Amplifying 10-30mV into at least 5V

Question

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?

Answer 1

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?

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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.

Answer 2

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.