0
\$\begingroup\$

I use a 9V 800mAh lipo Okcell battery to power my ESP8266 ESP-01 with a voltage regulator to change it to 3.3V. I use the ESP to detect the voltage peak from a piezoelectric when someone is running to measure the force and determine the time someone on air from the voltage graph.

The current consumption of my ESP8266 is about 75mA. From my understanding the battery can power my ESP8266 for about 8 hours because of the 20% power loss from the voltage regulator.

So for my project, can I use the piezoelectric to power my battery while im not using it. The piezoelectric I'm using is PZT type. I want to a use a force amplification frame for the piezoelectric so I use ANSYS simulation to simulate it.

the frame design

The stress on the piezo

The data generated by ANSYS

Of course the data is not 100% correct but I still cannot measure the voltage and current generated from the piezoelectric as I still did not fabricate it yet.

So is there any way to use the piezoelectric to power the battery theoretically speaking. It doesn't matter if it will take 1000 hours or not, my supervisor just want to know if it still can do.

\$\endgroup\$
4
  • 1
    \$\begingroup\$ Not realistically, no. Especially not without starting over with the most efficient components available for the load, which yours are orders of magnitude away from. \$\endgroup\$ – Chris Stratton Dec 1 '19 at 17:26
  • \$\begingroup\$ Related: You power use calculation is flawed. You MUST NOT use a linear regulator to produce 3V3 from 9V nominal. Deal with that urgently. || IF you can convert the Piezo high high voltage (288V nominal) to Vbat then you can charge the battery. How fast depends on power / energy levels which you don't mention and which are crucial. Energy = per cycle Joules. Power = energy per time. \$\endgroup\$ – Russell McMahon Dec 2 '19 at 0:57
  • \$\begingroup\$ @RussellMcMahon I'm sorry but why we should not use linear regulator to convert 9V to 3.3V? Then what do you suggest i use? A switching regulator? \$\endgroup\$ – faiz Dec 2 '19 at 5:34
  • \$\begingroup\$ @faiz A linear regulator wastes ALL the input energy that does not appear as output. If you input 9V at say 100 mA to a linear regulator and output 3V3 at 100 mA to a load you waste (9-3v3) x 100 mA. Pin = 9 x 0.1 = 0.9 Watt, Pout = 3v3 x 0.1 = 0.3W. WASTED power = (9-3V3) x 0.1 = 0.57 Watt. Efficiency = 0.33/0.9 =~ 37% !!!!. Yes - use a switching regulator or a battery FAR closer in voltage to the load. eg a LiIon at say 4.2V down to 3.4V in and 3V3 out gives 3.3/4.2 = 71% at 4.2 Vin, 97% at 3.4 Vin, and at average of (4.2-3.4) = 3.8V gives 87% efficiency. \$\endgroup\$ – Russell McMahon Dec 2 '19 at 12:07
2
\$\begingroup\$

In principle, any source of voltage/current can be used to provide power. In the case of piezoelectrics, the devil is in the details: PZTs simply don't produce much.

Especially in this case, you are looking at a losing cause. Consider that your esp8266 draws 75 mA at 3.3 volts, or 247 mW. Meanwhile, according to your illustration, the PZT produces 5.36 mW. I'm sure you see the problem.

If you were to make a dedicated micropower controller, you could produce a unit which transforms the high-voltage/low current PZT output to a usable level. The controller would turn off the 8266 most of the time, and direct the transformed PZT output to a battery charger. At regular intervals, the controller would turn the 8266 on and allow it to make a measurement, then turn it off again. You would, of course, need to account for the power draw of the controller, but dedicated simple logic can be built in CMOS with very low power draw.

In the best possible case, the duty cycle of the 8266 would be about 5.36/247, or about 2%. This would assume perfect efficiency both in the transformation/charging process, and in the 3.3 regulator. And you won't get that in either case. Regulators can operate in the 90% efficiency range if you know what you're doing, but battery charging is notoriously inefficient, especially as you approach full charge. I'd estimate that you might hope for something on the order of a factor of 5 - 10 worse than the ideal.

\$\endgroup\$
6
  • \$\begingroup\$ one more question is what happen if I connect two circuit to the piezoelectric?, one is ESP8266 and one more is energy harversting circuit. \$\endgroup\$ – faiz Dec 2 '19 at 5:41
  • \$\begingroup\$ That is essentially what I have suggested. \$\endgroup\$ – WhatRoughBeast Dec 3 '19 at 18:02
  • \$\begingroup\$ I'm sorry, what i mean is what happen to the power, does it split between the circuit and the 8266? Or do i only can do measurement and charging one at a time? I think that the 8266 can do the deep sleep mode and the current drawn based on the quiescent current if i'm not mistaken. \$\endgroup\$ – faiz Dec 3 '19 at 19:48
  • \$\begingroup\$ Please pay attention. The 8266 needs 247 mW. The PZT can produce (for the situation which you have shown data) 5.36 mW. As long as the input network (some sort of voltage divider) draws less power than the PZT supplies, there is power left over to drive the charging circuit. However, trying to power the 8266 via the PZT is simply impossible - the PZT can only provide about 2% of the power the 8266 requires. Also, be aware that PZTs cannot provide power when being driven by a DC mechanical load. You MUST AC-excite them in order to draw power. \$\endgroup\$ – WhatRoughBeast Dec 4 '19 at 2:07
  • \$\begingroup\$ I’m not trying to directly power the 8266 but I’m using it to power the battery. I’m sorry but my supervisor is really stubborn and she wants me to calculate the amount of time/steps it would take to charge the battery even by using assumption. \$\endgroup\$ – faiz Dec 4 '19 at 4:23

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.