# 5V Power Supply using 7805 Voltage Regulator with Design

## 5V Power Supply using 7805 Voltage Regulator with Design

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.

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

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.

# Circuit Diagram

• rusman

Hi,

thanks for the post.

when calculating R it should not be from secondary transformer since during discharge current direction is into the load.

so R = Vc / load into current via 7805.

do you agree ?

March 23, 2014 at 10:26 am
• arjun shinde

hello
thanks for giving specification of ICs 7805

July 28, 2015 at 2:35 pm
• Johanan Prime

hi there. im trying to design this in the Multisim program but i am unable to obtain the desired 5V output. instead, I am getting a constant 0.926pV output. I’ve tried changing the component values but I’m unable to obtain the expected result.

November 16, 2015 at 10:06 am
• 7805 is a proven voltage regulator. You will get it for sure. Try in hardware.

November 21, 2015 at 2:27 pm
• Nidheesh

Whether the output will be 5V 1A DC?

March 21, 2016 at 12:22 am
• Yes, 7805 maximum output capacity is 1A. So you should always use a current below it. And the current rating of the transformer also matters.

April 15, 2016 at 7:32 am
• giridhar

where have u used the 18k resistor?

June 8, 2016 at 3:13 pm
• giridhar

18ohm, sorry

June 8, 2016 at 3:15 pm
• It is the assumed load resistance.

August 27, 2016 at 12:15 pm
• zeineb

and if i want output max 5V 2A? what should i addto the circuit plz?

April 11, 2017 at 8:21 pm
• zeineb

and if i want output max 5V 2A? what should i add to the circuit?

April 11, 2017 at 8:25 pm
• Ravindran DVL

As per the data sheet of7805 the min input voltage should be 7.2 volts. Here in the circuit diagram it is shown 6 volts. Please clarify.

April 12, 2017 at 1:18 pm
• Shivanand Ganji

great article. simple for power supply design. transformer and other components actually makes it bulky.

power supply should be in Component form. like smps.

September 26, 2017 at 6:27 pm
• Sneha Mary Reddy Thumma

can you specify what is y and how u got that formula ???

July 5, 2018 at 12:23 pm
• Mohammed Mairajuddin Musharaf

Vr/2√3=0.4/2√3=0.115
But you got 0.3464p, please explain or correct ASAP

November 3, 2018 at 7:23 pm
• Balaji

(Vr/2)×√3=0.3464

December 9, 2018 at 2:29 pm