My thesis title is "Powerbank with sound to eletrical energy converter" meaning I'll be using sound to charge the powerbank, and I decide to used dynamic microphone instead of condenser microphone since dynamic microphone dont need any external voltage in order to work unlike condenser microphone.

I tried to measure the voltage output of the Dynamic Microphone under oscilloscope, every time I make some noise the reading gives 10-30mVAC.

What amplifier should I use in order to amplify the induce voltage coming fron the dynamic microphone which is (10-30mV) into 5V enough to charge a battery?

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    \$\begingroup\$ You're not going to get enough energy from a dynamic mic to charge a battery. Even if you could boost the voltage, maybe with a transformer, the resulting power would not even be enough to power the charger control circuitry let alone charge a power bank battery. \$\endgroup\$ – John D Nov 30 '17 at 23:07
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    \$\begingroup\$ 1 Pascal is getting close to a nearby jack-hammer. If you can work out the area of your microphone's diaphragm and the likely average stroke distance the diaphragm might make, you could work out the Joules per stroke. Multiply this by the frequency of motion and you have the theoretical watts available. Even with a jack-hammer, it won't be a large number. However, if you are serious about experimenting, I'd recommend looking at Energy Harvesting, to get a leg up on it. \$\endgroup\$ – jonk Nov 30 '17 at 23:23
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    \$\begingroup\$ What is the knowledge area in which you'll try to graduate? \$\endgroup\$ – mguima Dec 1 '17 at 0:38
  • \$\begingroup\$ I think your thesis is finished, after seeing Transistor's answer . \$\endgroup\$ – Mitu Raj Dec 1 '17 at 1:27
  • \$\begingroup\$ Of course you can amplify 10mV to 5V. But that amplifier needs a power supply. So you'd be better charging your battery directly from that power supply instead. The question title has nothing to do wirh energy harvesting; the other answers have nailed that one. \$\endgroup\$ – Brian Drummond Dec 1 '17 at 10:41

You are asking an xy question. The question is not, "How do I charge a powerbank with a microphone?", but rather "Can I charge a powerbank with a microphone?"

The first thing you should have done before you got yourself into this mess was to look at the basics of the question.

Let's assume that your sound emitter is producing 1 watt of acoustic power, broadcast with a hemispherical radiation pattern. A microphone with an collecting area 1 cm x 1 cm is located 3 meters from the sound source. How much power can the microphone collect?

Well, that's easy. The microphone has an area of 10^-4 square meters. The watt from the emitter is spread over an area 2 pi r^2, so the available power P is $$P = \frac{10^{-4}}{2 \pi r^2} = \frac{10^{-4}}{56.5} = 1.7 \mu W $$ Of course, your numbers may be different, so you should do your own calculations - but frankly, they're not going to get all that much better. 1 watt of power is pretty loud, especially compared to normal conversation levels.

If you compare this to the power needed to charge a powerbank, you will see that it's going to take a really, really long time to do the job. To make matters worse, using the numbers given and 5 volts as the battery voltage, the charge current will be on the order of 0.3 microamps. Now you need to find a fascinating number for your powerbank - the self-discharge current. I suspect that, for a commercially-available battery that number will be more than your charging current, which means that your charger will not actually be able to increase the amount of capacity lost during non-operation, merely slow the process down.

I am, frankly, astonished that a professor would be willing to commit to supervising your thesis - it is almost certainly a waste of time, except as a lesson in asking the wrong question. If you can, I suggest that you attempt to find a new thesis subject.

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    \$\begingroup\$ Is it possible that the professor that agreed to this thesis proposal was either asleep or not from engineering / physics and did not have a sanity check by one that was ... \$\endgroup\$ – Solar Mike Dec 1 '17 at 5:57

Understanding power will be key to this project. In electrical circuits power, voltage and current are related by the equation \$ P = VI \$ or, if we know the resistance, \$ P = \frac {V^2}{R} \$ or \$ P = I^2R \$.

You have a voltage reading but you have taken this with no load resistor so you don't know what current will be supplied when you add a load. Your next step is to find out what the microphone impedance (Z) is. It is probably 600 Ω (lo-Z) or about 10 kΩ (hi-Z).

Lets assume that it is 600 Ω. The Maximum Power Transfer theorem tells us that we can draw maximum power from the source when the load impedance = source impedance. The voltage will fall to half at this point. We can now work out the power using 15 mV as the voltage obtainable:

\$ P = \frac {V^2}{R} = \frac {0.015^2}{600} = 0.000,000,375\ \mathrm W \$.

Now lets look at a battery. I have on my desk a 1.2 V, 800 mAh, NiMH rechargeable cell. We can work out the energy storage in this as \$ 1.2 \times 0.8 = 0.96 \ \mathrm{Wh} \$ (watt-hours).

Now we can calcuate how long you will have to sing for to charge the battery:

\$ t = \frac {capacity}{charge\ rate} = \frac {0.96}{0.000000375} = 2,560,000 \ \mathrm{hours} \$. This is about 292 years.

Do you think you could sing a little louder to speed it up?

From your deleted comment:

But if I'll going to amplify it using op-amp or voltage amplifier, will it be possible?


simulate this circuit – Schematic created using CircuitLab

Figure 1. Amplifying the microphone signal requires power from somewhere else.

No. Amplifiers use a small signal to control power from somewhere else. In Figure 1 the power comes from the battery. The amplifier would waste about half of the power so you would be more efficient just using the battery.


simulate this circuit

Figure 2. A microphone transformer.

Just in case you are thinking of it, a transformer can step up voltage or current but not both. If you step up the voltage the available current decreases by the same ratio.

For a 100% efficient transformer \$ P_{OUT} = P_{IN} \$. Since \$ P = VI \$ we can rewrite this as $$ V_{OUT}I_{OUT} = V_{IN}I_{IN} $$

You can't get more power out than you put in.

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    \$\begingroup\$ I hope his degree doesn't depend on getting this to work. \$\endgroup\$ – Steve G Nov 30 '17 at 23:23
  • \$\begingroup\$ Oh yea sensible thought ! +1 :D \$\endgroup\$ – Mitu Raj Dec 1 '17 at 1:25
  • \$\begingroup\$ Bulk price for throat lozenges anyone :) \$\endgroup\$ – Solar Mike Dec 1 '17 at 5:58
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    \$\begingroup\$ @SteveG I kind of hope it does. \$\endgroup\$ – pericynthion Dec 2 '17 at 7:07
  • \$\begingroup\$ Front row at a Van Halen concert (Roth not Hagar), the charging time drops from 292 years to about 18 seconds. Due to the Evh coefficient. \$\endgroup\$ – Wossname Dec 2 '17 at 13:26

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