How much energy stored in 350 farad cap?

[Eric R Snow]

I don't know much about electronics but do remember when even a 1
farad capacitor was pretty big. Now, in the july edition of Nuts and
Volts there is a news blurb about a 350 farad, 2.5 volt capacitor. It
comes in a "D" cell size package. When charged to the max how much
juice does this hold expressed in watt hours? Is that the right term?
Maybe amp hours at 2.5 volts would be better?
Thank You,
Eric R Snow

 

[John Larkin]

I don't know much about electronics but do remember when even a 1
farad capacitor was pretty big. Now, in the july edition of Nuts and
Volts there is a news blurb about a 350 farad, 2.5 volt capacitor. It
comes in a "D" cell size package. When charged to the max how much
juice does this hold expressed in watt hours? Is that the right term?
Maybe amp hours at 2.5 volts would be better?
Thank You,
Eric R Snow

energy = 1/2 * c * v^2. That's 1100 watt-seconds if fully charged, 0.3
watt-hours. Even supercaps don't store much energy.

John

 

[Bob Masta]

energy = 1/2 * c * v^2. That's 1100 watt-seconds if fully charged, 0.3
watt-hours. Even supercaps don't store much energy.

And worse yet, the output voltage drops exponentially
when you try to use that energy!


Bob Masta
dqatechATdaqartaDOTcom

D A Q A R T A
Data AcQuisition And Real-Time Analysis
www.daqarta.com

 

[Eric R Snow]

energy = 1/2 * c * v^2. That's 1100 watt-seconds if fully charged, 0.3
watt-hours. Even supercaps don't store much energy.

John

So that means that if it could dump all it's energy in one second it
would be 1100 watts?Tthat would be 440 amps at 2.5 volts. This cap is
rated at 20 amps charge and discharge though. At that rate it would
take 55 seconds to discharge. Of course it probably can't do that in
the real world. With an ESR of 3.2 milliohms and 20 amp discharge
current how long would it take to discharge this cap?
Eric

 

[Terry Pinnell]

So that means that if it could dump all it's energy in one second it
would be 1100 watts?Tthat would be 440 amps at 2.5 volts. This cap is
rated at 20 amps charge and discharge though. At that rate it would
take 55 seconds to discharge. Of course it probably can't do that in
the real world. With an ESR of 3.2 milliohms and 20 amp discharge
current how long would it take to discharge this cap?

43.75 s, precisely.

 

[AC/DCdude17]

X-No-Archive: Yes

On Tue, 29 Jun 2004 20:04:21 -0700, John Larkin



And worse yet, the output voltage drops exponentially
when you try to use that energy!

Why do you say the voltage drops exponentially?

Assuming constant capacitance, E=V^2

Stored energy drops exponentially if the voltage drops linearly.

 

[Bob Masta]

X-No-Archive: Yes



Why do you say the voltage drops exponentially?

Assuming constant capacitance, E=V^2

Stored energy drops exponentially if the voltage drops linearly.

The problem is that to get the voltage to drop linearly,
you need a rising load resistance (ie, a falling current
drain). For a constant load resistance, the voltage drops
exponentially.







Bob Masta
dqatechATdaqartaDOTcom

D A Q A R T A
Data AcQuisition And Real-Time Analysis
www.daqarta.com

 

[AC/DCdude17]

X-No-Archive: Yes

The problem is that to get the voltage to drop linearly,
you need a rising load resistance (ie, a falling current
drain). For a constant load resistance, the voltage drops
exponentially.

I don't understand your reasoning. Voltage does not drop exponentially
as a function of stored energy.


The function looks like voltage = square root of stored energy.

 

[John Larkin]

The problem is that to get the voltage to drop linearly,
you need a rising load resistance (ie, a falling current
drain).

Backwards. Need falling resistance to get constant current to get
linear voltage droop. (But that's still not constant power.)

John

 

[Robert C Monsen]

X-No-Archive: Yes



I don't understand your reasoning. Voltage does not drop exponentially
as a function of stored energy.


The function looks like voltage = square root of stored energy.

Assume you have a cap which is being discharged through a resistor.
The voltage across it at any time is

v(t) = Vo * e^(-t/RC)

where Vo is the initial voltage at t=0.

The energy U across the cap as a function of v is

U(v) = C * v^2 / 2

Thus,

U(t) = C*(Vo*e^(-t/RC))^2/2
= C*Vo^2*e^(-2t/RC)/2

Clearly, as time goes by, both voltage and energy across the cap drops
exponentially. Since current is voltage over resistance, the current
through the resistor also drops exponentially.

Instantaneous power, P(t) = U(t)/t, so its dropping exponentially with
t as well.

P(t) = U(t)/t

I'm guessing thats what Bob meant.

Regards,
Bob Monsen

 

[Ratch]

X-No-Archive: Yes

I don't understand your reasoning. Voltage does not drop exponentially
as a function of stored energy.


The function looks like voltage = square root of stored energy.

From the Random House Webster's Unabridged Dictionary:

ex·po·nen·tial (ekÅsp$ nenÆshÃl, -spÃ-), adj.
1. of or pertaining to an exponent or exponents.
2. Math.
a. of or pertaining to the constant e.
b. (of an equation) having one or more unknown variables in one or more
exponents.
-n.
3. Math.
a. the constant e raised to the power equal to a given expression, as e^3x,
which is the exponential of 3x.
b. any positive constant raised to a power.
[1695-1705; EXPONENT + -IAL]

Since the capacitor voltage is proportional to the square root of its
stored energy, its voltage has a exponential relationship with the
aforementioned energy and vice versa. Certainly it is not a linear
relationship. Ratch