# Linear aproximation of an inductor in a switch circuit

Most of DC-DC switching circuit analysis uses linear approximation of an inductor voltage and current according to graphs below.

If we use basic formula for the inductor,

v = L di/dt

It is clear that linear changes in current produces constant voltage so from mathematical point of view it is clear.

However, I have trouble to fit this concept into the physical picture of the inductor, knowing that the transient changes on a inductor are described with an i(t) = I(1-e(-t/Tau)) and i(t) = Ie(-t/Tau).

What troubles me is the constant voltage of the inductor given in a approximation.

When the inductor is connected on DC power supply it will build magnetic field, which induces voltage that opposes to the source voltage, which than produces current that opposes the original current. So the original current through the inductor cannot raise nor decline instantly. Induced voltage will be "temporary" present and will tend to reach almost 0 after magnetic field is completely formed and reaches saturation. So approximation of the constant voltage is not quite clear to me from a physical point of view. How can this approximation be physically interpreted?

I think the assumption is that the the filter cap is large enough and the switching frequency high enough such that the voltage ripple across the output filter capacitor is small enough that you can can neglect dynamics in the inductor voltage, in the interest of keeping things simple.

Your current ripple can go all the way from zero to some $$\I_{Lmax}\$$ depending on mode of operation, but since it's a voltage regulator, your design goal is probably going to be a fairly constant voltage across your output capacitor (i.e. minimal ripple). If your Vin is also assumed to be constant then the only variations in the voltage across your inductor, outside of switching would be due to the voltage ripple across the output cap.

You're mixing models here -- you need to get down to the actual behavior of an inductor.

When the inductor is connected on DC power supply it will build magnetic field which induces voltage that opposes to the source voltage, which than produces current that opposes the original current. So the original current through the inductor...

This is a handwavy way of saying that an inductor self-induces a voltage. Yes, all that happens. As a consequence, the current on a perfect inductor follows the equation $$v(t) = L \frac{d}{dt} i(t)$$

That means that for a perfect inductor, and for a constant voltage, the current will rise in a perfect ramp. You can solve the above differential equation and see that.

However, I have trouble to fit this concept into the physical picture of the inductor knowing that the transient changes on a inductor are described with an $$\i(t) = I(1-e^{-t/Tau})\$$ and $$\i(t) = Ie^{-t/Tau}\$$.

That is true for an inductor in series with a resistance with a constant voltage applied. In the case of switching supply analysis, the resistance of a typical coil is small enough compared to the product of the switching time and the inductance that it doesn't matter much to the analysis.

In a well-constructed supply, long before you need to consider the effects of resistivity on the coil's current "shape", you need to consider non-ideal (and nonlinear) effects in the inductor core. At that point you're done drawing straight lines on paper, and you need to take your (hopefully very) educated guess about the power supply design and simulate it.

• Thank you for your time and for your answer.
– moci
Nov 4, 2020 at 8:39
• But still a bit confusing that the voltage remains constant across the inductor during transient period. I have a picture that it actually can pretty rapidly change to build magnetic filed or to release energy stored in a inductor during transient. But for a fast switches in a short period of time linearity should work.
– moci
Nov 4, 2020 at 8:57