[qstarin]

I am wondering if it is reasonable to consider these specs at face value, or do they change significantly during actual operating conditions?


Amplifier Efficiency specs:
86% @ 4-Ohm
85% @ 2-Ohm
68% @ 1-Ohm



Currently, I am running at a 1-ohm load and think I have considerable more than enough RMS watts for the speakers I'm driving (2 @ 500RMS).

I look at this and think, "Geez, if I went to dual 2 ohm VC subs instead of dual 4 ohm, I could run a 2-ohm overall load, have enough power, and reduce current draw at max RMS output by about 35 A".


But - is that too simplistic?

I mean, doesn't the resistance seen at the speaker terminals in operation vary with the frequency of the signal? If so, would that mean the efficiency specs listed above are more of a "worst case" - and that my current savings would be way, way less in reality?


Also, would the efficiency numbers likely change from full output to partial output (assuming for the better)?



Oh, and in case anyone really wants to know (but I don't think it matters really for this question), my setup is:

Car: 2001 Subaru Impreza RS

HU: Alpine iDA-x305 + H100 Imprint

FS: Alpine SPR-17S 6 & 1/2" components
RS: Alpine SPR-17C 6 & 1/2" coaxials (considering removing or switching to a "rear fill" signal)
4ch amp: Alpine PDX-4.150

Subs: 2 x Alpine SWR-1242D 12" dual 4ohm VC (considering switching to something else)
mono amp: Rockford Fosgate T1500-1bd

140A alternator, all wiring upgraded.

(I swear, I'm not an Alpine fanboi - just turned out that way )

 

[qstarin]

No one? I figured half you guys could answer this right off the top of your heads?


Is 1 ohm operation efficiency as bad as the specs make it look?

 

[ChrisB]

At 1 ohm, you generally have heat as a trade off, hence the reason your efficiency drops. Also, the components inside of the amplifier have to work really hard as they near a theoretical short.

 

[falkenbd]

Run it at 2 ohms, or maybe even 4 ohms.

You could run 4 ohms with 2 D4 subs (it sounds like that is what you have)

How much power does the amp put out in each mode?

Giving the same setup half the power only loses 3db - and you will barely notice that on music.

 

[Oliver]

If they are run full out, the efficiency will be high
quote>
Also, would the efficiency numbers likely change from full output to partial output (assuming for the better)?
quote>
Once your volts drop and the current increases [ U might blow fuses ]

 

[MidnightG35X]

I mean, doesn't the resistance seen at the speaker terminals in operation vary with the frequency of the signal? If so, would that mean the efficiency specs listed above are more of a "worst case" - and that my current savings would be way, way less in reality?


Also, would the efficiency numbers likely change from full output to partial output (assuming for the better)?

Resistance/Impedance does change across frequency. As a point of advertising, I would say the efficiency numbers posted were best case, not worst case. Judging by your efficiency numbers, you are using a class D amplifier also. A lot of times the efficiency numbers are also just using the amplifier efficiency without taking into account the power supply efficiencies. I'd say most modern MOSFET power supplies are 80-85% efficient.

Efficiency generally gets better as you drive the amplifier harder.

 

[qstarin]

Run it at 2 ohms, or maybe even 4 ohms.

You could run 4 ohms with 2 D4 subs (it sounds like that is what you have)

How much power does the amp put out in each mode?

Giving the same setup half the power only loses 3db - and you will barely notice that on music.

(rms watts @ 14.4v according to manual - incl. birth sheet stated 1803 @ 1ohm)
1500 @ 1ohm
1000 @ 2ohm
500 @ 4ohm


I think 500 watts split between the two subs would be a quite noticeable change.

1000 watts however vs. 1500 - well, I've already reduced the gains so that I should be putting about 1200 out (~35v, 50Hz 0dB test tone) at full HU volume anyway (found more didn't increase volume really, just audible distortion).


I would have to replace my subs with dual 2 ohm subs (and quite probably a different make/model, since I'm a little disappointed with the Type-R's, despite them being a good value for the price, IMO).

But, I have to consider if I switch subs, getting something that can utilize a little more power and stay running 1ohm.

Efficiency generally gets better as you drive the amplifier harder.

Really?? Cause I assumed the opposite. And when you say "harder" do you just mean more output watts (i.e. volume up), or is the lower ohm load considered "harder" - cause the manual's specs obviously seem to say the opposite if we're talking about ohm load.

 

[qstarin]

Maybe I should ask it this way, too:

Would it be less stressful on my electrical system to run 2 ohms with the gains up more to output 1000w rms or to run 1 ohm with the gain down more to still output 1000w rms?


I'm thinking it would be less stressful (considerably) to run at 2ohms, but I'm not sure if just having the gain dialed back at 1 ohm leaves me in about the same spot?

 

[Heartless]

Heat.

 

[Oliver]

Maybe I should ask it this way, too:

Would it be less stressful on my electrical system to run 2 ohms with the gains up more to output 1000w rms or to run 1 ohm with the gain down more to still output 1000w rms?


I'm thinking it would be less stressful (considerably) to run at 2ohms, but I'm not sure if just having the gain dialed back at 1 ohm leaves me in about the same spot?

