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Design >> Analog Design >> Switching power in cap
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Message started by mks on Mar 9^th, 2020, 8:16pm

Title: Switching power in cap
Post by mks on Mar 9^th, 2020, 8:16pm

If I have a step from 0 to V volts across a cap there is power dissipation of 0.5*CV^2.
Is it true that if I go from 0 to V/2 and then V/2 to V then the power dissipation would be 0.5*C(V/2)^2 + 0.5*C(V - V/2)^2 i.e. .5*CV^2/2

Which is half of the earlier case? I am not able to figure out where I am getting this wrong as the final charge on the cap is the same in both cases.

It will be great help if someone could help me understand this
Thanks

Title: Re: Switching power in cap
Post by rajasekhar on Mar 10^th, 2020, 7:32am

Your calculations are correct. The smaller the voltage charged to the cap, the lesser the energy lost in the resistor. You can consider each step charging voltage is V/N, replete this N time. When N-->inf, you would see no energy loss. Another simple way is to check the current waveform. In your example-charging directly to V, would result in impulse current of VC. If you do the two steps charging you would see CV/2 impulses two times. I hope it is helpful.

Thanks,
Raj.

Title: Re: Switching power in cap
Post by mks on Mar 10^th, 2020, 8:17am

Hi Raj,
Thanks for the clarification.
With infinite steps, if I have something like sine wave you meant I will have negligible consumption?
Thanks

Title: Re: Switching power in cap
Post by rajasekhar on Mar 12^th, 2020, 4:51am

I mean to say-first time V/n voltage, next 2V/n and so on until V..in, In this case, the Energy waste in the resistor is Zero when N tends to infinite. its good exercise..Let me know if you can't.. I will post the answer...

Thanks,
Raj.