Capacitance Calculation for Powering Pulsating Load: 100mA for 10mS at 3.3V

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To power a pulsating load of 100 mA for 10 ms at 3.3V, a capacitor value must be calculated to ensure it can deliver the required current. The power supply provides only 2 mA, necessitating the capacitor to store sufficient charge to meet the load's demands. The relevant equation for this calculation is i(t) = C * (Δv/Δt), which relates current, capacitance, and voltage change over time. Additionally, determining the time required to charge the capacitor is crucial to ensure it can handle the current surge effectively. Proper calculations will enable the design of a circuit that meets the pulsating load requirements.
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I want to build a small circuit which will power a pulsating load. I got a power supply which can provide 3.3V and 2 mA. Now my load requirement is 100 micro amp average and 100 mA for 10 mSec(all at 3.3V). I want to calculate

1. Value of capacitor which will provide 100mA for 10 mS
2. Time duration to sufficiently charge the capacitor(to determine after how much time load current surge can be adjusted)

Can anybody help me to solve this problem?
 
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Neo11 said:
I want to build a small circuit which will power a pulsating load. I got a power supply which can provide 3.3V and 2 mA. Now my load requirement is 100 micro amp average and 100 mA for 10 mSec(all at 3.3V). I want to calculate

1. Value of capacitor which will provide 100mA for 10 mS
2. Time duration to sufficiently charge the capacitor(to determine after how much time load current surge can be adjusted)

Can anybody help me to solve this problem?

The equation you want to use is:

i(t) = C \frac{dv(t)}{dt}

Or in simpler form (if you haven't studied calculus):

i(t) = C \frac{\Delta v(t)}{\Delta t}

or,

\frac{i(t)}{C} = \frac{\Delta v(t)}{\Delta t}

This means that the droop in voltage with respect to time is equal to the current out of the cap, divided by the size of the cap.
 
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