Monday, August 2, 2010

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.

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