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

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SUMMARY

The discussion focuses on calculating the capacitance required to power a pulsating load of 100mA for 10ms at a voltage of 3.3V. The user has a power supply capable of delivering 3.3V and 2mA, while the load averages 100µA. The key equation for determining the capacitance is i(t) = C * (dv(t)/dt), which can be simplified to i(t)/C = Δv(t)/Δt. This relationship allows for the calculation of the necessary capacitor value and the time required for charging.

PREREQUISITES
  • Understanding of basic electrical concepts such as voltage, current, and capacitance
  • Familiarity with the equation i(t) = C * (dv(t)/dt)
  • Basic knowledge of pulsating loads and their characteristics
  • Ability to perform calculations involving microamperes and milliseconds
NEXT STEPS
  • Calculate the required capacitance using the formula C = i(t) * Δt / Δv(t)
  • Research capacitor charging time and its implications on circuit design
  • Explore the effects of load characteristics on capacitor selection
  • Learn about power supply limitations and their impact on circuit performance
USEFUL FOR

Electronics engineers, hobbyists designing small circuits, and anyone involved in power management for pulsating loads.

Neo11
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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:

[tex]i(t) = C \frac{dv(t)}{dt}[/tex]

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

[tex]i(t) = C \frac{\Delta v(t)}{\Delta t}[/tex]

or,

[tex]\frac{i(t)}{C} = \frac{\Delta v(t)}{\Delta t}[/tex]

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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