Back EMF (C-EMF)

[Kane2oo2]

Hey gang!
I'll try and describe this as best i can.

If you take a capacitor (used as a battery) and connect it to a buzzer (like you find in a doorbell - the electromagnetic type).

Then connect a second capacitor & diode across the electromagnet to collect the counter emf or back emf.

Now is there any sort of formula to work out how much power/voltage would be collected in this cap? Ive found some formula which show how to calculate the C-EMF in a coil with an increasing current. But i dont think this applies to calculating the C-EMF created by a collapsing magnetic field of an inductor .... or does it?
examples welcomed

And secondly, by using a diode the voltage/current stored in the cap will leak back into the coil. Is there some simple way to stop this happening?
That way i can compare the actual charge stored, compared to the theoretical figures.

Ive attached an image of the proposed simple setup, is this correct or should the capacitor/diode be connected the other way round?

Thanks alot
Kane

 

[ljcox]

To calculate the capacitor voltage requires the solution of a second order differential equation.

The "Ohm's Law" of inductance is V = L di/dt where V = the voltage across the inductance, i the current through it and t = time.

For capacitance, it is i = C dV/dt where i = current through capacitor, V = voltage across the capacitor and t = time.

The answer also depends upon whether the circuit is over damped, under damped or critically damped. This depends upon the L, C and R values.

It also assumes that there is enough charge in C1 to keep the voltage approximately constant for the time of interest.

If not, then the solution becomes even more complex.

The charge on C2 won't leak back into the buzzer coils as the diode is reverse biassed.

Len