Power draw question for JBL PRX 612

[Telecruiser]

I contacted JBL and was told that this speaker draws 510w (4.25A). I am assuming that this is when the two class D's are at their top output of 1000w combined. How much does this vary as the power output (volume) is reduced or do the class D amps draw the same power no matter what the volume.

[dboomer]

Amplifiers don't make power, they just redistribute it. So when (and if) it puts out 1000w it will have to draw more than 1000w (probably 1500w if it is class D). You also have to remember that when it puts out 1000w it will only do that for a fraction of a second at a time. And when it is putting out 1000w peaks it is probably only putting out 125w of average power.

[Telecruiser]

So what is the power consumption figure that JBL gave me based on? Is it a minimum?

[dboomer]

Usually it's based on operation at 1/8 of the continuous power. That's usually about the point when the limiters start kicking in.

[Telecruiser]

Thanks. I'll give JBL another call.

[the_big_e]

What I did is get a short extension lead (a 6 inch one, comes in packs of 3, from walmart). split the cable into it's separate leads (live, neutral,earth) then used a clamp on amp meter on one of the leads while driving the amp. Not dead accurate but it will give you an idea of the current consumption at various volume levels.

If you are doing this to evaluate your needs for your battery/inverter setup, it was my experience that the end result was way off. EG. 4 amps at 120 v = 480 watts so in theory to get 480 whats out of the inverter, you'd need at least 480 watts going in. The input is 12v so that means a current of 40 amps.

If you assume that's peak than at the sort of volume you'll probably be using, let's assume a quarter of that so 120 watts out = 120 watts in = 10 amps in (at 12 volts). I never see anything like 10 amps going out of my batteries, more like 4 or 5 and that's including powering my mixer (large brick transformer) and pedals (5 wall warts on a strip inside my pedalboard).

 

I'm not an electrical engineer but I assume the disparity is because sine wave amps are not the same as DC amps (from the battery).

The end result as far as I am concerned though is that I ended up using far less from the battery than the 'calculations' and 'research' said I would.

 

That's just my experience, your mileage may vary.

 

[wesg]

I'm not an EE either. But AFAIK "sine wave amps" has no meaning. "Sine wave" refers to the shape of the voltage waveform. Amps refers to the rate of electron flow, without regard to voltage potential difference. I prefer to think of power consumption in terms of power, which is measured in Watts. In DC, P=IV (power equals current times volts). In AC, P=PF(IV) (power equals power factor times current times volts). I think you can safely assume PF is very close to 1.0 for this type of equipment.

 

So, your results are correct, multiply/divide by 10 when changing between 12VDC and 120VAC systems under these circumstances.

 

The reason you don't see the maximum possible current draw is because you're not drawing the maximum possible load for the sampling time of whatever device you are using to measure. Try generating a max-volume 60Hz tone for several seconds, that should peg your needle.

 

A good rule of thumb is that music through your PA system will use about 1/8th of the rated power times the inefficiency of the amp at full volume. For the purposes of napkin math, I figure most modern gear is about 75% efficient. The key word here is music. White noise with lots of LF would be another story entirely.