Circuitboard 3

Components:
-3 x 1kohm carbon film resistors
-1x 470 ohm resistor
-2 x 410 resistors
-3 x 1.7v LEDs (1 green, 1 yellow, 1 red)
-1 x LM324 op-amp
-2 x capacitors
-1 x 9v1 zener diode
-1 x10kohm resistor
-3 x 1N4007 diodes

Circuitboard 2

Components:
-2 x 1N4007 diodes
-1 x 2v2 zener diode
-1 x LM317 regulator
-180ohm carbon film resistor
-265ohm   "        "
-820ohm      "            "
-2 x 25v 330uF capacitors

Calculations:

                     Resistors R3 & R2
                                    Vout=Vref ( 1+R3/R2)
                                       5v/1.25 = (1+R3/R2)
                                                  4= (1+R3/R2)
                                              -1+4=1+R3/R2-1
                                                    3=R3/R2
                                              R2x3=R3/R2xR2
                                                  3R2=R3
                                           3x240=720
                                             R2=240ohms   R3=720
the calculation above is a ratio 3:1 (240:720)
Note# as the 240 and 720 not available i had to grab the closest ones above them (265ohm and a 820ohm and this is still the 3:1 ratio so it will not effect the opperetion of the circuit)


How the circuit works:
As the Raw 12v supply goes through the circuit it is being regulated by the zener diode which anything above the zener voltage is passed on to ground. When it passes through the regulator the out put voltage should be 5v because of the 3:1 ratio of R2 and R3. The capacitors job is to prevent voltage spikes by allowing any high voltage to pass through them to ground. The LED is there to indicate that the circuit is working, it does by lighting up when the 5v out put is coming out of the regulator.



Test procedure:
to test if 5v is coming out of the regulator i did an availible voltage test at the 5v output and i got 5v which means the circuit is regulating the raw 12v supply into a 5v output.





Problems:
One problem i had was at first when i completed the circuit the output voltage was 9 volts not 5 which indicated something wasnt working. Just by doing a visual check i noticed one of the 1N4007 diodes was around the wrong way, i De-soldered and re soldered it back in the right way and got an output voltage of 5v

Reflection:

things i could of done better:

-dont stretch the legs of some of the components so far to make the circuit look better.

Circuitboard 1

Component list:
-2xNPN C547 transistors
-2x1.6v LEDs (1 red and 1 green)

-2x 560ohms carbon film resistors

Calculations

-2x 1kohm carbon film resistors

Calculations NOTE# According to the data sheet .2 milli amps is not enough to fully saturate the transistors so instead i needed to use 5mA, therefore had to change the original resistors to suit this change.(the ones above in the component list)

How the circuit works

Relating to the circuit i drew up on lochmaster shown on the right:
when the 12v supply is applied the LEDs do not operate, but when the 1st %v supply is applied the green LED is turned on. This 5v supply is what switches the transistor on which completes the green LEDs 12v circuit which therefore allows the voltage go through the transistor to switch the green LED on because that circuit is now grounded. When the 2nd 5v supply is switched on the same thing happens to this circuit as the 1st LED circuit (12v supply is still on).
Note# the 560ohm resistors are there only to protect the LEDs.

Test procedure
A simple test to see if the circuit is working is a visual test, are the LEDs working when the 5v is supplied?
but to see if the circuit is working "properly" various test can be made with a multimeter. The 1st test i did was a availible voltage test at the LEDs and got 12v which is normal, another test i made was a voltage drop test across the LEDs, the legs of the transistors and across the resistors, all where normal voltages.

Problems
A problem which i encounted was that i had a short somewhere in the circuit, i dignosed this with a multimeter by doing various voltage drop tests. i got 12v across the emitter and the collector which indicated there was a shot somewhere. so to solve this i checked the tracks on the circuit board to find twotracks touching because of some solder, i cleared it with a cutter.


2nd 5v supply applied



both 5v supplys added

Reflection
to improve this circuit i could have placed the components closer together to make the circuit tidier.another thing was my soldering wasnt the best so i would get some practice before i made it.

BJT Transistors

Testing if a BJT transistor is an NPN type or PNP type

To test if a transistor is NPN or PNP you need a multimeter and put it on the diode test function.

to test for an NPN you place the red lead on the base (B) and the black lead on the collector (C) and you should get O/L when you swap the leads around you should get a reading because in a transistor the collector and emmiter are back to back diodes therefore you should not get a reading in the reverse direction (base to collector or emmiter). In a PNP transistor it is the exact opposite, you should get a reading from base to collector because noe the diodes are going the opposite way to an NPN.                                                                                                                

this table above represents the readings i got when i measured the different legs to see if the transistor was PNP or NPN and also to see if the transistor was functional.

Experiment 1: Transistor as a switch 

components 1 small NPN transistor, 2 resistors

exercise is to build this circuit on a breadboard

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Transistor number	Vbe	Veb	Vbc	Vcb	Vce	Vec
NPN	0.719	Ol	0.722	Ol	Ol	Ol
PNP	Ol	0..722	Ol	0.718	Ol	ol

when i connected the multimeter between the base and emmiter i got a reading of 0.7v which indicates that the diode within the transistor from the base to the emitter is operating and using 0.7v.

when i connected the multimeter between collecter and emmiter i got a reading of 1.4v which means the transistor is on.

there are 3 main regions that a transistor will operate in these are the active region, cut off region and the saturated region. During the active ragion the transistor is in between aturated and cut off which means it is on. when the transistor is in the cut off region it is off because it does not have enough mA to pass through the diodes. when the transistor is in its saturated region it is fully on and fully saturated.

 to work out the power dissipation for a transistor you use this formula Pd=Vce x IB

