Basic electronics

Electronics is the branch of science and technology which makes use of the controlled motion of electrons through different media and vacuum. In this course you will be able to make electronic devices and components. An AC DC Power Supply is one of the devices that we had as our project. 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. 




Terms and materials to remember:

*Rectifier-is an electrical device that converts alternating current (AC) to direct current (DC), a process known as rectification. 
*Capacitor- is a passive electronic component consisting of a pair of conductors separated by a dielectric (insulator).
*Transformer- is a device that transfers electrical energy from one circuit to another through inductively coupled conductors—the transformer's coils.


This is 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 diode such as 1N4004 types. The principal advantage of a bridge rectifier is you do not need a center 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 (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. 






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