transistors

First thing i did for transistors did a diode test on both of the diferent transistors the pnp and the npn the pnp is positive negative positive and npn is negative ositive negative to find out wich one is wich put positive on midle one of resistor and negative on one of the other and if geting reading its a npn and if not its a pnp. i took two transistors a npn and a pnp and identifyd all the legs and mesured them in difernt ways for example I measured vbe wich is the voltage thtough the trasistor from the base to the emitter and got a result of .720V for npn and for pnp i got a result of ol open circuit.The reason for the open circuit is because it was measured positive on base nd negative on emitter wich is the wrong way for a pnp transistor.I did this repetely until it was measured in all ways posible everytime that there was a result it was 0.720V.I created a circuit on my breadboard made out of a npn transistor and two resistors with a 15v supply I measured the voltage between the base and the emitter of the transistor and got a result of .720v because its in paralel and not losing its volts.Then i connected the multimeter between the collector and the emmiter and got a result of 0.56 because its losing more voltage through the collector than it does through the base because ther is a resistor on the rcollector side in series so its takin some of the voltage

transistor chart:
In this chart the arow marked A shows to the saturated are and B is the cut off area the saturated are has a high amprege flow and a low voltage drop wich gives a good flow were the cutoff area has high voltage drop and a low amprege flow which have high resistance and causes heat wich makes it bad flow.using this graph the power disipayion could be worked out by multiplying vce by Ic for example at 3vce it would be 3x13 that gives a result of 39watt. The beta could also be worked out by using the formula B=Ic/Ib for example at vce 2 it would be 15/0.6=25 and at vce 3 it would be 10/0.4=25 and for vce 4 it is 5/0.2=25 wich shows us that beta is 25 and stays the same at al volts acept at 5vce it would be 0.
I made a circuit using 470Ohm resistor a 5v supply a led and a npn transistror then i had to pick five resistors between 2k2 and 1M the reason for this is to get a range of resistor that wil alow you to see vce when the transistor is in the saturated region and when it is in the active region.The circuit stays the sae trhoughout the experiment acept for the resistor on the base side of the resistor that changes with the five resistors i had to pick.the five resistors i picked was 47k,220k,270k,330k and 1M. I changed the resistors in order from smalest to bigest so I used the 47k first.The first thing i had to test was vbe the voltag between the base and the emitter of the transistor and I got a result of .720Vthesecond test i made was vce the voltage between the colector and emitter of the transistor and got a result of  .11V the next test was Ib the curent at the base the way to do this is to pull the one side of the resistor out and put the black terminal of the multymeter on that side and the red on the base side of the transistor and i got a result of  18.96mA the last test was Ic this is measured the same way as Ib but wit the other resistor and the collector of the transistor and got a result of 8.96 mA.
I did the same for the other circuits after i changed the resistors and got results of:
resistors      vbeV       vceV      ibmA         icmA
220k          .680       .540        18.20          5.3
270k          .670       1.07        15.8   .        4.26
330k          .671       1.47        12.8            3.51
1M            .635       2.52         4.3              1.20
in the experiment the voltage of vce got more as i put biger resistors in the circuit because thhe biger resistors is after vce.In the circuit vbe did not have alot of change stayd in the same range because the emiter goes to earth so it has a low voltage.In Ib the curent got less beacuse the bigger the resistor the less current because the more resistance the slower it flows.In Ic curent got lower for the same reason as in Ib.

The picture above is a of a graph i drew it is the same kind of graph as the previos one the left side shows the values of vc the right side shows the value of Ib and the bottom shows vce the midle part indicates the active region top indicates the saturated area and the botom indicates the cut off region.The transistor works when it gets up to the active region where it switches and works in the saturated are the transistor does not work in the cut of region because the value is to low for the transistor to be able to switch over.

capacitor

this is my 1st one

I built 4 diferent circuits on a breadbord with difernt size resistors then used a ciliscope to worke out the charging time and get a graph of the charging time.I took some pictures of the graphs i got on the ciliscope.

the first circuit had a 1uf capisitor with a 1kohm resistor.I calculated the time in ms and got a result of 500ms when observed time was the same also 500ms.i used a wire that i conected to the negative side of the diode and then touch the positive with the wire to switch it on and of and then you can see on the ciliscope what happens when you charge and discharge the circuit and at the botom of the screan it shows on the left the voltage and at the right it shows the time in ms.you can also change the volts and the seconds.if it is set to fast it is hard to read and pause.it is important to have the seting set right to get an acurate result.

The picture on the right is a photo of the result on the ciliscope I got for the second circuit I made on the breadboard.The circuit was made of a 100uf and a 0.1KOhm resistor and a 12vsupply.The time i calculated was 50ms.I calculated the time by taking capacitance x resistance x5.The observed time was 70ms the reason for that could be because of the setings being a bit wrong to fast and made the result inacurate and hard to read acuratly.

