:
When the AC supply fails, D3 conducts, D4 is reverse biassed, and the
clock continues to operate from the battery with no loss of time.
However, since R2 is no longer able to keep micro input PD2 (pin 6) high,
the micro detects power failure and displays a "BATT" message. The battery
can be a small alkaline 9V type if power failures are rare, and will last
normal shelf life. If power failures are frequent or the clock is used as a
portable time source, a 6-cell 500mAH or better NiCd or NiMH battery pack is
recommended. R1 can be added to slowly charge the battery from the AC supply.

Schematic:
http://www.qsl.net/zl1bpu/MICRO/CLOCK/CLOCK2.GIF

* * * *

Best regards,

Luciano

[sayzer]

...

Schematic:
http://www.qsl.net/zl1bpu/MICRO/CLOCK/CLOCK2.GIF
...

Luciano,

9V battery in that schematic is indeed a good idea.

Thanks.