I have a 2.2V @ 20mA LED, that is driven by a 5V PSU, and therefore is connected to a 140Ω resistor.
I'd like to use to purchase a capacitor, that i'd be able to charge the capacitor to 5V, and that when I connect the capacitor to the LED circuit, as the PSU of this circuit, the LED's volume of illumination will not decline from 90% for 5 seconds.
I treated the LED as 2.2V/20mA=110Ω resistor, and did the next math:
4.5V = 5V * e^[-5sec/(250Ω*C)]
I reached C = 190mF.
Was it ok to treat the LED as a 110Ω resistor?
Is there any other way to reach this answer, reaching it using Q of the capacitor?
I'd like to use to purchase a capacitor, that i'd be able to charge the capacitor to 5V, and that when I connect the capacitor to the LED circuit, as the PSU of this circuit, the LED's volume of illumination will not decline from 90% for 5 seconds.
I treated the LED as 2.2V/20mA=110Ω resistor, and did the next math:
4.5V = 5V * e^[-5sec/(250Ω*C)]
I reached C = 190mF.
Was it ok to treat the LED as a 110Ω resistor?
Is there any other way to reach this answer, reaching it using Q of the capacitor?