# Electronic – Amplifying 10-30mV into at least 5V

amplifierenergy-harvesting

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?

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.