## 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.

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

**Component List :**

- Step down transformer
- Voltage regulator
- Capacitors
- 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 I
_{c = }1A - Output voltage range V
_{Max=5.2V ,}V_{Min=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/I_{c }

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=V_{ac-rms}/V_{dc}

V_{ac-rms} = V_{r}/2√3

V_{dc= }V_{Max}-(V_{r}/2)

V_{r=} V_{Max-} V_{Min}

- V
_{r}= 5.2-4.8 =0. 4V - V
_{ac-rms = .3464V} - V
_{dc = 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.

## Comments (16)

rusmanHi,

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 ?

arjun shindehello

thanks for giving specification of ICs 7805

Johanan Primehi 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.

Ligo George7805 is a proven voltage regulator. You will get it for sure. Try in hardware.

NidheeshWhether the output will be 5V 1A DC?

Ligo GeorgeYes, 7805 maximum output capacity is 1A. So you should always use a current below it. And the current rating of the transformer also matters.

giridharwhere have u used the 18k resistor?

giridhar18ohm, sorry

Ligo GeorgeIt is the assumed load resistance.

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

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

plz answer me!

Ravindran DVLAs 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.

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

power supply should be in Component form. like smps.

Sneha Mary Reddy Thummacan you specify what is y and how u got that formula ???

Mohammed Mairajuddin MusharafVr/2√3=0.4/2√3=0.115

But you got 0.3464p, please explain or correct ASAP

Balaji(Vr/2)×√3=0.3464