TTEC 4825-Vehicle Safety, Transmission and CAN Systems

Injectors Circuit Board Write-Up

Regan’s Injector Circuit Board Write-Up

Component List:
• Positive/Negative Inputs
• Channel One and Two Inputs
• Jumper Wires
• 1 Yellow LED
• 1 Clear LED
• 4x 1KΩ Resistors
• 2x NPN BC547 Small Signal Transistors

Calculations:
 • 2x 1KΩ LED Resistors (R1, R2)
(Vs-V Led)/I
(12-1.8)/0.01=1020Ω
Preferred Values of 1KΩ for R1, R2
• 2x 1KΩ Transistor Resistors (R3, R4)

I decided to use the same resistors as the LED resistors to keep the circuit simple and to ensure that my transistor will not be destroyed over time.


How It Works: First the circuit must be connected; so12V is run into the positive input and a second supply of 5V is needed to be attached to the channel one 5V input. The earths of both supplies are placed upon the negative input. With these connections in place the yellow (channel one) LED will emit light as an indicator that channel one is operational. Channel one works as follows; along the positive rail my 1KΩ LED resistor is placed in series before my yellow (channel one) LED. The Cathode leg has a jumper going from it in order to connect the Cathode leg to the collector of the NPN transistor. At the 5V channel one input rail, there is my 1KΩ resistor running into the base leg (switching circuit) of the transistor. The emitter is connected by use of a jumper to the negative rail. This is the discrete circuit channel one. Channel two works the same, except for the minor difference of my second 1KΩ resistor leading into my second NPN transistor on the channel two 5V input rail. Also at the positive rail my channel two 1KΩ LED resistor and channel two clear LED have to be separated from the channel one components by use of a drill hole after the yellow (channel one) LED and a jumper placed from before my 1KΩ channel one resistor to my 1KΩ channel two LED resistor. This all allows my circuit to run two channels individually.


Test Procedure: To test this circuit you must have a 12V supply and a 5V supply. The Negatives from both power supplies are to be connected to the negative input. The 12V positive is connected to the positive input. The 5V positive is placed on either the channel one or two input of the circuit based on which channel you wish to activate. The yellow LED should light up when channel one is connected and the clear LED for channel two.


Problems: No problems were encountered when making this circuit. I placed in all the components and tested the circuit to find that it worked as it should.

Reflections: Seeing as my circuit worked on my first attempt I would not change how I made or arranged my circuit as there is no need when the circuit works at first attempt.

12-5V Voltage Regulator Circuit Board Write-Up

Regan’s 12-5V Voltage Regulator Circuit Board Write-Up

Components:
 • Input/output Wires
• Jumper Wires
• 2x 1N Diodes
• 1x Zener Diode
• 2x 33uF 25V Capacitors
• 1x 317T 12- 5V Voltage Regulator
• 1x Red LED
• 1x 150Ω Led Resistor (1W Carbon) (R3)
• 1x 330Ω Resistor (1W Carbon) (R2)
• 1x 1000Ω Resistor (1W Carbon) (R1)

Calculations:
150Ω Led Resistor (R3)
(Vs-V led)/I
(12-1.8)/0.07=145Ω
Preferred Value of 150Ω for R3

For my R2 and R1 resistors used, I chose them based on the rule that the R1 is to be three times larger than R2. According to the LM317 resistor values, with 330Ω and 1000Ω I should obtain 5.04V. This is very close to my output which is 5.14V.


How It Works: My circuit is designed to have an input of 12V which is then regulated down to 5V by use of a 317T Voltage Regulator, with a 12V bypass wire. The complete run through as to how 12V goes in and 5V comes out is as follows. A 12V input voltage is run through the positive rail and the earth wire is run through the negative rail. The 12V first runs into a jumper wire leading up to the 12V bypass rail, ending at an output wire. The 12V also continues past the first jumper and through a 1N diode and onto a Zener diode which is also bridged to the negative rail. Alongside the Zener diode is a 33uF capacitor, also bridged down to the negative rail. Now the 12V will lead into the input leg of the Voltage Regulator therefore jumping my positive rail down to the output leg of the Voltage Regulator. This point in the circuit will now show 5V as the voltage has now been stepped down by use of the regulator. The Regulator still has a third leg, the adjust, which is bridged to a resistor (R1) which is bridged to earth. The resistor (R2) is bridged to R1 except it is jumped at the other end back up to positive rail. These two resistors provide current reduction for the Voltage Regulator. R1 and R2 can be any sized resistors as long as R1 is three times larger than R2. After the resistor (R2), back on the positive rail, now flows onto another 33uF capacitor (used as a filter), then onto a jumper which connects a diode, in reverse bias, from the positive rail after the regulator back around to the positive rail before the regulator. This is used to stop the 12V getting from the input to the output rail.
Back at the positive rail after the regulator and capacitor, a resistor (R3) is placed before a red Led, and then bridged to earth. This Led is used as an indicator to us that the circuit has power to it at that point. After R3 is the output wire which is measured at 5.14V, very close to the 5V rating.