Can you go to Basic car audio and Electronics ?

suggested reading for you

 

[falkenbd]

(rms watts @ 14.4v according to manual - incl. birth sheet stated 1803 @ 1ohm)
1500 @ 1ohm
1000 @ 2ohm
500 @ 4ohm


I think 500 watts split between the two subs would be a quite noticeable change.

You can predict the amount you'll lose. For every doubling of power, there is a theoretical 3db difference.

so you'd be off 6 db (but probably less than that due to other factors)

I changed wiring configs on my subs and was surprised... I had 1200 watts @ 1 ohm bridged between the 2, and admittadly the same amp at 4 ohms (500 watts) wasn't much different.

I changed amps all together and was blown away, now my amps are rated at 5 watts @ 4 ohms and I'm getting plenty of bass with "5 watts" per sub (altho the subs are 2 ohms so its more like "10 watts")

Moral of the story, try wiring your subs to 4 ohms and using the same amp and see how you like (IT'S FREE AFTERALL) then decide if you still want to spend the money...

 

[MidnightG35X]

Really?? Cause I assumed the opposite. And when you say "harder" do you just mean more output watts (i.e. volume up), or is the lower ohm load considered "harder" - cause the manual's specs obviously seem to say the opposite if we're talking about ohm load.

I was talking more output watts. Lower ohm loads will lower the efficiency due to the extra heat the amp will be producing due to much higher currents.

Amps produce greater efficiency at higher power levels because the losses of the biasing elements and transistors stay roughly constant (or increase at a slower rate than output power increases), but the output power is still increasing. When you say efficiency=(power out)/(power in) it should become apparent that as output power increases, the power in requirements do not increase at the same rate because the lossy elements have already used the power they need. I know that isn't a great explanation, but I am having a hard time trying to translate why this happens into just a couple sentences.

 

[qstarin]

Can you go to Basic car audio and Electronics ?

suggested reading for you

I get where you're coming from - really. But, I've spent at least a dozen hours reading just that site alone, including 4 straight hours yesterday before I posted this. I also have a fair amount of experience with simple electronics circuit calculations.

It just doesn't seem to be clicking. I'll keep re-reading, though.


But, I mean, its pretty simple - current draw is determined by power output and supply voltage and efficiency percentage, right? I = P/E/Efficiency, where P=1000w & E=whatever voltage the car's electrical system can maintain (12-ish)

I'm only trying to figure out if the listed efficiency specs are really applicable under the actual range of operating conditions?

It seems that a lot of the specs involved with audio equipment are measured under very specific conditions and tend to change a lot in actual operation.


If, for example, the efficiency at 2 ohms is only 85% very near the speakers natural resonant frequency (as an example that could be totally wrong), and drops down significantly as the frequency moves away from that point - then I would know that the difference in current draw from 1 ohm to 2 ohm isn't as great as it appears at first glance - and I would give that factor less importance when I make equipment decisions.


I guess nothing I've learned over on Basic Car Audio and Electronics made it clear to me how to draw my own conclusion on this point.




I mean, if output is 1000w, voltage is 31.6v @ 1 ohm & 44.7 @ 2 ohms, which makes current 31.6A @ 1 ohm & 22.35A @ 2 ohms.


In fact, I notice that for 1, 2, or 4 ohms, at the rated max power, voltage always ends up right around 44v - so I'm supposing this is related to the absolute maximum output voltage the amp can support.

Lowering the ohm load increases the current and therefore power. (I guess I should revisit the equations for how magnetic field strength is affected by voltage and/or current, since I'm not recalling off hand)


I'm gathering from the responses that driving it closer to its max voltage is actually more efficient?


I'm just trying to understand what's involved in balancing all these different things in a setup - and I have to work with what I already have, to a point. Sorry to be such a noob here, but I'm trying to ask reasonably thought-out questions.


I thought the answer would be as simple as, "Because of how class D amps work, that difference in efficiency is likely to be similar across its whole operating range." or "Because of how class D amps work, that efficiency difference likely only exists at max power in a narrow frequency band - outside of that it will be much more similar." Can it not be that simple?

Moral of the story, try wiring your subs to 4 ohms and using the same amp and see how you like (IT'S FREE AFTERALL) then decide if you still want to spend the money...

True. I guess I'm expecting a big difference going all the way to 4 ohm (if I attenuate the sub channel -6dB its certainly very noticeable, but like you say its free and i can test that within a couple minutes.

 

[qstarin]

I was talking more output watts. Lower ohm loads will lower the efficiency due to the extra heat the amp will be producing due to much higher currents.

Amps produce greater efficiency at higher power levels because the losses of the biasing elements and transistors stay roughly constant (or increase at a slower rate than output power increases), but the output power is still increasing. When you say efficiency=(power out)/(power in) it should become apparent that as output power increases, the power in requirements do not increase at the same rate because the lossy elements have already used the power they need. I know that isn't a great explanation, but I am having a hard time trying to translate why this happens into just a couple sentences.

I get you on the second part, thanks. I understand electronics to a fair level (mild hobby when younger, completed half of a computer engineering degree at a major university).

