5V Power Supply using 7805

Circuit Diagram

5V Power Supply Circuit using 7805 Voltage Regulator

In most of our electronic products or projects we need a power supply for converting mains AC voltage to a regulated DC voltage. For making a power supply designing of each and every component is essential. Here I’m going to discuss the designing of  regulated 5V Power Supply.

Let’s start with very basic things the choosing of components

Component List :

1. Step down transformer
2. Voltage regulator
3. Capacitors
4. Diodes

Let’s get into detail of rating of the devices :

Voltage regulator :

As we require a 5V we need LM7805 Voltage Regulator IC.

7805 IC Rating :

• Input voltage range 7V- 35V
• Current rating Ic = 1A
• Output voltage range   VMax=5.2V ,VMin=4.8V 

LM7805 – Pin Diagram

Transformer :

Selecting a suitable transformer is of great importance. The current rating and the secondary voltage of the transformer is a crucial factor.

• The current rating of the transformer depends upon the current required for the load to be driven.
• The input voltage to the 7805 IC should be at least 2V greater than the required 2V output, therefore it requires an input voltage at least close to 7V.
• So I chose a 6-0-6 transformer with current rating 500mA (Since 6*√2 = 8.4V).

NOTE : Any transformer which supplies secondary peak voltage up to 35V can be used but as the voltage increases size of the transformer and power dissipation across regulator increases.

Rectifying circuit :

The best is using a full wave rectifier

• Its advantage is DC saturation is less as in both cycle diodes conduct.
• Higher Transformer Utilization Factor (TUF).
• 1N4007 diodes are used as its is capable of withstanding a higher reverse voltage of 1000v whereas 1N4001 is 50V

Capacitors :

Knowledge of Ripple factor is essential while designing the values of capacitors

It is given by

• Y=1/(4√3fRC)  (as the capacitor filter is used)

1. f= frequency of AC ( 50 Hz)

2. R=resistance calculated

R= V/Ic   

V= secondary voltage of transformer

•      V=6√2=8. 4
•      R=8.45/500mA=16.9Ω standard 18Ω chosen

3. C= filtering capacitance

We have to determine this capacitance for filtering

Y=Vac-rms/Vdc

Vac-rms = Vr/2√3

Vdc= VMax-(Vr/2)

Vr= VMax- VMin

•   Vr = 5.2-4.8 =0. 4V
•    Vac-rms = .3464V
•    Vdc = 5V
•     Y=0 .06928

Hence the capacitor value is found out by substituting the ripple factor in Y=1/(4√3fRC)

Thus, C= 2314 µF and  standard 2200µF is chosen

Datasheet of 7805 prescribes to use a 0.01μF capacitor at the output side to avoid transient changes in the voltages due to changes in load and a 0.33μF at the input side of regulator to avoid ripples if the filtering is far away from regulator.