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tamita Normally the maximum power that can be taken from one output stage: (in case of complementary) Pmax = Usup on the square divided by 8 times the load resistance this is an ideal value, hence the loss of efficiency. Where Pmax = max power Usup = supply voltage Rload = speaker resistance (1kHz) So: 5x5 div. 8x4 = 0.781, this results in a loss of efficiency of about 75% = approx. 0.586 watts (voltage in volts, load in ohms) Given that in the present case the device is advertised as 50 watts, I could only determine the desired value by measurement. tamita

2020-10-15 04:46:58 Hasznos (0)
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