Gain Setting

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mtsooner

Member - Posts: 14

Joined: January 28, 2005

Posted: March 07, 2005 at 5:17 PM / IP Logged

I asked a similar question in another post, but I'd like to clarify something. I have a PG Xenon 200.4 amp. I'm going to be running my CDT Cl-61a components off of it and my Infinity Reference subs off of it. The amp has (2) input sensitivities....one for the front one for the rear. I've read DYohn's tutorial on setting the gain for optimum amp performance, but here's what I'm searching for. The amp's input sensitivities go from 8 volts on the left to .2 millivolts on the right (I now know that I should start from 8 volt/left side). My CL-61a's are rated at 90rms....if I set my amp to optimum performance, wouldn't I be kicking out WAY more than 90 rms to those speakers, hence taking the chance on blowing them! I think I should leave the gain set at 8 volts just to be safe, or am I being a little overly cautious on this? By the way, my HU is 4 volt outs.

DYohn

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Joined: April 22, 2003

Location: Arizona, United States

Posted: March 07, 2005 at 5:30 PM / IP Logged

To set your amplifier output based on wattage, you can use the voltmeter method.

Disconnect the componets from the amp channles and set the gain for that side to minimum. Set your HU to the max unclipped output level and play a 1000Hz test tone. Connect an AC voltmeter to the speaker outputs. Slowly increase the gain setting until you read 18.9 volts. This corresponds to 90 watts into a 4-ohm load and will set the amplifier gain to hit your speaker's max at the HU max. Shut it down and reconnect the speakers and test them. If you need to you can back the gain off from this level to achieve the proper balance.

Make sense?

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Hornshockey

Silver - Posts: 520

Joined: January 31, 2005

Location: Texas, United States

Posted: March 08, 2005 at 12:47 AM / IP Logged

DYohn, would the same hold true for setting the gain for subs?

Life moves pretty fast; if you don't stop and look around once in a while; you could miss it.

Ravendarat

Platinum - Posts: 2,806

Joined: February 23, 2004

Location: Canada

Posted: March 08, 2005 at 1:38 AM / IP Logged

The exact same procedure can be used to set any car amp for any purpose. Just substitute the 18.9 volts with whatever volltage corospondes to the amount of wattage you desire

double-secret reverse-osmosis speaker-cone-induced high-level interference distortion, Its a killer

Hornshockey

Silver - Posts: 520

Joined: January 31, 2005

Location: Texas, United States

Posted: March 08, 2005 at 1:51 AM / IP Logged

Would I still use a 1kHz test tone?

Life moves pretty fast; if you don't stop and look around once in a while; you could miss it.

mtsooner

Member - Posts: 14

Joined: January 28, 2005

Posted: March 08, 2005 at 8:23 AM / IP Logged

Thanks DYohn! I think I just had an epiphany! Ok, so I'll burn a 1000hz test tone to a CD and use a 60hz test tone for the subs....you made it easy to understand.

mtsooner

Member - Posts: 14

Joined: January 28, 2005

Posted: March 08, 2005 at 8:50 AM / IP Logged

Well, I just thought of one more question. If I have (2)-Infinity Reference 1032w subs rated at 250rms each bridged off my amp (400rms x 1 and 8-2 ohms). In order to set the output wattage correctly, do I use 250rms x 4 or 500rms x 4 to calculate the voltage?

DYohn

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Joined: April 22, 2003

Location: Arizona, United States

Posted: March 08, 2005 at 11:01 AM / IP Logged

Set the amplifier for its rating, not the subs. If the amplifier is capable of 400 watts into the load presented by the sub, then that's what you use.

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Poormanq45

Silver - Posts: 597

Joined: October 27, 2004

Location: United States

Posted: March 08, 2005 at 4:24 PM / IP Logged

DYohn: What is the formula that you used to calculate taht 18.9 volts = "90 watts into a 4-ohm load " . I have been searching for this formula, but I haven't been able to find it :(

Are there any situations which this formula does not work?

Thank you

-Brien

DYohn

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Joined: April 22, 2003

Location: Arizona, United States

Posted: March 08, 2005 at 4:30 PM / IP Logged

It's Ohm's Law. Voltage = square root of (wattage times resistance). 90 X 4 = 360, sqrt = 18.97

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