Help with piezoelectric circuit

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I connected three piezoelectric sensors to a rectifier which was further connected to a capacitor (100µF). I kept hitting the sensors for some time, and measured the voltage across the capacitor using a multi-meter. It displayed the voltage to be around 6 volts. Then, I used a breadboard to connect the capacitor, a resistor (100Ω ± 5%) and an LED bulb. The bulb didn’t light up even for a fraction of a second. I measured the voltage across the capacitor again and found out that it had been discharged.

Any suggestions on how I can fix this?ledcapacitorpiezoelectricitypiezoelectric-effectShareEditFollowFlagasked 3 hours agoMr.Brawler111 bronze badge New contributor

• 1how to fix what? – jsotola 3 hours ago
• 1What you have learned is that a 100uF capacitor doesn’t store a lot of charge, and a piezoelectric sensor doesn’t generate a lot of current/power. So if you want more usable amounts of electricity, you could connect hundreds of piezos to something that constantly vibrates and power a few small LEDs that way. What are you hoping for? If you want to make a light light up on impact you can amplify the signal from the sensor. Generating power with a piezo is more of a problem or it would be commonly done. – K H 3 hours ago
• Ah, you are telling us too many things at a time. Me IQ97 lost count at three. Now I suggest to eat the big elephant byte by byte in three or four bytes, starting the with the piezo. (1) So you hit the poor little piezo sensor with a hammer, and the piezo gets excited, generating electrons like a charged up capacitor. (2) Now the charged up piezo, say negative end, is connected to a capacitor through a rectifier, which lets electrons pass through but not allow them go back. (3) Now you use a multimeter and finds the capacitor charged up to 6V. So far so good eh? / to continue, … – tlfong01 3 hours ago   
• (4) Now let us look at the cap. If I remember correctly, the cap, after collecting the electrons (charges), should have an energy of 1/2 * (C * (V** 2)).= (100 * 10**-6) * (6V * 6V) (my always dodgy calculation is not proofread. :)) = (100 * 6 * 6 * 6) * (10**-6) ~= 21,000 micro Joule. (5) Now you connect the cap to a LED, which our stupid human eyes should see it growing red, if there is, say, current of 5mA, passing through it. (6) Now the starting discharging current should be big, something like V/R, where V = 6V, and R the “initial” resistance of the LED. / to continue, … – tlfong01 2 hours ago   
• (7) If the initial resistance R of the LED is too big, then the current passing through, by Ohme’s Law, I = V/R might be smaller than 5mA, so the LED wount bother to light up. (8) And even the initial current is bigger than 5mA, the cap might run out of steam, I mean, electrons very soon, perhaps in 10 micro seconds. In other words, even the LED lights up for 10uS, how can we stupid eyes notice any flick of light so smaller than 1/25 second? So my answer to your question is that my/your stupid human eyes are so very slow to see anything. / to continue, … – tlfong01 2 hours ago   
• (8) There are a couple of workarounds:(a) use a tight string, say guitar string to hit the piezo, then you might see LED blinking, as the string repeatedly hitting the piezo, refilling it with new electrons. (9) Use a digital storage oscilloscope, eg my Rigol DS1064, 50MHz, 1G samples per second, to store the cap discharge waveform, and display back in slow motion. (10) Actually I did this piezo hitting, current storing/displaying game a while ago. Perhaps I should search my lab log and show you. Ah locking down supper time. See you late this evening or tomorrow. – tlfong01 2 hours ago   
• Ha, I quickly found my old posts: (1) “Protecting circuit from piezoelectric disc voltage spike – Asked 3 months ago Active 3 months ago Viewed 222 times”: electronics.stackexchange.com/questions/528419/…. (2) Piezo sensor picking up audio instrument signal. raspberrypi.stackexchange.com/questions/103868/…. Have a great locking down weekend. Cheers. – tlfong01 2 hours ago    
• 1@tlfong01, please don’t answer questions in the comments. (But you know this already!) – Transistor 1 hour ago
• @Transistor. Ah sorry about that. So I edited my comments to an answer. Cheers. – tlfong01 30 secs ago   Edit   

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2 Answers

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The bulb didn’t light up even for a fraction of a second.

It probably did but it would have just been a short time. For a capacitor, we can say this: -Q=C⋅VQ=C⋅V

And if we differentiate Q to become rate of change of Q we get current and rate of change of voltage: -i=C⋅dvdti=C⋅dvdt

So, if the LED is taking about 30 mA, the rate at which the voltage collapses on the charged capacitor is: -dvdt=30 mA100 μF=300 volts per seconddvdt=30 mA100 μF=300 volts per second

So if the LED needs at least 2 volts to operate and the capacitor is charged up initially to 6 volts, then the LED lights for about 13 ms (give or take a few milliseconds).

You could try making the 100 Ω resistor a lot bigger (say 1 kΩ) so that there will only be about 3 mA flowing (on average) into the LED and it should still light but, it will light for maybe 130 ms. Turn the lights down to see it.ShareEditFollowFlaganswered 2 hours agoAndy aka324k1818 gold badges257257 silver badges565565 bronze badgesadd a comment0

Question

Why a 100uF cap charged up by a repeated stressed piezo through a rectifier can not light up an LED?

Answer

This is a bit complicated and needs a long answer.

(1) So you hit the piezo sensor repeatedly, and the piezo becomes a charged capacitor.

(2) Now the charged up piezo, say negative end, is connected to a capacitor through a rectifier, which lets electrons pass through but not allow them go back.

(3) Now you use a multi-meter and finds the capacitor charged up to 6V.

(4) Now let us look at the capacitor. The cap should have an energy of 1/2 * (C * (V** 2)).= (100 * 10-6) * (6V * 6V) = (100 * 6 * 6 * 6) * (10-6) ~= 21,000 micro Joule.

(5) Now you connect the cap to a LED, which should growing red, if there is, say, current of 5mA, passing through it.

(6) Now the starting discharging current should be big, something like V/R, where V = 6V, and R the “initial” resistance of the LED.

(7) If the initial resistance R of the LED is too big, then the current passing through, by Ohme’s Law, I = V/R might be smaller than 5mA, so the LED won’t light up.

(8) And even the initial current is bigger than 5mA, the cap might run out of electrons pretty soon, perhaps in 10 micro seconds.

In other words, if the LED lights up for 10 uS, human eyes won’t notice the flick of light

(8) There are a couple of workarounds:(a) use a tight string, say guitar string to hit the piezo, then you might see LED blinking, as the string repeatedly hitting the piezo, charging and recharging it.

(9) Use a oscilloscope, to store the discharge waveform, and display back in slow motion.

(10) Protecting circuit from piezoelectric disc voltage spike.

(11) Piezo sensor picking up audio instrument signal.

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