Inductors and Practical Magnetics

• Inductance resists current changeThe dual of a capacitor. Current through an inductor cannot change instantaneously; trying to interrupt it produces a large voltage spike. Everything painful and everything useful about inductors follows from this one fact.V = L x dI/dt. Energy stored = 0.5 x L x I^2.
• The LR time constantAn inductor driven from a voltage source through a resistance ramps its current exponentially with tau = L/R. In switching converters the on-time is short compared to this, so the current ramp looks essentially linear — which is what makes buck converter maths simple.tau = L / R. During a short on-time, dI = V x t / L.
• Flyback and freewheelingSwitch off a relay coil, motor or solenoid and the collapsing field drives current through whatever path it can find, usually destroying your transistor. A reverse-biased diode across the coil gives that current a legal path. This is non-negotiable on every inductive load you will ever switch.Without a flyback diode, V_spike = L x dI/dt can reach hundreds of volts from a 12 V coil.
• Saturation currentPush enough current through an inductor and the core saturates: inductance collapses, current spikes, and the converter it lives in can destroy itself in microseconds. Datasheets give Isat (often at a -20% or -30% inductance drop) separately from Irms, which is a thermal limit.Choose Isat above the peak current including ripple, not above the average current.
• DCR and core lossThe winding has resistance (DCR) which costs efficiency and heat; the core has frequency-dependent loss. A physically small, cheap inductor usually buys its size with higher DCR. Efficiency targets are often really inductor-size targets.
• Ferrite beads are not inductorsA ferrite bead is deliberately lossy: it converts high-frequency energy to heat rather than storing it. Its spec is an impedance at 100 MHz, not an inductance. Using one to filter a supply rail that draws real DC current will cost you voltage and can resonate with the bulk capacitor.

ProblemA relay coil is 12 V, 30 mA, with 200 mH of inductance. You switch it low-side with a MOSFET rated 60 V. What happens without a flyback diode, and what diode do you need?

1. 1
   Coil resistance R = 12 V / 30 mA = 400 ohm. LR time constant = 0.2 / 400 = 500 us.
2. 2
   Energy stored at full current: 0.5 x 0.2 x 0.03^2 = 90 uJ. Small, but the voltage is the problem, not the energy.
3. 3
   The MOSFET turns off in, say, 100 ns. If the current were forced to zero in that time: V = L x dI/dt = 0.2 x 0.03 / 100e-9 = 60,000 V.
4. 4
   In reality the MOSFET avalanches at its 60 V rating and absorbs the energy itself. Repeated avalanche shortens its life and is outside its datasheet's guaranteed operating area.
5. 5
   Add a diode across the coil, cathode to +12 V, anode to the drain. When the MOSFET opens, coil current freewheels through the diode.
6. 6
   Diode requirements: reverse voltage > 12 V (choose 100 V for margin), forward current >= the coil current of 30 mA continuous during decay. A 1N4148 (100 V, 200 mA) is more than adequate; a 1N4007 also works.
7. 7
   Decay time with the diode: current decays with tau = L/(R_coil + R_diode_dynamic) ~ 500 us, so the relay releases in a few milliseconds. If faster release is needed, add a Zener in series with the diode to clamp higher and dissipate faster — at the cost of a higher voltage on the MOSFET.

Answer1N4148 across the coil, cathode to +12 V. Without it the MOSFET is driven into repeated avalanche at 60 V. With a Zener-plus-diode clamp you trade MOSFET voltage stress for faster relay release.

Tool: ngspice transient analysis with an inductive load

• Create flyback.cir:

  Relay flyback
  V1 vcc 0 DC 12
  L1 vcc d 200m
  R1 vcc d 400
  S1 d 0 g 0 SW
  Vg g 0 PULSE(0 5 0 1n 1n 5m 10m)
  .model SW SW(Ron=0.1 Roff=1e9 Vt=2.5)
  .tran 1u 12m
  .end

• Note R1 is in parallel with L1 to model the coil's own winding resistance.
• Run it and: plot v(d) — you will see the drain voltage spike far above 12 V at turn-off. Read the peak.
• Add the flyback diode: insert 'D1 d vcc DMOD' and '.model DMOD D(Is=1e-14 N=1.8 BV=100)'.
• Re-run and plot v(d) again. The spike is now clamped to roughly 12 V plus one diode drop.
• Plot the inductor current with: plot i(L1) — observe the linear-ish ramp on and the exponential decay through the diode after turn-off.
• Measure the decay: how long until i(L1) falls below 10% of its final value? Compare to 2.3 x L/R.

• AC Electrical Circuit Analysis (Fiore) - Inductance chapteropen licencereading~90 minInductance and the RL transient. Same open textbook as the capacitor skill.
• All About Circuits - Inductors chapterfreereading~45 minPlain-language treatment of magnetic field storage and the LR curve.
• Wurth Elektronik - Trilogy of Magnetics (free chapters) and app notesfreereference~45 minManufacturer knowledge base. Read the material on saturation, DCR and ferrite bead impedance curves. Free, no purchase needed.
• Any real inductor datasheet, e.g. a Coilcraft XAL series partfreedatasheet~20 minOpen a real datasheet and locate: L nominal, tolerance, Isat (and its inductance-drop definition), Irms, DCR typ/max, SRF. Do this before the project.

Specify the protection and the inductor for two inductive loads: a 12 V/30 mA relay switched by a MOSFET at 1 Hz, and a 5 V brushed DC motor drawing 800 mA stall, switched by PWM at 20 kHz.

Deliverables

• For the relay: a full flyback calculation as in the worked example, a specific diode part number with its reverse voltage and current ratings quoted from its datasheet, and a stated release-time consequence.
• For the motor: the same analysis, but justify why a Schottky is preferred over a signal diode at 20 kHz, with reverse-recovery time quoted from two real datasheets you compare.
• An ngspice simulation of the relay case, with and without the diode, and the measured peak drain voltage in each case.
• An inductor selection exercise: given a 5 V to 3.3 V buck at 500 mA and 1 MHz, compute the required inductance for 30% ripple current, compute the peak current, then pick a real part from a distributor and show that its Isat and Irms both clear your numbers with margin.
• One paragraph on what would happen if you chose that inductor by inductance value alone and it saturated at 400 mA.

Acceptance criteria — how you know it is good enough

• Peak flyback voltage without protection is calculated, not merely described as 'large'.
• Both diode choices name a real part and quote real datasheet numbers.
• The reverse-recovery argument for the 20 kHz case is supported by numbers from two datasheets.
• The inductor is selected against peak current (average + half ripple), not average current.
• The simulation shows a clamped waveform and the clamp level is explained.

If you want to push further

• Add a Zener-plus-diode clamp to the relay simulation and quantify the release-time improvement versus the extra MOSFET voltage stress.
• Model the inductor's DCR in the buck calculation and estimate the conduction loss and the resulting efficiency hit.

• I can state the inductor V-I relationship and explain why interrupting inductor current is dangerous.
• I can calculate an LR time constant and a current ramp over a short on-time.
• I never design a switched inductive load without a freewheeling path, and I can size the diode.
• I can read Isat, Irms, DCR and SRF off a real inductor datasheet and say what each constrains.
• I can explain why a ferrite bead is not an inductor and where it is the wrong part.
• I can select an inductor for a buck converter against peak, not average, current.