Power Circuit

Components list

R3= 330R (0.6W)

R2=100R (0.6W)

R1=100R (0.6W)

C15 and C16= 33 micro farad. (25V)

D13and D12= Diodes (1n4001) (1A, 50V)

D15= Zenor diodes

Regulator LM317.

Red l.e.d (30mA, 1.8V)

To calculate R1 i used the formula R=(Vs-Vled) / I

= 5-1.8/0.030= 106.6 Ohms. I used the preferred value of 100 ohm.

To calculate R2 and R3 i rearranged the formula Vout=Vref (1+r3/r2)

Vref= 1.25V, Vout=5V

To Vout/Vref -1 = R3/R2.

So 5/1.25=4.

4-1=3.

3= R3/R2

3 is the ratio. R3 must be 3x larger than R2. This is why i used 330ohm and 100 ohm resistors.

The circuit is a voltage regulator with 12V entering the circuit and 5V leaving. R3 and R2 adjust the regulator voltage because they connect to the adjust leg of the regulator. The l.e.d will emit light if there is voltage leaving the regulator. There should be 5v after the output leg of the regulator. The l.e.d shows if the output circuit is working. The capacitors are filters. D12 is to protect the regulator. It protects the regulator by stopping the 12v connecting to the output. The zenor diode is in reversed bios. This will drop voltage to X volts of the zenor before the input leg of the regulator.. All the circuit needs to work is the regulator and the resistors the rest of the components are optional. This circuit could be used at any time when you have a 12v input and require 5v for a component or circuit.

Test procedure.

Check if l.e.d is emitting light. There should be 12 at the input of the regulator and 5v at the output. There should be also 5v at the wire before the l.e.d There should be 12v at the 12v output wire. Test the output wires they should read around 5V. My circuit ended up having 5.2V across the output which is acceptable for this power circuit.

Problem solving.

First problem i had was it was shorting. I think this was because i didn't have my holes drilled correctly. I ended up taking the circuit apart and re drawing it on lochmaster. It then worked.I had 5.2V on the output. If i was to do the circuit again in the future i would double check everything before i soldered it to the board. Also i would be more careful when cutting the board I ended up with a little crack in the board. I did end up moving a resistor from the above drawing.

Injector Circuit

Components used:

• R14 and R15= 180R (0.6W)
• R13 and R16= 390R (0.6W)
• 1x White l.e.d (30mA 1.8V)
• 1x Yellow l.e.d (30mA 1.8V)
• 2x Transistors. (C547 w64 NPN)
  

First i calculated the values for R15 and R14 the resistors that run off the L.E.Ds.

12-1.8/0.030= 340 ohms.

I used 2x 390 Ohms as it was the closest preferred value.

For the load resistors R13 and R16 i calculated them by.

5/0.030= 166.6 Ohms.

I used 2x180 Ohm resistors as it was the closest preferred value.

I calculated the resistors incorrectly and also used the incorrect ones. The circuit still worked. I needed to use the the current across the base to calculate the resistor. I used the IC max of 100mA to work out IB

Ib= IC/B. I used 100/110= 0.90mA.

4.3/0.009 = 477.7 Ohm

I could of used 2x470R

The circuit is actually two different circuits on a single circuit board. Each circuit can be used individually by having the 12V on the positive and negative rails and 5v to the required circuit. If both circuits have a 5V input both l.e.d.s will light. R15 and R14 resistors are there to protect the l.e.d and R13 and R16 are the load of the transistor. This limits the current to the base. The transistors is acting as the switch in this circuit. No l.e.d.s will light if there is no 5v input because the collector and emitter are shorted because there is no base connection. so VCE would be 10.2V with no 5V input. (12-1.8v LED). The transistor is said to be in its cut off zone. when the 5v input is connected the transistor is considored saturated The circuits can be easily switched between each other. It is called an injector circuit because it works similar to EFI circuit because the circuits can changed between each other quickly with the help of an ECU or a controlling circuit

Test procedure

Check the l.e.d.s are active when the 5v input has been connected. Check VBE is 0.7V this is when the transistor is saturated. There should be 1.8 Volts across each l.e.d. Check all drillings have been made under the resistors.

I didn't have any problems with this circuit it worked first attempt.If i was to remake the circuit in the future i would probably use jumper wires instead of stretching component legs and have all the components flat on the circuit board.