## How to project circuits of Power Supply using integrated circuit of the family 78xx

With the integrated circuits of the family 78xx it is possible to set up a circuit of power supply with easiness and efficiency.

The circuits integrated tension regulators are a class of integrated circuits, quite used, those circuits present in your internal structure, circuits for the reference supply, error amplifier, control device and overload protection, everything in an only package.

The regulators of tension of three positive terminals, are available in several packages, being the most common and found with more means TO-92 (78Lxx max. 70MA) AND TO-220 (78XX max. 1A). The fixed values of tension are understood from 5 to 24 volts.

##### Get new posts by email:

Those integrated circuits ally some factors for a good acting: limitation interns of current, area holds of compensation and thermal stoppage, tornado that quite reliable component when the tolerance demanded by the circuit it goes same + / – 4%.

what means that the entrance tension should be larger than the tension of exit of the integrated circuit. As several manufacturers of those circuit exist integrated we can find those components with several suffixes, like KIA7805, LM7805, ua7805, etc.

Below a table with the numbers of the ic and the tension of minimum entrance:

 Number of the ic Regulated tension Tension entered low 7805 + 5 V 7,3 V 7806 + 6 V 8,35 V 7808 + 8 V 10,5 V 7810 + 10 V 12,5 V 7812 + 12 V 14,6 V 7815 + 15 V 17,7 V 7818 + 18 V 21 V 7824 + 24 V 27,1 V

### Regulator circuit using the zener diode reference

In that case Vout = Vout(IC)+Vz.

I1 â‰¥ Vout(ic)/R1. I1 â‰¥ 5mA.

Then when projecting circuit of regulated supply using that schematic, the exit tension will be the tension of the integrated circuit + the tension of work of the diode zener, like this when we want a supply of certain tension and we didn’t find the ic regulator with the tension that we want, we can use that schematic above, reminding that when calculating the circuit the current i1 should be larger or equal to 5mA and larger I1 or equal the vout(ic)/R1.

#### Circuit using resistor as boost

A circuit same to the previous schematic however now using a resistor as reference, the formula is Vout=Vout(ic)(1+(R2/R1) + R2*IB.

it gives like this for extending the tension of exit of the circuit integrated regulator.

To do those calculations it establishes the value of the resistor R1 to find the other values.
IB is the current quiescent of the integrated circuit = 5mA and vout (ic) it is the tension of work co regulator.

#### Circuit of regulator supply using system of additional protection

Even if the own ic regulator presents protection circuit against short and overload, in that supply schematic is used additional components the diode zener in the entrance it is for protection of entrance on tension, remember to use fuse, because the zener will enter in short if the tension arises.

the tension of the diode zener owes larger than the entrance tension.

in the exit, so that the capacitor is not discharged in the points of drop current of the integrated circuit damaging him/it.

Bibliography: Circuits based on Semiconductor Technical it Dates it goes three terminal positive voltage regulators it dates of Korea Electronic corp. ltd.

Click on a star to rate it!

Average rating / 5. Vote count:

No votes so far! Be the first to rate this post.

As you found this post useful...

Share on the social networks!

We are sorry that this post was not useful for you!

Let us improve this post!

Tell us how we can improve this post?