For a vast number of applications, a simple unregulated power supply is sufficient. This type of power supplies are normally used to supply power to current-hungry loads, for instance:

- Amplifiers

- Control circuits

- Irrigation systems

- Alarm circuit

Although not regulated or stabilized, when designed correctly, they can be a cheap alternative to regulated power supplies.

As can be seen, the power supply consist of only a few components:

- Power switch
- Protection fuse
- Transformer
- Rectifier
- Capacitor

**How it works **

The power switch S1, via the fuse F1, gives power to the transformer TR1. This will normally be 110V AC or 220V AC, depending on your location. The transformer perform two tasks:

- It converts the utility voltage from a high voltage to a safe working voltage of your power supply
- It gives electrical isolation between the utility network and your power supply output.

Rectifier B1 converts the AC voltage to a DC voltage.

Lastly, capacitor C1 is used to filter out the 50/60Hz components present on the DC output.

## Step 1: Waveforms of the Power Supply

**Waveform 1**

This shows the AC output of the transformer.

**Waveform 2**

This shows how the bridge rectifier changes the AC voltage into a DC voltage. However, the DC voltage fluctuates from minimum to maximum voltage. This is not suitable for most applications.

**Waveform 3**

The addition of the capacitor reduces the ripple of the DC output voltage. This ripple will change depending on the capacitor size and load current.

## Step 2: Designing the Power Supply

For each application, the specifications will be different. We will need the following information to start with the design:

- Normal output voltage
- Maximum output current
- Maximum output ripple (amount of change on the output voltage)
- Line frequency (50Hz or 60Hz)
- System voltage

**Example**

This Instructable will show you how to design a 12 Volt, 2 Amp unregulated power supply:

- Input Voltage: 220V 50Hz
- Output Voltage: 12V
- Output Current: 2A
- Maximum Ripple: 5%

## Step 3: Calculations for Capacitor

In the above waveform, the output can be seen behaving almost like a wave. This change in output voltage is called the Ripple. Capacitor C1 needs to limit this ripple to 5% of the output voltage.

**Calculating C1**

Vc x C = I x t

where:

Vc = voltage change allowed on output (ripple, given as 5%)

I = maximum current (given as 2A)

t = time period between 2 peaks (50Hz given)

Vc = ( % ripple / 100 ) x Voutput

Vc = ( 5 / 100 ) x 12V

**Vc = 0.6V**

**I = 2A, given**

t = 1 / ( 2 x frequency )

t = 1 / ( 2 x 50 )

**t = 0.01 second**

Now:

Vc x C = I x t

0.6 x C = 2 x 0.01

C = ( 2 x 0.01 ) / 0.6

C = 0.033 333 Farad

**C = 33 000 uF**

**Capacitor Specification**

The capacitor voltage rating must be about twice the maximum output voltage, so a good choice will be to use a 25V capacitor.

__Thus, C1 is 33 000uF 25V__

## Step 4: Calculating of Transformer Output Voltage

Vc = ripple voltage

Vc = ( 5/100 ) * 12V

**Vc = 0.6V**

The output voltage was given as 12V, which is the average output voltage.

The ripple was given as 5%, or 0.6V as calculated in previous step.

This means that our output voltage can change by 0.6V. This is TOTAL voltage change, lower or higher. So the average voltage of 12V, plus half of the ripple, will be the maximum output voltage, and 12V minus half of the ripple will be the minimum output voltage.

The maximum DC voltage for the power supply after the diodes then becomes:

Vmax = Voutput + ( 0.5 of ripple voltage)

Vmax = 12 + ( 0.5 x 0.6V)

Vmax = 12 + 0.3V

**Vmax = 12.3V**

But we must also take into account the voltage drop across the diodes. Two diodes will conduct in any given power cycle. Each diode will have around 0.7V voltage drop across them.

Vin = Vmax + ( 2 x diode forward voltage drop)

Vin = 12.3V + ( 2 x 0.7V )

Vin = 12.3V + 1.4V

**Vin = 13.7V **

Vin is the maximum output voltage of the transformer. We now need to calculate the RMS voltage of the transformer:

Vrms = Vin / 1.414 (1.4114 is the square root of 2)

Vrms = 13.7V / 1.414

**Vrms = 9.68V AC**

Typical, a 10V transformer will be available.

**Thus Vseconary will be 10V AC**

Transformers ratings are normally specified in VA, which is the amount of current the transformer can withstand.

Power = V * I

Power = Vin * Iload

Power = 13.7V * 2A

**Power = 27.4VA**

**Transformer Specification:**

*Vprimary: 220V AC*

*Vsecondary: 10V AC*

*Frequency: 50Hz*

*Power : 28VA*

## Step 5: Calculating Diode Bridge Ratings

We must now calculate the minimum voltage and current ratings of the diode bridge.

**Voltage Rating**

During any given time, two diodes will be conducting, while the other two diodes will have a reverse voltage applied across them. This reverse voltage is the minimum voltage that the diodes should be able to withstand.

Our transformer output voltage was calculated at 10V. This voltage is RMS voltage, and not the peak output voltage of the transformer.

Vpeak = Vrms x 1.414 (square root of 2)

Vpeak = 10V x 1.414

Vpeak = 14.141V

But the diodes must withstand twice this voltage

Vd_reverse = Vpeak x 2

Vd_reverse = 14.141V x 2

Vd_reverse = 28.3V

This is the minimum reverse voltage. A good practice is to rate the diodes for at least twice the calculated rating:

Vd_reverse = Vd_reverse x 2

Vd_reverse = 28.3V x 2

Vd_reverse = 56.6V, or 60V

**Current rating**

The power supply is rated at 2A contiuous. This is the minimum current rating of the diodes. A good practice is to rate the diodes for at least twice the calculated rating:

Idiode = Imax x 2

Idiode = 2A x 2

Idiode = 4A

**Diode Bridge Ratings**

Idiode = 4Amp

Vreverse = 60V

## Step 6: Calculating Fuse Rating

To protect the power supply and load in the case of an fault, the HV side of the power supply is fitted with a protection fuse.

The power supply is designed for 2 Amps continuously, but there will be times when the current will exceed 2 Amp. For instance, connecting a load with large supply line capacitors. The power supply itself will also require more than 2 Amps when switched on, to charge up the capacitor.

In these cases, we do not want the fuse to blow. A good rule is to allow for a 100% overload on the transformer;

Imax_lv = Imax x 2

Imax_lv = 2A x 2

Imax_lv = 4A

But the fuse is on the HV side of the transformer, so

Imax_hv = Imax_lv x ( Vlv / Vhv )

Imax_hv = 4A x ( 10V / 220V )

Imax_hv = 0.182A, or 0.2A

Increasing the current rating of the fuse alone, will prevent it from blowing in most cases, However, some loads will take a couple of milliseconds to reduce their peak current. For this reason, the fuse must also have a timing element to allow for short periods of overload. For this reason, the fuse must be a slow-blow fuse, and not a fast-blow fuse.

**Fuse Rating**

This example is for a 220V power supply, so the fuse must be able to break the current at this voltage. The fuse rating is now:

220V, 0.2A slow-blow

## Step 7: Floating Versus Grounded Output

The circuit diagram indicates that the output of the power supply can be connected to the mains earth (the dotted line).

When the 0V line is not grounded, the output voltage will float, and can produce around 24V - 75V with respect to earth.This is by no means unsafe, but can give of a little bite if the output is touched. However, this method totally isolate the power supply from any other electrical circuits.

This is purely up to the user how this must be configured, and will not change the behavior of the power supply.