Power regulator circuit

Components needed:
2 x Rectifier diodes, 1N4007
1x L.E.D 5mm any colour
1x Zener diode
2x 33uF capacitors
1x Voltage regulator, LM317T
1x 180ohm resistor
1x 270ohm resistor
1x 820ohm resistor

Calculations:

R1= Vs - VD@L.E.D / ione
(5 - 1.8 / 0.02)
=  160ohm


Vref = 1.25v
Vout = 5v
Vout is = to Vref(1+R3/R2)
5v = 1.25(1+R3/R2
5v/1.25v = 1+R3/R2
4 = 1+R3/R2
3 = R3/R2

R3 is 3 x larger than R2 so I chose to use a 270ohm resistor for R2 because 820/3 = 273 and the closest resistor to value was a 270ohm.

How it works?
This circuit is parallel. It has a 12v in for the Vs, it has a jumper wire connecting both the 12v in and the 12v raw together. From the point this meets, the current will flow through the rectifier diode to the zener diode to ensure the current is flowing one way only. After that it goes to the 1st capacitor which stores the charge. From here, the charge goes through the IN port on the regulator, Also if you look at D3 it is a bypass for the regulator for any extra current that may occur from voltage spikes, this ensures that there is no damage done to the regulator. From here the current flows through the OUT port regulated down to a lower voltage, from there it flows through the 270ohm resistor which links to the ADJ port
then it flows to the 2nd capacitor which stores the charge, then flows to the 180ohm resistor where it meets the 5v signal in supply which then flows to the 5mm L.E.D, then all the left over voltage is used up by the 820ohm resistor.

Testing to see if circuit is correct:
Vs is 12v in, there is a 0.68 VD across D1(rectifier diode), there is a 11.4 VD in the capacitor, but this VD is not constant because the charge is release once the capacitor fills. There is 11.2v flowing through the zener diode to earth. 11.2v at the IN port of the regulator. Out port of the regulator is 5.1, VD at capacitor 2 is 5.1 but again is not constant, R1 = 4.9v  R2 = 2.1 R3 = 2.9v The VD at the L.E.D is 2.3. These tests prove that the circuit is functioning proper for the purpose it was designed for.

Problems found?
One problem that I came across whilst constructing the circuit was that some of my jumper wires in the design were not placed in the correct areas, so my finished product had a couple of slight improvements compared to my template designed on lochmaster.

Reflection:
If I had the chance to rebuild this circuit, I would look at trying to again improve on my soldering, making it a lot more neat to stop potential shortages in the circuit, and having the regulator set up in a different place because the way it was designed made it quite hard to get it soldered on properly, aswell as having it in the place that it is it meant the circuit had to be spread out using unnecessary space.

Sorry, another bad quality picture.

Injector circuit write up

Components needed for this board:

2x 560ohm resistors
2x 1kohm resistors
2x PNP transistors
2x 5mm L.E.D

Calculations to figure out what resistors were needed:
We needed to figure out the resistance for R1, R2, R3 and R4 to get the correct resistors

The data needed to calculate this was found on a datasheet I found online:
http://www.datasheetcatalog.com/datasheets_pdf/B/C/5/4/BC547.shtml

R1 = (V-L.E.D/I) 12-1.8/0.20 = 510ohm/560ohm resistor
R3 = same as R1's calculation.
R2 = (signal-V.D/Ibe) 5-.7/0.005 = 860ohm/1kohm resistor
R4 = same as R2's calculation

How it works:
This circuit has a 12v in, which runs through each 560ohm resistor which are both set up in a parallel circuit. From the resistors the current flows through the L.E.D's into the collectors. A 5v signal current is then sent through the 1kohm resistors which are also in a parallel circuit, this 5v signal current then flows to the base allowing the transistors to saturate meaning that the current can flow from the collector to the emitter, so therefore the L.E.D's are supplied with an earth as well as a Vin current which allows the L.E.D's to emit light.

Testing to see if circuit is correct:
First I started by testing 12v in voltage to see if i was getting enough current, my Vs reading was 12.2v so i adjusted this to 12v at power supply because that was the specified voltage. After Vs was tested, i tested the volt drop across both the 560ohm resistors and Vd reading was 9.7v, then from there i tested the volt drop across the L.E.D's and they were using the left over voltage which was 2.3v, which is correct because according to the data sheet for these particular L.E.D's the max power dissipation without damage was 2.5v. I then tested both the signal in's for the base port, the volt drop between them was slightly to much at 4.8 volt so I decided to change the resistor to a lower value resistor to allow the the base to saturate. After this was changed I tested the volt drop again and the result was alot better being 4.4v. After this change it allowed the base to saturate resulting in the L.E.D's lighting up.

