Experiment #8 Common Emitter

Pick five resistors between 2K2 and 1M for Rb. You want a range of resistors that allow you to see Vce when the trnsistor is the saturated switch region and when it is in the active amplifier region. I used 47K, 220k, 330k and 1M, but this can vary depending on your transistor. Some may need to use 2k2. Put one resisitor in place, and measure and record voltage drop across Vce and Vbe. Also measure and record the current for Ic and Ib. Then change the Rb resistor and do all the measurements and record the new readings. Do this for each of the resistor values above.

Rb: 2K2
Vbe: .778V
Vce: 26.5mV
Ib: 1.94mA
Ic: 6.19mA

Rb: 10KR
Vbe: .738V
Vce: 45.5mV
Ib: 0.42mA
Ic: 6.15mA

Rb: 47KR
Vbe: .709V
Vce: 90.5mV
Ib: 90uA
Ic: 6.06mA

Rb: 220KR
Vbe: .684V
Vce: .525V
Ib: 19.4uA
Ic: 5.29mA

Rb: 270KR
Vbe: .678V
Vce: .921V
Ib: 16.4uA
Ic: 4.49mA

Discuss what happened for Vce during this experiment. What change took place, and what caused the change?
: As resistance is increased the voltage for Vce increases.

Discuss what happened for Vbe during this experiment. What change took place if any, and what caused this change?
: Because the transistor is fully on with each resistor Vbe will be close to or around 0.7V as was shown by the results.

Discuss what happened for Ib during this experiment. What change took place, and what caused the change?
: As resistance increases Ib gradually gets smaller and smaller to the point where it gets into uA's.

Discuss what happened for Ic during this experiment. What change took place, and what caused the change?
: As with Ib, the more resistance that is added to the circuit the less current that will be sent to the collector.

Experiment #7 Transistor Switch

Connect the multimeter between the base and emitter.
Note the voltage reading and explain what this reading is indicating.
: 0.730V, this indicates a saturated transistor that is fully on.

Connect the multimeter between collector and emitter.
Note the voltage reading and explain what this reading is indicating.
: 50mV, this indicates again a saturated transistor as Vce will be very close to 0V when fully on.



A = Saturated (fully on)
B= Cut-Off (fully off)

How does a transistor work in these regions? Explain in detail:
When fully Saturated Vce will be very small but be putting out a high current.
When fully Cut-Off Vce will be small or large but be putting out very small current because the transistor becomes reverse biased.

What is the power dissipated by the transistor at Vce of 3V?
: Pd = Vce x Ic
3V x 12.5mA
37.5V Correction: 37.5W

What is the Beta of this transistor at Vce 2, 3, & 4 volts?
: B = Ib / Ic

2V: 20mA / 0.8mA = 25
3V: 12.5mA / 0.5mA = 25
4V: 5mA / 0.2mA = 25

This is because the beta will always be the same as it is taken off the gradient line you see in the graph.

Experiment #6 Meter Check

Identify the legs of your transistor with your multimeter.
In order to do this I used the diode test function of my multimeter by switching it to the point where a picture of a diode is displayed.

NPN

Vbe: .532V
Veb: OL
Vbc: .541V
Vcb: OL
Vce: OL
Vec: OL

PNP

Vbe: .583V
Veb: OL
Vbc: .587V
Vcb: OL
Vce: OL
Vec: OL

Experiment #5 The Capacitor

T = R x C x 5

Calculated Times:

0.0001 x 1000 x 5 = 0.5s or 500ms

0.0001 x 100 x 5 = 0.05s or 50ms

0.0001 x 470 x 5 = 0.235s or 235ms

0.0003 x 1000 x 5 = 1.5s

27/10/10 Correction for 330uF: 0.000330 x 1000 x 5 = 1.650s or 1650ms

Observed Times: 1: 0.5s
2: 0.2s
3: 0.3s
4: 1.8s

How does changes in the resistor affect the charging time?: As resistance increases the time taken to reach maximum charge increases also.

How does changes in the capacitor affect the charging time?: Like increased resistance, the larger capacitor will take longer to charge and also because of the larger capacity will take longer to discharge

Experiment #4 Zener Circuit 2

Explain what is happening here and why: The Voltage for V1 and V3 is at or around 5V because the Zener diode is acting as a 5V regulator. V2 is reading as it should as the diode will drop 0.7V.
V4 is highest at 15V because its coming in before the diode and so takes the most of the voltage.

27/10/10 Correction for V4: Increasing the voltage will therefore increase the resistance at V4 resulting in a higher voltage being read at the resistor.

Experiment #3 Zener Circuit

For R=100 ohms and RL=100 ohms, Vs=12V

What is the value of Vz?: 4.98V

Vary Vs from 10V to 15V
What is the value of Vz?: 4.74V - 5.06V

Explain what is happening here: At all stages of voltage, Vz stays at around 5V because the Zener is regulating the voltage to a maximum of 5.1V

What could this be used for?: As i stated above this circuit would be used as a voltage regulator.

