[Xenolith]

Hi I'm new!

Is it possible to damage earphones by playing a square wave on it? I was using test tones for equalization but then someone warned me that playing square waves on speakers can damage it. Personally I think it's ridiculous but then my OCD kicked in.

 

[beerguy0]

Quote:

Originally Posted by Xenolith /img/forum/go_quote.gif Hi I'm new! Is it possible to damage earphones by playing a square wave on it? I was using test tones for equalization but then someone warned me that playing square waves on speakers can damage it. Personally I think it's ridiculous but then my OCD kicked in.

I'm not sure if a square wave would actually damage headphones, but it's probably not a good idea. Square waves are composed of a series of odd harmonics, and hence have a lot of high frequency energy, beyond the audible range. I've modeled square waves in Mathcad, and you need to go out to at least the 19th harmonic to get a decent looking square wave. That means a square wave, at 1 KHz, will have significant energy at, say, 19 KHz and up. A real square wave generator will have energy at frequencies much higher than that. A spectrum analyzer shows this clearly.

Too much HF content can definitely damage tweeters, but I'm not sure about full-range drivers such as those found in headphones. (I once cooked a tweeter with the Mobile Fidelity recording, "The Power and the Majesty". I was listening to the thunderstorm side at low level, and during a hailstorm sequence, a tweeter gave out. Not much power, but enormous HF content).

Sine waves are a much better alternative, as the energy is at a single frequency.

 

[threepointone]

I think there's an easy passive way to convert pwm to sine waves--might want to look it up

 

[tangent]

Quote:

Originally Posted by Xenolith /img/forum/go_quote.gif Is it possible to damage earphones by playing a square wave on it?

I suppose it depends on the fundamental frequency, how loud you play it, and what headphones you use. Theoretically, yes, I believe it's possible. I just don't think it's guaranteed.

If it's not important that you use actual headphones, I recommend building a set of dummy headphones. You can use these for any test that requires the load of a headphone, where you risk damaging real headphones. It's also a better baseline for comparison, since you don't have to worry about the nonlinearities of a particular headphone affecting the test.

Quote:

I was using test tones for equalization

Square waves

for equalization?

beerguy0 is right: a perfect square wave has infinite decaying harmonics out from the fundamental. I can't see how that can possibly be useful for equalization.

If you wanted to do broad-band equalization with just one test signal, use white noise instead. That's equal in voltage at all frequencies, so you can measure what actually gets through to the output and EQ out the differences.

 

[Xenolith]

Thanks for the replies.

I did a bit of my own research and this is what I found out. Please correct me if there are any mistakes in my comprehension.

When I say square waves, I'm referring to square waves recorded in wave or mp3 files.

I don't understand why square waves can be damaging at all. A square wave is composed of many sine waves. By using a steep low pass filter, one may observe the fundamental frequency of the waveform. It is a sine wave and it has the same frequency as the square wave. Also, the power level of the of the fundamental is drastically lowered upon reaching the third overtone.

If square waves are damaging to speakers, then music must also be as damaging.

When a speaker is fully saturated with sound signals, like a transistor going full on when the signal goes to maximum, they cannot be overworked because unlike transistors, you can reach to the volume control to change their loudness and wattage.

It seems that damage from square waves can only happen when you play it at a very very very loud volume. Play any music at that volume and the result will be same. It doesn't matter how the signal is generated.

It looks like the main source of confusion here is how high you have to turn the volume to for any damage to be acquired, since loudness is quite subjective and the level of abuse tolerance varies among speakers or in my case earphones.

Please point out flaws in my understanding. Has this myth been busted?

 

[DaKi][er]

That's it, if it is coming from a digital source, they wont be able to produce frequencies much above 20khz by design, so don't go too loud and you'll be right

 

[tangent]

You may be right, it may be safe. But why are you trying to talk yourself into doing it if you aren't dead certain?

One mitigating factor I should have mentioned: your average audio player isn't going to be able to produce the higher harmonics. This is both because of design limits, and actual design features. Digital players, for instance, often have low-pass filters in them, which would stomp on the higher harmonics. The caution against testing speakers is aimed more against people using dedicated test equipment, which can operate at much higher frequencies.

 

[Xenolith]

Tangent you are right, that is an extremely mitigating factor you just brought up.

I should keep this in mind when I set up my monitor speakers.

Thanks!

 

[GaryJW]

I strongly doubt that a square wave will damage single driver headphones.

Tweeters on speakers can be damaged by a square wave or distortion since any clipped signal contains a significant amount of high frequency component which is then routed to the tweeter by the speaker crossover. Tweeters are not normally rated to take much power and under these conditions can be overdriven/burnt out.

The same mechansim just does not exist in most headphones which have a single driver (with the exception of some of the multi driver IEMS).

 

[cobaltmute]

I may be quite wrong, but doesn't a perfect square wave consist of two states, high and low? And in a perfect wave, aren't those states high and low DC? And doesn't DC make a mess of a driver?

