# How to calculate the Back-EMF energy on an Inductor?

The Back-EMF generated by an inductor in Volts is: $V = -L\dfrac{di}{dt}$

Where $L$ is the self-inductance and $\dfrac{di}{dt}$ the rate of current change.

Now, let's say I want to store those Volts on a capacitor $(F, V_0)$ where $F$ is the capacitor capacity in Farads and $V_0$ is the initial voltage on the capacitor.

How can I find out the final voltage on the capacitor? How much energy stored on the Back-EMF field?

• Inductor is current source and capacitor is voltage source. So an energy transformation is necessary. This transformation done by the equivalent resistor of both. – GR Tech Dec 18 '14 at 15:38

The instantaneous energy stored in an inductor is

$$E = \frac{1}{2}L I^2$$

The energy stored in a capacitor is

$$E = \frac{1}{2}C V^2$$

You can see that there's a tradeoff between the capacitance value and the voltage required to store a particular amount of energy.

Assume there is a perfect diode between L and C, and that C is large.

Switch off the current source, and $$\frac{dI}{dT}$$ will be negative, so the EMF V will be positive.

At V=V0, the diode turns on, so $$\frac{dI}{dT} = \frac{-V_0}{L}$$.
(approximately constant because we assumed C is large)

At I=0 there is no more energy stored in the inductor so the diode turns off, occurring at $t = I_0 \cdot L/V_0$.

Charge transferred Q is the integral of the current transferred over time,

$$Q = \frac{I_0 \cdot t}{2}$$

$$Q = \frac{I_0^2 \cdot L}{ (2 \cdot V_0)}$$

which will add dV = Q/C to the capacitor voltage.
(if dV is large, our assumption about C was incorrect)

The energy transferred in this cycle is simply $$E = Q \cdot V_0$$ or as Dave Tweed says, $$E = \frac{I_0^2 \cdot L}{2}$$