# Coil abrupt discharge BEMF amplitude

My goal it to understand and attain the highest voltage possible for an amateur experiment.

Can't find specs on BEMF properties, only that it's a "dangerous spike" on the instant a switch opens. This is a very poor description of a phenomena Tesla was very interested on.

Normal voltage formula is $$[*1]: v = L * \frac {di} {dt}$$

Does this still applies on abrupt discharge? (simulators suck at producing consistent results, and results depend on simulation speed).

Ideally on abrupt discharges $$[*2]:\frac {di} {dt} \to \infty$$ Which makes v infinite as well.

In reality, i won't be instantaneously zero due to:

• Air gap conductivity/spark (Relay, manual switch).
• Fall time (10-20ns) (MOSFET).
• i's absolute value running through the coil the instant switch opens.

Simulated current rise, and it's resulting voltage. Abrupt discharge (max resolution). Circuit schematics (per @andy-aka request): I manually waited to Amps to get flat, paused, opened the switch, resumed simulation, to create above readings:

• Voltage across the inductor
• Amps in the inductor.

## Assumptions:

All this must be implicitly related to the magnetic field storing energy in the coil, its sudden collapse becoming the voltage ring at natural frequency (NF) (**Also haven't seen anyone explaining why or how the coil transforms the current variation in voltage, it's just a give **) of the coil for a cycle (limiting the decay of i [*2] (from max to zero) to the duration of the collapse of the magnetic field). This field has a collapsing speed as well (1/2 period of natural frequency of the coil?).

## Questions:

• If so, what causes the oscillation in a grounded coil (without a capacitor), is it the inertia of the current generated in the 1st half period?

• If the magnitude of the voltage impulse (BEMF) is related to [*1], but happens as a ringing at NF, at the field collapse speed, what adjustments should be made to have $$v \to \infty$$ in real life?

• Is a capacitor the ideal source of max $$\frac {di} {dt}$$ in case relying on the current variation of the coil itself (decaying magnetic field) is too slow compared with a capacitor discharge on the coil?

• How do different cores (air, stainless steel, ferrite, etc) influence magnetic field collapse speed? (is that accounted for implicitly in Inductance unit H?)

• Any better idea based on considerations I didn't even think about to achieve the goal? (The Tesla magnifier relies on the primary and secondary tuning, but it escapes me ATM why he said that any desired voltage can be achieved.)

Note: in the scope, with only the signal generator (max 25V about 15ns fall time) I only see a very small voltage rise. Probably because while the signal has high precision, it has low current. It also has issues falling to zero with DC offsets, or maybe it's the scope centering the image but it looks like the square wave always gets to a negative value when I want zero-max.

#### Edit:

Will perform some tests and load values on this spreadsheet over time. Hopefully correcting/accounting for parasitic values as explained in comments.

• Yes, V = L di/dt is the proper way to understand this. There is always stray or parasitic capacitance. In fact coils have a self-resonant frequency (SRF) due to stray capacitance. This is why you get an oscillation. I am not sure I am the best person to fully answer your question but I thought I would try to shed light on that part. Commented Oct 27, 2022 at 16:30
• I see, well, Tesla bifilar pancake coils embrace built in capacitance :) But I tend to suspect we are hiding the current inertia contribution behind the (also present) concept of parasitic capacitance. And I'm not talking charged particles (which I understand are slow, and some argue of their existence altogether), but dielectric field. Commented Oct 27, 2022 at 16:34
• In addition to the inductor's self-capacitance that @mkeith mentions, the switching MOSfet adds significant capacitance to that (in parallel). To achieve high pulse height, these capacitances should be minimized. Commented Oct 27, 2022 at 17:08
• A perhaps obvious point. But when you ask a MOSFET to turn off current through an inductor, the MOSFET will suddenly experience a very high voltage which (if you are trying to get maximum voltage) will probably cause the MOSFET to go into avalanche conduction. This will defeat your purpose and also stress the MOSFET. If the energy in the inductor is large enough, then it will destroy the MOSFET permanently. Automotive ignition uses, essentially, a transformer to generate the ignition spark. So the voltage spike at the primary is survivable for the electronic switch. (Not an expert on ignition) Commented Oct 27, 2022 at 17:50
• Show a circuit and show the waveform and show where you measure it (simulation or real circuit, either). Commented Oct 27, 2022 at 18:15