|
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. |
