v= (L) di/dt assistance

I've asked inductor questions before and never fully understood the answers.

First of all, I took calculus and learned derviatives as: X^3 d/dx = 3X^2.

The voltage across an inductor is given by: v= (L) di/dt

I'm unsure how to take the derivative of that formula because there is a di in the numerator.

I understand it can be said as delta i / delta t. But this would be in the case of t approaching 0 or a linear change in current; i.e. a linear ramp.

What if I had a poteniometer that was changing at an acceleration rate instead of a linear rate? Then my i would not be linear, it would be curved.

What if I wanted to know the current at exactly X seconds not using a delta assumption?

Any calculus assistance will be appreciated.

p.s. I am aware this formula can be used in Diff EQ and a formula will show the voltage at every point in time.

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Certainly a better way of writing this is: d/dX (X^3) = 3x^2.

Your misunderstanding is mostly of the physics. The mathematical aspect is not so important. In uyour context, I presume, you need to have a specific functional form for i. Not to be original, suppose i(t) = t^3 and L = 2 henries, what would v be as a function of time?


Reply to
Salmon Egg

Not sure why you want the derivative of it in the first place. That would give you dv/dt on left and d^2i/dt (second derivative of i) on the right.

To understand the voltage/current relationship, suppose the applied voltage is v = ksin(wt). Substitution would give you...

ksin(wt) = (L) di/dt k/L sin(wt) = di/dt

Now, *integrate* both sides....

k/L ((1/w)-cos(wt)) = i (ignoring the integration constant for now...)

Note how i is a negative cosine function while we started with v as a sin function. This shows the current is 90 degrees out of phase with the voltage (current lagging voltage).

If the applied voltage is something more complex than a simple sinusoid function, the integration can get a bit more difficult, but the principle is the same.

Or, if you want you can use a current source driving current through the inductor. Let us say the current source is a sin function such that i = ksin(wt). We find the derivative of i...

di/dt = d(ksin(wt) / dt = kw cos(wt)

Substituting in the formula for the inductor....

v=(L) di/dt = (L) kw cos(wt)

Again, the shift from sin to cosine shows the voltage out of phase by 90 degrees from the current (voltage leading current). And again, if the injected current is something other than a sinusoid, the principle is the same, but you may have more work to do to arrange so that i can be reduced to di/dt.

Hope this helps...


Reply to

"daestrom" wrote in news:47c73d94$0$6509$ snipped-for-privacy@roadrunner.com:

Question about your answer. When you took the integral of both sides, does the i come out as a constant because d/dt disappeared.

That example is fine if I'm applying a sine wave, but what if I have a DC source? Using deltas will be a good approximation, however, it's not exact because the current is di/dt and not delta i / delta t.

I believe I'm confusing this too much. :)

Reply to

Your questions may require more math than I can muster. But you can only use di/dt and its integrals on smooth continuous functions, not step functions. So for DC with a step change on/off, I don't think you can use smooth integrals.

Where.... k/L sin(wt) = di/dt

integrated to

k/L ((1/w)-cos(wt)) = i

No, 'i' is not a constant . The right hand side is the instantaneous current, which happens to be a function of time (say, f(t) = i ). Reversing things and taking the derivative...

k/L ((1/w)-cos(wt)) = f(t) = i

k/l sin(wt) = f'(t) = di/dt

The sine function works out nicely in part because the function is continuous and we can ignore any transient parts by only looking at the 'steady-state'. If you replace the applied voltage with something like a step change at time 0, where at all times less than zero the voltage and current are zero and then voltage steps to some fixed positive value, then you have....

v=0, i=0 (t0)

Notice that there is no 'upper limit' to current, it just keeps growing forever (at the rate di/dt = v/L). This is because there is no resistance in this 'theoretical' circuit. Of course a real circuit has a resistance, which means we apply Kirchoff's law to get....

0=v - Ri - Ldi/dt (t>0)

Solving this R-L circuit can be found in many calculus textbooks. If you ignore the transient part, obviously i=v/R for the steady-state where di/dt = 0 (kind of the definition of 'steady-state' in this case). But the transient part can be 'interesting' and will show the current rise from zero up to the steady-state is a function of e^(-R/Lt). Hence the 'time constant' for such a circuit is TC=R/L

Does that help at all??


