# Why does a diode placed in series before an inductor not protect from surge?

I checked some posts that explain quite well the concept of a freewheeling diode, but all of them seem to place the diode in parallel with the inductor.

Why doesn't placing the diode right before the inductor in series (see picture below) protect the circuit from the back surge?

• For the same reason you can't put a brick wall and expect it to stop a car while not having damage incurred to the both the wall and the car (presumably you want everything to be able to continue working rather than just doing the job once). The force in the car will go somewhere just like the inductor current will. All you can do is spread it out over time or reroute it through a less damaging path. Commented Nov 14, 2022 at 15:28
• Sometimes, not very often, you do see a circuit with an inductor and diode in series. But that is to accomplish a different goal. Commented Nov 14, 2022 at 16:48
• there is no before the inductor ... the diode could be anywhere in that circuit Commented Nov 14, 2022 at 18:57
• Why would that protect anything? Commented Nov 15, 2022 at 19:27

Usually we want the free-wheeling diode to protect the switch, which is typically a semiconductor, from high voltage when the switch opens.

In this configuration, the diode won't protect the switch because the current will continue to flow in the same direction through the diode (forward biased) so the voltage across the switch will still become very large (infinitely large ideally).

The "trick" to understanding this, is to understand what the inductor "wants" to do.

The inductor "wants" to maintain the same current through it at all times. It tries to maintain the same current by changing the voltage.

When the button is pressed, current starts flowing through the diode and into the inductor. At first, the inductor is seen as barely there - like an open circuit, dropping all of the available voltage across it. This current flows in one direction, "charging" the inductor in that direction. The inductor draws little current at first, but this increases with time.

After some milliseconds, this reaches a maximum value, and the inductor has the maximum stored charge (as a magnetic field.) It now appears like a short, dropping little voltage across it.

When the button is released, there is nothing supporting the magnetic field anymore, so it almost instantly collapses. The current (now a magnetic field) has a polarity, and the polarity cannot change. So without support, the voltage across the inductor must invert and increase rapidly, trying to keep the same current flowing in it. It inverts (goes below zero volts) because that must happen to keep the current flowing in the same direction without support. If this voltage is not stopped or "clamped", it can reach a very high voltage in a very short amount of time.

This is why the diode must be placed across the inductor, and not in series with it. This diode is placed anti-parallel with the normal current flow, so that the diode is only "active" when the inductor is trying to dump it's stored current. Such a diode is termed a "free-wheeling", "kickback", or "clamp" diode.

It can also work to use a diode from the inductor to ground, or some other assortment of components (snubber, transient voltage suppressor) - anything that prevents the inductor's negative voltage spike from getting too large.

Here is a simulation. Before time = 1 s the switch is depressed (allowing current to pass through the inductor), and after 1 s the switch is released:

As others have said, the inductor wants to maintain the same current flowing through it. To do so when the button is released, the inductor increases it's voltage. As you can see, the voltage drop across the switch is gigantic (input - output voltage). The simulation overestimates the spike amplitude because a real inductor has a finite saturation current and non-zero parallel capacitance.

What happens if you put a random diode in series?

That doesn't help! Why? Because just after the button is released, current is momentarily flowing in a clockwise direction because the inductor is trying to maintain it's current, and the diode is facing the wrong way to stop the current.

If you put the diode in parallel with the inductor, the voltage drop across the switch is limited:

The diode now dissipates the spike through self heating.

I usually think like this: The current through an inductor does not "like to" change and it adapts its voltage to achieve the least change in current.

In your case it will produce a large negative voltage on the diode node (relative to the resistor node) in an attempt to keep the current flowing in the inductor and the diode. The diode will therefore just stay forward biased and to no use for protecting the remaining circuit.

Inductor act like current sources when charging discharging so the flyback diode is put in parallel to the wires connected to the battery so it provides a path of smaller resistance for current to flow in order not to flow inside the battery.

Let's finally generalize all these specific explanations in a "current source philosophy".

The problem is how to interrupt (stop) the current produced by an (almost) ideal current source, which here is a charged coil.

The answer is the current cannot be stopped by breaking (opening a series-connected switch); current can be stopped only by diverting (closing a parallel-connected switch). In short, the rule is:

Current is stopped not by interrupting but by diverting.

The "current steering" phenomenon observed in transistor differential amplifier stages is a typical application of this rule.