Keep AB or Upgrade To D class?

Hi, I have a Crossfire CFA602 and I plan on using a 15" Audiobahn AW151T rated at 600rms bridged@4ohms.  Should I keep this amp or invest in a Class D amp that's rated at the same watts? I'm just not sure if the AB would pull more current than the Class D because of the circuit design.   I may purchase a Power Acoustik A1800D rated at 600rms@4ohms.  Thanks for any info.

Why are you concerned about how much current the amp is "pulling"?

The only difference between the ab and d (aside from power depending on the models rating) is the class D amps are typically dedicated sub amps and mono, so you can actually connect a 2ohm (or 1 ohm again depending on the amps rating) load to it, with the 2 channel you can only connect a 4ohm load to it when bridging. All else being equal I doubt you would hear much difference between the 2.

As far as which will pull more current that depends on the efficiency of the specific amp, your ohms load, it's max output, and how much you crank the volume. Your main concern should be to not overpower the sub and blow it.

If you have flexibilty with your ohms load configuration you should be able to use a stereo or mono amp without issues.

If you are thinking about replacing amplifiers because of their potential demand on the system resources, you should do the appropriate math that will give you the answer you need:

Class A/B is approx. 60% efficient, with a factor of 40% as waste.
Class D is approx. 80% effiicient, with a waste factor of 20%.

Now, plug in your numbers.  You said 600 watts, so we'll use that as maximum RMS power.  If your amp is going to output 600 watts, it needs more than that much from the car's system resourcess because of the waste factor:

Finding the waste factor is a bit tricky for me as I am math-deficient.  We want to know what the draw on the system in watts has to be, in total, that will result in 600 watts after applying the waste factor.  I would first look at 600 watts / 1000 = 60% (apparent because 600 equates with 60%).  I'll have to use some trial-and-error to find the result for the class D amp, though, as 80% isn't as apparent as the 60% with these numbers.

A/B:  1000 * 60% efficiency = 600 watts.
D:  750 * 80% efficiency = 600 watts.
(Anyone who knows how to apply this math without resorting to trial-and-error, please post it.)

Looking at these numbers, now use Ohm's Law to find how that relates to amperage draw from the car's electrical resources.  Here's an Ohm's Law chart for reference.

To use Ohm's Law, if you have two known values you can find the third value.  You are considering power draw so you will use the system's resources as the known voltage.  With the engine running, voltage (E) can be considered to be 14.4 volts.  Power (P) is the demanded power we have found above.

I = P/E 
I = 1000 watts / 14.4
I = 69 amps (current)
This is the determined draw for the class A/B amp.

I = P/ E
I = 750 watts / 14.4
I = 52 amps
This is the determined draw for the class D amp.

You see that at maximum draw, there is a difference of approx. 17 amps needed.

Let's also look at it in typical terms rather than maximum.  At 200 watts normal output rather than 600 watts, the difference in amperage draw (do the math as above) amounts to 7 amps.  You can look at the answer to this problem as being a difference ranging from 7 to 17 amps.

When you are contemplating major changes like this for the reason you stated, you are looking at saving 7 to 17 amps draw in total.  If your car's electrical system can't provide enough for the A/B amp and showing that in dimming lights, etc,, you need to upgrade the electrical system.  See "Big 3" and "h.o. alt".  The difference in power draw when changing to a class D amp is unlikely to fix the problem as that difference in current draw is relatively minor when you consider everything.

You also have to examine the quality of the gear you are thinking about replacing the amp with.  Are the specs known to be accurate with the given manufacturer?  What trade-offs will you incur?  But as you can see, using class D for subs is the smart way to go;  you just have to determine whether switching, for you, is the smart thing to do.

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Build the box so that it performs well in the worst case scenario and, in return, it will reward you at all times.

Steve's answer is what really matters in terms of changing amplifier class.  Thanks, Steve!

