Voltage Regulator

Component List


Components I have used in this voltage regulator.

• D12 = lN4001 Diode

• D13 = lN4001 Diode

• D15 = Zener Diode

• C15 = 25v 33uF capacitor

• C16 = 25v 33uF capacitor

• R1 = 150ohm resistor

• R2 = 270ohm resistor

• R3 = 820ohm resistor

• LM317 regulator

• High Density L.E.D



Calculations used to work out the resistors in the voltage regulator circuit


Vref = 1.25

Vout must = 5v

5/1.25 = 4

4 = 1+ R3/R2

4 – 1 = R3/R2

3 = R3/R2



R2 = 750ohm (closest = 820)

R3 = 250ohm (closest = 270)



R1 = (5v – vled) / I

R1 = (5 – 1.8) / 30

R1 = 3.2 / .03

R1 = 106ohm (closest = 100ohm)

Problems Encountered.

I found myself trying to sort out a couple of problems in this circuit both of them to be small silly mistakes, the result of these problems is a low voltage output and my L.E.D not lighting up. The reason for my led not lighting up is that I had two of my resistors the wrong way round so there was too much resistance for my L.E.D to light up.

The other problem with the output voltage at 1.8v had me stumped for quite some time, then was told that i had my 5v output wire in series with the resistor and led and not in series.


Test procedure

To test this circuit all I had to was connect it to a 12v power supply and make sure the led is on. Then I had to measure the output voltage which had to be 5v and take voltage drop readings across the components.

Output = 5.09v

D12 = 6.26v
D13 = 0.73v
This show the the diode is in foward bias and this is the required voltage for it to work
D15 = 11.36v
C15 = 11.36v
C16 = 5.09v
R1 = 3.17v
R2 = 1.24v
R3 = 3.84v
LED1 = 1.91v
The voltage drop across the led is the required voltage for the led to work.

Reflection.

Threre is really nothing i would do differently when making this circuit, but next time I will make sure I have all my components in the right place and the wires in the right place.

Circuit design



 
 
 
 
 
 
 
 
 
I made only a few small changes to my circuit my moving some of the componets across the regulator so it would work.