- #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?
(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?