This amp breaks physics?! (Solved)

[PixelGeek]

   In my truck, (yes, not home theatre, but this is the closest sub-forum) I use a Sound Storm EV2.1200 Evolution. This amplifier claims to output 1200 watts of pure power when bridged, however, there is a 25 amp fuse on it.

 

   I'm not an electrical engineer, but the fuse should blow at only 300 watts. 

12v * 25A = 300w

And I'm pretty sure 300w is considerably less than 1200w. 

 

So, how can it output 1200 watts? 

[TH3R34P3R]

It could be performing some sort of electrickery as in where its doing something internally to make it be able to live up to the 1200W name without ever really reaching 1200W. 

 

[PixelGeek]

4 minutes ago, TH3R34P3R said:

It could be performing some sort of electrickery as in where its doing something internally to make it be able to live up to the 1200W name without ever really reaching 1200W. 

 

   I like the term "electrickery", but there isn't really too much you can do. 1200 watts is 1200 watts. If you put in an amount of power, the circuit can only ever spit out that much power back. What you put in is what you get out. Mayhaps, the fuse is on the output end of the circuit. (I can't find a wiring diagram (and I don't want to open the amp)) If the fuse is on the amplifier output, and the amp is putting something like 48v to the sub, then a 25A fuse would work. 

 

48v * 25A = 1200w.

 

But, I do not think it's putting 48v to the speaker. I could be totally wrong, though. 

[Dedayog]

20 minutes ago, PixelGeek said:

   I like the term "electrickery", but there isn't really too much you can do. 1200 watts is 1200 watts. If you put in an amount of power, the circuit can only ever spit out that much power back. What you put in is what you get out. Mayhaps, the fuse is on the output end of the circuit. (I can't find a wiring diagram (and I don't want to open the amp)) If the fuse is on the amplifier output, and the amp is putting something like 48v to the sub, then a 25A fuse would work. 

 

48v * 25A = 1200w.

 

But, I do not think it's putting 48v to the speaker. I could be totally wrong, though. 

The fuse should be on the input side of the amp, so that it doesn't get shorted by too much juice coming in.  The output doesn't have anything to do with the input side.  It's how amps work, they boost the input by the gain they're built to produce.

[TheFlyingTraut]

Because it's a cheap amp with box marketing. Not a chance it hits anywhere close to that.

[keskparane]

Also it's designed to have a 35A fuse https://www.manualslib.com/manual/587584/Ssl-Ev-2-1000.html?page=8#manual

[PixelGeek]

3 hours ago, jstudrawa said:

The fuse should be on the input side of the amp, so that it doesn't get shorted by too much juice coming in.  The output doesn't have anything to do with the input side.  It's how amps work, they boost the input by the gain they're built to produce.

The amp is externally fused, 100amps on the input,  and it says on the external wiring diagram to externally fuse it. So, I'm doubting that the included fuse connects to the input. 

[PixelGeek]

3 hours ago, keskparane said:

Also it's designed to have a 35A fuse https://www.manualslib.com/manual/587584/Ssl-Ev-2-1000.html?page=8#manual

Reguardless, 35A is a far cry from the 100A I would expect it to have. 

[PixelGeek]

3 hours ago, TheFlyingTraut said:

Because it's a cheap amp with box marketing. Not a chance it hits anywhere close to that.

Possible. I need to put a Multimeter on it. 

[BabaGanuche]

5 hours ago, PixelGeek said:

So, how can it output 1200 watts? 

The short answer it is cannot. The longer answer it can but only for very short periods of time, and assuming the average power is under the rating for the fuse.

 

The fact the people like big number means they will stick the larger instantaneous number, rather than the real continuous number.

[TheFlyingTraut]

No it is a cheap amp. It won't output 1200 rms. guaranteed. It probably puts out 120rms. Maybe. 

[tikker]

Is it this one? https://www.soundstormlab.com/product/car-audio-sound-system-auto-2-channel-amplifier-sound-storm-ev2-1200/

Quote

1200 W MAX Power 2 Channel
450 W X 2 RMS @ 2 ohm
225 W X 2 RMS @ 4 ohm
900 W X 1 RMS Bridged @ 4 ohm

Max power is different form RMS power, the latter being what it may normally output sustained while the former is what it could maximally handle for a split second.

[keskparane]

Using a different Ohm's Law

I = (W / R)₂

I = (1200 W / 2 ohms)₂ = 24.49 A

[PixelGeek]

9 hours ago, keskparane said:

Using a different Ohm's Law

I = (W / R)₂

I = (1200 W / 2 ohms)₂ = 24.49 A

 I have never seen this form of Ohm's law. Thank you, this was driving me 'round the bend. 

[PixelGeek]

13 hours ago, tikker said:

Is it this one? https://www.soundstormlab.com/product/car-audio-sound-system-auto-2-channel-amplifier-sound-storm-ev2-1200/

Max power is different form RMS power, the latter being what it may normally output sustained while the former is what it could maximally handle for a split second.

@keskparane figured it out. While yes, RMS is different from Maximum output. A 25A fuse would still limit me to 300w. Still considerably less than the 450w minimum RMS listed, according to the equation I was familiar with. I was not accounting for the 4 ohm resistance of the speaker bringing my required apms down considerably. 

[PixelGeek]

18 hours ago, BabaGanuche said:

The short answer it is cannot. The longer answer it can but only for very short periods of time, and assuming the average power is under the rating for the fuse.

 

The fact the people like big number means they will stick the larger instantaneous number, rather than the real continuous number.

Thanks for the reply. @keskparane figured it out. I was not accounting for the resistance in my equation. The math works out when I do. 

[tikker]

9 hours ago, PixelGeek said:

 I have never seen this form of Ohm's law. Thank you, this was driving me 'round the bend. 

Using P=U*I for power, R for resistance, U for voltage and I for current it's a simple rewrite:

Ohms law: U / I = R

=> I = U / R

=> I*I = (U * I) / R

=> I = sqrt( P / R)

 

9 hours ago, PixelGeek said:

I was not accounting for the 4 ohm resistance

Yep, as much as wel like ideal circuits in theory, there's (practically, apart from some exotic things) always resistance in real life