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?
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.
To calculate the power dissipation by the transistor in a Series Op-Amp Regulator, we need to find the current passing through the transistor and then multiply it by the voltage across it.
In a Series Op-Amp Regulator, the current through the transistor (IB) is given by:
IB = IL * (1 + (R2/R1))
Given that IL = 260mA, R1 = 4.0kΩ, and R2 = 10kΩ, we can substitute these values and calculate IB:
IB = 0.26A * (1 + (10kΩ / 4.0kΩ))
= 0.26A * (1 + 2.5)
= 0.26A * 3.5
= 0.91A
Next, we need to calculate the voltage across the transistor (VBE). It is given by:
VBE = VIN - VZ
Given that VIN = 14V and VZ = 3.6V, we can substitute these values and calculate VBE:
VBE = 14V - 3.6V
= 10.4V
Now, we can calculate the power dissipation by the transistor using the formula:
Power Dissipation = IB * VBE
Substituting the values we calculated, we get:
Power Dissipation = 0.91A * 10.4V
= 9.44W
Therefore, the power dissipation by the transistor in the Series Op-Amp Regulator is approximately 9.44 Watts.