Calculating Final Voltage in Capacitor Discharge: What Could be Wrong?

In summary, the conversation discusses the analysis of capacitor discharge, with a focus on the minimum capacitance needed for the voltage to remain above a certain level after a given time. The problem lies in taking the square root of both sides of the equation, which is incorrect. The conversation also touches on using RC time constant for discharge calculations. Font size changes in the middle of the message are also mentioned.
  • #1
Corey Spruit
5
0
Hi All,

(First post, be nice).

I'm analysing the discharge of a capacitor. Starting from the energy in a capacitor:
[tex]E = \frac{1}{2} CV^2[/tex]
This can be represented as follows (seen this elsewhere, there is nothing wrong with this):
[tex] \Delta E = P \Delta t = \frac{1}{2} C(V^2_s-V^2_f)[/tex]
Where [itex]\Delta t[/itex] is the time of energy transfer, [itex]V_s[/itex] is start voltage (DC) across the capacitor and [itex]V_f[/itex] is final voltage. It's an inverse square drop off of the voltage across the capacitor, assuming [itex]P[/itex] is constant and a resistive load.

For example, [itex]V_s[/itex] = 100V, [itex]V_f[/itex] = 50V, [itex]\Delta t[/itex] = 0.1sec and [itex]P[/itex] = 100W, then [itex]C_{min} [/itex] = 2667 [itex]\mu F[/itex].

Rearranging the above (also seen this elsewhere, there is nothing wrong with this):
[tex]C_{min} = \frac{2P\Delta t}{V^2_s-V^2_f}[/tex]
This gives the minimum capacitance needed for its voltage to remain above [itex]V_f[/itex] after time [itex]\Delta t[/itex].

Now the issue I have is using this equiation to calculate the final voltage after a given time given a value of capacitor. The results I'm getting are not correct. That is, if I plug in the same values, the final voltage I get is way out.

I have rearranged the above equation to the following:
[tex]V^2_s-V^2_f = \frac{2P\Delta t}{C_{min}}[/tex]
Then:
[tex]V_f = V_s - \sqrt{\frac{2P\Delta t}{C_{min}}}[/tex]
Is this not correct?

With the above example values and using 2667 [itex]\mu F[/itex], I'm calculating [itex]V_f[/itex] = 13.4V, it should be 50V. What is wrong here? o_O
 
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  • #2
The problem is with your square root. sqrt (a^2 - b^2) is not a-b. (a-b)^2 is a^2 - 2ab +b^2. Some day I'll learn Latex.
 
  • #3
Your problem is the last step: You can't just take the square root of both sides.
 
  • #4
Thanks guys. Ah, this is the trouble with being an engineer, sometimes you use high-school maths and you end up looking like a 6th grader..
meBigGuy said:
The problem is with your square root. sqrt (a^2 - b^2) is not a-b. (a-b)^2 is a^2 - 2ab +b^2. Some day I'll learn Latex.
This is not relevant, I went off and tried this, and ended up solving a quadratic equation (for fun, digging up old high-school maths again). But [itex]x^2-y^2 \neq (x-y)^2[/itex].

EverGreen1231 was correct, I can't just square root both sides. Interesting discussion on why on Stack Exchange if anyone is interested.

I left the equation at:
[tex]V_s = \sqrt{ \frac{2P\Delta t}{C_{min}} -V^2_f}[/tex]
 
  • #5
You assume energy is removed at a fixed rate which would require a switching regulator.

If on the other hand, if the regulator is linear, then the charge will be removed at a constant current …

You know that C = Q / V and that Q = I * t. Then; C = I * t / V.
So; Cmin = Imax * dt / dV
 
  • #6
Baluncore said:
You assume energy is removed at a fixed rate which would require a switching regulator.

If on the other hand, if the regulator is linear, then the charge will be removed at a constant current …

You know that C = Q / V and that Q = I * t. Then; C = I * t / V.
So; Cmin = Imax * dt / dV

Yes, totally understand that. I'm intending on this cap to feed a DC-DC switching regulator which will compensate the current drawn due to the drooping voltage, but I might verify with an experiment. Thanks.
 
  • #7
Corey Spruit said:
This is not relevant, I went off and tried this, and ended up solving a quadratic equation (for fun, digging up old high-school maths again). But [itex]x^2-y^2 \neq (x-y)^2[/itex].

EverGreen1231 was correct, I can't just square root both sides. Interesting discussion on why on Stack Exchange if anyone is interested.

