Kaushik Duari

Introduction

Today, most consumer electronic products require a steady DC voltage supply in order to function. The transmission of electrical power happens in AC as it is more efficient to transfer with minimal power loss. It is the job of the design engineers to somehow use this AC voltage and convert it to a stable DC voltage that the electronic products can use.

This process requires that we first make all of the sinusoidal wave positive i.e. rectify it and then make the drops from the positive peaks to 0 less abrupt, i.e. filter the signal. After this, we must bring the voltage down to a required output i.e. regulate it.

This is what we are trying to achieve in this project over the three phases.

Phase 1

Abstract

We will be designing a power supply system in this project with the following schematic.

For a specifically chosen output, we will work with three different power supply design choices and make use of specific components which will match the required design specifications.

The three power supply system choices are as follows:

1. Based on single ended half wave rectifier

2. Based on single ended full wave rectifier

3. Based on center tapped full wave rectifier

For each of the above three design options, we will be determining the following values:

1. Rs -> Resistance across filter output

2. Rz -> Max zener diode impedance

3. Vripple ->Voltage ripple at filter capacitor

4. C -> Capacitance of the filter capacitor

5. Maximum reverse voltage across rectifier diode

We will also make component choices based on these values and determine the total cost of implementation of each design option.

Theory

Zener diode (regulator):

The zener diode with an output of 10 V across it and 20 mA of test current (Iz min) through it has an impedance of 17.

Filter output and Rs:

From the figure,

Is = Io + Iz (By KCL)

When Io = 0, Is = Iz, Iz here is Izmax. Thus, Is = Izmax.

Rs = (Vpeak – Vd – Vout) / (Iz + Io) (From webpage)

Based on the rectifying circuit used Vd might be 2*Vd (Center-tapped) as 2 diodes are used or 4*Vd (Single-ended full wave) as 4 diodes are used.

Output ripple and voltage ripple

Allowed output ripple is 2% of Vout.

We can use the output ripple value as the output voltage across Rz.

Then we can apply voltage divider rule with Vripple as input voltage to find Vripple:

Output ripple = Vripple * (Rz) / (Rs + Rz)

Theory of maximum reverse voltage

Half-wave rectifier:

Here, the peak voltage from the transformer is 30 * (2) ^ ½

The largest possible reverse voltage across the diode is if there is an output of peak voltage from the transformer and there is the opposite polarity if the same value going in to the filter. This puts our maximum reverse voltage at 2*Vpeak.

Single ended full wave rectifier:

Here, the top arm of the transformer (with L2) outputs only 15 V rms.

Therefore, Vpeak = 15 * (2) ^ ½

The maximum reverse voltage across any diode from the same theory as the half-wave rectifier = 2 * Vpeak.

Center tapped full-wave rectifier:

Here again the 2 arms of the transformer (with L2 and L3 respectively), each output 15 V rms.

Therefore, the peak value of voltage = 15 * (2) ^ ½

Again from the same theory as used in the half-wave rectifier as well as the single ended full wave rectifier, the maximum reverse voltage across the diodes is 2*Vpeak.

In summary, we need to know the maximum reverse voltage across the rectifying diodes because diodes have only a finite reverse voltage tolerance before they break down and it is essential that the rectifying diodes don’t break down.

Application of theory and results

Appropriate choice of zener diode

We chose to have an output voltage of 10 V. The zener diode which gives us this output is the DIN5240.

We are allowed to use 75% of the maximum power output the zener diode