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Discussion Starter · #1 ·
hi i was wondering if i can use a 200W rms Amp for a 100W rms subwoofer. amp is Kenwood KAC-6104D 600 Watt Max Power Class D Mono Power Amplifier, and the subwoofer is Kicker 10C84. Thanks
 

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technically you could use any amp for any speaker, but if under/over powered you run the risk of ruining/blowing it. that said, if u do hook it up to that amp make sure u dont turn the gains up to high.
 

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I don't think there will be a problem, the sub should still only draw about 100 watts because it has the same resistance regardless, so you just wouldn't be using your amp's full potential.
 

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but if under/over powered you run the risk of ruining/blowing it. that said, if u do hook it up to that amp make sure u dont turn the gains up to high.
Not exactly.

Underpowering doesn't blow speakers, clipping does.

Same goes for overpowering..esp subs

It would take a lot more than rated power to toast a sub.

Unless it's some 2000w 59 dollar one from ebay.
 

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Even if your amp is rated at 5000 watts, the case is the same.

Power = (Voltage)^2 / (Resistance)

Because the resistance of your sub will always be relatively constant, it will dissipate the same power as long as it has ~12 volts running through it. The amp's higher power rating means it can pump out about 17 amps at about 12 volts (ignoring the math behind taking the root mean square). Just because the amp can dish out 17 amps at 12 volts, doesn't mean the sub will draw 17 amps at 12 volts. In other words, the sub chooses the current, not the amp, make sense?
 

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Makes sense

Speakers and subs rarely if ever see full power anyway since music is dynamic.

Now if OP was playing a 20hz test tone at full tilt, he'll toast the sub.
 

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Even if your amp is rated at 5000 watts, the case is the same.

Power = (Voltage)^2 / (Resistance)

Because the resistance of your sub will always be relatively constant, it will dissipate the same power as long as it has ~12 volts running through it. The amp's higher power rating means it can pump out about 17 amps at about 12 volts (ignoring the math behind taking the root mean square). Just because the amp can dish out 17 amps at 12 volts, doesn't mean the sub will draw 17 amps at 12 volts. In other words, the sub chooses the current, not the amp, make sense?
Wrong. The resistance is the OHMs of the subwoofer, not it's RMS power.

OP - You will be fine with 200W on a 100W RMS sub. Just don't crank the gain up.
 

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Wrong. The resistance is the OHMs of the subwoofer, not it's RMS power.

OP - You will be fine with 200W on a 100W RMS sub. Just don't crank the gain up.
You obviously didn't read the post correctly.

I stated that the amp's power rating meant it could pump out 17 amps at 12 volts.

17(or 16.66)A * 12v= 200w (note: FOR THE AMP)

I ALSO stated that just because the AMP can put out 17 amps, doesn't mean the subwoofer will draw the same current. The resistance of the sub will determine the amount of current, not the subwoofer.


I may have not made my point clearly, but at NO POINT did I equate RESISTANCE with POWER.


Please read a post before you just flag something as wrong. Some may find it rude and take it as a sign of your ignorance.
 

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I don't think there will be a problem, the sub should still only draw about 100 watts because it has the same resistance regardless, so you just wouldn't be using your amp's full potential.
Even if your amp is rated at 5000 watts, the case is the same.

Power = (Voltage)^2 / (Resistance)

Because the resistance of your sub will always be relatively constant, it will dissipate the same power as long as it has ~12 volts running through it. The amp's higher power rating means it can pump out about 17 amps at about 12 volts (ignoring the math behind taking the root mean square). Just because the amp can dish out 17 amps at 12 volts, doesn't mean the sub will draw 17 amps at 12 volts. In other words, the sub chooses the current, not the amp, make sense?
This stuff is so wrong, I'm not sure I would even know where to start :lmao:
 

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Discussion Starter · #11 ·
You guys have been extremely helpful. I'm going to be installing my first subwoofer in my car, thanks again. :D
 

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well, P = V^2/R is right ... the rest is way out there

Sub impedance (AC resistance) varies WILDLY with frequency from 2-3 ohms up to 50 ohms or more in the 20-80 Hz range of a sub. The box can help substantially to flatten that out. The amount of power the amp can put out is NOT a simple function of the speakers impedance - otherwise EVERY amp would put out the same power to a 4 ohm speaker.

