4
\$\begingroup\$

I'm planning on a little experiment with a friend. We're gonna be using 2 Xbee series 1 modules at 3.3V and 50mA. One of the Xbee modules will be sealed in a music instrument and I was wondering: is it possible to power it using energy harvesting?

The instrument will be in constant vibration while in use, frequency will be from 80Hz to 350Hz, concerning the amplitude I don't have a clue yet as I'm building it, but it isn't constant for sure as it changes according to the music. The instrument is about the size of a guitare with a slightly different shape,it is made of wood.

The power supply needs to allow a range of action from 5 to 10 meters to allow freedom with the instrument.

The radio can't go to sleep while in use , and needs to reach 250kbps in order to tansmit sound data with a quality near mp3.

What kind of device or electrical circuit could I use to do so? Is it even possible to have 3.3V with energy harvesting ?

\$\endgroup\$
  • \$\begingroup\$ You might be better to start by having a look at some of the low power modes of the module and work out the average current requirement. I suspect getting 165mW continuously over that distance may be a bit more challenging than you think. \$\endgroup\$ – PeterJ Apr 25 '14 at 12:16
  • \$\begingroup\$ When in use the module will be constantly on with a high rate of data transmission, possibly 250kbps. I can't put it to sleep and get it back again. After some investigation it seems you're right about wireless power see my edit \$\endgroup\$ – Morendo Apr 25 '14 at 12:54
  • \$\begingroup\$ You might expose better your setup: maybe there's a solution that does not involve a "piezo generator" but some sort of other energy harvesting system. \$\endgroup\$ – Vladimir Cravero Apr 28 '14 at 11:38
  • \$\begingroup\$ What is the frequency and amplitude of vibration? If you don't know, can you at least describe it qualitatively? \$\endgroup\$ – gwideman Apr 28 '14 at 21:44
  • \$\begingroup\$ see my edits concerning frequency and amplitude \$\endgroup\$ – Morendo Apr 29 '14 at 11:14
4
+50
\$\begingroup\$

I think you'd be hard pressed to get the required 150mW or so, although it isn't impossible.

Looking into piezo-based energy harvesting systems, one product claims about 7mW constant power. I would assume this scales reasonable well though. This is using a 3x7cm bendable piezo element and some electronics to store and convert the energy. There's a lot of info on piezo.com.

Using the device above you'd need 20-25 of these elements. As they flex quite a bit and obviously should not touch each other, you end up needing quite a large box.

If, however, you are able to design the radio so it only transmits at a low duty-cycle and goes into some deep-sleep power saving state if it doesn't transmit, this seems feasible.

Although he isn't using XBees (but RFM12B radios), have a look at the jeelabs.net website - there's some examples on low-power use of microcontroller driven radios.


With the added detail that the frequency of the vibration will be between 80-260Hz, have a look at this device. The specs state it can generate 7mW at 50Hz - they're not cheap though at $600 for a kit. And you'll be needing something in the order of 25 of them.


Two bonus links: Linear has done some research on this (you've probably seen this already), using this sensor. Quote:

energy harvesting can produce about 1mW–10mW, where the active sensor-transmitter combination may need 100mW–250mW.

\$\endgroup\$
  • \$\begingroup\$ What about systems like this one ? linear.com/product/LTC3588-1 it seems they are almost there or am I misreading this ? \$\endgroup\$ – Morendo Apr 28 '14 at 11:48
  • \$\begingroup\$ The power supply isn't the problem - the issue is how much power you get from a piezo device. \$\endgroup\$ – RJR Apr 28 '14 at 12:11
  • 1
    \$\begingroup\$ If the vibration is only in 1-axis and it moves enough it might be possible to use a magnet in a tube with a coil around it (and some energy harvesting IC) instead of a piezo element. \$\endgroup\$ – kerblogglobel Apr 30 '14 at 15:29
4
\$\begingroup\$

You could easily provide far more useful information than you have. If you are serious about getting good answers you can help yourself by giving as complete and accurate a description of the environment as possible subject to secrecy constraints. If secrecy is not an issue then tell us what you are doing.

In the comments you say "5 to 10 meters" but there is no indication that you needed that sort of separation in the question.

Why do you need contactless power transfer?

Sounds like a quad (or other) copter. If so, you could use downwash from the props to drive a "wind turbine". If it is electrically powered why not use the main supply?
If "gas" powered then an alternator (magnets moving past coils) could easily be driven from a motor.

Piezo tends to be hard to get substantial power from BUT a magnet vibrating in a coil can easily meet your needs. Mechanical resonance can be employed to achieve high amplitude (with the system moving into resonance as the driving frequency moves, as long as it does not move too fast.

