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I need to make a speaker that is capable of creating several tones for a school project, so nothing too big nor expensive.

I'm using a micro controller sending a frequency to a transistor and the transistor is connected to 12v and the speaker.

The sound level must be a minimum of 80 dba in a 2 meter radius. I could only achieve 70 dba with a 3W 8 ohm speaker salvaged from an old radio.

How can I make it louder? Should I change the type of speaker?

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    \$\begingroup\$ What makes you think that with 3W you can get arbitrarily high output volumes? \$\endgroup\$ – PlasmaHH Oct 3 '17 at 20:16
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    \$\begingroup\$ "... with a 3 W 8 ohms speaker savaged from a old radio." I'm trying to picture you savaging a radio. I think you mean "salvaged". \$\endgroup\$ – Transistor Oct 3 '17 at 20:27
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    \$\begingroup\$ @Transistor I mean, depending on how the OP took the radio apart, both terms might be applicable. \$\endgroup\$ – duskwuff -inactive- Oct 3 '17 at 20:29
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    \$\begingroup\$ (1) Add in a schematic of your amplifier. There is an easy-to-use app built into the editor toolbar. (2) If it doesn't have to radiate in every direction you can add a "horn" to your loudspeaker. This will help improve the volume in front of the horn. \$\endgroup\$ – Transistor Oct 3 '17 at 20:30
  • \$\begingroup\$ Easiest way I know of is to get a cheap powered speaker as is commonly used for PCs. \$\endgroup\$ – WhatRoughBeast Oct 4 '17 at 1:32
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A horn can increase the loudness remarkably. It does it in 2 ways:

  • better acoustical matching between the air and the speaker cone increases the power of the soundwave

  • the directivity concentrates the acoustic power to smaller sector. The in-beam dB levels can be much higher than the average or off-beam levels at the same distance.

The horn is that effective only if it's designed for quite narrow frequency range. A single frequency horn can be much more effective than one designed for example reproducing speech understandably.

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  • \$\begingroup\$ Since they mention "80 dba in a 2 meter radius", I'm thinking that they want it to have that loudness in every direction. Hard to tell without further information but OP should at least consider that. \$\endgroup\$ – pipe Oct 3 '17 at 21:36
  • \$\begingroup\$ @pipe This is both omnidirectional, or perhaps radial, but for normal use your listeners are standing on the ground, and a horn: duevel.com/Schnell/BLSchnitt.jpg \$\endgroup\$ – winny Nov 5 '17 at 22:33
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Nice to see that you actually have a clear, objective definition of "louder"!

From an electrical engineering point of view, loudspeakers have a frequency response which says basically in what frequencies the device performs better. Just increasing volume isn't going to help that much because you might be just on a situation that the speaker is not only distorting the audio signal but attenuating it.

Another trick is to match the impedances of the output circuit (whatever it is) with the input impedance of the speaker. This increases the power that can be transfered from the circuit to the speaker. It is really hard to help without having some more detail but that's what I can tell for now.

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    \$\begingroup\$ You can't just blindly apply the maximum power transfer theorem to everything you see. Think about what happens if you have an amplifier with a maximum peak-to-peak output voltage driving an 8 ohm speaker, and now connect an 8 ohm resistor in series. \$\endgroup\$ – pipe Oct 3 '17 at 21:30
  • \$\begingroup\$ Agreed. Anyway, the Theorem still holds (it is a Theorem after all). The 8ohm resistor in series'd form a resistence divisor with the input impedance of this particular speaker. But this is a new circuit, for which the load impedance should be matched again. Thinking more carefully, even if the input power of the load will drop (by a factor related to the resistance divisor), the matching impedance would still be the same, right? (these are legitimate questions, really). Is this your point: the series resistor would break the theorem even though it doesn't change the matching impedance? \$\endgroup\$ – daniel.franzini Oct 3 '17 at 22:56
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    \$\begingroup\$ @daniel.franzini pipe is telling you to imagine a 0.0001-ohm amplifier connected to a 8-ohm speaker, and then to imagine adding a resistor to make it an 8-ohm supply connected to an 8-ohm speaker. In the second case you get less power transfer. The MPTT only tells you that if you have a 0.0001-ohm amplifier then having a speaker lower than 0.0001 ohms will decrease your power transfer, and if you have an 8-ohm amplifier then having a speaker lower than 8 ohms will decrease your power transfer. \$\endgroup\$ – user253751 Oct 3 '17 at 23:28

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