Alright, so, you've got an inductor. You know it makes a magnetic field, but beyond that, so what!? Well I'll tell you what...

No, not really- the worst thing I have is a tinge of differential calculus. Hey wait, come back here, it's a lot easier than you think! The fundamental inductor equation is:

If you're much familiar with capacitors, you'll note the striking resemblance to the capacitor equation, I = C * dV/dt. Indeed, the capacitor and inductor are mirror images of each other (not opposites, for reasons I'll get to later).

How to apply this equation? Well, when V and L are relatively constant, you can change dees to deltas and use it to find how much current rises in some period of time, when you apply a voltage. If you feel like solving differential equations, you can do all sorts of stuff, too.

Take an inductor, ground one end and apply a square wave to the loose end. If this is what voltage looks like, this is what current will look like. Where for a capacitor you charge it with a constant current to get a voltage slope, when you charge an inductor with a constant voltage, you get a current slope.

Note that, when voltage reverses, current stays put. It starts moving, but it doesn't change position. Compare to a resistor, where current is always in the direction of voltage. This can cause trouble for circuits not designed to handle reverse currents, such as audio amplifiers. Actually, the average audio amplifier, probably using an emitter follower output, wouldn't handle it too badly, but as flyback current decays, the amp has to "catch" it in time as the current flow reverses polarity. I have an example of this from my induction heater drive transformer:

If you check the page I originally posted this image on, you'll see I didn't know what the bumps were after the transitions. It turns out the inductor equation describes it perfectly:

- When voltage rises, finite primary inductance causes a dI/dt.
- When voltage falls to zero, primary current holds constant (no change -- dI/dt = 0). (Remember, an inductor holds its charge at 0V terminal voltage, just as a capacitor holds its charge (i.e., dV/dt = 0) at 0A terminal current!)
- When voltage goes negative, current remains positive for a moment. The class C emitter follower driving this is conducting through the NPN transistor (at a relatively large cost of power, since it has full supply voltage across it).
- As "flyback" current decays, less current flows through the NPN. Gradually, the PNP is turned on instead. At this point, the reverse of the cycle starts.

The crunchy part about this is, to turn on MOSFETs I need a quick burst of dV/dt. When the driver circuit starts a pulse, the forward transistor might conduct briefly due to the capacitance, but then as the voltage comes to a stop, the flyback current takes over and the opposite transistor is needed. Bipolar transistors take time to shut off, so this could create a situation where shoot-through occurs, seriously blasting the dissipation. The original circuit I built, using TIP31 output transistors, almost certainly suffered from this, as well as general slowness of the transistors.

Let's say you have an inductor laying around, and you want to know what value it is. You can estimate current rating by the wire gauge (if you can see it) and you can find DC resistance with an ohmmeter. Together you can estimate how much copper wire is on the thing, and if you can see the geometry, even calculate a ballpark inductance. But that's not accurate, and all bets are off if a core is used.

Here's what I use to measure inductors. You'll need a floating signal generator or 'scope since they have 20V or so between grounds. Alternately, you could use a current transformer instead of the resistor, but that's perhaps another web page. Anyway, when the MOSFET is switched on (remember to drive it with around +10V / 0V square wave, and start with a low duty cycle in case things want to burn!), voltage appears on the inductor and current rises blah blah. Magnetic field is proportional to current. Magnetic materials can only help out so much, so if there is a ferrite core, you can test saturation current by measuring at what current the inductance starts to increase. This is labeled I_{sat} on the diagram. You may want to check voltages in the circuit, power supply sag for instance, so you know exactly how much voltage is on the inductor. Since you know voltage, peak current and what time it took to reach that current, you can use V = L * ΔI/Δt, er, L = V * Δt/ΔI to find inductance.

You can also find inductance from the falling (flyback) curve (which is actually rising, because the voltage *was* negative, but you know what I mean), but voltage is usually more variable so I prefer to take it from the front slope. A note: it helps to have an extra source of current for the flyback load. That way, inductor current falls to zero. As shown, current won't actually go to zero -- duty cycle mostly determines flyback voltage and current will be continuous, which can make ugly displays.

Here's a passable equivalent circuit. The capacitor C represents the parasitic capacitance between every turn. Resistor is the DC resistance. You could add a series RC to represent loss at HF, but I don't know that that would be accurate. L of course represents inductance. Although this is a parallel resonant circuit, it looks very inductive for low frequencies and impedances, which is the point. However, when you turn off the transistor, the impedance goes through the roof, and you can observe the ringing at the end of the cycle. In fact, since you calculated inductance from the first half of the cycle, you can use this opportunity to measure the capacitance of the inductor. The ringing will proceed around the resonant frequency, which can be measured by counting divs, and used to calculate C = 1 / (4*π^{2}*L*F^{2}). Just don't forget to subtract MOSFET drain capacitance!