Reply to

"daestrom" wrote in news:47cb26a1$0$6138$ snipped-for-privacy@roadrunner.com:

Well yes, a switch would be consider linear. I'm thinking more of say a poteniometer where I'm changing the rate of current from 1 amp / second,

2 amps / second, then 4 amps / second, then 8 amps / second.

Of course this would be difficult to achieve but in that example a linear ramp would clearly not be accurate. So using delta i / delta t would be incorrect.

Now using that example, the current would be changing on a curve.

I think I was pampered with d/dx (x^3) = 3x^2.

Now I'm having problems trying to solve (basic) electronic equations.

Reply to

Principles are the same, just the math can get ugly.

Suppose you have....

di/dt = t^2

Integrating (assuming i = 0 at t=0)...

i = 1/3 t^3 + 0


v = L di/dt = L t^2

Instead, using your example....

t=0 di/dt = 1A/s t=1 di/dt = 2 A/s t=2 di/dt = 4 A/s t=3 di/dt = 8 A/s

This is a 'doubling' function so it's exponential. This one I can see from inspection....

di/dt = 2^(t)


v = L (2^(t))

Integrating this one is a bit harder...

di/dt = 2^(t) di/dt = e^(t*ln2)

i = int(e^(t*ln2)) dt

Now, I *think* that you can integrate this term by setting u=(t*ln2) and du=ln2dt. Then knowing int(e^u du) = e^u

i=1/ln2*int(e^u du) i=1/ln2*e^u i= 1/ln2 *e^(t*ln2)

-or- i = 1/ln2 * 2^(t)

But I could be mistaken here.

Well, if it was *easy*, you wouldn't have to go to school to learn it now, would ya? :-)

The 'trick' is if you can find a function that describes di/dt in the first place. The more complicated that function is, the harder it is to integrate it to find 'i'.


Reply to

You take the derivative of i with respect to t because i is actually some function of t.

In other words, i changes as t changes (or di/dt = 0).

In order to make use of the above relationship, you have to know what i(t) is or know what v(t) is and work backwards to find i(t)

Reply to
Paul Hovnanian P.E.

First, instead of X^3 d/dx = 3X^2, the correct form is d/dX X^3 = 3X^2. That is, you take the derivative with respect to X 'of' X^3. This dictates the order of the operator in the equation.

Second, the function v = L di/dt can simply be interpretted as saying that the voltage across an inductor is proportional to the rate of change of current through the inductor. (The constant of proportionality is L.) So, if the current changes faster the voltage developed across the inductor will be larger. If a current that drives an inductor is suddenly switched off, for example, then di/dt will be very large, so v will be very large. This accounts for the term "inductive kickback."

Let's take the case you describe that causes the variation of the current to accelerate. For example, let i(t) = K t^2 where K is a constant.

Since v = L di/dt, then v = L d(K t^2)/dt = 2 L K t

What does the result mean? It means that the voltage across the inductor continues to increase linearly with time. The "scale factor" is 2LK. So at t = 0, v = 0. At t = 1, v = 2LK. At t = 2, v = 4LK. And so on.

Reply to
Bob Penoyer

"daestrom" wrote in news:47cde496$0$22823$ snipped-for-privacy@roadrunner.com:

After reading your explainations and playing with some circuits in MultiSim, I believe I have a better understanding.

I was trying to understand the formula (v = L di/dt) without understanding what I was doing or inputting a function.

If I understand everything correctly:

  • Any linear change such as a continous increase (or decrease) in current over time will be linear and thus making the formula delta i / delta t.

  • If I'm applying some function to a circuit, then I need to solve for it such as what you did.

I did have some issues when simulating. I was unable to obtain a higher voltage than my input. Adjusting the input frequency only caused the voltage across the inductor to equal my input voltage but only for a different time period (obviously equal to the inductor formula).

It would have been nice to generate 100 volts with 5 volts input. Most online HV circuits I found online contained step-up transformers and any inductor sites beats the inductor formula to death without explaining it.

Another problem I saw: I had a 1H in series with a 100 ohm resistor and opened the circuit (via a switch) and the voltage across the inductor was a triangle wave (above and below ground) that appeared to never hault.