I think you've got the first part a little backwards, though.

Lower ohm loads will lower the efficiency due to the extra heat the amp will be producing due to much higher currents.

Isn't the heat generated because the amp is dissipating more power internally (less efficient)? Rather than the other way around (I mean, I know rising temperature is also going to further effect the operation of lots of the components, but we don't say heat is the cause of inefficiency, rather inefficiency is the cause of heat).

I just want to know if the values they list are likely fair at face value, or like most other specs - only applies in narrow conditions and don't translate to 17% less current draw all the time, as it would seem at first glance.

 

[Oliver]

With a clamp-on ammeter you can determine what is actually happening.

The reality is I will say what I have to to sell things.

If anyone tested it, I'd dispute their ability as far as testing is concerned.

In the real world as things heat up they should be derated [ they will become increasingly less efficient ]

Your comparison is equal to 6 of one and a 1/2 dozen of the other[ meaningless as far as I can determine ].

Just connect it, run it, Check It, and form your opinion

 

[qstarin]

Your comparison is equal to 6 of one and a 1/2 dozen of the other[ meaningless as far as I can determine ].

I suppose that could make it hard to get a good answer.

Just connect it, run it, Check It, and form your opinion

It always comes down to that, doesn't it.

 

[MidnightG35X]

I get you on the second part, thanks. I understand electronics to a fair level (mild hobby when younger, completed half of a computer engineering degree at a major university).

I think you've got the first part a little backwards, though.



Isn't the heat generated because the amp is dissipating more power internally (less efficient)? Rather than the other way around (I mean, I know rising temperature is also going to further effect the operation of lots of the components, but we don't say heat is the cause of inefficiency, rather inefficiency is the cause of heat).

I just want to know if the values they list are likely fair at face value, or like most other specs - only applies in narrow conditions and don't translate to 17% less current draw all the time, as it would seem at first glance.

I DO say heat is the cause of inefficiency. The transistors in the amplifier are forced to provide more current with lower impedance loads. This increased current causes heat. This heat causes the transistors to drop in efficiency. On top of that, other components inside the amp heat up and are not able to conduct as efficiently as they could if they were cooler. This leads to even more loss.

I would also say the numbers published by a manufacturer are best case, under ideal conditions. I'm not sure I've seen a graph from a car audio manufacturer that lists efficiency across frequency with different loads.

 

[MidnightG35X]

I'm only trying to figure out if the listed efficiency specs are really applicable under the actual range of operating conditions?

It seems that a lot of the specs involved with audio equipment are measured under very specific conditions and tend to change a lot in actual operation.

I say they are best case, so I would agree that it changes in operation.

If, for example, the efficiency at 2 ohms is only 85% very near the speakers natural resonant frequency (as an example that could be totally wrong), and drops down significantly as the frequency moves away from that point - then I would know that the difference in current draw from 1 ohm to 2 ohm isn't as great as it appears at first glance - and I would give that factor less importance when I make equipment decisions.

I find a couple things puzzling in that statement. First, (series) resonance implies 0 ohm impedance. You can't have 2 ohm and 0 ohm at a specific frequency. Impedance does change across frequency. The only way to find out exactly what the impedance across frequency is would be to use a network analyzer or impedance analyzer. The difference between 1 ohm and 2 ohm current wise is quite large. The whole argument if an amp is 1/2 ohm, 1 ohm, 2 ohm, whatever stable comes down to how much current the amplifier can supply. I would keep it as one of the factors you look for in equipment decisions

Lowering the ohm load increases the current and therefore power. (I guess I should revisit the equations for how magnetic field strength is affected by voltage and/or current, since I'm not recalling off hand)

Maxwell's equations need not apply here... way too high level of discussion

I thought the answer would be as simple as, "Because of how class D amps work, that difference in efficiency is likely to be similar across its whole operating range." or "Because of how class D amps work, that efficiency difference likely only exists at max power in a narrow frequency band - outside of that it will be much more similar." Can it not be that simple?

Like most things... the answer is not always that simple. There are too many factors to answer that question with a simple yes or no. I say the best answer is if you want high efficiency, go with a class D amp and run it at the highest possible impedance for the power level you want.

 

[qstarin]

Okay, so simply switching from coils and speakers both in parallel (1 ohm nominal) to coils in parallel and speakers in series (4 ohm nominal) is HORRIBLE.

Huge loss in output.


I know that doesn't say a lot, but it says something. What exactly? I'm still digesting that.

 

[br85]

If 500 Watts aren't enough to power 2 subs and still give a decent "pound":

* your subs are horribly inefficient - get new subs (those type R's are 85 dB at 1 volt, so that COULD be a reason)
* your subs are in the back of a sedan or something, and can't properly get sound into the cabin - fix the install to make it work
* you are a basshead, get a 3000 Watt amp and a couple of 15's
* you have the subs turned down too far on your HU or the gains are set too low
* your subs are out of phase (although not likely given that they worked fine before)

Moral of the story: 500W should be PLENTY of power for any sub system if you're into sound quality. Something else must be wrong.