Confusing speaker specifications

Thread Starter

Gump

Joined Jun 7, 2010
57
Hello,

I have bought these speakers for a small project, but I'm struggling with the technical details - particularly how much voltage is actually needed and how much current is drawn.

1) The speakers require 4 AA batteries, but can use instead a 5V power adapter. So, is it more likely that the speakers actually only needs 5 volts rather than the 6 volts provided by the batteries and is probably using a voltage regulator for the batteries?

2) The rated input power is 3 watts, so wouldn't this mean that the current is 0.6 amps (I=P/V = 3/5), if so, why have they stated a DC power adapter of 1 amp?

Thank you.
Gump.
 

MrChips

Joined Oct 2, 2009
19,384
The power drawn from the power supply or battery is not the same as the audio power delivered to the speakers.

Most audio amplifiers do not use a voltage regulator because it is not needed. A voltage regulator would just waste electrical power.

You can get a rough estimate of the delivered power to the speakers by assuming the full battery voltage is applied to the speaker. For this you need to know the impedance of the speaker. Use the RMS voltage.

So if the voltage is 6V and the speaker is 4 ohms, the RMS wattage would be 3 x 3 / 2 / 4 = 1.1W.

Now some manufacturers cheat and quote peak power instead = 3 x 3 / 4 = 2.2W.

And remember to multiply by 2 for two speakers.

The adapter has to exceed this power rating otherwise the signals will be clipped resulting in distortion.
Most adapters for applications such as audio speakers do not require regulated output. A 5V adapter will usually have a no-load output as high as 7 or 8 volts.
 
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Adjuster

Joined Dec 26, 2010
2,148
I would very much doubt that the "6V" battery would require any regulation down to 5V. For a start, over much of its life the battery voltage will be less than 6V, running down to 5V or less before the battery is considered discharged. A practical battery amplifier therefore has to cope with this range of variation.

As for the current drain, no amplifier is 100% efficient, so even on that basis it would be normal to expect the input power required to be greater than the output power.

In addition, it is a good idea to allow for a "safety factor" or margin for error, so that you can be sure that the power supply will not be overloaded, and preferably will run at less than its full rating. This is likely to make it more reliable, and allow it a longer lifespan.
 

Thread Starter

Gump

Joined Jun 7, 2010
57
Hello,

Thanks for the replies.

You can get a rough estimate of the delivered power to the speakers by assuming the full battery voltage is applied to the speaker. For this you need to know the impedance of the speaker. Use the RMS voltage.
Ah I'm sorry, I think I've confused things then. I'm actually struggling with how much current the speakers would pull from the batteries.

Thanks,
Gump.
 

MrChips

Joined Oct 2, 2009
19,384
It depends on how loud you are going to play your music. 2W + 2W is a lot of acoustic power. At normal listening level, 1/10 of that should do, 0.2W + 0.2W.

So I am guessing about 100 - 200mA.
 
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Thread Starter

Gump

Joined Jun 7, 2010
57
It depends on how loud you are going to play your music. 2W + 2W is a lot of acoustic power. At normal listening level, 1/10 of that should do, 0.2W + 0.2W.

So I am guessing about 100 - 200mA.
Hm, perhaps I've ordered speakers that are too big... Is it possible for me to work out the maximum possible current? I've made an attempt:

Current = Power / Voltage
0.5A = 3W / 6V

The 3 watts is the Rated Input Power, and voltage is the full voltage of the batteries.

Thank you.
 

crutschow

Joined Mar 14, 2008
23,508
..................
You can get a rough estimate of the delivered power to the speakers by assuming the full battery voltage is applied to the speaker. For this you need to know the impedance of the speaker. Use the RMS voltage.

So if the voltage is 6V and the speaker is 4 ohms, the RMS wattage would be 3 x 3 / 2 / 4 = 1.1W.
...................
That's for a single-ended output. If the amp has a bridge output then the peak output voltage is 6V (assuming the amp has no voltage drop at the voltage peaks) and the RMS power would be 6 x 6 / 2 / 4 = 4.5W.
 

crutschow

Joined Mar 14, 2008
23,508
Hm, perhaps I've ordered speakers that are too big... Is it possible for me to work out the maximum possible current? I've made an attempt:

Current = Power / Voltage
0.5A = 3W / 6V

The 3 watts is the Rated Input Power, and voltage is the full voltage of the batteries.
If 3 watts is the input power and not the speaker power then your calculations are correct.

The reason they state you should use a 1 amp adapter is likely just as a design margin.
 

Audioguru

Joined Dec 20, 2007
11,251
Will you play continuous tones at full blast? Then you need a continuous input of 3W.

Or do you play normal music that has occasional peaks but averages 1/10th the peak power? Then you need 3W for the duration of the peaks and only 300mW for most of the time.
 
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