CIRCUIT TWO ( USING LINEAR REGULATOR)

circuit Liner Regulator

This adjustable liner regulator is used to support the input voltage and the otput voltage.
In this circuit the supply voltage is 12V and various componints are used to for the purpose of voltage drop and LED to physically see if the circuit is working (warning light) and capacitors are used to store energy and release back when switch off.
The purpose of this circuit is to find out the value of voltage out as we did in our calculation which was 5 Volts , this gain value depends on the value of risistors R3 and R2 from the formula.

Calculatioin:
Calculate of the values R3, R2
Vref= 1.25         Vout must = 5V
Use Vout= Vref (1+R3/R2)
Vout = vref (1+R3/R2)
5V= 1.25 (1+R3/R2)
5V/1.25= 1 + R3/R2
4 - = R3/R2
3<R2 = R3
R1 = Vs - VLed/I= 5v - 1.80VLed/ 0.02= 160 ohms
R1= 160 ohms
R2= 240 ohms ( from data sheet)
R3 = 3>R2 ( 3*240= 720 ohms)
R3= 720 ohms.

Designed of circuit on Loc Master 03

I have started to builed the circuit used these components:
One LM317 regulator, three risistors R1180 ohms, R2= 280 ohms R3= 860 0hms. one 1.8V LED, two 25V 47uf capacitors, two 1N401 diodes and one zener diode

Test

First checked the voltage out of cicuit 1, 5.2V out , 2 0V , 3 12Vout.

Checked availble voltage in varios points, the result was positive, available voltage after D13 , 11. 35V which is voltage drope 0.65V, 11V for both zener and C15, avilable voltage before R3 on adj 3. 82V, after R3 availble voltge 0V , across R1 volage drope 3. 01V , and across LED 2V and out is exactky 5.2V that which we got on our calculation.

Problems;
in this crcuit which i buled i had no problem, because my calculation was correct and choesed the correct componint for this circuit and also works perfect which i expected.
Reflection:
Itry to make the circuit simple as possible and used the componint mathematically approved .
next time i hoppe i will make batter circuit than this one .