I left the equation at:
[tex]V_s = \sqrt{ \frac{2P\Delta t}{C_{min}} -V^2_f}[/tex]

1. I never said [itex]x^2-y^2 = (x-y)^2[/itex] but that was what you said originally. I pointed out it was wrong, and was the source of your error. That mistake is the source of your original error.

I guess there is confusion about "take the square root of both sides". You can always take the square root of both sides of an equation. If you are somehow thinking that a - b somehow has two sides, then you need to adjust your vocabulary.

2. In the final equation above, you correctly took the square root of both sides of the equation but the equation is STILL wrong. You should have ADDED [itex]V^2_f[/itex] to both sides before taking the square root of both sides.
 
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  • #8
meBigGuy: Yeah, it's clear now. Sorry my response was a bit vague too, I was referring to "(a-b)^2 is a^2 - 2ab +b^2."

Having looked at it again just now, I instantly noticed the -ve should be a +ve; I guess I hadn't reviewed what I entered that closely. When I went off and did the calcs for the actual reason I wanted this for, I had it correctly as a +ve and got the result I was after.

Thanks.
 
  • #9
EEs often use the RC time constant for this type of calculation. R times C gives the time needed to discharge a capacitor by ≈ 63.2%. That is often a nice model for figuring capacitor discharges which skips much of the math.
 
  • #10
Jeff Rosenbury said:
EEs often use the RC time constant for this type of calculation. R times C gives the time needed to discharge a capacitor by ≈ 63.2%. That is often a nice model for figuring capacitor discharges which skips much of the math.
That RC decay effect is quite different and inapplicable to both the constant current of a linear regulator and the constant power of a switching converter.
 
  • #11
RC is often used in ripple voltage approximations, which is what this sounds like to me.

Vripple(pp)=Vmax/120RC, for small values of ripple. R is of course determined by Ohm's law. V can be approximated as Vmax for small Vripple or Vmax - Vripple(pp) for larger ripples. This doesn't give an exact solution, but is good enough for capacitor selection.

Now why did my font size change in the middle of my message? :oldconfused:
 
  • #12
Jeff Rosenbury said:
RC is often used in ripple voltage approximations, which is what this sounds like to me.

Vripple(pp)=Vmax/120RC, for small values of ripple. R is of course determined by Ohm's law. V can be approximated as Vmax for small Vripple or Vmax - Vripple(pp) for larger ripples. This doesn't give an exact solution, but is good enough for capacitor selection.

Now why did my font size change in the middle of my message? :oldconfused:

I am trying to work out a value of capacitance required to 'ride out' interruptions in the incoming supply before it is fed into a DC-DC Regulator. So the Reg has a min/max input voltage range, looking at how to maintain this in case of a dropout in the supply.

That's a handy equation too when designing simple rectifiers. It should be noted that it is applicable to 60Hz systems, it's discussed on the Wikipedia page; the "120" is actually [itex]2f[/itex], so it would be 100 in 50Hz systems.
 
  • #13
How long do you need it to provide power?

Also this sounds like a job for supercapacitors. Supercapacitors are misnamed IMO. They are more like lame batteries. Still they have niche uses and this might be one of them. They last much longer than batteries. (I'm not sure they ever wear out.) But they can't discharge quickly like a real capacitor. You seem to want a slower discharge, so maybe that's not a problem.

A clever person could make the ripple equation work by treating R as a function, but perhaps not as easily as the true equation. I'm sure it doesn't work with supercapacitors though.

Good luck.
 

1. What is the formula for calculating capacitor discharge time?

The formula for calculating capacitor discharge time is t = RC, where t is the discharge time in seconds, R is the resistance in ohms, and C is the capacitance in farads.

2. How is the voltage across a capacitor related to time during discharge?

The voltage across a capacitor is exponentially related to time during discharge, following the equation V = V0e-t/RC, where V0 is the initial voltage across the capacitor.

3. What is the difference between charging and discharging a capacitor?

Charging a capacitor involves applying a voltage across the capacitor to store energy, while discharging involves releasing that stored energy. Charging is a process that increases the voltage across the capacitor, while discharging decreases the voltage.

4. What is the purpose of a resistor in a capacitor discharge circuit?

The purpose of a resistor in a capacitor discharge circuit is to limit the amount of current flowing through the circuit and prevent damage to the capacitor. It also controls the rate at which the capacitor discharges.

5. Can the capacitance or resistance values affect the discharge time of a capacitor?

Yes, both capacitance and resistance values can affect the discharge time of a capacitor. Higher capacitance or resistance values will result in a longer discharge time, while lower values will result in a shorter discharge time.

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