How much current in an Outlaw?

jrlouie

Hey guys,
I emailed Outlaw asking about their 755. I asked them if they could tell me how much current it could output and I also asked why they don't post these specifications. Here's their response...

Thanks for contacting the Outlaws!

There is no standardized test for measuring "current" availability.
Unfortunately, we do not know how to measure this in any meaningful way
to
compare to other amplifiers. Once a standardized procedure for this is
established we will be happy to publish this statistic.

Please let me know if you have any further questions.

Thanks and regards,

So does anyone know how much current these babies can put out?

bikezappa

That's said!

I assume you want to know the current to your speakers at full power.

Watts=Current times Voltage

or

W=IE

Also Ohms law is:

Voltage = Current times resistance

E=IR

Substituting for E=IR in the equaltion for watts gives us:

W=IIR

Solving for I:

I= the square root of (W/R)

Assume you have 4 ohm speakers and a 100 watt amplifier, then

I= the square root of (100/4)=5 amperes of current

Hope this helps.

Peter

bikezappa

That is 5 amperes to one speaker.

jrlouie

Thanks bikezappa. That does help. That's much more than I knew
The main reason I ask is because I read forums talking about the ability for an amp to supply the needed current for certain speakers. Obviously, Lsi's like current, which I have.
It almost sounds like the formula listed defines the current to your speakers based upon some given assumptions.

Assume you have 4 ohm speakers and a 100 watt amplifier, then

I= the square root of (100/4)=5 amperes of current

So based on that, it appears this formula states all 100 watt amps will provide 5 amperes of current. So why do people compare current (amperes) output between two amps of the same wattage?

Sorry for the questioning. I really don't know this stuff and I'm just trying to learn a little.

Thanks!

bikezappa

The amount of undistorted current an amplifier supplies is based on the, maximum power from the amplifier, the resistance of the speaker and the volume control setting.

If you had 8 ohm speaker with a 100 watt amplifier you would have the following maximum current:

I= square root of (W/R) = square root of 100/8 = 3.5 amperes.

The 100 wattts is based on the maxinum power the specific amplifier can deliver. The real power the amplifier delivers is based on the volume control, 0 to 100. Other amplifiers can deliver more or or less power or Watts. Watts and power are the same thing.

What I think you were asking was what is the maximum current to my speakers. The maximum current is a function of maximum power in your ampilifier and the resistance of your speakers.

There are many ways to cheat with these power formulars:
Is the power rating of the amplifier continuous?
What is the distortion at the maximum power setting?
What is the frequency range at maximum power?
.....




Hope this helps.

heiney9

Originally posted by jrlouie
Thanks bikezappa. That does help. That's much more than I knew
The main reason I ask is because I read forums talking about the ability for an amp to supply the needed current for certain speakers. Obviously, Lsi's like current, which I have.
It almost sounds like the formula listed defines the current to your speakers based upon some given assumptions.


So based on that, it appears this formula states all 100 watt amps will provide 5 amperes of current. So why do people compare current (amperes) output between two amps of the same wattage?

Sorry for the questioning. I really don't know this stuff and I'm just trying to learn a little.

Thanks!

It sure would be nice if THIS were that simple, but it's not. Think of it as calculating horsepower ratings for a combustion engine. The key difference being measuring vs. calculating. Lot's of other variables involved. From a pure mathematical perspective Ohm's Law is great. In practical use (end results) the outcome varies widely. I suspect your question is more about producing peak current output, this varies widely depending on the amp design and components used. That's where the diff is. Ohms Law simply allows you to calculate continuous current delivery, all other things being equal (in real life they never are). Also be aware that wattage is partly derived from the rail voltage in a power amp. Rail voltage is heavily reliant on the power supply, etc, etc...gets complicated when you consider each stage is dependent on each other stage. If you are truely interested in understanding read the links below. A bit technical, but they do a good job explaining things step by step.

See Links for an indepth explaination.

http://www.rocketroberts.com/techart/amp.htm

http://www.rocketroberts.com/techart/powerart_a.htm

Have Fun

H9

jrlouie

Okay, cool. Thanks.

Outlaw's figures are respectable. So for my M200 (or I suppose their 755 also).....with .05% or less distortion at 300 watts for a 4 ohm load (my Lsi's), we could say she's puttin out....

300(watts)/4 = 75 amperes!

That's damn respectable. Wow.



Ooops, heiney got a post in there. I didn't read it before posting

bikezappa

Wrong

I= the square root of (watts/resistance)= 8.6 amperes.

As I said I think this is what you were asking.

Rail voltages and instantnious power stuff is interesting but not what he was asking.

I like to make it simple first than it may get complicated.

heiney9

Originally posted by bikezappa
Rail voltages and instantnious power stuff is interesting but not what he was asking.

I like to make it simple first than it may get complicated.

If this is all he's asking about then all 300 watt amps @ 4 ohms = 8.6 amperes. How does that help one in deciding what amp to choose? I was just giving him a resource to allow him to see the BIG picture.

Ohm's law can be used all day to calculate things, doesn't give any insight to how a particular piece will perform or how it's been designed.

Same as my ex. for horsepower that can be calculated all day long; pure figures don't tell much about the big picture.

Take my answer FWIW

H9