Experiment 8

As we increase resistor value at the base, transistor starts to close and voltage drop increases at the Vce (emitter-collector)

Vbe require at least 0.7 volts for the transistor to work, so it doesn't change much.

With increasing the resistor at the base we we make transistor work less efficient so less (Ib) current flowing through the base and that effect the current flow through the collector-emitter (Ic), its also decrease.

Calculate Beta of this graph, Vce 1.5 volts

B=Ic/Ib=3.3miliAmps/55microAmps=0.0033/0.000055=60

At this graph i can see that 0.089volts at Vce transistor fully open (saturated), at 1.374volts at Vce its in active region, and at 2.35volts at Vce its cut off.

Experiment 7

o.8 Volts volt drop across base and emitter of transistor indicates that its working and in saturated condition as it requre at least 0.7 volts to work.

54.1 mVolts voltage drop across collector and emitter means that transistor is fully open.

This graph shows working areas od BJT transistor. Area marked with A is a saturated region, which means that BJT fully on, and area marked B is a cut off area, which means it fully off.

The more voltage drop at the base of transistor, the less voltage drop at emitter-collector of transistor, it means that if more " fully on".

Power dissipation at 3 volts Vce.

P=Ic*Vce=3volts*0.014=o.o42 watts= 42 mWatts.

Calculate Beta of this transistor at Vce 2,3,4 volts

B=Ic/Ib

At 2 volts Vce B=21/0.75=28

At 3 volts Vce B=12.5/0.52=26

At 4 volts Vce B=5.5/0.22=24

Beta is how much Ic signal amplified compare to Ib.

Experiment 6

Emitter-base always have slightly higher forward voltage drop then collecter-base, so can be identified easy with multimeter.

NPN transistor:

PNP transistor