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amp and speaker calculations ohms law


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todubd 
Member - Posts: 7
Member spacespace
Joined: January 17, 2015
Location: Florida, United States
Posted: January 18, 2015 at 12:08 AM / IP Logged  
Hey, all! Newbi here, first post. And I'm sure you get a lot of these, but I really can use the help and it's all very much appreciated. Especially from anyone audio mathematical savvy.
First up, my amp:
Alpine MRV-M500
-mono subwoofer amplifier
-300 watts RMS x 1 at 4 ohms or 500 watts RMS x 1 at 2 ohms
Second, the speakers:
Kenwood KFC-W3013PS
12" 4-ohm subwoofer
power range: 50-400 watts RMS
"peak power handling: 1,200 watts"
frequency response: 34-300 Hz
Alright. So my plan is to have this amp powering two speakers with the possibility to just use one speaker depending on what will be my best option sound wise running really clean.
Also, I'm going to tune a 9 AC volt single stock speaker signal down to 5 AC volts with a pac LP7-2R... "hopefully"... I know everyone hates these LOC's because they use transformers or what not... and sound like garbage. But my other option is to just pull the signal and plug it into one channel on the amps signal level input. We'll see what ya'll think.
What I really need to know is:
I'm calculating that with one mono amp Y'd off "parallel" to two 4 ohm single voice subs, then amp is going to think that it's running at 2 ohms. Therefore, as stated in the specs: 500 watts RMS x 1 at 2 ohms.
So using the (V=Sqrt P*R)
Ac Volts = Sqrt 500(watts RMS) * 2(ohms RMS)
Ac Volts = 31.62
31.62 is what the gain should be set at running two 4 ohm subs.
Or if I only wanted to use one 4 ohm sub, it would be:
300 watts RMS x 1 at 4 ohms
AC volts = Sqrt 300 * 4
AC volts = 34.64 (gain level)
Or do I have this all wrong. And the calculation is:
Ac Volts = Sqrt 400(speaker RMS) * 4(ohms amp)? or 2 (ohms amp)
Cause then Ac volts = 44.72 or 28.28... which is a big difference.
No matter if I use one or two speakers...?
If anyone can put this and explain it to me in a "setting your subs up for dummies" kinda terms, I'd really really appreciate it.
I know you can just plug it all up and do all this by ear, but I'm trying to really get this figured out so I don't damage anything.
Thanks for any real good and simple help!
Tdub
oldspark 
Gold - Posts: 4,913
Gold spacespace
Joined: November 03, 2008
Location: Australia
Posted: January 18, 2015 at 12:41 AM / IP Logged  
I know nothing about setting gains - that's the domain of other gurus hereon - but you got the two parallel 4R = 2R correct (where R = Ohms).
I'd expect your gains to be set for the max output situation (500W @ 2R; ie 2 x 250W) tho maybe it is set for voltage - ie, single 300W @ 2R...? (C'mon Grus!) But either way IMO it should be a set & forget thereafter - irrespective of the 1 or 2 subs. Then again, for critical installations (sound comps; pedantic listeners)...
I'm amused by the speaker specs - eg, 50-400 watts RMS ie, a rating of 400W; a peak power handling or 1200W... who cares? wth does that mean - do they define the conditions (not that it matters) or is it simply 3x RMS because 2xRMS is called "Music Power"?
I don't know what that 9V to 5V (AC is about. If it's a mere signal drop I'd probably use a resistive network or trimpot. If it's a speaker rating noting that they are all AC and usually defined by a Wattage and Nominal Impedance (Ohms), then it's power conversion that may involve impedance matching/changing (ie, 4R input to 2R output), but assuming the max power involved is 9W (ie, 9V into a 1R speaker) maybe mere signal conversion products will handle it.
BTW = from V=IR & P=VI comes P=VV/R = IIR - ie, voltage-squared or current-squared (div or times R respectively). That avoids all AC and root-2 conversions etc.
todubd 
Member - Posts: 7
Member spacespace
Joined: January 17, 2015
Location: Florida, United States
Posted: January 18, 2015 at 12:55 AM / IP Logged  
oldspark wrote:
I know nothing about setting gains - that's the domain of other gurus hereon - but you got the two parallel 4R = 2R correct (where R = Ohms).
I'd expect your gains to be set for the max output situation (500W @ 2R; ie 2 x 250W) tho maybe it is set for voltage - ie, single 300W @ 2R...? (C'mon Grus!) But either way it should be a set & forget.
I'm amused by the speaker specs - eg, 50-400 watts RMS ie, a rating of 400W; a peak power handling or 1200W... who cares? wth does that mean - do they define the conditions (not that it matters) or is it simply 3x RMS because 2xRMS is called "Music Power" ?
Thanks. So if I'm understanding this Ohm thing, the lower the ohm, the harder the amp is working (or thinks it's working), so you set the gain lower for more power with two speakers? Then if I only had one speaker, I can push out more power (AC volt) at a higher 4 ohm load?
So the speakers max RMS is 400.
So it's either
two speakers:
AC volts = Sqrt 400 * 2
or
One speaker
Ac volts = Sqrt 400 * 4
IDK why it seems backwards or I'm just not grasping something.
I guess let's see what the gurus have to say.
