I have 0.047 F Capacitor charged to 5V, now does this capactitor have an Ampere-Hour Rating? If yes, how can i find it...

Please note that with time the charge on capacitor is exponentially decreasing when current is drawn, so current supply also decays, so how can we find ampere-hour rating in this non-linear environment?

 

Responding to H. Shazad's query:

The Amp-hour "capacity" is determined by integrating the current-time curve of the current transient... in other words, the area between the current curve and the horizontal or 't' axis.

Note that such a curve is related to the discharge resistor, R, connected in series with capacitor, C. Hence, A-hr capacity must be expressed for a given R value.

Let me know if you need additional detail.

Regards, Phil Corso, PE Boca Raton, FL [[email protected]] ([email protected]) {[email protected]}

 

Convert to coulombs. Q = IT so 1 Ah should = 3600 coulombs. Q = E*C or in your case roughly 5*.05 or .25 coulombs So you have about 1/14400 Ah. That's if you have .047 F rather than uF. I don't have a calculator handy. It's easy to see why batteries prevail.

Regards

cww

 

Ampere-hour ratings normally apply to batteries. They assume a useful battery life down to a specific working voltage. I suppose an analogy could be drawn. Let us suppose that the capacitor is supplying useful power from 5 volts down to 3 volts. Suppose further that the load is continually adjusted so as to make the voltage go down at a fixed rate of 2 volts per hour. Then over that hour the current would be

I = C * (dV/dt)
= .047 * (2/60/60)
= .000026 amps

So this capacitor would have a .000026 amp-hour rating as a "battery". Note that the amp-hour rating would be cut in half if you define the working voltage as only going down to 4 volts instead of 3 volts.

-Robert Scott
Real-Time Specialties
Embedded Systems Consulting

 

Phil, I'm going to mildly disagree. Ampere hours is a measure of potential energy. While in theory they should be equal, in the real world we know the full amount of potential energy is never recovered as
kinetic energy. You have correctly stated the case for dynamic energy. I would give the proper equations for deriving ampere hours (mah ?) but I assume everyone except our student already knows these
equations. I don't want to deprive the student the satisfaction of deriving these equations. Besides, I doubt Shazad would want share the fruits of his grade with me.

Fred Townsend

 

Phil,

You should know better than that - that A-hr is NOT related to the particular resistance used (assuming an ideal capacitor = no self discharge, zero internal effective series resistance). Moreover, the integral of an natural exponential function is an natural exponential function and vice versa. The real question is the voltage drop threshold - namely between 5 V and which lower voltage the A-hr counts ? Meir

 

Hi All,

OTOH, H. Shazad could be intending to replace a battery with a capacitor to power some item of equipment and wants to know what size capacitor to use. In this case the Amp-hour rating depends on initial charge voltage and minimum usable charge voltage which could be fairly similar. If the allowable volt droop is small (say 0.5 volt) then the usable amp-hour capacity is much smaller than the total amp-hours (or energy) stored in the capacitor. (The calculation is left to the student).

In this case also, usable amp-hour capacity depends on leakage resistance (or self discharge rate) particularly if the capacitor is required to hold a charge for days or weeks rather than seconds.

The usable amp-hour capacity is related to not just the load resistance but also the internal resistance of the capacitor. If the internal resistance is zero (perfect capacitor)then the amp-hour capacity is independent of the load resistance. If the discharge rate is very high then internal resistance can be a significant source of energy loss. Hence the use of low ESR capacitors.

All of which just goes to show that amp-hour capacity of a capacitor is a very seldom used concept and why in general terms a capacitor does not have an amp-hour capacity rating in the way that a battery does.

Regards

Peter whalley

 

well the solution i came up is this:

1A = 1C/1sec
1C = 1A x 1 sec
then for an hour
3600 C = 1A x 1 hr
now...
capacitor is charged tp 5V, this means, Q=CV
0.235 C = 0.047F x 5V

so then, i can safely say that....
0.235 / 3600 = 65.2 uA / hr

so ampere hour ratting will come to 65.2 uA-hr, provided the current draw is constant.

Please guide me if its okay,

 

Paraphrasing the not too famous Kennedy, George, "Gentlemen, what we have is a case of simple mis-communication":

Fred: The "potential" energy stored by the capacitor is 1/2xCxV^2, in Joules. When the capacitor is series connected to a resistor, R, then the potential energy is converted to "thermal" energy, or joule heat, or I^2xR loss, in Watts. R, then, affects the rate, J/s, at which the energy is converted.

Meir: Your comment that the integral of a natural exponential function is another exponential is correct, but only when the exponent "constant", call it A, is unity. If otherwise then the exponential function is multiplied by A.

Peter: There is nothing inherently wrong with replacing a battery with a capacitor. I have used, but not in great quantity, capacitors rated 1 Farad @ 5 Volts.

H. Shazad: Since the 'haves" won't willingly extend helpful knowledge to the "have nots", then contact me directly!

Regards, Phil Corso, PE Boca Raton, FL [[email protected]] ([email protected]) {[email protected]}