Reverse the polarity of the Zener Diode
What is the value of Vz? Make a short comment why: .863V, this has happened because the Zener in reverse bias becomes like a regular foward bias diode.

Experiment #1 Resistors

Obtain 6 Resistors of different values. You are then going to determine their value two ways: use the colour code, Include the maximum and minimum tolerance value of each resistor then measure the resistor value with an ohmmeter.
Record your values in a chart

Choose 2 resistors and record their individual ohm resistance value measured with an ohmmeter.
R1: 996K ohm
R2: 468K ohm

Calculated Series: 1464K ohm
Measured Series: 1461K ohm

Calculated Parallel: 318.4K ohm
Measured Parallel: 319.6K ohm

What principles of electricity have you demonstrated with this? Explain.

With these readings and calculations I have shown the principles of resistors in a circuit.
When connected in series, 2 resistors are added together to get a total resistance.
When connected in parallel, total resistance of 2 resistors is given by the formula Rt = R1xR2/R1+R2

Oxygen Sensor Write-Up

Parts:

• 1 x 14 pin LM324 Op-Amp.
• Resistors = 1KR 0.5W (R2,R3,R4), 390R 0.5W (R5), 10KR 0.5W(R6), 270R 0.5W (R7) and 470R 0.5W (R8).
• 2 x Capacitors (C1 + C2, Monolythic Dip C5K 1uF)
• 3 x 1.8V LED (Green, Yellow and Red)
• 3 x 1N4001 0.7V Diodes (D2, D3 and D4)
  
• Zener Diode 9V1 (D1)
• Vero Board 19x25
• Jumper Wires
• Power Wires = 12V Red, Earth Black and 1V Red

What it Does: Works as a tester for an oxygen sensor.

Calculations: Resistor Calculations. R2, R3, R4 = 12V-1.8V(LED) - 0.7V(Diode) = 9.5V/0.0095A(LED) = 1KR
R5 = 12-9.1V(Zener)/0.0095A = 390R (prefered value. answer was 305 ohms)
R6 = 10KR as stated in diagram (we use this to find the current for R7 and R8: 9.1V/10K ohms =
0.00091A)
R7 = 0.4V / 0.00091 = 470R prefered value (440 ohms)
R8 = 0.23V / 0.00091 = 270R prefered value (250 ohms)


How it Works: Connect the positive of a 12V Vs to the 12V Red wire at the top right of the board and the negative wire of the Vs to the Earth Black wire of the board. Also connect a seperate 1V Vs to the 1V Red wire at the bottom right corner of the board and the negative to the Earth Black wire. Pin 4 of the IC is connected to 12V and Pin 11 is Earthed. Pins 2, 5, 10 and 13 are connected inbetween resistors R6 and R7 leading to Earth.

At first the current passes through the Zener Diode, Capacitor and R5 resistor limiting voltage to 9.1V at the positive rail. This then goes down to each LED having passed through 3 more resistors limiting current at the IC. These all lead to each Output pin of the IC. A Sensor Input that can be adjusted to delover 0V -1V is connected to terminals 3, 6, 9 and 12. When initially hooked up the Green LED will light up. When adjusted to 0.3V the Yellow LED lights up. Then when 0.6V is put through, the Red LED lights up.

Testing the Circuit: Using a Voltmeter.
Each LED should read 9.9V when off.
Pins 2 and 5 should read 0.64V
Pins 10 and 13 should read 0.23V
Pin 11 should read near 0V as it is Earth.
Pin 4 should read 11.2V

Problems: As with everyone that everntually got this circuit to work I had not bridged Pins 2 and 5, and Pins 10 and 13 properly and not at the right points inbetween resistors. Once this problem was rectified my circuit still would not work apart from the LED's lighting up but not in correct sequence. This was because a jumper wire I had hooked up was touching another jumper wire. When I made sure both were no longer touching my circuit then worked as it was supposed to.

Reflection: If I were to repeat this task I would be much more careful with my jumper wire placement (less would be good aswell) and pay closer attention to the diagram for my wire placement with the resistors in mind.

Power Circuit Write-Up

Parts:

• D1 1N4001 1A 100V
  
• D2 1N4001 1A 100V
  
• Zener Diode 5V1 400mW
  
• 25V 33uF Capacitors x2
• Voltage Reg. LM317T
• R1 resistor 1KR 0.5W
  
• R2 resistor 330R 0.5W
  
• R3 resistor 100R 0.5W
  
• LED 2.V 30mA
  
• Jumper Wires x3
• Red Wire 12V
  
• White Wire 5V
• Black Wire Earth
• Vero Board
  

Whats its used for: Anything requiring a 5V Vs when only a 12V Vs is available up to 30V.

How it Works: This circuit only needs a voltage regulator and 3 resistors to work which means that everything else is there for protection.

