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Am i calculating my volts right? Please take a glance. Trying to set m

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Ive got an alpine mrp m350 amp that does 350 watts @ 2 ohm and a 4 ohm dvc ssa dcon 10". Im trying to run the sub at around 300-350 watts in parallel at 2 ohms. I want to accurately set my gain and found the equation that you are suppossed to use to find the optimal volts.

Voltage = sqrt(rms x impedence)

So is this what i would do for my 4 ohm dvc woofer?

Voltage = sqrt(300x4ohm)

Voltage = sqrt(1,200)

Voltage = 35

Or is it this since im running the amp at 2 ohm to a 4 ohm dvc woofer

Voltage = sqrt(300x2 ohm)

Voltage = sqrt(600)

Voltage = 24.5

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I believe it would be the later since that's the actual impedance at the amp, but setting the gain using that formula is just a rough guestimate at best.

Use your ears instead, turn the gain all the way down, turn your head unit volume up to max listening level.

Turn gain up until it sounds distorted, you may have to take a step back from the vehicle to tell, then back it down til it sounds clear.

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26.5v is the expected AC voltage to obtain 350w@2ohms...

However.... It would be safe to probably go over this number a little because voltage typically increases a little with impedance due to the stress it puts on the power supply.

Notice on many big amps how they may be rated as follows-

1000w @4ohm

1800w @2ohm

3000w @1ohm

When u see an amp such as this-

1000w @4ohm

2000w @2ohm

4000w @1ohm

You should expect to be disappointed...honestly.

So, a slight increase in voltage with impedance rise is what usually happens.

Now, instead of getting all technical on where the amp should be.. just do what stefanhinote said.

You are not on a dying level to have the amp set in such a way to get every single 0.1v out of it..

If you have to ask, you are not competing, therefore nothing crucial is necessary.

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I heard by ear is a guestimate?

Your formula doesn't take into account amplifier efficiency, and your voltage will vary from frequency to frequency, and music is dynamic.

You could set the gain to that target voltage and give it a listen and do some tweaking by ear.

It's not as hard as everyone makes it out to be...

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Amplifier efficiency has nothing to do with that formula. It's just Ohms Law.

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Amplifier efficiency has nothing to do with that formula. It's just Ohms Law.

heh, I was thinking of the power input to the amp like it takes xxx amp at 14v to produce xxx watts but since the efficiency of the amp is not 100%, it would require more amperage.

Instead of the output power, being final imp at amp with a target wattage(more of a guess) to find voltage.

So yeah, makes sense. :P

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