V(be)= 0.8vThis is the voltage drop in-between the emitter and the base of a c547 type transistor, it shows that it has a knee voltage of just 0.8v
V(ce)= 0.05v
This is the voltage drop in-between the collector and the emitter, the voltage drop here is 0.05v this meaning the current is flowing freely from the collector to the emitter.
In the plot on the left the two regions labelled A and B represent two different things A being the saturated zone where the transistor is fully open and B being the "cut off" zone where no power can pass through working almost as a switch.
It takes O.7v for the transistor to be fully open(Saturated) and anything under that, is considered as the active zone where some flow is let through, but not none at all.
I had to calculate how much power is dissipated by this transistor at Vce of 3v using the graph above.
Calculations:
P=IxV
P=15x3
15exp-3x3 = 0.045ma
Power dissipation at 3v was (0.045ma)
I also had to find the Beta of this transistor at Vce 2,3 & 4 volts
Calculations:
Beta= IC/IB
2v: 20ma/0.8ma = 25
3v: 13ma/0.4ma = 32.5
4v: 5ma/0.2ma = 25
good explanation However BETA does NOT have a unit
ReplyDeleteit is just the gain or difference between IB and IC
Power is expressed as watts not Amps and remember that a transistor is a current operated devise eg once .7 volts at the base is reached the current at Ib controls how much the transistor is opened
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