With a 12V Vs connect the positive wire to the Red wire on the Vero Board and the Earth wire of the Vs to the Black wire on the Vero Board. Current will pass through the first 1N4001 Diode ,which has a forward bias voltage of 0.7V giving us 11.3V, a Zener Diode and a 33uF Capacitor , both of which function as protectors for the Voltage Regulator, current then hits the IN terminal of the voltage regulator. There is also an additional 1N4001 Diode in Reverse Bias providing extra protection to the Voltage Regulator.

In order to calculate the resistors needed I remembered that because the outut voltage was going to be 5V I could use any Resistors as long as their ohm values had a ratio of 1:3 for R1 and R2 and used the formula Vout = Vref ( 1 + R3/R2 ) and ended up with 1KR and 330R resistors for my ADJ circuit. Now getting 5V out of the OUT Terminal of the Voltage Regulator, current goes through a 33uF capacitor then my Red LED protected by a 100R Resistor ( R3 ) will light up. At the end of the circuit is 3 wires. My 12V out, my 5V out and my Earth coloured Red, White and Black.

Testing: With a Voltmeter. As long as you are getting 5V out of the 5V output and 12V out of your 12V output your circuit is working as it should.

Problems: At first I realised my 1N4001 Diode in reverse bias didnt have a drill hole underneath it which I promptly included. Secondly I was only getting 3.3V from my OUT terminal of the VReg. After much fault finding and help from Carl it was discovered that I was using the wrong resistors for R1 and R2 in the wrong format. After switching to a 1KR for R1 and using the 330R for R2 i rearranged them in the format you can see in the picture. At this point i was still getting 3.3V. For a reason unknown to me the placement of my 5V out wire was affecting my readings as when I relocated it to right in front of the VReg I got 5V and my LED lit up.

Reflection: If I was to do this again I would double check my resistor calculation so as to not have the problems I had and after having learnt how to fault find properly double check the R1 and R2 resistor layout to make sure it would work properly.

Injector Circuit Write-Up

Parts:

• Vero Board
• Jumper wires x5
• R1+R2 resistor 470R Correction : 0.5W
  
• R13+R16 resistor 100R Correction : 0.5W
  
• BJT Transistor NPN type BC 547 x2
• LED 1 2V 5mm 30mA Yellow
• LED 2 2V 5mm 30mA Red

What it does: ECU operated, switches injectors at high speeds.

How it Works: First of all the circuit must be connected to a Vs of 12V. This sends current through the Resistors and LED's to the Collector terminals each BJT.
In order for the LED's to operate current must also be sent to the Base terminal of either BJT from either 5V Vs'. This switches the BJT fully on (saturated) and thus lights up the LED. This 5V supply can be switched to either BJT so as to turn either LED simultaneously. To complete each circuit each Emitter terminal of the BJT's is Earthed at the 0V rail of the Vero Board.

Test Procedure: To test this circuit you must have a 12V supply and a 5V supply. The negative wires from both power supplies need to be connected to the negative wire on the board. The 12V positive wire is connected to the positive wire on the board. The 5V positive wire is placed on either the channel one or two wires of the circuit depending on whether channel you wish to activate the Yellow or Red LED. The yellow LED should light up when channel one is connected and when channel two is connected the Red will light up.

Calculations: Finding the values of my resistors.

R1 = 12/0.03mA = 400 ohms (470R)

R2 = 5 - 0.7 = 4.3/0.09A = 47 ohms (100R)

Problems: No problems or faults were encountered whilst making this circuit.

Reflection: If I were to do this circuit a second time I would not do it any differently apart from seeing if at all the amount of space on the board could be reduced and not drill my holes as large as I did.

Experiment #2: Diodes

In experiment #2 I was asked to identify the anode and cathode of my LED and 1N4007 Diode using my multimeter.

My readings were LED: Anode - OL Forward Bias
Cathode - OL Reverse Bias

1N4007: Anode - 695mV Forward Bias
Cathode - OL Reverse Bias

In able to identify these without a multimeter, on the LED, I would have looked for the longest stem, it being the Anode, and the shorter being the Cathode. For the 1N4007 the end of the Diode which has the gray stripe would be the Cathode and the opposite being the Anode.


Calculate the first value of current flowing through the diode, now measure and check your answer.

Calculated: 5V/1000R = 5mA
Measured: 4.78mA

Is the reading as you had expected?: Yes, I expected to get 5mA if not just slightly less than 5mA.

Calculate the voltage drop across the diode, now measure and check your answer.

Calculated: 695mV ( taken from previous question )
Measured: 665mV

Using the data sheet, What is the maximum value of the current that can flow through the given diode?: 1A

For R = 1KR. What is the maximum value of Vs so that the diode operates in a safe region?
: 1000V

Replace the diode by an LED & calculate the current, then measure and check your answer.

Calculated: 5V/1KR = 5mA
Measured: 3.41mA

What have you observed?: Because the voltage has remained constant and the current has decreased, the Resistance must have increased when I switch the Diode with the LED.

Notes

Here are some pictures of notes taken during lessons and some from my workbook and two of my home power supply that I made in class.