Re: YD7377 4x10W radio amplifier, how can it output 4x25 W?
"Let's use Behemot's setup here as an example. You have four 2-Ohm speakers, and you are using the IC in single-ended mode to drive each channel. With a 14.4V supply and ignoring voltage drops in the amp IC, the maximum output power is 25.92 Watts RMS per channel into a 2-Ohm speaker"
Post 17 by fzabkar:
P = (Vrms x Vrms) / R = (Vpeak x Vpeak / 2) / R
= (14.4^2 / 2) / 4 = 25.92W
That's assuming that there is no distortion over the full peak-to-peak voltage swing for a bridged configuration.
So Per formula post 17: R= 4 Ohms, V supply is 14.4V, Bridged mode configuration
P = (Vrms x Vrms) / R = (Vpeak x Vpeak / 2) / R
In the example shows = (14.4^2 / 2) / 4 = 25.92W
The 14.4 is the peak Voltage due to bridge mode.
So in single end mode, the Vpeak will be only 14.4Vp-p/2 =7.2Vp
So using formula in post 17 with Vp of 7.2V due to the supply Voltage is 14.4 V and in single ended mode = (7.2^2 / 2)/2 Ohms = 12.96 W R.M.S. So where does 25.92 Watts RMS into 2 Ohms in single ended mode come from?
Am I missing something? From the Radio Amplifier manufacturer product spec sheet? Which will not be surprised.
"Let's use Behemot's setup here as an example. You have four 2-Ohm speakers, and you are using the IC in single-ended mode to drive each channel. With a 14.4V supply and ignoring voltage drops in the amp IC, the maximum output power is 25.92 Watts RMS per channel into a 2-Ohm speaker"
Post 17 by fzabkar:
P = (Vrms x Vrms) / R = (Vpeak x Vpeak / 2) / R
= (14.4^2 / 2) / 4 = 25.92W
That's assuming that there is no distortion over the full peak-to-peak voltage swing for a bridged configuration.
So Per formula post 17: R= 4 Ohms, V supply is 14.4V, Bridged mode configuration
P = (Vrms x Vrms) / R = (Vpeak x Vpeak / 2) / R
In the example shows = (14.4^2 / 2) / 4 = 25.92W
The 14.4 is the peak Voltage due to bridge mode.
So in single end mode, the Vpeak will be only 14.4Vp-p/2 =7.2Vp
So using formula in post 17 with Vp of 7.2V due to the supply Voltage is 14.4 V and in single ended mode = (7.2^2 / 2)/2 Ohms = 12.96 W R.M.S. So where does 25.92 Watts RMS into 2 Ohms in single ended mode come from?
Am I missing something? From the Radio Amplifier manufacturer product spec sheet? Which will not be surprised.
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