Dot Physics

This comes from Buzz Out Loud Episode 865 which got the story from Slashdot regarding a possible new technology that would use piezoelectric devices to charge cell phones while you talk. The original article the slashdot story pointed to talked mostly about the advances in piezoelectric devices, but I want to look at the possibility that sound could charge a phone.

First for the basic physics. How do you make sound and what is it? Sound is a compression wave in the air. To make a sound you need something to push the air (yes, I simplified this quite a bit). When that something pushes the air, it will have a force exerted over a distance. This means it takes energy to make sound. Recall the work definition:

i-de1d6578776c9c582bb4cbb90a56ee82-networkw-23.jpg

The reverse can happen when this sound hits something (like a microphone). The air will push the device and move it – thus work will be done on the device. This leads to a change in energy of the device:

i-9d6dd8738d429b701d5af20bf43a0d2c-work-energy-2.jpg

So, it takes energy to make sound and you can get energy from sound. The best way to look at this is with the intensity. Hyper Physics (basically an online textbook) has a good description of this. The intensity of sound is the power per square meter.

i-699a1a02dacf47afef04011d22bf96ea-powerintensity.jpg

If I have a sound source that makes sound uniformly in all directions, then the farther a receiver is from a source, the lower the intensity. You can think of this sound as an expanding sphere. When the sphere expands the energy in a given amount of time (the power) is “spread” over the whole surface area of the sphere. If the original source has a power output of P, the intensity (I) will vary as the distance with:

i-071470a6b566116edd45b6a8188d4daf-powerdistance-1.jpg

I think that is enough to proceed with the calculation.

The question is: how much power could you get from talking to a phone? Well – how much power could the PHONE get? How much power do you output when talking? The typical value for talking is that normal speech is around 60 decibels. Human ears are kind of awesome in that they don’t not interpret the intensity. If they did, how would your brain comprehend a wide range of intensities. To compensate, our ears (or brain – not sure) work on a log scale such that:

i-6ee5f4b8eb040be9e3634866a0690de0-loudness.jpg

So, I need to convert human speech from human perceived loudness to real power per area.

i-566ac58341dc279e87a72242f0235af2-loud-133.jpg

Actually, I should write this in general terms of L (instead of 60 dB) so it can be more useful.

i-0024b49410cbe5bfc89e733abeaa28ec-intensity-q-3.jpg

Where L is the loudness in decibels. Now I can use this to do some calculations. Remember, I already stated that I assumed the sound from the speaker was uniform in all directions (obviously not true). I will also assume that the piezoelectric can convert 100% of the power from sound to electrical energy. In this calculation, I will use:

  • Speech of loudness L.
  • Piezoelectric device is a square with a width of d.
  • The phone is a distance r from the mouth.

So to calculate this power, let me look at a normal conversation. Wikipedia lists a normal conversation from 40-60 dB at 1 meter away. Clearly, someone would not hold the phone 1 meter away. I want the intensity at r distance away. First, I would find the intensity (see above). That is the power per square meter for a sphere that is 1 meter in radius. The total power would be the same if it was a sphere of radius r, but it would give an intensity of:

i-e5dc9663988e350aa56aaa47f9d77556-intensity343.jpg

Where I1 is the intensity of speaking at 1 meter. If r is less than 1 meter, then the intensity will be greater. The power delivered to the cell phone will be the intensity times the area of the cell phone (d2). Putting this all together, I get:

i-80933f0523b1c7b7dc10ecaacf30c3be-finalpower.jpg

Now, what values do I put in? I will put

  • L = 60 dB (as stated before)
  • d = 2 cm = 0.02 m (really just guessing here).
  • r – 2 cm = 0.02 m (another wild guess).

If I plug these numbers in to the previous formula, this works out nice. The d2 cancels with the r2 and I get:

i-af7642bae36623d8feab22a1df751968-total-watts.jpg

How much power does a phone need? To run, I am not sure. I would guess it would be much greater than 10-6 watts. When transmitting, they use perhaps on the order of 1 watt. Looking at amazon for cell phone batteries – it looks like 1000 mAh is a reasonable guess for the energy stored in a battery. Yes, this would be 1 amp-hour. If I talked into this phone, how long would it take to charge 1 amp hours? If this is a 3.7 volt battery with 1 amp-hour of charge stored, then this would be 3.7 Joules of energy. How long would it take a power supply of 10-6 watts to get this amount of energy?

i-a015b5579c3a53a5e627b8329a06e951-tottime.jpg

That is a long time. Yes, I made some assumptions – but it would STILL be a long time even if some things were changed. Also, this is essentially the same conclusion that they came to on the Slashdot discussion.

Comments

  1. #1 Tom
    December 8, 2008

    There’s a saying that you’d have to yell for about 8.5 years to generate as much energy as it takes to heat up a cup of coffee (8 years, 7 months and 6 days is often quoted). I remember running the numbers, and the claim is reasonably accurate.

  2. #2 at&t refurbished cell phones
    September 28, 2009

    When did I wind up in high school ap calculus. These are some crazy formulas. Some of it made sense, than again I have been out of high school for the past ten years. You obviously kow what your doing and talking about. Great post.

  3. #3 calin
    June 12, 2011

    Well, the time needed to charge the phone will definitely be acceptable if you go on a football stadium (117 dB), or near an airplane taking off (140 dB), or near a baby crying (110 dB) or use an ambulance siren (120 dB), or near fireworks (163 dB) at 3 feet, etc. For example using a shotgun (170 dB) could charge the phone quite fast. But of course there is the problem of fast can the device absorb that power.