You seem to think that the amp puts out 12V because it is connected to a 12V power system (the car's). This is nowhere near what is happening. The 12V and ground from the cars electrical system goes thru a DC-DC power convertor that makes rails of +/- voltage of 30V or 45V or whatever the design calls for. The transistors take the relatively small input from the headunit and amplify the signal up to a max of the level of the rails. So if you have 40V rails = 80V peak to peak = appx 27V RMS. At full output into a nominal 4 ohm speaker the amp will make 27^2/4 or 182 watts RMS. The sub at this point is a passive particiapant, it accepts whatever the amp places accross it voice coils. The amount of power the amp puts out is directly proportional to the signal level at the input of the amp. Think of the example of the FET transistor, it is like a section of hose pipe that you have a pair of pliers on. The water comes from the source side and flows out thru the drain side - the amount of water coming out of the drain side is completely dependent on how hard you squeeze the pliers. Having a varyinging or pulsing water output means you are squeezing and letting up on the pliers in a rythmic motion.

I hope I didn't lose too many people on this explanation.

P.S. If you actually ran the 12V DC thru a speaker, you would fry it immediately
 

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PaulD is correct, this is wrong on so many levels.

Though you have the correct formula, you grossly misapplied the meaning.

Audio signals are AC signals, not DC. And they typcially swing plus/minus voltage relative to ground. In the old days before high power amps, they had to bias the speaker up to ~6 volts so the AC signal would swing from 0 - 12 volts. That's why there used to be a turn-on thump when the radio first turned on.

Anyway, to address the 200 watt question here:

Assume the speaker load is nominally 4 ohms. Speaker impedance is not "nearly constant". It varies as a function of frequency. The 'resistance' is just the real portion of the complex impedance. The imaginary (or reactive) portion (either capacitive or inductive) is what varies as a function of frequency.

To create a 'clean' 200 watt signal into 4 ohm impedance.

V^2 = power * impedance

So V = sqrt (200 * 4) = +/- 28 volts.

The amp must create a +/- 28 volt RMS (AC) signal to the speaker. Also note this is RMS voltage, the peak-to-peak requirements are factor of ~1.414 higher (sqrt(2)), so a peak-to-peak voltage of ~40 volts is required. And depending on how much headroom the amplifier designer wanted, he/she may want another 30% on the internal power supply rails of the amp. So now we're up to +/ 52 volts.

So how do you create a +/- 52 volt power supply with a single 12 volt input? It's called a switching power supply. This is needed because you can not transform DC into a higher (or lower for that matter) voltage. You need to convert the incoming DC into an AC signal, transform that, then filter that to create a dual voltage power supply (same principle as your cars voltage coil to create the spark on your spark plugs).

It's impossible to address 4 years of engineering classes into a short forum reply, so some shortcuts had to me made... :)
 

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PaulD is correct, this is wrong on so many levels.

Though you have the correct formula, you grossly misapplied the meaning.

Audio signals are AC signals, not DC. And they typcially swing plus/minus voltage relative to ground. In the old days before high power amps, they had to bias the speaker up to ~6 volts so the AC signal would swing from 0 - 12 volts. That's why there used to be a turn-on thump when the radio first turned on.

Anyway, to address the 200 watt question here:

Assume the speaker load is nominally 4 ohms. Speaker impedance is not "nearly constant". It varies as a function of frequency. The 'resistance' is just the real portion of the complex impedance. The imaginary (or reactive) portion (either capacitive or inductive) is what varies as a function of frequency.

To create a 'clean' 200 watt signal into 4 ohm impedance.

V^2 = power * impedance

So V = sqrt (200 * 4) = +/- 28 volts.

The amp must create a +/- 28 volt RMS (AC) signal to the speaker. Also note this is RMS voltage, the peak-to-peak requirements are factor of ~1.414 higher (sqrt(2)), so a peak-to-peak voltage of ~40 volts is required. And depending on how much headroom the amplifier designer wanted, he/she may want another 30% on the internal power supply rails of the amp. So now we're up to +/ 52 volts.

