Capacitor discharge

[Soulkeeper]

It's been over a decade since i've sat in a math class.
I can't even remember how to use my graphing calculator.

After searching for formulas on capacitor discharge rate I find myself confused by the symbols (some multi-character). "legends" or "keys" are seldom provided in formula explanations when I search google.

How would I go about calculating the capacitance needed to operate a 208ma load at 5v for 1 second ?
It seems i'd need to plot the discharge and then find an intercept for a lower "threshold" voltage say 3v ?

 

[Soulkeeper]

1*208mA/2v = C

is this right ?
104mF ?

 

[sm625]

Assuming a constant load of 208mA during the entire discharge period from 5V to 3V, yes that is correct. In a real circuit, your load current is probably going to change significantly as your voltage goes from 5V to 3V and on down to zero. Most likely load current will also go down, which means you could get away with less capacitance. But if load current increases as voltage drops then you will need more capacitance or else you wont last for an entire second.

 

[Hitman928]

1*208mA/2v = C

is this right ?
104mF ?

That's a linear approximation of the discharge rate of a capacitor, it will be accurate at your initial voltage but less and less accurate as capacitor discharges and as sm625 noted, this assumes a constant current which may or may not be very accurate depending on the load.

With all that said, the linear equation and non-constant load will probably give you a larger capacitor than you need but that won't hurt for what you need on the discharge side and it's not like capacitors are super expensive. The question though is if other aspects of your system are sensitive to capacitor size (charging time, phase, all that stuff).

 

[DeathRayLoveMachine]

Isn't it better to use multiple capacitors in series for this kind of thing?

ETA: nope, forget this

 

[ScottMac]

Try this RC time calculator.

You'd want ~ 24 Ohms for the series resistor, as a current limiter, and ~50000mfd which should deliver voltage for a bit over a second. If you charge it to five volts, you'll get five volts (minus some small losses).

Tweak the values to suit your need.

http://easycalculation.com/engineering/electrical/capacitor-energy-rc-time-constant.php

 

[rifken2]

I don't have anything add except that I think I recall the discharge rate for a cap is something like 7 cycles or some such... it has been to long since I took AC/DC and semiconductors in college... If you don't use it you lose it...

I will echo Hitman and say that you should probably be more concerned with the other end of the equation and figure out the charge cycle and impact on the opposite side of the cap and not so much on the discharge rate.

 

[Soulkeeper]

5v ---C-----------------------------------------
             \   \   \   \   \   \   \   \
             R   R   R   R   R   R   R   R
              \   \   \   \   \   \   \   \ 
              D   D   D   D   D   D   D   D
               \   \   \   \   \   \   \   \  
----------------------------------------------------

C = capacitor
R = resistor
D = blue led

This is what I had in mind.
I'm thinking I won't get carried away 100uF should be fine for such a simple thing.

Thanks everyone

 

[Abwx]

Discharge is an exponential curve , voltage across the load being
U(t) = V.exp(-t/RC).

t is time in seconds , V is the initial voltage ,
R is the load impedance , C is the capacitor in Farad.

In your case the load is R = 5/0.208 = 24 ohms ,


If you want the capacitor to still provide 50%
of the initial voltage after 1 second you equal
the equation to 2.5 (V) for t = 1 (second) :

U(1) = 5.exp(-1/24C) = 2.5

Solving to extract C :

5.exp(-1/24C) = 2.5

exp(-1/24C) = 0.5

C = -1/24.ln(0.5)

C = 0.06 Farad

That s quite a big capacitor...

 

[rifken2]

What's the breakover voltage on a blue LED? When I was in school we didn't have the fancy LEDs these kids have now... we had germanium and silicon which where .3v and .7v if memory serves. I thought the blue LEDs took well over a volt, close to 1.2v if I recall, maybe more.

That being said, I think that impacts his circuit on when the lights "flip" on and not the discharge rate... the reason I bring it up is that he might be able to get away with 25% voltage instead of 50% after 1s. That should make the cap much smaller I would think...