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The power dissipation of a BRT is the sum of the power dissipated by the internal transistor (Q) and the built-in bias resistors (R1 and R2).
Here, let’s calculate the power dissipated by the BRT shown in Figure 1 when it is in the “on” state.
Suppose that the BRT is the RN1402 (R1=R2=10 kΩ), the input voltage (VI) is 10 V, the collector-emitter voltage (VCE) is 0.2 V, the base-emitter voltage (Vbe) is 0.7 V, and the operating current gain (hFE) of the internal transistor is 10.
Then, IB, IR2, Ib, and IC are calculated as follows:
IB = ( VI – Vbe ) / R1 = ( 10 – 0.7 ) / 10 = 0.93 mA
IR2 = Vbe / R2 = 0.7 / 10 = 0.07 mA
Ib = IB – IR2 = 0.93 – 0.07 = 0.86 mA
IC = Ib * hFE = 0.86 * 10 = 8.60 mA
Let the power dissipation of R1 and R2 be PR1 and PR2 respectively and the collector-emitter and base-emitter power dissipation of the transistor be PCE and Pbe respectively.
R1: PR1 = ( VI – Vbe ) * IR1 = ( 10 – 0.7) * 0.93 = 8.65 mW
R2: PR2 = Vbe * IR2 = 0.7 * 0.007 = 0.05 mW
VCE: PCE = VCE * IC = 0.2 * 8.6 = 1.72 mW
Vbe: Pbe = Vbe * Ib = 0.7 * 0.86 = 0.60 mW
Hence, the power dissipation (P) of the BRT is calculated as:
P = PR1 + PR2 + PCE + Pbe = 11.02 mW