# How to work this out?

Discussion in 'General Electronics Chat' started by RadioRental, Jan 11, 2010.

Sep 3, 2009
3
0
Hi All,

How do i go about working out how long i can power something from a capacitor. i have always wanted to know this and it just so happens a relevant project has come up where i get to try it out.

I want to try and run a circuit for atleast 5 seconds in the event of a power failure. The circuit runs off a DC PSU

Basically, when the power fails the capacitor provides the power for as long as it can (5 sec). The caps i have chosen are super caps.

I know my circuit uses 12V and consumes 1.5A when running. if i choose a 1F capacitor, how do i work out the legth of time it will take before it dies?

I know caps have an exponensial discharge curve but i also know my circuit operates anywhere from 12V to 9V. Therefore this is the range i wish to keep the circuit powered.

The hard part is working out how long it can provide 1.5A. i can work out how long it takes to fall from 12V to 9V but what about the current?

Can anyone help me with this or provide a formula.

Thanks

2. ### Mike33 AAC Fanatic!

Feb 4, 2005
349
25
Ouch, I'd like to see the answer to this, too!!
As you stated, you can figure out your time constants, right?

You can't really add a series resistance to slow down the cap discharge without causing your load (I'm using 8 ohms in LT Spice to look at this) to increase and therefore your load sees less current in operation...

So I tried a resistor in series with the shunt capacitor. Using 1F, 12V and an 8 ohm load, I get the current dropping to 100mA after 5s with a 100 ohm resistor in series with the shunt cap. 535mA w/10 ohms, and 997mA with no resistor at all. So, maybe you just have to rely on the cap value of 1F? And that would get you down to about an ampere after 5 secs. Depends on where your load will 'drop out', too....
Someone else will chime in on this one, I'm sure!

3. ### thyristor Active Member

Dec 27, 2009
94
0
The discharge equation for a capacitor is:

Vc = V'exp(-t/RC)

where:
Vc = the voltage on the capacitor at time t
V' = the fully charged capacitor voltage
R = the resistance into which the capacitor discharges

So if t = 0 Vc = V', ie: the capacitor is fully charged.

Your circuit draws 1.5A @ 12v so it looks, in simple terms, like an 8 ohm resistance (12/1.5). Also V' = 12v and you state you have a 1F capacitor.

Therefore, if we look at the voltage on the capacitor after each second, using the formula above, we obtain:

t = 0 secs; Vc = 12v
t = 1 sec; Vc = 10.6v
t = 2 secs Vc = 9.4v
t = 3 secs; Vc = 8.3v

So after 3 seconds your circuit voltage will fall below the requisite 9v.

A better way to use the equation would be to determine the value of capacitance needed to maintain at least 9v after 5 seconds.

Rearranging the equation to solve for C, we obtain

C = -t/[R ln(Vc/V')] where "ln" is the natural logarithm

So, if we substitute R = 8 ohms, t = 5 seconds, V' = 12v and Vc = 9v, we obtain:

C = -5/[8 ln(9/12)] = 2.2 farads

Last edited: Jan 11, 2010

Sep 3, 2009
3
0
Excellent, i knew there was an equation i could use.

I have been thinking about how the current changes in all this. I know the voltage discharges at an exponential rate but what does the current do?

My device requires a solid 1.5A continuous between 12V and 9V. If the voltage is changing and the resistance (8R) is fixed then surely the current is also changing and not continuously 1.5A.

I am confussed

5. ### thyristor Active Member

Dec 27, 2009
94
0
You are correct, the current will decrease as the capacitor's voltage drops. The equation is i = Cdv/dt

where:
i = the current
C = the capacitance (2.2F in this case)
dv/dt = change of voltage over the change of time

So, for example, the change of voltage over the first second is 0.7v. This gives us 2.2 x 0.7 / 1 = 1.5A and so on as tabulated below

@ t= 1 sec, i = 1.5A
@ t= 2 sec, i = 1.4A
@ t= 3 sec, i = 1.3A
@ t= 4 sec, i = 1.2A
@ t= 5 sec, i = 1.2A
@ t= 6 sec, i = 1.1A

6. ### nomurphy AAC Fanatic!

Aug 8, 2005
567
12
A little algebra based on the above will give you: dt = C*(dv/i)

Although it is not a constant-current circuit, by using the worst-case current load of 1.5A you will see that:

dt = 3F*(12V-9V/1.5A)
= 3*(3/1.5)
= 6.0 sec minimum

Which is more than the 5 sec. minimum you desire. The above estimate allows for circuit discrepencies and provides some fudge factor or "headroom" -- also be cognizant of capacitor tolerances.