You can calculate the approximate mass and distance needed from a vibrating energy harvester (regardless of how it is then turned to electrical power) by calculating the energy required to accelerate the mass from rest to peak velocity and back again 2F times per second.

At frequency \$f\$ one cycle takes \$1/f\$ second = \$t\$.

The mass accelerates from rest to \$V\$ peak in \$t/4 = 1/4f\$.

If total stroke is \$d\$ the it accelerates as above over \$d/2\$. For linear velocity ramp (which you will not have) mean velocity is \$distance/time\$.

$$V_{mean} = \dfrac{d/2}{1/4f} = 2fd$$

Peak velocity will be double the mean (for linear ramp) = \$4fd\$.

Energy in an accelerated mass will be

$$\dfrac{1}{2} \times m \times V^2$$ $$ = 8 \times f^2 \times d^2 \times m $$

That's a half cycle so you get two per cycle and f per second so

$$Power = \text{Energy per 1/2 cycle} \times 2f$$ $$Power = 16 \times f^3 \times d^2 \times m$$

E&OE

You can be ABSOLUTELY CERTAIN that this is wrong! Either because of a bad assumption and/or a calculation error or my having missed something. But it gives you a starting point to think on.

The above formula if correct (and it's not) gives the power delivered to a vibrating mass by a linear motion ramp and which is then removed by braking (alternator). Even as I write this I'm raising mental objections to what I've done but it gives you a starting point

Let's plug in some figures "just for fun". Say 200 Hz, 10mm full stroke. What mass do we need to get say 200 mW?

$$Power_{wrong} = 16 \times f^3 \times d^2 \times m$$ $$m = \frac{Power}{16 \times f^3 \times d^2}$$
$$ = \frac{0.200}{16 \times 200^3 \times 0.01^2}$$ $$ = \text{about 16 micro gram (too small)}$$

So there's something wrong above - but is it one of the rough assumptions or a straight error?


Light

150 mW needed.

At 15% overall efficiency you need about 1 Watt of light energy to be converted by a PV cell.

Full sunlight has a power density of \$\cong 1000 W/m^2\$ so using sunlight you'd need

$$1/1000 \, m^2 = 10 cm^2 = 1000 mm^2$$

That's about \$30 \times 30 mm\$.

A modern LED can produce 25% + light output/ DC in.

So a 5 Watt LED focused on whatever area of PV cell you choose to use would produce 150 mW+. Whether that is feasible over the distance concerned depends on factors as yet undisclosed to us.


Inductive power transfer

How large a device is the target? How large may a target coil be?

You can achieve 10's of Watts at 10 metres at around 10% - 20 efficiency with larger coils than you probably want to use.

MIT 2007

enter image description here

Link to image

Image source

Another account here

MIT technology review - A wireless powered lightbulb - part 1 and part 2

Note the distance versus efficiency curve at left in the photo.

enter image description here

Cornell paper

Georgia Tech report


Agilent Wireless power transfer

\$\endgroup\$
1
\$\begingroup\$

Whether what you want to do is feasible (as opposed to possible) depends on several factors including the efficiency of each piezo energy harvester, which is another way of saying, How much energy will you have to expend applying vibrational energy to the transducers in order to produce a given number of mW at the output of the harvester? (Some harvesters also have a narrow frequency bandwith which is outside your proposed range of 80-350 Hz.)

Consider a loudspeaker which is a cone with an armature winding that slides back and forth along a magnetic core. Applying amplified audio to the armature generates a magnetic field which forces the cone back and forth, producing sound waves in the air. A loudspeaker's cone will also move back and forth in response to sound waves, producing a magnetic field and ultimately electrical energy -- in other words a loudspeaker can always be used as a microphone of sorts. A loudspeaker is an inefficient microphone and, while you might get enough audio-frequency electrical power from the loudspeaker to be useful if amplified, this is an inefficient way to acquire audio-frequency electrical power compared to the use of a microphone. Even leaving aside considerations of frequency response and signal-to-noise ratios, it's just the wrong tool for the job.

Attempting to generate power in useful amounts from a piezo-energy harvester may also be an inefficient, or less cost-effective, method of obtaining power. While attempting to produce electrical power piezoelectrically from mechanical vibration doesn't exactly compare to using a loudspeaker as a microphone, it seems to me that the two methods are similarly inefficient when it comes to generating usable, cost-effective amounts of electrical energy.

That is to say, it's easier and more cost-effective to get sound from electricity using a loudspeaker than electricity from sound; and it's easier and more cost-effective to get vibration or other mechanical motion from electricity than vice versa.

Also, one of the app notes (Barbehenn) described a piezo-harvester system that would power a sensor that drew, at most, 60-80 microamps. Your Xbees draw 50 ma, in other words several hundred times as much power.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.