In this project, I will show you a 12V to 5V Buck Converter Circuit that can efficiently convert 12V DC to 5V DC with a maximum 50mA output current. Let's make it!

## Components List

The following components are required for this 12v to 5v converter.
1. 5V Zener Diode (1 pcs)
2. 150R 1/2W Resistor (1 pcs)
3. 100nF Capacitor (1 pcs)
4. 3mm LED (1 pcs)
5. 1K 1/4W Resistor (1 pcs)
6. 2 Pin Terminal connector (2 pcs)

## Zener Diode Based 12V to 5V Buck Converter Circuit Diagram

The schematic of 12v to 5v buck converter circuit using Zener diode is shown below. This circuit design is for 50mA Load.

## How to Calculate the Buck Converter Circuit for 5V 50mA Load?

To get the DC 5V, 50mA from 12V dc source, we need the Current through resistor (R1) constant at all times. Even the load current will change from 0 mA to the maximum value (50 mA), and the Zener diode current needs to change to keep the voltage across the output at 5V.

### Calculations

Power of the Zener diode is,
PZ = VZ x IZ
[Where, PZ: Power of the Zener diode, VZ: Voltage through Zener diode, and IZ: Current through Zener diode.]
= 5V x 50mA
= 0.25 watts
Here, we use 5V 0.5W Zener diode.

Current limit Resistor R1 Value is,
R1 = (Vin – VZ)/ IR
[Where, R1: Resistor, Vin: Input Voltage, and VZ: Zener diode voltage.]
= (12V – 5V)/ 50mA
= 140 ohms or about 150 ohms.

Current limit Resistor R1 Power is,
PR = VR x IR
[Where, PR: Power of R1, VR: Voltage through R1, and VZ: Zener diode voltage.]
= (12V – 5V) x 50mA
= 7V x 50mA
= 0.35 watts or use 0.5 watts.

The filtering capacitor value, C1 = 100nF.

## Working principle of Zener Diode based 12V to 5V Buck Converter Circuit

The main component of this buck converter circuit is a 5V Zener diode. The working principle of the 12V to 5V step-down converter is easy.

When this buck converter circuit gets power from a 12V DC source, a input power indicator light indicates. A suitable current limiting resistor R1 drops the input higher current to 50mA current. This sufficient current drive the Zener diode safely to produce a stabilised 5V 50mA output with low ripple under varying Load current conditions. A filtering capacitor C1 is used to remove the ripples from the output. Thus the buck converter circuit converts 12V to 5V.