Question

An inductor in a switch-mode power supply that operates at a peak current of 8 A needs to store 80 mJ of energy, thereafter release it to a 10 ? load resistor. Suppose this inductor had to be custom designed and built as a solenoid coil wound on a pencil (10 mm diameter) with no magnetic properties. Only one layer of turns is allowed on the pencil’s outer surface, and 0.1-mm diameter enamel coated wire has to be used. Answer the following carefully:

(a) Calculate the required inductance to store the energy.

(b) Draw an isometric sketch of the air-core inductor with only one layer of turns, showing its dimensions and a few turns of wire. Also draw some flux lines.

(c) Calculate the effective cross-sectional flux area of the air-core inductor.

(d) Do 3 iterative calculations to solve the number of turns (tightly wound next to each other) required to achieve the inductance in (a). Comment on your result after each iteration.

Answer 1

(a). I=8Amps, E =80 mJ

We know that, energy stored in the inductor, E = (1/2)LI² ;

Calculating the value of inductance, L using above relation and the given values of E and I.

Solving above we get, L=2.5mH

(b). Find the solution in the attached figure:

(c). Diameter of pencil =10mm

Radius of wire== 0.05mm

Effective radius of the inductor is the distance of the mid of flux path to the mid point of pencil , which is ,r=(5+0.05)mm =5.05mm

Effective Area of air core inductor=

(d). If number of turns = N and the length of inductor =l ;

then l=0.1N 10^-3 m= (10^-4 N)metres (1)

where, 0.1 = width/diameter of the wire

also, for the solenoid, (2)

where A= effective area of cross section

Substituting the values in the above relation:

L=0.0025H,

A=80.0810^-6 m²

= 1.25710^-6H/m

Put values in (2):

0.0025=1.25710^-6 N²80.0810^-6/l

N²/l =2.4810⁸ m^-1 -(3)

Substituting the value of l from (1) in (3):

N=2.4810⁴ =24800

Substituting the value of N in (1):

l= 10^-4N =10^-424800 =2.48m

Hope this was helpful.