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Power Supply Design

By Engineering Knowledge administrator Rev1.0

Many of electronics hobbyist have built there own 5v and 12v PSU ("Power Supply UNIT") by them self, in order to power up the electronics experimental; which require a little bit of power. However, if the load requires too much amps, you have to calculate all the parameters to build a stable power supply. After you read this tutorial, you will be able to ("DIY, Do It Yourself") your efficient power supply!

How Power Supply Works?

The main function of power supply is to convert AC to DC; as shown in fig 1, the first stage is to make full wave rectifier to the AC signals by using bridge rectifier, filter the rectified wave by using filtering capacitor and finally select the appropriate voltage regulator to generate pure DC signal.

Current rating for bridge rectifier should be suitable with load current, also; the designer should consider the voltage drop across each diode, which is normally equal to 0.7v. Note that only two diodes are required when using center-tapped transformer.

The following formula is used to calculate the capacitance value for the filtering capacitor:

. . . (1)

C: Capacitor value.
Vp: Peak voltage. ("Bridge output max voltage")
f: Frequency of the AC supply.
i: Load current.

Note that the above equation for 10% ripple voltage.

We will use in this project LD1084#12 voltage regulator, which can provide up to 5Amps ("guaranteed") with only 1.3v dropout. Input voltage must be around 15v. It is highly recommended that you follow this value exactly, if the input voltage is more, the regulator temperature will increase more.

Tip:

It is possible to build your power supply without adding a regulator, when you want to power up an electrical motor, a small ripple in voltage will not affect the performance, unlike the electronic circuits and devices!

Designing of +12v 5A Power Supply:

To calculate the voltage required and the transformer secondary winding we first determine the input voltage for the regulator, which is 15v, plus a 10% of this value for ripple. For a regular transformer we have to consider a bridge rectifier, as a result; we will add 1.4v. So the secondary winding should be 15+1.5+1.4=18.9 lets say 18v @ 5 Amps. Now we will calculate the capacity of the filtering capacitor. By using equation number 1 and assuming that f=60Hz, we will get C=5 x 5 / ((18-1.4) x f) =25,100µF

Tip:

If C value is too large, connect more than one capacitor in parallel in order to reach the desired value.

To be sure what you will get exactly, we can simulate the circuit by using Simulink® ("below is the screenshot of the model"), download the model and see how the capacitor will act. Parameters are changeable, so you can adjust it as per your project requirements.

Invitation: Share your knowledge with others.

If you know how to build the circuit by another simulation software, build it and send it to contact@engknowledge.com . All credit will be yours.

As mentioned before LD1084#12 regulator will drop the output voltage by 1.3v when the load require 5 Amps, to go around this, we will use two regulators connected in parallel in order to get high quality and stable output voltage. A proper heat sink should be used with regulators.