To calculate the power dissipation of the transistor, we first need to calculate the output voltage of the regulator.
The output voltage is given by:
Vout = VZ * (1 + R2/R1) + IL * R3
Substituting the given values, we get:
Vout = 3.6 * (1 + 10/4) + 0.26 * 4.9
Vout = 12.84V
The voltage drop across the transistor is VIN - Vout = 14 - 12.84 = 1.16V
The power dissipation of the transistor is:
P = V * I = 1.16 * 0.26
P = 0.3016W or 301.6mW
Therefore, the power dissipation by the transistor is 301.6mW.
In a Series Op-Amp Regulator, given R1=4.0kΩ, R2=10kΩ, R3=4.9kΩ, VZ=3.6V, VIN=14V.
Assume IL=260mA, what is the power dissipation by transistor?
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