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# Inductor kickback (2 of 2)

We give the inductor's kickback current a path to flow by adding a diode. Created by Willy McAllister.

## Want to join the conversation?

• Under ideal conditions would the current in the diode, inductor circuit remain unchanged forever? Or at least until the button is pressed again?
• Hello Hstrathman,

Correct.

An ideal diode has zero voltage drop when forward biased and zero leakage current when reversed biased. By these definitions it has zero resistance.

An ideal inductor has zero resistance.

If we have current flowing in this ideal inductor and then disconnect it the current will continue to flow through the fly-back diode. Since neither part has resistance the current would continue on forever.

In the real world this is not possible and the inductor's energy will be "burned" in the resistor.

There are some interesting application related to this concept. Suppose you have a solenoid such as the injector for a diesel engine. Ref: https://www.youtube.com/watch?v=NUvWnOd5lFw

Here it is important to turn off the solenoid as quickly as possible to stop the flow of fuel. Know that the solenoid will remain energized as long as current flows. Once again the trick is to dissipate the inductors energy as fast as possible. This can be done with resistors or other more complex methods.

Regards,

APD
• Is there somewhere where there is effectively an endless amount of circuit problems and videos that walks through the process of how to solve certain circuits? (RC, LC, RLC, Op-amp, etc.? )
• Hey Willy, then can we replace the diode with a capacitor for the same result? If yes, does the connected polarity of the capacitor matters? Thank you in advance.
• No. A capacitor can't be used in place of the diode. The action of the circuit needs to have the forward conducting path through the diode in order to protect the switch. A capacitor does not have a conducting path.
• Is there a similar effect with capacitors...?
(1 vote)
• Capacitors are the "dual" of inductors. Exchange terms: C for L, v for i, and short for open...

If you build up a current in an inductor and then suddenly open the circuit, you get a sudden jump in voltage.

If you build up a voltage in a capacitor (storing charge) and then suddenly short the leads of the capacitor, you get a sudden jump in the current (all that stored charge comes out in a rush).

Both events can damage nearby components.

The capacitor effect does not have a special name like 'kick back'. Usually it's just called 'Dang it! Stop doing that!".
• sir,I am confused a little bit on the issue that when we close the circuit inductor should oppose current and voltage should be less than 3?please help me if i am missing something.
• why is the stored energy around the conductor preventing or opposing the current that create this stored energy? is the reason behind that the principle of energy conservation?
• Hey Willy, thanks for tutorial. I learnt alot. However, I have some questions and hope you or anyone could help me with it.

1) At , you mentioned that the current is headed up. At , why is there a current flowing down through the inductor when the circuit is still open?

2) As there is a new current path to flow, would the 3.6V keep going up to 100,000V as calculated before?

3) Would the inductor be activated again(even though the circuit is open) as the current is going in a loop as described at , flowing through the inductor in a forward biased manner?

(1 vote)
• 1. There is a current flowing in the inductor when the circuit is open because the inductor says so. You cannot force the current in an inductor to instantly go to zero. It's like trying to stop a spinning bicycle wheel with your hand. The momentum of the wheel will force your hand to move, even if you don't want to.

2. No. The voltage stays "clamped" at 3.6V by the diode. The inductor current is happy because it has a new path to flow, so there is a much smaller change of current (di/dt) and therefore a much smaller voltage = L di/dt.

3. The current flowing up through the diode comes around and flows down through the inductor. This loop current continues until the it eventually falls to zero. When we talk about inductors we don't need terms like activated or forward biased. All you need is current, slope of current, and voltage.
(1 vote)
• Hey Willy,

I have a feeling that I may have been confused with "Back EMF" and "Inductive Kickback" generated by inductor(s) in the circuit.

Further search and reading on Google didn't help to distinctively differentiate them, except with common suggestions to "always use capacitor to help regulate voltage ripple in the circuit caused by inductor(s)".

1) But how are those voltage ripples different from the inductive kickback mentioned here?

2) And if they are different, how was it created(differently) compared to the inductive kickback taught here?

Hope you could shed some light on the matter pls.
(1 vote)
• Will the low voltage generated from a diode enough to create a spark to actuate the switch that is supposed to be actuated by a 100,000V?
(1 vote)
• The switch is only actuated by pushing it closed. The spark is created when there's both a small gap while the switch is being opened, and a high voltage from the inductor kickback (ie the 100,000V). Because the diode stops the high voltage from being made, there won't be enough voltage to make a spark across the air gap.
(1 vote)
• Hi !

At , when the button is pushed down, what stops the current to flow into the diode after it leaves the inductor ? Is it because there is no voltage difference over the diode ?
Thank you.
(1 vote)
• When the switch is closed there is a voltage difference across the diode. 3V on top, 0v on the bottom. But, if you look carefully at the diode direction, its symbol is pointing up, so it is reverse biased. Pretty much zero current flows in the diode. All the current is in the inductor.
(1 vote)