E.g. 3 x 0.005 =0.015 therfoe the answer is 15mW.

capacitors

Capacitors

• stores electrical charge
• consists of 2 metal plates close together seperated by an insulator
• a power souce flows into the negative plates to charge up the capacitor
• when the power source is removed the capacitor stays charged
• capacitance of the capacitor is measured in Farads (F)

Identifying capacitor size
by using this table i can work out the size of a capacitor by looking on the capacitor for the EIA code

μF	nF	pF	EIA code
0.00001	0.01	10	100
0.0001	0.1	100	101
0.001	1.0(1n0	1000	102
0.01	10	10000	103
0.1	100	100000	104
1.0	1000	100000	105
10	10000	1000000	106
100	100000	10000000	107

Expirement 1: compnents: 1 resistor, 1 capacitor, 1 switch



Circuit number	Capacitance (uF)	Resistance (Kohms)	Calculated time (ms)	Observed time (ms)
1	100	1	50	50
2	100	0.1	5	5
3	100	0.47	23.5	23.5
4	330	1	165	165

First i had to calculate the time the capacitor will take to charge up for each circuit (as above). To calculate the time first you have to put each unit in its base units (F,ohms,S) and then the formula is Capacitance x resistance x 5 e.g. 0.0001Fx1000ohmsx5=0.5seconds (answer to circuit 1) but because the calulated time has to be in ms you just convert it from seconds to milli seconds therefore the answer is 50ms.

observed time on an occiliscope circuit 1

circuit 2

circuit 3

Diodes

^                                ^
    Cathode "C           Anode "A"

Task 1# Identifying the Anode and the Cathode of a Diode
by simply looking at the diode you can instantly tell which is the anode and the cathode, in this case the grey band on the diode indicate that that side is the cathode which leaves the othersode to be the anode.

another method to identify the anode and cathode is to use a multimeter on the ohms funtion. The task was to use a multimeter to identify the anode and cathode of a diode and an LED using a multimeter.
first i measured the voltage drop of the diode in the foward biased direction (anode to cathode) with the red lead on the anode side and the black lead on the cathode and got a measurement of 1.9m ohms, this shows that the side with the red test lead is the anode because voltage can pass through diode from anode to cathode not in opposite.

next i measured the LED the same way as the diode in the 1st test except a different reading of 1.1m ohms which also identifies that the anode is the side with the red test lead, in the reverse biased direction you shouldd get an infinite reading both for the LED and Diode.
also another way to identifying the anode and cathode on an LED is to remember the flat side is the cathode and another way i remember which side is positive and which is negative is to think of the longer leg of the LED is positively longer.

Next task was to take 1 1k resistor, 1 1N4007 diode and have a voltage supply of 5v and put it in this circuit on a breadboard

  R=1kohm   D=1N4007 Vs=5V

Now I had to work out the current:  Calculated= I=V/R                   Measured= 4.2mA

                                                                           =5-0.7/1000

                                                                           =4.3mA

By simply using ohms law i calculated the current by dividing the voltage by the resistance, to get the correct voltage i had to minus the voltage drop over the diode which was 0.7v because according to the data sheet of a 1N4007 the voltage drop across it is 0.7V.

to measur it i used a multimeter on the Amps function and made the multimeter part of the circuit to get the reading of 4.2mA. The calculated and the measured amperage readings are only 0.1 mA out so therefore i can assume that the circuit is working properly.

Next was to replace the diode with an LED and then work out the current

calculated= I=V/R                          Measured= 3.3mA

                   =5-1.7/1000

                   =3.3mA

Even though the diode was replaced with a LED the ohms law is still used to work out the current, the only thing which changed was the voltage drop over the LED which was 1.7v because in this case this is how much the LEd needs to emit light.

The 3rd experiment was to obtain a breadboard, and grab the following components and build them in the circuit below : 2 resistors, 1 5v1 400mW Zener diode

                                                

           R=100ohm  RL=100ohm  Vs=12V
First i measured the volt drop of the Zener diode(Vz) and got a reading of 4.9v and when i varied the supply voltage from 12 to 15v the value of Vz remained 4.9v, what is happening here is this circuit is being a voltage regulator. what is regulating the voltage is the zener diode because of its properties, the zener diode has a zener voltage of 5.1 voltage which means when the voltage gets above 5.1v the zener will open and alow the rest of the voltage to go to ground which therefore keeps the voltage at Vz around 5.1v. Tis circuit can be used as a voltage regulator.

When i reversed the polarity of the zener diode the value of Vz decreased to 0.8v, because it is in reverse it just acts as a a normal diode which only needs 0.6v-0.7v to break through the diode which explains the small voltage of Vz in this polarity.  

the 4th experiment

componets: breadboard, 1 resistor, 1 5v1 400mW Zener diode, 1 diode 1N4007

Vs=10 & 15v, R=1k ohms

Volt drops:  10V          15V

-V1=             4.6          4.7

-V2=             0.63        0.67

-V3=             5.26        5.48

-V4=             3.75        7.48

Calculated

Current (A)    6.2mA     9.2mA

the voltage drop across the zener (V1) is measuring the zener voltage which is the voltage which passes through at a certain voltage in this case 4.6v with a supply of 10v and 4.7v at 15v supply.

the voltage drop across the 1N4007 diode (V2) is measuring how much the diode uses to breakthrough which in this case is 0.63 for 10v and 0.63v for 15v supply.

V3 is the combine voltage drop of the zener and the 1N4007 diode which is the 4.6 going through the zener and the 0.6 through the other diode (for 10v supply).

as the supply voltage changes from 10v to 15v the amperage in the circuit increases because the resistance in the circuit increases.