 

[ruZZ.il]

In my limited understanding, all digital players always have low pass filters on them since the sampling itself replicates all present frequencies around harmonics of the sample frequency. Also, your analog chain probably has some high pass filters in it too.
Remember, its not just the single frequencies that are the square wave, its the sum of them. They compliment each other where needed, and subtract where needed. Even though a system can handle all the frequencies, it doesn't often naturally happen at such timing that they create step waves... Anyway, I personally wouldn't want to expose my headphones to what still manages to be rather close to DC much...

 

[GaryJW]

A lot of misunderstanding here. A square wave is not DC. A square wave can be decomposed down a series of sine waves; specifically one at the fundamental frequency and then an infinite series of higher frequencies (the odd harmonics (i.e 3.f, 5.f, 7.f etc) each divided by the harmonic multiplier;

e.g. sin(2.pi.f.t) - sin(6.pi.f.t)/3 + sin(10.pi.f.t)/5 ..........

(t=time, pi=3.14159.., f=frequency)
(check on wikipaedia).

The way the driver is going to behave with this depends on the frequency components, of which none are lower than the base frequency. The reason that DC is (often) bad for drivers is that drivers are reactive (i.e. a substantial portion of the impedance is in reaction to a changing signal) and as such can be very low impedance to DC, leading to high currents and overheating.

To summarise; A square wave will not damage single driver headphones and does not behave as short segments of DC. It is in fact the opposite and behaves as the fundamental + an infinite sum of higher frequency harmonics.

(And just before somebody tries to prove me wrong, the above is correct for all frequencies, but the driver safety part depends on the value of the frequency; i.e. feeding a susceptible driver with a large enough 1Hz signal (whether sine or square wave) will have much the same impact as DC).

Quote:

Originally Posted by cobaltmute /img/forum/go_quote.gif I may be quite wrong, but doesn't a perfect square wave consist of two states, high and low? And in a perfect wave, aren't those states high and low DC? And doesn't DC make a mess of a driver?

 

[cobaltmute]

Okay. I get that a square wave can be theoretically decomposed to a series of sine waves. No issue for me there. I do understand the additive nature.

Where I have a problem is the transition from theory to reality. A clock signal in a DAC is a square wave. It has two states - high and low. The device that generates those two states only generates those two DC levels. There is rise/fall time as switching is not infinitely fast, but each goal of each output state is DC. So you are now telling me that it is actually outputting summed sine waves?

 

[sanderx]

Quote:

Originally Posted by cobaltmute /img/forum/go_quote.gif Okay. I get that a square wave can be theoretically decomposed to a series of sine waves. No issue for me there. I do understand the additive nature. Where I have a problem is the transition from theory to reality. A clock signal in a DAC is a square wave. It has two states - high and low. The device that generates those two states only generates those two DC levels. There is rise/fall time as switching is not infinitely fast, but each goal of each output state is DC. So you are now telling me that it is actually outputting summed sine waves?

Outputing more than one level of "dc" really means outputing ac - as soon as the "signal" varies over time, you have ac. Look at it this way - batteries only seem to output dc, what they really output is ac with frequency on the order of microherzes to nanohertzes depending current draw. Because the voltage gradualy changes. never mind that they (be design) usually never complete a single cycle

In comparison, your dac clock changes the voltage many millions of time per second and hence outputs very high frequency ac.

 

[cobaltmute]

Quote:

Originally Posted by sanderx /img/forum/go_quote.gif Outputing more than one level of "dc" really means outputing ac - as soon as the "signal" varies over time, you have ac.

Ok that I can get.

So let's take at 1Hz square wave that goes from 0V to 1V. For ~0.5 seconds, it will be outputting 1V DC. In reality, there will be a brief rise and fall time around the 1V state. Taken over time this will then be an AC signal modeled by that formula.

Now let's take a look at that effect on a driver. To my understanding, DC is bad for drivers as it tends to heat voice coils. The issue is that when the voice coils gets too hot, the glue holding the coil wires in place melts and wires can short.

So during the "on" state of the square wave, we are applying DC to the voice-coil. This generates heat and a rise in temperature of the coil. We then transition to the off state. This provides a cooling period and therefore decrease in temperature for the coil. If the amount of heat generated during the on period is greater than the amount of heat dissipated during the off period, we end up with a situation on a long enough term that will be an issue as the eventual thermal rise will cause the aforementioned voice coil issues.

Of course the amount of the heating effect is dependent on voice coil design as well as the amplitude, frequency and duty cycle of the square wave. Speakers designed to handle low frequencies tend to have more robust voice coils. (e.g. car subwoofers) than those designed to handle high frequencies (i.e. tweeters). In fact, the heating effect is why they put cooling fluid into tweeters.

So a square wave can damage a speaker. It does need to have a certain set of characteristics to do so however. In fact isn't that what a clipped signal is - a square wave with a really "bad" rise and fall time?