Problems found:
There was one problem found, the resistors I used for the base port were of to higher value so they weren't allowing the base to saturate, I then changed those for a lower value resistor resulting in the base to then start saturating.

Reflection:
If I was to re built this circuit, I would make my soldering alot neater than it is, I would also try to make it more compact to save board space. But other than that I am quite happy with the result I have gotten out of this circuit.

Sorry about the bad quality photo, but it is of the circuit board working

TTEC 4284 exp 8

Results:

27k  rb: 4.10   vbe: .72   vce: .10   ib: 4.08   ic: 4.70

67k  rb: 4.11   vbe: .71   vce: .15   ib: 4.10   ic: 4.67

180k  rb: 4.14   vbe: .69   vce: .18   ib: 4.15   ic: 4.54

560k  rb: 4.17   vbe: .65   vce: .24   ib: 4.18   ic: 4.22

1m  rb: 4.24   vbe: 0.64   vce: .35 ib: 4.24   ic: 1.00

Discuss what happend to VCE during this experiment. What change took place, and what caused the change?
Whilst testing VCE, when the resistors were replaced with higher resistors the value of VCE raised due to the transistor acting as an amplifier, the higher the resistance is the more the transistor tries to make up for the volt drop.

Discuss what happend to VBE during this experiment. What change took place, and what caused the change?
When we were testing VBE the value decreased as higher resistors were added to the circuit. The reason for this is that the collector circuit is using more of the current supplied.

Discuss what happend to IB during this experiment. What change took place, and what caused the change?
When we tested Ib after attaching higher value resistors, the result we got was lower each time due to the increase of resistance at the collector, the base didnt get as much current.

Discuss what happend to IC during this experiment. What change took place, and what caused the change?
The value of IC kept decreasing due to there being less current provided to the base port because of the higher resistance at the collector port.

B = Ic/Ib

B = 4.24/0.1 = 42.4
B = 42.4


Explain what the load line graph is telling you. Discuss the regions of the graph where the transistor is Saturated, Cut-off, or in the Active area:
The load line graph indicates that most of the readings given are either just about in the cut-off region if not in the cut-off region, but one of our readings, the one that ranges up to 1.1mA on the Ib scale, shows that it is in the saturation area of this transistor.

TTEC 4284 exp 7

Transistor as a switch:

Components: 1 x Small Signal NPN transistor, 2 resistors

Exercise: Connect the circuit as shown in Fig 12 and switch on the power supply.

Connect the multimeter between base and the emitter. Note the voltage reading and explain what this reading is indicating:
The reading that was shown on the multimeter was 0.71v , what this reading is indicating is that the voltage given to us by the multimeter is the voltage needed to let the current flow across the base and emitter ports.

Connect the multimeter between collector and emitter. Note the voltage reading and explain what this reading is indicating:
The reading that we got from the multimeter was 4.87v. This indicates that the transistor is still in Active Region but near cut off point, so therefore there is a slight amount of amperage flow in the circuit but not alot due to it being so close to it's cut off point.

In the plot given, what are the regions indicated by the arrows A & B?

Plot A: Saturation Area

Plot B: This is the cut off point














How doe's a transistor work in these regions? Explain:
The term "Saturation" means that the transistor is at full working capacity and is giving the most amount of amperage to the circuit as it possibly can, this also means that the V at the base is higher than the collector and emitter.

When the term "Cut off point" is used, this means the circuit will not be getting its full potential of amperage, this also means the collector is higher than the base and emitter, which means that the circuit will not be recieving amperage.

What is the power dissipated by the transistor at Vce of 3 volts?
Pd = IxV
Ic = 13
V = 3
Pd = 0.013x3 = 0.039 power of 3.
Pd = 3.9v

What is the Beta of this transistor at Vce 2,3 & 4 volts?
Beta = Ic/Ib
2v = 19/0.7 = B(27.142)
3v = 13/0.5 = B(26)
4v = 5/0.2 = B(25) 

TTEC 4842 exp 6

Identify the legs of your transistor with a multimeter.

Diode test (V) meter readings	Table Cell	Table Cell	Table Cell	Table Cell	Table Cell	Table Cell
Transistor number	Vbe	Veb	Vbc	Vcb	Vce	Vec
NPN	O.L	O.L	0.834	O.L	O.L	O.L
PNP	O.L	0.841	O.L	0.84	O.L	O.L

HTML Tables


With this experiment what we were trying to do was to figure out which way current will flow on an NPN transistor, and a PNP transistor. Our results show that an NPN transistor, current will flow from emitter to collector once a base current is supplied. The results for the PNP transistor show that current will flow from the collector to the emitter once a base current is supplied.