Test Procedure: This circuit has a simple test procedure to see if it is working. A 12V supply with a positive and negative lead is attached to the positive and negative input wires. Then a multimeter is used at the output wires, with the black lead on the negative output, and the red lead on the 12V bypass output. The reading on my multimeter was as expected, 11.97V. Whilst leaving the black lead on the negative output change the red lead to the 5V output wire. My reading at that point was 5.14V which is very close to the 5V mark and acceptable as a finished 12-5V Voltage Regulator.

Problems: The only problem that I encountered was after I had completed my circuit; I tested the voltage with my voltmeter at the 5V output and obtained a reading of 11.97V. I removed my 317T Regulator only to find that it was in working condition. After visually inspecting my board further, I noticed that my diode going across the voltage regulator was placed along the same rail without having a drill hole placed between it. Meaning that it was allowing the 12V from the input to link up with the output rail, resulting in my 12V reading I obtained at my 5V output. To fix this problem I drilled a hole under the diode and retested my 5V output to get a reading of 5.14V

Reflection: If I had the opportunity to do this circuit again, I would place the diodes across the rails so that I do not have to drill holes under them, also to use the diodes as jumpers at the same time. Simply to limit the jumper wires as much as possible and keep the board clean.

TTEC 4824-Automotive Electronics

Experiment No. 1

Identifying, Testing and Troubleshooting Semiconductor Components.

6 resistors were given to me with the aim to determine there values two ways:

• Using the colour code chart (as shown)
• Using an ohmmeter across the resistors
• Also the max and min tolerances were asked for

Resistor 1:

• Measured value = 986Ω
• Calculated value = Brown (1), Black (0), Red (2), Gold (5%) =1KΩ
• Tolerance Level = (950-1050)Ω

Resistor 2:

• Measured Value = 46700Ω
• Calculated Value = Yellow (4), Violet (7), Orange (3), Gold (5%) = 47000Ω
• Tolerance Level = (44650-49350)Ω

Resistor 3:

• Measured Value = 99.7Ω
• Calculated Value = Brown (1), Black (0), Brown (1), Gold (5%) = 100Ω
• Tolerance Level = (95-105)Ω

Resistor 4:

• Measured Value = 99.3Ω
• Calculated Value = Brown (1), Black (0), Brown (1), Gold (5%) = 100Ω
• Tolerance Level = (95-105)Ω

Resistor 5:

• Measured Value = 2170Ω
• Calculated Value = Red (2), Red (2), Red (2), Gold (5%) = 2200Ω
• Tolerance Level = (2090-2310)Ω

Resistor 6:

• Measured Value = 465Ω
• Calculated Value = Yellow (4), Violet (7), Brown (1), Gold (5%) = 470Ω
• Tolerance Level = (446.5-493.5)Ω

Choose Two Resistors and measure there values with an ohmmeter:

• Resistor 1 = 2170Ω
• Resistor 2 = 465Ω

I placed the resistors in series with each other then calculated and measured them for readings:

• Calculated resistance of resistors 1 and 2 in series = 2670Ω (Add them together)
  
• Measured resistance of resistors 1 and 2 in series = 2640Ω

I also placed the resistors in parallel with each other then calculated and measured them for readings:

• Calculated resistance of resistors 1 and 2 in parallel = 940Ω = (2200x470/2200+470)
  
• Measured resistance of resistors 1 and 2 in parallel = 383Ω

The Principles of electricity demonstrated here are that in series Rt (total resistance) is the sum of each resistive unit. (add them all together).
When in parallel Rt (total resistance) is less than any one branches resistance.
Rt = R1xR2/R1+R2 (Formula is for two resistors only).

or for three resistors:

Rt=R1xR2/R1+R2=R3

Rt=R2xR3/R2+R3