This is a photo of the third circuit result I got on the ciliscope.This circuit was made out of a 100uf capacitor and 0.47KOhm resistor and 12v power supply.The result of the time I calculated was 235ms and the observed time was 250 ms.this is a good result even though it is not a perfect result.

The picture shown here(the bottom one) is the 4th result shown on a ciliscope.The circuit was made out of a 12vsupply a 330uf capacitor and a 1KOhm resistor i calculated the time and got a result of 1650ms the observed time had a perfect result of 1650ms the reason my result was perfect is because all the setting was set correctly to give a more accurate result.
It is very important when working with capisitor to put it the corect way in the circuit negative of capasitor with negative on the power supply and positive on capasitor with the positive in the supply,if put incorectli in the circuit capasitor could explode and be dangerous.

diodes

I used a multimeter to identify the anode and cathode of the diode and the Led
I measured the voltage in forward biased direction
Led:1.77v
diode:0.56v
measured voltage in reverse direction
Led and diode were open because n type meterial is more negative and when connected to positive side of battery it attracts increasing the size of the completion zone,same prosec happens with p type meterial on positive side of the battery.depolition zone increases to such a size that it stops the flow of electricity
the cathode can be identified without a multimeter.On diodes there is a stripe on the cathode side and on Leds the flat edge is the cathode.It is realy important to be able to identify the anode and cathode of a diode
I built a circuit on a breadboard the circuit has a 5v power supply and a 1kOhm resistor with a diode of 1n4007.After I made my circuit i calculated the value of current 5-0.7/1000=4.3mA 5is because of the voltage suply 0.7 is the diode and 1000 is the Ohms.I measured this with a multimeter and got 4.8mA
I expected this result because my voltage was a bit high wich means my curent would also measure a bit high.The picture shown here is a picture i took of my breadbord.

I calculated the voltage drop across the diode and got a result of 0.6-0.7v and then i mesured it with my multimeter and got a result of 0.64v so the result i calculated was within range of measured with multimeter.
After i made the circuit on the breadboard with a diode I replaced the diode with a Led.Icalculated the curent and got a result of 3.6mA I worked out by using ohms law A=V/R 3.6/1000=.0036A= 3.6mA 3.6 is the volts i got and 1000 is the resistance. i mesaured the current with a multimeter and got a result of 2mA the big diferance in measured and calculated can be cause by alot of thing but in this case i think it has to do with the multimeter because it had a blown fuse on mA so i measured it with amps.
I created a circuit with 2 100Ohmresistors and a 400mWzener diode with a 12v power supply.I mesaured the Vz and got a result of 4.9V then put a 15V supply on it and got a result of 5.15V because the more voltage you ad the more gets pushed through the zener diode.this circuit could be used as a voltage regulator because of the voltage amount it carries.After this i put the zener diode in reverse polarity and got a reading of 0.8V because the negative was on the negative side of the diode and the positive on the positive side of the diode.
The 4th circuit i built on the breadbord was a 1kresistor+400mWzener diode and a 1n4007 diode and mesured it once on 10V supply other on 15V supply  got result of:
                   10V                                                15V
zenner diode:4.65V                                            4.7V
diode:0.65V                                                       0.67V
both:5.32V                                                         5.46V
resistor:5.7V                                                      9.58V
current:5.7mA                                                    9.5mA
shows that the higher the voltage supply the more voltage flows throug the diodes and resistor.

Resistors

I am going to calculate the resistors
I toock 6 resistors with diferent values and calculated them manualy and then with a multimeter to confirm my result.the resistors have colour codes and every coulure has a value,the first two bands is the numbers you write down and the next band is the multiplier
the colour values are:
black:0
brown:1
red:2
orange:3

yellow:4                                                                      
green:5
blue:6
violet:7
grey:8
white:9
The picture on the top is a good example of how to work out the values of the coulors.I got this picture from www.reuk.co.uk
The first resistor i calculated was red,red,yellow,gold.I got a result of 220kohm because red is2 and the yellow is 4 wich is the amount of zeros and a 1000=1kohm and with the multimeter it was 212.9kohm and i have a 5%tolerance wich i stayd in.because I can go as low as 209ohm
the other results was as folowing
value(colour codes)                                                  value(multimeter)
brown,black,brown,gold               
100ohm,5%tolerance=95-105Ohm                                  98.4ohm
brown,black,green,gold
1mega ohm,5%tolerance=.950-1050MOhm                      .993mega ohm
brown,black,orange,gold
10kohm,5%tolerance=9.5-10.5KOhm                               9.98kohm
brown,black,red,gold                              
1kohm,5%tolerance=.950-1.05KOhm                               .992kohm
yellow,violet,black,black,brown
470ohm 1%tolerance=446.5-493.5Ohm                            470ohm

all my results were within tolerance givin so was done corectly
refrences:www.reuk.co.uk