So how do you create a +/- 52 volt power supply with a single 12 volt input? It's called a switching power supply. This is needed because you can not transform DC into a higher (or lower for that matter) voltage. You need to convert the incoming DC into an AC signal, transform that, then filter that to create a dual voltage power supply (same principle as your cars voltage coil to create the spark on your spark plugs).

It's impossible to address 4 years of engineering classes into a short forum reply, so some shortcuts had to me made... :)
First off, I was avoiding considering the subwoofer as a phasor to simplify my efforts of making a simple analogy.

Second, I wasn't aware that A class D amp had a switching power supply to up the voltage. In all the diagrams I've seen, the signal is turned into a PWM signal, then sent to MOSFETS with a constant input voltage (which I assumed to be 12v) before a passive low-pass. It was the assumption of a constant 12v driving voltage (pk-pk) that led me to assume the impedence of the load determined the power consumption. Most of what I know deals with small scale (um) class A/AB amplifiers with a certain Vcc controlling most of the circuit, hence the root of my misunderstanding.

I apologize for coming off terse, I get really tired of people who simply say "you're wrong" without a justification.I really appreciate your input, and in an effort to not hijack the thread please PM me so I might learn a bit more about what's going on. Feel free to include the 4 years of classes (and references to texts).
 

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keeping it in PM's will rob other people of being informed ... they might have the same questions or misinterpretations. A class D amp just uses PWM in the output devices. This is a pretty good article on audio amps http://en.wikipedia.org/wiki/Electronic_amplifier
 

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keeping it in PM's will rob other people of being informed ... they might have the same questions or misinterpretations. A class D amp just uses PWM in the output devices. This is a pretty good article on audio amps http://en.wikipedia.org/wiki/Electronic_amplifier
Yes, but nowhere in there does it talk about shifting the peak-peak voltage to achieve a 200W output, which was the source of the misinterpretation.
 

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I apologize if I came across a bit flippant. I didn't mean to, so for that, I'm sorry.

I don't want to highjack this thread either, so I'll try and keep this short.

I was typing in my original response when I noticed PaulD also replied. So in essence, we independently came to the same conclusions.

"I don't think there will be a problem, the sub should still only draw about 100 watts because it has the same resistance regardless"
Subs/speakers don't choose how much power to draw, they have to absorb whatever the amp puts out.


"Because the resistance of your sub will always be relatively constant..."
As noted before, this is not correct.


"it will dissipate the same power as long as it has ~12 volts running through it. The amp's higher power rating means it can pump out about 17 amps at about 12 volts (ignoring the math behind taking the root mean square). Just because the amp can dish out 17 amps at 12 volts, doesn't mean the sub will draw 17 amps at 12 volts. In other words, the sub chooses the current, not the amp, make sense?"
As noted before, this is not correct.


"I apologize for coming off terse, I get really tired of people who simply say "you're wrong" without a justification."
I guess I'm not sure what you meant. People read these threads to get information. If something is clearly wrong, then I think it is justified to correct any wrong/misleading information.
 

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Yes, but nowhere in there does it talk about shifting the peak-peak voltage to achieve a 200W output, which was the source of the misinterpretation.
The psuedo schematic in the link is a single ended example (0V to V+). From previous replies, to have 200W RMS across a 4 ohm load, the voltage swing need to be ~80 V (+/- 28 V RMS, +/- 40 V pk-to-pk, or 80 V total). So V+ had to be at least 80V in this trivial example.

The output will need to biased up to approx V/2. And a DC bias on speakers is bad, so they have the DC blocking capacitor on the output.

A more common approach is to create a dual supply, +/- V. That way the "midpoint" can be 0V (ground) and you don't need the output capacitor any more.
 

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I guess I'm not sure what you meant. People read these threads to get information. If something is clearly wrong, then I think it is justified to correct any wrong/misleading information.
Exactly, you never actually corrected anything until much later, you just said it was wrong. People, including myself, are looking for *information* not just "you're wrong." When you show *why* something is wrong like in your later posts, then cool, I learned something. If you just say "you're wrong" without any justification, it does absolutely nothing.
 
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