5 VOLT REGULATED POWER SUPPLY DIAGRAM USED IN MICROCONTROLLER BASED PROJECTS

                  REGULATED POWER SUPPLY

A variable regulated power supply, also called a variable bench power supply, is one where you can continuously adjust the output voltage to your requirements. Varying the output of the power supply is the recommended way to test a project after having double checked parts placement against circuit drawings and the parts placement guide. This type of regulation is ideal for having a simple variable bench power supply. Actually this is quite important because one of the first projects a hobbyist should undertake is the construction of a variable regulated power supply. While a dedicated supply is quite handy e.g. 5V or 12V, it's much handier to have a variable supply on hand, especially for testing. Most digital logic circuits and processors need a 5 volt power supply. To use these parts we need to build a regulated 5 volt source. Usually you start with an unregulated power supply ranging from 9 volts to 24 volts DC . To make a 5 volt power supply, we use a LM7805 voltage regulator IC .

The LM7805 is simple to use. You simply connect the positive lead of your unregulated DC power supply (anything from 9VDC to 24VDC) to the Input pin, connect the negative lead to the Common pin and then when you turn on the power, you get a 5 volt supply from the Output pin.

CIRCUIT FEATURES

Brief description of operation: Gives out well regulated +5V output, output current capability of 100 mA

• Circuit protection: Built-in overheating protection shuts down output when regulator IC gets too hot
• Circuit complexity: Very simple and easy to build
• Circuit performance: Very stable +5V output voltage, reliable operation
• Availability of components: Easy to get, uses only very common basic components
• Design testing: Based on datasheet example circuit, I have used this circuit successfully as part of many electronics projects
• Applications: Part of electronics devices, small laboratory power supply
• Power supply voltage: Unregulated DC 8-18V power supply
• Power supply current: Needed output current + 5 mA
• Component costs: Few dollars for the electronics components + the input transformer cost

BLOCK DIAGRAM

CIRCUIT DIAGRAM

BASIC POWER SUPPLY CIRCUIT

            Above is the circuit of a basic unregulated dc power supply. A bridge rectifier D1 to D4 rectifies the ac from the transformer secondary, 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.

                   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 (I_L) 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) 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.

                  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).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 –

                 In our example above we were taking 0.5A out of the Vsec of 10V. The VA required is 10 X 0.5A = 5VA. 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.

THEY MUST BE HIGH VOLTAGE TYPES RATED FOR A.C USE

                  

                     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

                   The whole project MUST be enclosed in a suitable box. The main switch (preferably double pole) must be rated at 240V or 120V at the current rating. All exposed parts within the box MUST be fully insulated, preferably with heat shrink tubing.