Cuircuitz

Circuit 1:

Before wiring up the circuit board I made it on "loch master" to get an idea of how it would look like on the cuiruit board...

When i first wired up the circuit the 12V connect, lead was huked up to the positive on the left hand side power supply (pic below). The second Leed below was connected to the 5v positive turminal on the rite hand side powersupply. The connector below was the 5v negetive and the last lead was the negetive (Ground).

Components:
Resistor: 2 X 510Ω, 2 X 884Ω
LEDs: 2 X 5mm

Calculations:
LED ÷ ILED1
R14 = 12 – 1.8 /0.20
R14 = 10.2 /0.20/8
R14 = 510Ω the closest one 560Ω

R15 = V12 – VLED / ILED1
R15 = 12 – 1.8 / 0.20
R15 = 10.2 / 0.20
R15 = 510Ω the closest one was 560Ω

R13 = VPWM0-1 – Vbe / Ibe
R13 = 5 – .7 / 0.005
R13 = 4.3 / 0.005
R13 = 860Ω the closest was 1000Ω

R16= VPWM1-1 – Vbe / Ibe
R16 = 5 – .7 / 0.005
R16 = 4.3 / 0.005
R16 = 860Ohms the closest one was 1000Ω

when first turned on the power switch it the cuircuit was not working (one light was working yoet not the other one).. Therefor i did a voltage droptest on the recistors and the capasitors and found out that the capacitors are the wrong way round. insted of power going from the colector to the emmitor the current was flowing from the emitor to collector.

So i took them apart and changed them the correct way and i wired it up i got proper volt drop reading.

were the 12 volt supply sends the current flow to the cuircuit the 5volt supply gives the current flow to the bace wich opens the colector and the emitor to fow the 12volts threw.

How does this Circuit Board work?
This is a parallel circuit. The voltage comes from V12 supply, goes through the resistor (R14 and R15) and LED to the collector . Then 5V (PWM0-1 and PWM1-1) goes through R13 and R16 to the Base which give enough saturation to open the path flow between Collector and Emitter to let the current flow to earth.

Test procedure:

Vcc is 12V. There is 9.5V voltage drop between R1, R2 and 1.8V in LEDz. The datasheet says if the LEDs have 2.5V it would work safely.

There was 5V at PWM0-1 and PWM1-1. The voltage drop between R3, R4 is 3.8V

Cuircuit board 2..

Recisters.. working:

R1 = VOUT – VLED ÷ ILED1
R1 = 5 – 1.8 / 0.02
R1 = 3.2 / 0.02
R1 = 160Ohms .. Closest was 180Ohms

R2, R3 = VOUT / Vref (1 + R3 ÷ R2)
R2, R3 = 5 ÷ 1.25 – 1

R3 = 750 - the closest was 820Ω resistor
R2 250 -the closest was 280Ω resistor

COMPONENTS:
2 -Rectifier Diodes- 1N4007
1 - Zener Diode -1N4748-
2- 33uf capacitors
3- resistors 180ohms, 270Ohms, and 820Ohms
1- LED

1 -Voltage Regulator LM317T

THis cuircuit is also a parrelel circuit to some extent, the 12 volt supply ges threw the first diode to the next diode and threw the rest of the trackones it reaches the first diode the voltage drop stayes the same>

As it reaches the capacitor ,it charges and stores the energy and released as needed. From the jumpercables after the diode negetive . it jumps the voltage to the zena diode. the zena diode holds 5 volts back and lets the rest threw the track.

Voltage will flow from the OUT terminal of the voltage regulator. It will go through the 280 OHm resistor which is connected from the OUT terminal to ADJ terminal then to the ground going through the 820Ohme resistor. Voltage moves to the capacitor. The capacitor will charge and store Voltage. It will move on to the 180 Ohme resistor to the LED then to the ground. Voltag flow 5V wen tested...

VOLTAGE DROP READINGS.
Vcc is 12V.

0.7V voltage drop between the diodes.

Capacitor voltage drop is 11.6V.

Voltage regulator is 12.1V drop

R1 is 5V

LED voltage drop is 1.8V.

R3 Voltage drop is 3.4V

R2 is 1.24V.

Voltage regulator is 5.6V.

Cuircuitboard 3: