Designing a 5V 10mA DC Regulated Power Supply from 230V 50Hz AC Line

In this article, we will discuss the design of a 5V 10mA DC regulated power supply using a 230V 50Hz AC line supply. This project involves the use of a 9-0-9 transformer, bridge rectifier, capacitor filter, and a zener diode as a voltage regulator. Each component will be selected and designed based on the given ratings to ensure proper operation and safety.

Introduction to the Components

The power supply will be designed to provide a stable 5V DC output current of 10mA from an AC line supply of 230V 50Hz. The design will include the following components:

9-0-9 transformer with a 500mA or higher current rating. Bridge rectifier to convert AC to DC. Capacitor filter to smooth the rectified waveform. Zener diode as a voltage regulator to maintain a constant 5V output. A current limiting resistor in series with the zener diode. A load resistor to draw the required current from the power supply.

Designing the Zener Diode and Series Current Limiting Resistor

The zener diode will work in reverse bias, producing 5V across it when 5V is applied to its cathode-anode junction. The value of the zener current ( I_{z} ) is determined by the formula:

[ I_{z} I_{load} frac{9 times sqrt{2} - 0.6 - 5.1}{R_s} ]

Given the specifications for the 1N4733A zener diode:

Vz 5.1V Pz 500mW

Substituting the values, we get:

[ P_{z} V_{z} times I_{z} 500mW implies I_{z} frac{500 times 10^{-3}}{5.1} approx 0.098A 98mA ]

Thus, the total current ( I_{z} I_{load} 108mA ), and the series resistor ( R_s ) is calculated as:

[ R_s frac{9 times sqrt{2} - 0.6 - 5.1}{0.108} approx 65.5 Omega approx 66 Omega ]

The power dissipation of the series resistor can be calculated as:

[ P_{R_s} (I_{z} I_{load})^2 times R_s approx (0.108 times 10^3)^2 times 65.5 approx 65.5mW ]

Selecting the Rectifier Diodes D1 and D2

The rectifier diodes must have the ability to carry at least 108mA of current. In practice, it is better to use higher-rated diodes as a safety margin. The diodes should also be able to block a Peak Inverse Voltage (PIV) of at least 91.414V, which is approximately 13V. A Schottky diode with part number 726-BAS12504WH6327, which has a 25V blocking capacity and a 700mA current carrying capacity, can be used for this purpose.

Selecting the Filter Capacitor

A filter capacitor should have a voltage rating of at least 25V. The capacitance value is derived using the following formula:

[ C frac{V_m}{2fR_{load} Delta V_r} ]

Where ( Delta V_r ) is the ripple voltage, which is 2% of the average voltage. Given:

The ripple voltage ( Delta V_r 2 times 0.0251 0.102 )V The peak voltage ( V_m ) in the secondary side of the transformer is 9V

The required capacitance ( C ) is thus:

[ C frac{9V}{2 times 50 times R_{load} times 0.102} ]

Assuming a load resistor ( R_{load} 100 Omega ), we get:

[ C approx frac{9V}{2 times 50 times 100 times 0.102} approx 9 mu F ]

A 470μF/25V capacitor can be used as the filter capacitor.

Connecting the Zener Diode and Load Resistor

The zener diode should be connected in reverse bias, and a series resistor of 470 or 560 Ohms may be used. The zener diode of 5.1V, either 500mW or 1W rating, can be used as a 5V regulator. The load resistor can be connected across the output as required.

Conclusion

This design of a 5V 10mA regulated power supply involves careful component selection and proper circuit design to ensure stable and reliable operation. By incorporating a 9-0-9 transformer, a bridge rectifier, a capacitor filter, and a zener diode as a voltage regulator, a robust and efficient power supply can be achieved. Each component plays a crucial role in ensuring the proper functioning of the circuit, and the design considerations outlined in this article should be followed to achieve the desired output.