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.