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I'm experimenting with how inductors store and release energy into a circuit. I made the following circuit (I'm not sure of the actual inductance value because I'm using a "Snap Circuits" electromagnet part without the optional "iron core" inserted. I understand why the LED flashes when I open the switch. What confuses me is that the LED flashes very weakly also immediately each time I close the switch. I don't understand what is causing this. Can anyone help me?

schematic

simulate this circuit – Schematic created using CircuitLab

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    \$\begingroup\$ Are you familiar with the Laplace Transform? That'd nicely reduce the amount of math in an answer :) (still not an easy thing to answer, I'm not even sure what causes this exactly, but we can reasonably argue with math that there's energy oscillating in and out through some form of overshoot by the LED having a bit of capacitance, and the switch not being perfect, either, but doing this precisely is going to be .. uff.) \$\endgroup\$
    – Marcus Müller
    Apr 14 at 17:21
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    \$\begingroup\$ Could it be due to contact bounce? You close the switch, it makes and breaks several times and each interruption causes the inductor to push current through the LED. \$\endgroup\$
    – Transistor
    Apr 14 at 17:22
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    \$\begingroup\$ Remove the inductor. Does the LED light up? \$\endgroup\$
    – JRE
    Apr 14 at 17:24
  • \$\begingroup\$ I took the circuit apart and rebuilt it the same way and now the LED isn't flashing when I close the switch. Weird. I was flashing before every time I closed the switch. \$\endgroup\$
    – Annette Gates
    Apr 14 at 17:35
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    \$\begingroup\$ It looks like it's contact bounce. I tried several different switches and also touching the contacts with a jumper and I intermittently saw the little flash from the LED. That explains it. Thanks! \$\endgroup\$
    – Annette Gates
    Apr 14 at 19:06

3 Answers 3

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The switch "bounces" a little. The particular phenomenon is complex and include little sparks, vibrations and sometimes chemical processed over the switch contact surfaces.

So, it is never "off" and then simply "on".

It is "off", then the current starts flowing in pulses that leave less and less pauses and at some moment it is finally "on".

The whole process happens in a miliseconds timescale (when the switch is sane), but it has enough on-off transitions in order to power the LED.

The whole "bouncing" business is quite a hassle when one deals with switches.

A modern CPU can run thousands of instructions while e.g. the switch under the keyboard button "bounces". There is a separate term "debouncing" that deals with sanitizing a signal from a mechanical switch.

On the other hand, the same "bouncing" accelerates the degradation of contacts switching a significant electrical power (e.g. in power relays). Having an inductance in the circuit makes the problem worse. In some cases, additional elements (diodes, capacitors, etc...) are used to protect the mechanical contact.

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    \$\begingroup\$ Or additional elements such as a pool of mercury metal! Mercury-wetted relays aren't remotely common anymore, but they used to be used where contact bounce was unacceptable, and as far as I know they still see occasional use in high-reliability circuit design. When one of your contacts is liquid, bouncing doesn't really happen anymore. \$\endgroup\$
    – Hearth
    Apr 15 at 2:35
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Everything has resistance and here it is relevant.

The maximum current thru the Inductor must not exceed the Absolute Maximum Rating of the LED, otherwise, MTBF degrades rapidly. (even instantly)

What limits the current?

The battery has an effective series resistance ESR like e-caps measured by a short pulse low R shunt. (don't blow your DMM fuse) This is predictable for many chemistry types.

The inductor has a low DCR and will be very low for 1uH.

An LED will have an ESR or Rs bulk resistance inversely proportional to its power rating and somewhat with voltage. e.g. 5mm Red is ~ 10 Ohms, 5mm White is 15 Ohms at rated current. This rises exponentially at low voltage and also has some capacitance.

What limits the rise time and fall time?

\$V=L ~ dI/dt\$

Although this circuit when switched rapidly can boost voltage to create a steady DC to drive an LED it must be pulsed to perform as an unregulated switch mode supply.

The energy stored depends on E=0.5 L I^2 in Watt-sec [J] but the self-charge time constant depends on the L/R for series current.

So it is possible to blow the LED with excess current using a manual switch.

In order to raise R and L for longer visible manual operations and thus limit current for short momentary contacts, consider the example below and compute \$T_{\text{63%}}=L/R\$ and \$dI/dt= V/L\$ using the plots.

enter image description here

The LED leakage current of typ. 1~10 uA is modelled with a shunt R.

Also, most switches have contact bounce and depends on spring force and power rating. Typ 1 to 15 ms.

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There isn't any way to tell but it's most likely ESD from your body as you close the switch.

Another possibility, the flicker may also be some kind of resonance with the inductor and parasitic capacitance in the circuit.

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