LM2577s-12 Inductor selection

Hello Gurkan,

For a boost, you could calculate the inductance required as follows:
L = Vin*D/(deltaI*F). deltaI is the ripple current through the inductor and it is calculated as follows:
deltaI = 0.3*Iout/(1-D). D is the duty cycle and it is calculated as follows:
D = (Vout - Vin)/Vout.

After substituting the right parameters (Vin as 5V, F as 52KHz and Iout as 500mA) and calculating the L I get about 150uH as a target value. Further, the peak current through the inductor is calculated as:
Ipeak = 1.5*Iout/(1-D). This gives the peak current as 1.38A. 
Therefore I would suggest using a 150uH inductor rated for 1.7A or so. 

Using a 100uH inductor, you will get a larger ripple across the inductor. If your current is going to remain within the 500mA boundary, then you could use this value too. Only thing you should remember is to find an inductor with a relatively larger current rating. Therefore if you choose a 100uH inductor I would suggest rating it for about 2A. 

I hope this information is helpful. Please write back if you need more information.

Regards,
Akshay