I would like to better understand how LTspice works with the variable inductor defined by using the flux through it. I have some troubles to understand why according to the above post the formula of the flux as to be integrated… It is said that it is internally integrated? What does it mean? Does it mean that LTspice is observing dI and not i(t)?
Here is the formula according to the Faraday's Law and the relation between the voltage and the variation of current across an inductance:
$$d(Flux) = \frac{L}{N}dI$$
If we time integrate the above equation, we obtain:
$$Flux(t) = \frac{L}{N}i(t)$$
And that is the relation that we have to enter into LTspice. So in this case we have to integrate if we have a relation in function of dI.
According to me, I think that by observing the current $$i_{L}(t)$$ LTspice tries to adjust the coefficient $$\frac{L}{N}$$ in order to maintain the above relation equal.
So suppose I want to modelize an inductance. I know how the permeability change according to the excitation field H:
$$µ(H) = µ(H=0)*f(H)$$ where f is a function which I know
Then, $$B = uH$$
If I consider that the magnetic field is homogenous over the all surface of the inductance S, I can say that
$$Flux = B/S$$
Then If I do the same approximation about H around a wire but on a curve l:
$$H = NI/l$$
I can say that:
$$Flux = \frac{NI}{lS}*[µ(H=0)*f(\frac{NI}{l})]$$
In this case we do not have to integrate the relation for having the flux. Isn't it? What do you think?
Thank you and have a nice day!
Best Answer
The behavioural inductor accepts expressions that are
time-dependent. There is a keyword,x, signifying the current through the inductor, which is nothing but a shortcut forI(L1)(assuming someL1). For behavioural capacitors,xis the voltage across them, and in both cases the behavioural expressions are solved is by applying a partial derivative w.r.t. each variable (fromLTspice > Circuit Elements > C. Capacitor):This means that, in order for a behavioural expression to reflect the desired change, it needs to be integrated, first.
So, for example, if the inductance needs to vary according to the mathematical expression
$$f(x)=-2x^3+3x^2+1$$
(with
xbeingtime), then the behavioural expression needs to be integrated first, resulting in$$g(x)=-\dfrac12x^4+x^3+x$$
To test this, apply a unity current ramp (which results in the derivative of the current as 1) and read the voltage across the inductor (overlapping voltages have been shifted by a small amount):
Lahas the raw equation as its behavioural expression, andV(a)shows a parabola identical to whatB2shows,V(c2), which means the interpreted result is$$h(x)=-6x^2+6x=\dfrac{\text{d}}{\text{d}x}f(x)$$
Lbhas \$g(x)\$ as its behvioural expression, which means the voltage across it,V(b), is the same as the one supplied byB1,V(c). And, since the example implied the usage oftimeas the variablex,Lb2shows that the voltage across it,V(b2), reflects the integrated expression withtimeas the variable. If another quantity would have been needed instead ofx, such as a voltage, or a current, or any othertime-dependent expression, it would have to be integrated, first.Caution is needed when dealing with expressions, since they are, usually, within a bounded interval. Exceeding the limits may have very unpleasant consequences, and I don't mean the kind that are visible (voltages going sky-high, or similar), but the kind that are insiduous and give apparent good results, yet wrong -- the best debugging sessions. In this case,
x(and, thus,time) is assumed to vary between [0...1], which means that simulating with.tran 2will show software fireworks, but fortime≤1.5 the results will seem correct.Be careful when handling the mathematical expressions in LTspice:
-x**4will not give you what you expect (see the paragraph below the boolean table in the help:LTspice > Circuit Elements > B.). You will need to write them in the two ways shown: with appropriate parenthesis, or with a multiplier in front of it (dummy or not).I would have replied to your comment, but this information should be added here, better. While
xdoes signify the current through the inductor, usingI(La)as the current throughLais not a good idea, because of the following quote from the help (LTspice > Circuit Elements > B., the 3rd point):I(La)meets the above condition, so the proper, 40+ year old way to use the current through a device is to add a series, zero valued voltage source as a sensor, then use that current, in the same way a CCCS (F source) or CCVS (H source) would be used.Now, LTspice considers inductors as voltage sources, so it is possible to avoid the use of a series voltage source as a current sensor (e.g. a CCCS would have the value
La 1), but not in this case, because of the quote above. Therefore, usingI(La)may (most probably will) result in an error -- best usex.OTOH, if, for example, the inductance needs to vary according to some external voltage, then there is no need to use
x, but the expression still needs to be integrated. If the voltage varies according to a known mathematical function, the behavioural expression forFlux=can be derived and written, explicitely, otherwise something along these lines should be used, whereG1andC1act as a unity integrator:The same slight offset has been applied, like in the picture above, for the same reason. Note that, due to the internal derivative, there may be initial spikes of MV or similar, lasting some 1 ns, hence the 1 ms skipped time in
.tran 0 1 1m. Otherwise the response is right.