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

[Neo11]

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?

 

[berkeman]

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.

 

[oswaler]

Based on the given information, the capacitance required to provide 100mA for 10mS can be calculated using the formula C = I*t/V, where C is capacitance, I is current, t is time, and V is voltage. Plugging in the values, we get C = (100mA * 10mS)/3.3V = 0.303mF (or 303μF). This is the minimum capacitance required to sustain the load current for 10mS at 3.3V.

To determine the time duration for the capacitor to sufficiently charge, we can use the formula t = RC, where t is time, R is resistance, and C is capacitance. Since we are given the voltage and current, we can calculate the resistance using Ohm's Law (R = V/I). Plugging in the values, we get R = 3.3V/100μA = 33kΩ. Now, using the calculated capacitance of 303μF, we can calculate the time as t = (33kΩ * 303μF) = 9.99mS. This means that it will take approximately 10mS for the capacitor to reach its maximum charge and be able to sustain the load current of 100mA.

It is important to note that these calculations are based on ideal conditions and may vary in a real-world circuit due to factors such as internal resistance and capacitance of the power supply. It is always recommended to have a margin of safety when designing circuits, so it is advisable to use a slightly higher capacitance and allow for a longer charging time to ensure stable and reliable operation of the circuit.