Electrical characteristics of sound reinforcement system
1. Continuous power (rated power)
Continuous, long-term or RMS (effective value) power (the duration of the test signal lasts more than 1 hour), it is also called continuous pink noise power (when pink noise is used), long-term sine wave sweep power (when sine sweep ), The final measurement point gives the lowest safe power value.
2. Program power
Program or music power (test signal with music characteristics, about 1 second), it is also called continuous program power.
3. Peak power
Peak or instantaneous power (test signal is short-term, less than 0.1 second), this test gives the highest power value, which is 6dB higher than continuous / long-term / RMS power (four times the rated power).
CSR digital power amplifier
Using high-efficiency class D circuit, the power amplifier has small size, light weight, low heat generation, conversion rate up to 95%; low power consumption and high power utilization rate.
All adopt 1U height design; the power supply part adopts high-efficiency switching power supply, which improves the stability of the long-term work of the whole machine.
With perfect protection function, overload protection, short circuit protection, temperature protection, DC protection.
Built-in automatic limiter function to avoid excessive input and distortion of output.
Actual power consumption
1. Determine the type of power amplifier used (Class A, B, AB, H, D, TD, etc.).
2. The conversion efficiency of each type of power amplifier is as follows: Class A, B, AB: less than 50%; Class H: about 75%; TD: up to 90%; Class D: up to 95%.
3. Demand coefficient: The calculated total power can be multiplied by a demand coefficient, because the music signal is not continuous, so the sound system is not a constant consumption load for the power supply. The system of general language application can use coefficients of 0.3-0.5, the theater system is about 0.6, and the night field should be above 0.8.
For example: a night show project uses 20 TD-type 8-ohm 900W two-channel power amplifiers. How much power does it need to provide?
Solution: 900 × 2 × 20 = 36000 (Watts), 36000 ÷ 0.9 = 40,000 (Watts), 40000 × 0.8 = 32000 (Watts), 32000 ÷ 1000 = 32 (Kilowatts) The required power consumption is 32KW.
(Note: This calculation is already very conservative. Under normal circumstances, it is impossible for these amplifiers to work at full power at the same time.)