Question

Rₑ=1 Ω

Rₗ=8 Ω
V C C=15 V
βₙ=βₚ=770

(a) What is the required output swing voltage to deliver 5 W power to an 8Ohm load?

Answer 1

To deliver 5 W power to an 8 Ohm load, the required output swing voltage is approximately 17.8 V peak-to-peak.

Step-by-step explanation:

The required output swing voltage to deliver 5 W power to an 8 Ohm load can be determined by using the relationship between power (P), voltage (V), and resistance (R), which is P = V^2 / R. We need to solve for V, the voltage across the load, to ensure it can deliver 5 W power.

Substituting the given values into the equation:

5 W = V^2 / 8 Ω

This leads to:

V^2 = 5 W × 8 Ω

V^2 = 40 V^2

V = √(40 V^2)

V = 6.32 V (RMS)

Since we want the peak-to-peak output swing voltage (VP-P), we need to multiply the RMS voltage by 2√2, which gives us:

VP-P = 6.32 V × 2√2 ≈ 17.8 V

Thus, the required output swing voltage is approximately 17.8 V peak-to-peak to deliver 5 W power to the 8 Ohm load.