Experiment 7-4824

By using a voltmete accross the B to the E, I found a 0.8 Vd, this is indicating that the transistor uses 0.8 V to function. The 0.8 is its knee voltage.

The Vd accross the C to E is 0.05V. This V difference is very small. The more the transistor is on, the less the difference in Vd. The less its on the greater Vd accross the C to E.

The lower the amps at the base, the lower amps will be at the collector, therefore a high voltage drop. The more amps at C and B the more the transistor is on, so less Vd down to the 0.8 needed for it to operate.

So 'A' in the graph is in the region of Saturation and means it is fully working with minimal Vd, whereas 'B' is in the 'off' region and the transistor will not function.

The power dissipated by the transistor at Vce of 3Vd is:
Ic=13ma Vd=3
13x3=39W

The Beta of the transistor at Vce 2,3 and 4 is:

At 2Vd Ic=20ma ÷ Ib=0.8ma Beta=25

At 3Vd Ic=13ma ÷ Ib=0.5ma Beta=26

At 4Vd Ic=5ma ÷ Ib=0.2ma Beta=25