That being said, I would NEVER recommend replacing anything by Crossfire with anything from Power Acoustic.  That is a big downgrade in quality.  I would also not recommend purchasing Audiobahn.  There are far better subs out there in the same price range.

The sub is an single voice coil rated at 650rms 4ohms.   I will not be investing in a Audiobahn amp.

Thanks!

Thanks Steve, impressive as usual.

And you took my simple answer and turned it into a math assignment..lol

But I completely agree that no matter if using a class AB, or a class D, if the electrical system is being taxed due to excessive current demands, that issue needs to be addressed first.

Also note upgrading just the alternator is not always nessecary..and at the same time not always enough.

Read the thread on the "big 3" if you haven't already. The wimpy factory wiring/grounding is a prime offender that needs to be addressed once you get into high power systems.

I know thats the first thing I address and it has been working for me forever.

Stevdart - Here is your formula for efficiency(E is efficiency in decimal form)

(RMS * 2) / (E * 2) = Total RMS

You can then apply ohms law to calculate amp draw.

I have a question regarding efficiency.

Can you really be that general in saying an AB is 60% efficient and a D is 80% efficient ?

While researching the topic I was lead to believe they are both fairly inefficient..but better then a class A in comparison.

Wouldn't efficiency be amplifier specific as a rating , rather then generalizing? Or is it that those specs (60% / 80%) can be applied with decent accuracy regardless?

Thanks.

Master5 - Those efficiency numbers can be used with relatively good accuracy, but they do only represent the average and are not specific to all amplifiers. Some can be better or worse depending on the manufacturer.

Thanks vinspo, Thats pretty much what I though.

But using your formula above is not really a formula for efficiency. It is used to determine the total RMS if you know the efficiency to determine if the electrical/charging system/wires are up to task.

The link I posted for the calculator does all the math for you..and then determines the total power drawn from the alternator. However, you still need to know some specifics and the rockfords give the specific info with every amp they sell.

I wonder if anyone knows a way to determine the actual efficiency of a particular amp by taking measurments ?  If so please post.

What I have been doing is simply relying on the power specs of the amp depending on the ohms load. Then once I know the length of the wire I use an IASCA wire chart to determine the wire gauge and current demand. Seems to work for me so far.

Thanks everyone for all the info

BTW..if someone does now how to determine the efficency of a specific amp I do know the formula, it's basic division at the worst. But do you think a simple watt meter used to measure power in / power out would be accurate enough or is there other factors to be taken into consideration.

Thanks again.

Ok, so with this info rather then be too concerned with the specific efficiency...is it safe to "assume" that a class AB rated at the same RMS power of a class D could draw substancially more current?

And should we take this into account when deciding on system demand and wire gauge?

Iasca tells us to assume double the rated RMS power when using the chart to determine wire ga regardless of design to be on the safe side ...but my chart may be a bit dated.

Is there anything I am missing or am I on track?

]Findi wrote:

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g the waste factor is a bit tricky for me as I am math-deficient.  We want to know what the draw on the system in watts has to be, in total, that will result in 600 watts after applying the waste factor...  I'll have to use some trial-and-error to find the result...

A/B:  1000 * 60% efficiency = 600 watts.
D:  750 * 80% efficiency = 600 watts.

(Anyone who knows how to apply this math without resorting to trial-and-error, please post it.)

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vinspo wrote:

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Stevdart - Here is your formula for efficiency(E is efficiency in decimal form)

(RMS * 2) / (E * 2) = Total RMS

You can then apply ohms law to calculate amp draw.

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Thanks, vinspo.  Just what I was looking for.  So for the 600-watt class D amp with the rough estimate of 80% efficiency that I used in my post, the math would look like this:

(600 X 2) / (0.8 X 2)

1200  /  1.6  =  750 watts demanded.

I'll save that formula in my calculator files. 

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Build the box so that it performs well in the worst case scenario and, in return, it will reward you at all times.