Easy math question

WagBoss

Member
2013-01-28 1:27 am
I have an unbalanced miniDSP that can do 0.9V RMS MAX output. I have an ep4000 in bridged mono mode and an lms-r 15" set up as a 4 ohm load.

which math is correct:

4 ohm bridged mode:

1950 W = (V*2*50)^2/4

V = 0.88 V

When bridging an amplifier, the voltage gets doubled, which is where the *2 comes from. So from this math the ep2500/4000 only needs 0.88V to reach it's max output on 4 ohm bridged. The measured power value is from here Measuring Amplifiers

The *50 is from the 34 dB gain of the ep4000 amplifier. The /4 is from the 4 ohm load.


OR

1950 W = (V*50)^2/4

V = 1.73 V

where bridging the amp doesn't double the voltage.

Which way is the correct way?

Thanks
 
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WagBoss

Member
2013-01-28 1:27 am
No I know how bridging works, it just inverts the signal with a 180 degree phase shift so the voltage difference is twice as big. P is proportional to v^2 so a doubling in voltage quadruples the power. (I think you meant to say it does 4X the power, not 1X or 2X, or you meant the output voltage is doubled?).

What I'm asking is in my calculations for input sensitivity while bridging, do we look at the gain as 100* (because bridging doubles the voltage output) or as 50*.
 
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Mooly

Administrator
Paid Member
2007-09-15 8:14 am
I think you answered your own question.

For example, two amplifiers rated at 28 Vrms output (100 watt/8 ohm). Each has a voltage gain of say 28. So 1 volt in gives 28 volts out.

In bridge mode to get 28 vrms across the load would only require each amp to give 14 volts rms. So that's 0.5 volts input for the bridge amp to put out the "same" output as the single ended version would give.

So the input sensitivity is halved in bridge mode compared to the same output from the single ended version. In terms of gain its twice as much.

Is that what you meant ?