Reply to

Glad I could help...

This may be a limit of multisim (never used it), or the way you set it up, don't know.

Well 20:1 is asking a bit much, but it certainly is possible to use an inductor to 'pump' to a higher voltage. One simple circuit is to arrange the DC supply to feed through the inductor and return via a FET transistor. Between the inductor and FET, tap off with a diode to feed a capacitor. Now arrange a 'suitable oscillator' to turn on the FET so the inductor basically shorts the DC supply. When current builds up, suddenly switch off the FET. The inductor magnetic field will collapse and create a voltage spike that will momentarily forward bias the diode and push some charge into the capacitor. Once inductor current drops near zero, turn on the FET and start over again. This uses the strong magnetic field built up in the inductor when it's 'shorting' the power supply to 'pump' charge the capacitor. But as with all things, there are limits to how much load you can draw off the high-voltage capacitor because if you draw more charge off than the 'pump' circuit can put in, voltage on the capacitor drops.

(I'm sure someone here could devise a better circuit or tell me where I may have made a mistake with this, but the principle is used in a lot of DC-DC converters that are used for step-up)

That certainly is a multisim issue. What should happen is you get a very high voltage 'spike', but it has a very short duration. A 'real' switch would end up arcing momentarily from the high voltage spike and the energy in the inductor would be dissipated in the arc.


Reply to

--------- Part of the problem is just what circuit you are trying to simulate and the source involved (voltage or current) and it may be that the simulation software simply doesn't like the initial conditions given or that the model is not realistic (e.g. switches in series with an ideal current source would be a serious problem- a non-ideal source can be converted to a voltage source behind the switch). The circuits involved don't have the complexity to need a simulator so you should be able to get a good handle on it without one.

Reply to
Don Kelly

"Don Kelly" wrote in news:EAmCj.86204$w94.33585@pd7urf2no:

MultiSim allows you to set the initial conditions. Typically I leave it at the default (let the simulator determine initial conditons), however, this time I have it set to 'zero'.

I have a 1H inductor directly to ground; the 1H makes the math easier. The switch allows for a linear ramp instead of the FET; it was causing non-linearity most likely because of the internal capacitance.

I do, however, get double the voltage across the inductor than I calculate. As an example, my switch takes 9.9us to open (as measured in the program) and it's telling me the current is 7.4mA. That should be [7.4mA / 9.9us] * 1H = 747v. but my measured voltage is 1492v. I'm not sure if this is a fault of the program since I have an inductor directly to ground. The circuit turns into a Diff EQ if I add a resistor and makes the math tricky to deal with while "learning" the basics.

Reply to

I have no idea of what multisim is doing but it appears that you are calculating an average voltage over the switch opening. I do have a problem with the "measured" value -if it is the multisim result, it isn't actually measured -however, it does agree with the maximum value assuming such a ramp down. I don't know enough about multisim (having never used it) to say what it is actually doing. Is it trying to represent an ideal current source with a switch in series? If so, problems occur. If you are representing an ideal current source that ramps down from 7.4 to 0 in 9.9 microseconds- that is another animal and your value of 750V is good. Otherwise??? Again, exactly what circuit is it that you are trying to represent? A great deal of the problem appears to be that you are trying to model an unrealistic system. Resistance always exists so there is no such thing as a pure inductance. I would suggest that for true learning purposes, you ignore multisim which is a tool for those that have a firm hand on basics. It is not a substitute for the work involved in learning the basics (as you have found) , including the dread DEQ (1st and second order are sufficient at this stage and the solutions of such are very straightforward and simple - by the book .

Don Kelly snipped-for-privacy@shawcross.ca remove the X to answer

Reply to
Don Kelly

yes, i know this is late, nevertheless...

Steve wrote: [...]

that is accurate, and also true for non-linear change in current. the issue is, you cannot develop a voltage across a pure inductance without a change in current through the inductance.

use a current source, not a voltage source.

the time constant for L-R circuits is, naturally, L/R. your description suggests that you simulated an unrealistic condition -- you open switch, current in your inductor must drop to zero, but if you don't add another conductive path somewhere the delta-i has no place to "go". your simulator then does whatever it can do, and the result isn't necessarily realistic.

hth, hand.

Reply to
D. Ismay

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