hold up time or storage time for the electrolytic capaictor

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srevish

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Dear Sir,

In my SMPS I am using 470uF, 35V capacitor, the output voltage is 24Vdc.

This 24Vdc output is given to the load of 2.5W through this capacitor.

Input voltage of this SMPS is 72Vdc.

I want to know after switching off this SMPS, what will be the time taken by this electrolytic capacitor of discharge its energy.

Is there is any formula to calculate it.
 

The time constant of the RC circuit will decide how long the capacitor can supply the current to the load. As you are saying a 2.5W load with 24V operation, the resistive load for this would be (24^2)/ 2.5 = 230.4 ohm. Now with the capacitor of 470uF and 230.4 ohm load, one time constant will be RC = 108 msec. The capacitor will get fully discharged in 5 time constants = 541 msec.

As the capacitor discharges providing the load current, the voltage across the capacitor also drops exponentially. The instantaneous voltage can be calculated by the formula: Vb X exp(- t/RC). Where Vb will be 24V initial voltage, t is time instance, R is load resistance and C is capacitance.

For example the instantaneous voltages at 10msec, 15 msec, 100 mesc and 150 msec will be 21.9V, 15.1V, 9.53 and 6V.
 


Thanks for your reply,

Also please I have checked this link

http://www.aimtec.com/site/Aimtec/f...tionNotes/increasing hold up time in smps.pdf
 

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