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We have been using CAT4240TD-GT3 part for 24 V/40mA application for LCD backliting purpose.
Below is out inductor value calculation for Output Voltage=19.2V and LED Fwd Current=40mA. Our input Voltage is 5V.
D=(Vout - Vin)/Vout = (19.2-5)/19.2 = 0.739 (74% duty cycle).
IL=I out/1-D = 40m/ (1- 0.739) = 0.153Amp.
Ipk = IL(1+(ripple ratio /2)) = 0.153(1+(0.4/2)) = 0.1836Amp.
L= (Von*D)/(ripple*IL*f) = (5*0.739)/(0.4*0.153*(10^6)) = 60.3758uH (f = 1MHz switching frequency).
Final Inductor Value = 60uH.
We wish to confirm above values suitable for Vout=19.2 with 40 mA.
We also have input=5V, output Voltage=23.1V and Current=40mA for this part and used inductor value is 47uH where as calculated value is 53 uH.
But system is working ok. Please advise.
Regards
Yk
A 47 uH inductor is recommended for most of the CAT4240 applications. In cases where the efficiency is critical, inductances with lower series resistance are preferred. Inductors with current rating of 800 mA or higher are recommended for most applications.
Answered by: ON Semiconductor
2013-04-17 08:56:54.349