Thanks for the input... anything simple helps. lol
Tdub
todubd 
Member - Posts: 7
Member spacespace
Joined: January 17, 2015
Location: Florida, United States
Posted: January 18, 2015 at 12:59 AM / IP Logged  
Also, just something else stupid...
the frequency response on the speakers is 34-300 Hz.
So I typically just set the freq knob or "low pass filter" to the 300 area...
I hope that's about right also. lol thanks again!
Tdub
oldspark 
Gold - Posts: 4,913
Gold spacespace
Joined: November 03, 2008
Location: Australia
Posted: January 18, 2015 at 1:40 AM / IP Logged  
Noting that I edited my reply AFTER your following reply...
POST EDIT - AND missed your last reply...end POST EDIT
Working harder...? Hmmm - maybe it's easier for the amp to spill its entire guts... Tho technically since "work" means "force times distance" and if "harder" means bigger or quicker or more, then yes, things work harder into lower impedances (resistances).
I prefer to look at it in ITS simple basics... The lower an impedance, the easier it is pushing things thru it. EG - water in a fatter pipe (to use oft used electrical-water analogy) or current into a load.
You might drop the pressure (voltage) to keep the flow (current) the same as the larger impedance/resistance, or use the lower resistance to extract more current at a given voltage (I=V/R; if V is constant, I & R are inversely proportional - eg, 1/3rd the resistance means 3 times the current.
An easy trap is confusing loads with source, ie individual speaker power & amp output.
For example, assuming a constant source voltage (amp output voltage) the 4R speaker outputs the same power irrespective of whether the other 4R speaker is in parallel. However with 2 paralleled 4R speakers, the amp is outputting twice the current.
But other things have impacts - eg, if the amp output is limited to a certain current, or its gains are varied. (Amps reduce voltage to limit current - except f.ex PWM output amplifiers.)     
I can push out more power (AC volt) at a higher 4 ohm load?"...
A 4R needs twice the voltage of a 2R for the same power output.
But power output is independent of speaker impedance. It has wiring to handle whatever current and insulation to withstand a whatever (way over-rated) voltage (not that insulation is a "specification limiting" feature of a speaker). Its coil's wire length (at whatever Ohms/meter resistance) determines its resistance and its coiled etc winding determines its inductance which - along with other aspects including capacitance - determines its impedance. Its thermal else winding (current) or other physical things determines its power handling capability.
The ideal way to increase power output is to increase the output voltage - ie, minimise disadvantages and losses incurred using higher currents - but there are limits to voltage - eg, component voltage ratings; dangerous voltage levels. The latter is (I presume!) the reason car audio regularly uses 2R or 1R or lower speakers - they have to in order to push more power into (singe) speakers and avoid dangers or Licensing/Design issues with high voltage audio outputs. (Many seem limited to ~35V rails which accords nicely with some common safety limit - namely 70V - tho I think that's a DC voltage standard.)
I'll look at your rooted two's later. (But one speaker is limited to 300W, not 400W.)
todubd 
Member - Posts: 7
Member spacespace
Joined: January 17, 2015
Location: Florida, United States
Posted: January 20, 2015 at 1:55 PM / IP Logged  
Or is it 2x 400 RMS speakers at 2 ohms?
Ac volt = Sqrt 800 (total RMS) * 2 (ohms)??
Which is it pros? lol thanks!
Tdub
oldspark 
Gold - Posts: 4,913
Gold spacespace
Joined: November 03, 2008
Location: Australia
Posted: January 20, 2015 at 2:35 PM / IP Logged  
There is no 800W output into those 2x400W speakers. You'd need a different amp for that.
todubd 
Member - Posts: 7
Member spacespace
Joined: January 17, 2015
Location: Florida, United States
Posted: January 20, 2015 at 6:37 PM / IP Logged  
Ok, so it's:
AC Volts = Sqrt (Amp RMS) * (4 ohm speaker load)
AC Volts = Sqrt 500 * 4
AC Volts = 44.72
44.72 is where I set my gain to AC Volts on the voltmeter using this setup:
amp and speaker calculations ohms law - Last Post -- posted image.
Would there be any benefit to run it at 8 ohms or worse? There is no 8 ohm rating for the amp.
The wire wizard shows this as well:
amp and speaker calculations ohms law - Last Post -- posted image.
Tdub
todubd 
Member - Posts: 7
Member spacespace
Joined: January 17, 2015
Location: Florida, United States
Posted: January 20, 2015 at 6:46 PM / IP Logged  
todubd wrote:
Would there be any benefit to run it at 8 ohms or worse? There is no 8 ohm rating for the amp.
The wire wizard shows this as well:
amp and speaker calculations ohms law - Last Post -- posted image.
Cancel... nm, I'm dumb..
Tdub
oldspark 
Gold - Posts: 4,913
Gold spacespace
Joined: November 03, 2008
Location: Australia
Posted: January 21, 2015 at 2:03 AM / IP Logged  
You mean a temporary lapse or distraction due to other priorities, or if festive season, or I caused you temporary insanity.
Yeah - 8 Ohms means less power "received".   
Highest power delivery is with all speakers (loads) in parallel. Of course the amp (source) may not handle that configuration - f.ex if the combined impedance is lower than the min spec'd output resistance/impedance allowable for that amp.
P = VI = V*V/R = I*I*R (by substituting Ohm's Law ie V=IR and I=V/R respectively into P=VI).
V = the root of {P x R).

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