Piezo to Capacitor

[TBadertscher]

If I have a piezo electric actuator that produces 5 mW with one actuation (using the piezo as a reverse generator i.e. applying force creates the voltage rather than applying voltage to actuate) does that mean that 2 actuations would equate too 10 mW. If this is not true, why not?

Let me explain my train of thought so you can correct me where I am wrong. Is this still false when you connect a super cap/battery? I look at a Watt(unit of power) as a joule per second which is a form of energy. Since a piezoelectric device pulses the current(the current is not constant), there is a definitive amount of energy per acuation. Lets say 5 volts and 5mili amps per actuation which is 25 mW. When that energy is stored in the battery, and another actuation occurs; that does not equate to 50mW going into the battery/super capacitor? If not, how can I harvest the energy from each actuation to charge a super cap? Thanks in advance!

 

[Arouse1973]

Hello
If 5 mW is 5 mJ / second which is a rate of change of energy from one form to another. Then 5 mW *2 is just 5 mW *2 and not 10 mW. For it to be 10 mW this would need to be 10 mJ / Second. What I think you are referring to is the amount of energy stored in the electric field of a capacitor. This is 1/2*CV^2 or 1/2 * Q*V which ever you prefer. So let ignore the imperfections of a capacitor for a moment and say that every time you add energy into a capacitor by moving electrical charges, this energy adds to the existing stored energy in the capacitor. This is why the voltage increases because voltage is also energy (Joules/coulomb) which in this case is electrostatic potential energy.
Thanks
Adam

 

[TBadertscher]

Okay, so the energy from the 2 actuations will go into the capacitor along with the 3rd actuation and so on. It just will not be at a faster rate like I was implying?

 

[Arouse1973]

Okay, so the energy from the 2 actuations will go into the capacitor along with the 3rd actuation and so on. It just will not be at a faster rate like I was implying?

Yes correct. And you will also have to overcome the voltage that is increasing also. When your output voltage equals the capacitor voltage no more energy can be stored. And then you use the formula above to work out your energy.
Adam

 

[TBadertscher]

How can you increase the power? In series and parallel?

 

[Arouse1973]

How can you increase the power? In series and parallel?

Sorry you lost me. What do you mean?
Adam

 

[TBadertscher]

The power will increase if I add 2 or more piezoelectric actuators in series or parallel, correct?

 

[Arouse1973]

If you add two energy sources in parallel that dont interact with each other and remove energy from each other then yes..
Adam

 

[TBadertscher]

I know this may be a little different of a question but if I have 25 mW of power for 2ms (the power is not continuous) that energy will go into a capacitor regardless if it is constant? And after lets say 20 sec, I have another 25mW for 2ms, that will also be added into the super capacitor (ignoring leaking current from capacitors)?

I really appreciate all your help.
Tyler

 

[Arouse1973]

It the same as the question you asked before. What is it you dont understand about power, current, and voltage. There is something?
Adam

 

[TBadertscher]

I understand all of those principles, I'm just worried the super cap wont accept it or if it will self discharge to fast. I will just have to test for myself then to see the results. Thank you for your help

 

[Arouse1973]

Do you really? Anyway, if you put more energy into a capacitor than you take out of it then it will store energy. It does not matter the size.
Adam