TTEC 4842 exp 5

Components: 1 x resistor, 1 x capacitor, 1 x O/N switch
Exercise: First, calculate how much time it would take to charge up the capacitor. Then, connect the circuit as shown above. Measure the time taken by the capacitor to reach the applied voltage on an oscilloscope. Fill in the chart below. Also draw the observed waveforms in the graphs below, filling the details on each one.





            

Circuit #	Capacitance (uF)	Resistance (Kohm)	Calculated time (ms)	Observed time (ms)
1	100	1	600ms	500ms
2	100	0.1	60ms	50ms
3	100	0.47	280ms	250ms
4	330	1	2.4s	1.75s

HTML T

circuit 1

circuit 2

circuit 3

                                           circuit 4

How does changes in the resistor affect the charging time?
When a resistor is changed, the amount of voltage supplied to the capacitor changes so therefore that effects the rate of charge in the capacitor.


How does changes in the capacitor affect the charging time?
This effects the charging time because a change in capacitor size means a change in storage size, the more storage space the longer it takes to fill the capacitor, and the same rule applies to a smaller capacitor except there is less storage so therefore it takes less time to fill. The capacitor also has a curve in the oscilloscope graph because a capacitor releases voltage gradually, the release time (graph curve) fully relys on the specific specifications of the capacitor being used, so the release time curve will vary for different capacitors

TTEC 4842 exp 4

                 10volts                             15volts
Vd V1:      4.63                                4.80
Vd V2       0.66                                0.68
Vd V3       5.29                                5.48
Vd V4       4.72                                9.50
Calculated current 10v: 10mA     Calculated current 15v: 14.98mA

Describe what is happening and why you are getting these readings:
When the voltage increases all the components use the required amount of voltage to power them which leaves the remainder of the voltage to be consumed by the resistor.

TTEC 4842 exp 3

For R= 100ohm and RL= 100ohm. Vs= 12v

What is the value of Vz?
Vd= 4.78
Vary Vs from 10v to 15v:
What is the value of Vz: 10v= 4.78v  15v= 5.13v

Explain what is happening here?
The voltage flowing through the diode increases when the supply voltage is increased.

What could this circuit be used for?
To regulate voltage in a circuit

Reverse the polarity of the zener diode.

What is the value of Vz? Make a short comment why you had that reading:
The reading we got at Vz in reverse polarity was 0.8v, it was 0.8v because when the diode is in reverse polarity it has more resistance than the resistor in this circuit, so therefore all the voltage will flow through the resistor because electricity will always flow through the path of least resistance.

TTEC 4842 exp 2 (2

Components:
1 x resistor, 1 x diode, 1 x L.E.D
Exercise:
For Vs= 5v, R= 1Kohm, D= 1N007 build the following circuit on a breadboard:


Calculate first the value of the current flowing through the diode, now Measure and check your answer?show your working:
Calculated:
                  4.3/1000  = 0.0043/4.3mA

Measured: 4.60mA

Is the reading as you expected; explain why or why not?
The reading is pretty much exact as to what I expected it to be, it was slightly different due to the tolerance/resistance involved in the individual components involved in this experiment.
Calculate the voltage drop across the diode, now Measure and check your answer?

Calculated: 0.7v

Measured: 0.4v

Using the data sheet given in Table 1,

What is the maximum value of the current that can flow through the given diode?
1A @ 75 degrees c
For R = 1Kohm. What is the maximum value of Vs so that the diode operates in a safe region?
1000v

Replace the diode with an L.E.D & calculate the current, then measure and check your answer?

Calculated: 3.1mA

Measured: 3.17mA

What do you observe? Explain briefly: V = Voltage
The resistor has a voltage drop because it is a consumer of V and is in series so the over all value of V is reduced after being partly consumed by the resistor. The L.E.D has a voltage drop of 1.83v because every L.E.D needs at least that amount of V to let current flow through the circuit and the drop is also due to resistance in both components.

TTEC 4842 exp 2 (1

Components:
1 x Diode, 1 x L.E.D (light emitting diode)


Exercise:
Using a multimeter, identify the anode and cathode of the diode

Explain how you could identify the cathode without a multimeter:
The easiest way to figure out which side of an L.E.D is the cathode is that it will always be the short piece of wire out of the two wires built into the L.E.D. For a rectifying diode the cathode will always be the side which has the strip on it.