A power supply could be something as simple as a 9v battery or it could be as complex as a precision laboratory power supply. Here we deal in ac dc power supplies and present a basic unregulated dc power supply. The ac from the transformer secondary is rectified by a bridge rectifier then filtered.


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What is a power supply?

A power supply could be something as simple as a 9v battery or it could be as complex as a precision laboratory power supply. Here we deal in ac dc power supplies.

Below is the schematic of a very basic unregulated dc power supply. The ac from the transformer secondary is rectified by a bridge rectifier D1 to D4 which may also be a block rectifier such as WO4 or even four individual diodes such as 1N4004 types. (see later re rectifier ratings).

The principal advantage of a bridge rectifier is you do not need a centre tap on the secondary of the transformer. A further but significant advantage is that the ripple frequency at the output is twice the line frequency (i.e. 50 Hz or 60 Hz) and makes filtering somewhat easier.

This image is copyright © by Ian C. Purdie VK2TIP - schematic of basic power supply

Figure 1 - schematic of basic power supply

As a design example consider we wanted a small unregulated bench supply for our projects. Here we will go for a voltage of about 12 - 13V at a maximum output current (IL) of 500ma (0.5A). Maximum ripple will be 2.5% and load regulation is 5%.

Now the rms secondary voltage (primary is whatever is consistent with your area) for our power transformer T1 must be our desired output Vo PLUS the voltage drops across D2 and D4 ( 2 * 0.7V), all divided by 1.414.

This means that Vsec = [13V + 1.4V]  / 1.414 which equals about 10.2V. Depending on the VA rating of your transformer, the secondary voltage will vary considerably in accordance with the applied load. The secondary voltage on a transformer advertised as say 20VA will be much greater if the secondary is only lightly loaded.

If we accept the 2.5% ripple as adequate for our purposes then at 13V this becomes 13 * 0.025 = 0.325 Vrms.  The peak to peak value is 2.828 times this value. Vrip = 0.325V X 2.828 = 0.92 V and this value is required to calculate the value of C1. Also required for this calculation is the time interval for charging pulses. If you are on a 60Hz system it it 1 / (2 * 60 ) = 0.008333 which is 8.33 milliseconds. For a 50Hz system it is 0.01 sec or 10 milliseconds.

The formula for C1 is:

C1 (uF) = [ ( IL * t )  / Vrip ] X 10 6

C1 = [ ( 0.5A X 0.00833 ) / 0.92V ] X 10 6

C1 = 0.00453 X 10 6 = 4529 or 4700 uF

Remember the tolerance of the type of capacitor used here is very loose. The important thing to be aware of is the voltage rating should be at least 13V X 1.414 or 18.33. Here you would use at least the standard 25V or higher (absolutely not 16V).

Power Supply diode rectifier rating

With our rectifier diodes or bridge they should have a PIV rating of 2.828 times the Vsec or at least 29V. Don't search for this rating because it doesn't exist. Use the next highest standard or even higher. The current rating should be at least twice the load current maximum i.e. 2 X 0.5A or 1A. A good type to use would be 1N4004, 1N4006 or 1N4008 types. These are rated 1 Amp at 400PIV, 600PIV and 1000PIV respectively. Always be on the lookout for the higher voltage ones when they are on special.

Transformer Rating for Power Supply

In our example above we were taking 0.5A out of the Vsec of 10V. The VA required is 10 X 0.5A = 5VA. For our power supply this is a small PCB mount transformer available in Australia and probably elsewhere. This would be an absolute minimum and if you anticipated drawing the maximum current all the time then go to a higher VA rating.

The two capacitors in the primary side are small value types and if you don't know precisely and I mean precisely what you are doing then OMIT them. Their loss won't cause you heartache or terrible problems.


The fuse F1 must be able to carry the primary current but blow under excessive current, in this case we use the formula from the diagram. Here N = 240V / 10V or perhaps 120V / 10V. The fuse calculates in the first instance to [ 2 X 0.5A ] / [240 / 10] or .04A or 40 ma. In the second case .08A or 80 ma. The difficulty here is to find suitable fuses of that low a current and voltage rating. In practice you use the closest you can get (often 100 ma ). Don't take that too literal and use 1A or 5A fuses.

Construction of Power Supply

The whole power supply project MUST be enclosed in a suitable enclosure, preferably of the insulated variety. The main switch, also preferably double pole, must be rated at least 240V or 120V at the current rating. All exposed parts within the power supply box MUST be fully insulated, preferably with heatshrink tubing.

UPDATE - 20th January, 2002

One very sensible suggestion for housing your power supply project is to use the case of a defective or surplus Switched Mode Power Supply (SMPS) from a computer!

Think about it? A case, suitable socket - probably with a surplus cable, switch maybe, fuse maybe, fan if you need it etc.... Possibilities are endless. Just remove the SMPS guts and discard to the junk box... FWIW. Also I'll try and do some photos.

Power Supply Kits available through AMAZON.

BOOK - Power Supplies by David Lines

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