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The
base voltage Vb, is set
by potential divider 9K
and 1K = 1 volt
(ignoring
Ib)
The emitter voltage Ve,
is Vb - 0.6 = 0.4 volts.
The emitter current is
Ve / Re = 0.0008A
The voltage across the
5K load resistor is 5000
x .0008 = 4 volts
Therefore the collector
voltage is 10 volts - 4
volts = 6 volts
The power dissipated by
the transistor is the
voltage across the
transistor multiplied by
the current through it.
= 5.6 volts x 0.0008A =
0.00448 watts
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