Maximum Sound out of a Speaker

Question

I am new to electronics, so please excuse the noob question.

I have an 8 ohm 1/4W speaker. I would like to get the loudest sound out of it that I can. Based on my understanding, which may be entirely wrong, the maximum current that can go through it without damaging it is

I = Square root ( P / R ) = sqr ( 0.25W / 8ohms ) = 0.176A.

To get this current at 5V, I need an over all resistance of

R = V / I = 5V / 0.176A = 28.3ohms.

Since the speaker has 8ohms, I need a resistor in series of 28.3 - 8 = 20ohms. I have a 22ohm resistor that I am using.

Is this correct?

Answer 1

You want to put 250 mW into a 8 Ω speaker. You don't get a choice of either voltage or current. The Watts into a resistor is:

  W = A² Ω = V²/Ω

where A is the current in Amperes and V the voltage.

Flipping these around, you can solve for the voltage and the current:

  A = sqrt(W/Ω) = sqrt(250mW / 8Ω) = 177 mA

  V = sqrt(W*Ω) = sqrt(250mW * 8Ω) = 1.41 V

As a sanity check the voltage and current should be related by Ohm's law:

  Ω = V / A = (1.41 V)/(177 mA) = 8.0 Ω

Good, that checks out.

You seem to be asking what resistance you should put in series with this speaker if you have a 5 V audio source. The calculations you show are correct in themselves, but probably not what you actually want.

They are correct if you truly have a 5 V RMS source. Note that such a source would have peaks of ±7.1 V, for a peak-peak voltage of about 14 V. And that's just what the peaks need to be for a sine wave. Audio contains occasional peaks significantly higher than the average. To properly reproduce the sound, this amplifier would need to be capable of at least ±30 V, although even that's quite slim.

If you really only have a 5 V circuit producing this audio output thru a capacitor to remove the DC, then you actually have 5 Vpp, which is 2.5 Vp, which is 1.77 V RMS for sines. That's only slightly more than the 1.4 V RMS to get the 250 mW into your speaker. Most likely the 5 V circuit can't actually drive all the way to both rails effectively. Your speaker should be fine driven from this circuit.

Then the real issue is that putting a resistor in series with a speaker is not a good idea. It will change the frequency response since the speaker was designed to be driven from a low impedance.

All in all, don't worry about this. Just turn the volume down. If the sound level gets beyond what the speaker can handle, you will hear it clipping and creating other artifacts. Your speaker isn't going to magically blow up at 260 mW. The 250 mW spec is most likely the limit where the distortion specs are met, not a absolute maximum rating where the speaker will be damaged.

You are over thinking this. Just turn down the volume.

Answer 2

I assume you want to put something in series with the LSP besides that resistor? Otherwise you won't hear anything except the initial click. That anything will cause an additional voltage drop, affecting you calculation.

That anything should switch the current through the speaker on and off. Assuming 50% on, 50% off, that means that you can double the current during the on period.

This kind of current/no current driving is not the most friendly way to treat a speaker, but for making a lot of noise you can ignore that.

So assuming a 50% duty cycle, and 0.6V drop in the switching element, the peak current should be sqrt( 0.5 W / 8 Ohm ) = 0.25 A, the total resistance 4.4 V / 0.25 A = 16 Ohm, which leaves a series resistor of 8 Ohm.

I'd advise you to unclude a bleeder diode parallel to the speaker.