Experiment 7
By looking at the voltage drop between the base and the emitter V(BE) we can see that this transistor(C547) has a knee voltage of 0.8 V.
VCE = 0.05 V
And also by looking at the voltage drop between the collector and the emitter we can see that this transistor is in saturation , therefor working as a good switch.
Section A is the Saturated region which in other words mean the transistor is fully on
Section B is the cut off region and also in other words mean this transistor is fully off.
Power dissipated by this transistor when the Vce = 3 V
Power dissipated = Vce x Ic
by using the above graph we can get Ic which is 14 mA
Power dissipated = 3 x 0.014
= 42 mW
Caculating β = Ic/Ib
Vce = 2 V β = 20 / 0.8
β = 25
Vce = 3V β = 14 / 0.5
β = 28
Vce = 4 V β = 5 / 0.2
β = 25
Experiment 8:
As we change the resistance of Rb I've recorded the readings in the table below
and from this readings I have plotted the graph below:
Vce is the voltage drop between the collector and the emitter of the NPN BJT transistor. You can see in these results, as we increase Rb resistance ,so does Vce . This is showing that lower current at the base, due to using bigger resistor before the base terminal of the transistor, causes a higher voltage dop between the collector and emitter. so the transistor is more closed or more off.
Vbe is voltage drop between the base and emitter terminals and does not change much , therefor it shows in this transistor it more acts like a diode and a rectifier and 0.7 v is the knee voltage to operate this diode .Ib is current at the base terminal .As we increase Rb it reduces the Ib and ultimatly this results in lower current flow at the collector .
Ic is current at the collector terminal as we can see above as Rb decreases ,Ib increases so Ic also increases so we can see as we increase Ib ... Ic also increases in current flow at the collector
