In the lm3914's datasheet, it uses a 7.5 ohm resistor and a 2.2uF tantalum cap to cut heating in half, with 5v supply and 20mA through LED's. How do you calculate the resistance and capacitance?

 

In the lm3914's datasheet, it uses a 7.5 ohm resistor and a 2.2uF tantalum cap to cut heating in half, with 5v supply and 20mA through LED's. How do you calculate the resistance and capacitance?

Could you save us some bother and post the datasheet with the page where you got the info?

 

it's under application hints, on page 19.
LM3914 Datasheet by Texas Instruments | Digi-Key Electronics (digikey.com)

 

In the lm3914's datasheet, it uses a 7.5 ohm resistor and a 2.2uF tantalum cap to cut heating in half, with 5v supply and 20mA through LED's. How do you calculate the resistance and capacitance?

If you are concerned about power dissipation, use high efficiency LEDs that will be plenty bright with 1 mA current.
Also don't use bar mode. Use dot mode.

 

it's under application hints, on page 19.

They're assuming LED voltage is 2V. That leaves 3V across the driver. P = IV = 200mA * 3V = 600mW.

Since the driver is constant current, inserting a 7.5 ohm resistor will reduce dissipation in the driver by P = RI^2 = 7.5 ohms * 200mA^2 = 300mW.

The capacitor value isn't very critical, and you probably don't even need it because small variations in supply voltage won't likely affect LED brightness noticeably.