Buck Converter Explained

Understanding the Basic Concept

The difficulty here is that the numerical solution jumps directly to formulas without explaining what a buck converter and the inductor are doing physically.

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Step 1: What is a Step-Down (Buck) Converter?

A buck converter takes a higher DC voltage and produces a lower DC voltage by rapidly turning a switch ON and OFF.

• Input voltage = 230 V
• Desired average output voltage = 50 V

Since the switch is not ON all the time, the load sees an average voltage lower than 230 V.

The fraction of time the switch is ON is called the duty cycle (D).

V_o = D × V_s

Here:

D = 50 / 230 = 0.217

This means the switch is ON only about 21.7% of the time and OFF for 78.3% of the time.

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Step 2: Why Do We Need an Inductor?

Imagine connecting the load directly to the switching voltage:

230 V → ON
0 V   → OFF
230 V → ON
0 V   → OFF

The current would jump up and down violently.

An inductor resists sudden changes in current.

Think of an inductor as a flywheel for current.

• During ON time, current increases slowly.
• During OFF time, current decreases slowly.

Instead of large current jumps, you get a small ripple around the average value.

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Step 3: What Does "15% Ripple" Mean?

Average current:

I_avg = 100 A

Allowed ripple:

ΔI = 0.15 × 100 = 15 A

So the peak-to-peak ripple current should not exceed 15 A.

Example:

Maximum current = 107.5 A
Minimum current = 92.5 A

Ripple = 107.5 − 92.5 = 15 A

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Step 4: Why Does a Larger Inductor Reduce Ripple?

The basic inductor equation is:

V = L(di/dt)

or

di/dt = V/L

This tells us:

• Large L → Current changes slowly
• Small L → Current changes quickly

Large inductor → Small ripple
Small inductor → Large ripple

That is exactly why the question asks for the minimum value of L that keeps the ripple within 15%.

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Step 5: Where Does the Ripple Formula Come From?

During ON time:

Voltage across inductor = V_s − V_o

= 230 − 50 = 180 V

Current rises for:

t_on = D × T

where

T = 1/f = 1/400 = 2.5 ms

Thus:

t_on = 0.217 × 2.5 ms = 0.543 ms

Using the inductor law:

ΔI = (V_L × t) / L

Substituting the values:

15 = (180 × 0.543 × 10^-3) / L

L = 6.52 × 10^-3 H

L ≈ 6.5 mH

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