When winding your own coils, it helps to know what you're winding. For single layer solenoids, I use COIL.EXE, written by Brian Beezley, which is very accurate, from small coils to large and across the frequency spectrum. For larger or stockier inductances, though, I need a ferrite core, so what I do is put down say 20 turns of 18AWG on an ungapped core picked from my drawer and hook it up to this circuit. I determine inductance and, if resistance is low enough, saturation current as well. (Remember, the DCR of the inductor causes a voltage drop inside the wire itself which subtracts from the pure inductance's voltage). Let's say I found 50uH. Since inductance is proportional to turns squared, I divide inductance by square of the turns (400 in this case) to find my core does 125nH/T^{2}. That means one turn would be 125nH, two would be 0.5uH, 200 turns would be 5mH, and so on. Likewise, the magnetic field corresponds to the total rate of electrons spinning around the core, or amp-turns. If saturation happens at 5A, I have 5*20 = 100A·T saturation on this core. The maximum energy storage is E = ½*L*I^{2} (compare to the equation for a capacitor's stored energy, or Newtonian kinetic energy for that matter!), or 0.625mJ. If I find this is insufficient for my application, I might add paper or cardboard gaps to the core, measure again, and if it's still no good despite a huge gap, I'll just look for a different core (or come up with some creative combinations!). When you're doing this, remember inductance drops off roughly in proportion to the increase in amp-turns. Since amps are squared and L is linear, you can hold more energy with a greater gap, but since turns are higher for the same inductance, you need bigger wire for the same resistive loss. Pretty quickly you're fighting a losing battle as you discover you have pounds of wire wrapped around this poor little core that's gapped to nothingness. ;-)

HTML is probably going to suck at this, but meh. Let's say you connect an inductor L in series with a resistor R, and connect this to a power supply V. What happens? Well, first of all let's start with basics. Kirchoff says:

V_{R} + V_{L} = V so V_{R} = V - V_{L}

That is, the voltage across the resistor and inductor respectively sum to the supply voltage, and:

I_{R} = I_{L} = I,

I being the current in the loop, being that the circuit is only one loop.

By Ohm's law, V_{R} = I * R, and by the inductor equation, V_{L} = L * dI/dt.

So, I*R = V - L*dI/dt or dividing by R, I = V/R - L/R * dI/dt.

*(Note that I'm using the asterisk to indicate multiplication, and strategically placed spaces to group fractions and coefficients. Hopefully, this will remain clear!)*

What kind of function is that? Well let's break it down to its fundamental form.

Let: C = V/R, and τ = -L/R. Now it looks like:

I = τ*dI/dt + C,

that is to say, I equals the time derivative I, times a constant. I need a function where, when I take the derivative of it, I get itself times a constant. Remember way back in calculus class that d/dx * e^{k*x} = k*e^{k*x}, and it should be realized that this is eerily similar to the *current* situation (ha! ha!). So on a hunch, I'm going to let I = e^{τ*t}. Thus, dI/dt = τ * e^{τ*t} = τ * I. Substituting,

e^{τ*t} = τ^{2}*e^{τ*t} + C

...This is where I get stuck... I have yet to take differential equations. It does yield the familiar RL decay equation, I = V/R * e^{-t*R/L}.

Likewise, you can hook up an inductor and capacitor and get a differential equation.

For the capacitor, I = C*dV/dt, and for the inductor, V = L*dI/dt.

If I take the derivative of the C equation,

dI/dt = C*d^{2}V/dt^{2}, which can be substituted into the L equation:

V = L*C*d^{2}V/dt^{2}. (Note that I haven't used Kirchoff's rules, other than calling the dI/dt's equal. V should actually be negative!) The function that fits this equation is a trig function, simple, so that the second derivative equals the opposite of the function. Sine and cosine work, and for convienience I'll pick cosine. Something like V = Vi * cos(ω*t), where L and C determine ω (ω = 1 / √L*C), as L and R determine τ above. So that, at t = 0, V = Vi, which C is charged to before the switch is closed. The fun thing about this is you get a repeating function...funky. This implies, nay, describes resonance, significantly important for selecting frequencies and such.

Of course, if you include a resistor, either in series or in parallel, the oscillation decays. Everything real has resistance, so it's important and inescapable. But since I suck at differential equations, I'll leave it to other people to calculate it and present the transition from these differential equations to the equations of reactance, and some creative and very useful ways of using them.

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