Calculus Problem For Electronics Help - Page 1
Forum section: electronics / beginners

author: bostonman
author_rank: Super Contributor
timestamp: February 11, 2020, 09:20:51 pm
content: This formula (or formulas) was given to me for the melting point of a fuse, however, I don't understand it.Can someone explain it to me?

author: Renate
author_rank: Super Contributor
timestamp: February 12, 2020, 07:58:57 pm
content: Erm, well...They seem to be talking about an exponential pulse of current going through a fuse.They don't seem to be differentiating the resistance of the fuse itself and this ambiguous "R" which is setting the time constant.So you have an exponential decaying pulse of current, okay...So you take this current and square it, which gets you square amperes, not an especially useful unit.If you were to multiply that by the (forgotten) fuse resistance you would get the power dissipated in the fuse.If you integrate that you would get the energy dissipated in the fuse.Now, if you put that fuse in a vacuum (and a mirrored chamber), 1 mA of current would eventually raise it to the melting point.If you put the fuse in liquid nitrogen it could probably take 10,000 A without melting.Knowing the fuse's resistance, we can easily calculate how much power is going into heating.We don't know how much the ambiant cooling works or what the melting point is.A (real) recent case in point. I have a fuse.I just determined that the resistance is 0.33 milliohm.The fuse is a nominal 150 A.We take (150 A)^2*0.33milliohm = 7.425 W.So, apparently the manufacturer presumes that 7.5 W of heating in ambient air is not enough to cook the fuse.Of course to blow instantly, it would need a ton of current.

author: bostonman
author_rank: Super Contributor
timestamp: February 21, 2020, 07:55:22 pm
content: Sometimes the letters and stuff confuse me.To focus on one confusing issue, the left side of the equation: integral of I(t)^2 dtDoes this get treated as integrating the t in parenthesis equaling I(t)^3 / 3 or a constant leaving it as I(t) * t ?

author: T3sl4co1l
author_rank: Super Contributor
timestamp: February 22, 2020, 12:22:30 am
content: Neither: I is a function of t. It's not a multiplication process (despite parenthesis being used to indicate that as well).They are giving:1. I(t) is defined as a peak current Ip(they didn't format the subscript, I think) times a decaying exponential, presumably for t >= 0.2. They give the square of (1), to prepare for the next step.3. Fusing is defined by the integrated current-squared time passed through the fuse. "dt" indicates the variable we are integrating over, and the (early modern?-)English long-s indicates integration (a continuous, curvy 's', in analogy to the capital sigma Σ used for discrete summation -- integration is continuous summation!). Normally sub/superscripts are applied to the integral sign to indicate a range; without them, this is the improper integral, the result of which can be evaluated anywhere needed.So we must calculate this to find out what happens with this waveform. The right hand side has been substituted, and evaluated, but an error was made: the improper integral has a "+ C" ("plus a constant") at the end.Alternately, it should be written as integration over a range; in that case, the range indices are kept outside of brackets on the right-hand side, before simplifying further in the next step.4. The integration limits have been added, making this a proper integral, and we evaluate the right hand expression at the indices given. Note that infinity is not a number, and this actually means "take the limit as this index independently goes to infinity".* The values are substituted (and again the limit is implied), and simplifications are made -- in particular, the constant factors have been taken out (Ip^2 RC/2) leaving the dependent parts (exp(-t/RC)).5. exp(0) = 1 and exp(infty) = 0, making the parentheses almost trivial -- integrals can be scary, but integrals of exponents like this happen to be nice and easy -- and we get the expression we presumably were looking for, i.e. the fusing current in terms of an RC time constant and the initial peak current.6. An underlying circuit seems to be assumed here, presumably a capacitor C charged to some initial voltage for t <= 0, in series with a switch that turns on at t >= 0, in series with a resistor R, and the fuse (which isn't really a circuit element, just a current monitor which is sensitive to I^2 t). The initial current is Ip = V/R.*I use the particular language, because in the case of an integral from -infty to infty, you might reasonably ask: what if we take both limits at the same time? That is, integral from -a to a, limit a --> infty. There is such an integration method, but because the values of a and -a are arbitrary -- they could be general functions of a, not just +1 and -1 times as in this case -- you can see this is actually a very much more tricky route to follow, where the choice of functions can affect the result. (As it turns out, these pathological considerations do not apply to the function we are integrating, not that it's being integrated from multiple directions anyway.)We can, by the way, format this on the forum! Quote this post to see the code.\[ I(t) = I_p \, e^{-\frac{t}{RC}} \]\[ I^2(t) = {I_p}^2 \, e^{-\frac{2t}{RC}} \]\[ \int I^2(t) \, dt = {I_p}^2 \, e^{-\frac{2t}{RC}} + c_0 \]\[ \int_0^\infty I^2(t) \, dt = {I_p}^2 \, \frac{RC}{2} \left[ e^{-\frac{2t}{RC}} \right]_0^\infty \]\[ = {I_p}^2 \, \frac{RC}{2} \left[ e^0 - e^{-\frac{\infty}{RC}} \right] \]\[ = {I_p}^2 \, \frac{RC}{2} \]\[ = V^2 \, \frac{C}{2R} \]For the "plus a constant", I used lowercase c_0, to avoid confusion with C (capacitance?). Also note that:\[ e^{-\infty} = \lim_{a \rightarrow \infty} e^{-a} = 0 \]Tim

author: bostonman
author_rank: Super Contributor
timestamp: February 25, 2020, 08:02:48 pm
content: I appreciate the explanation, most of it was basic as I've already taken calculus. I just never understood how to apply it to real world problems and tend to get mixed up when many 'letters' exist.Maybe listing my questions in order would help:Why square everything in step one?How do you know to use dt when taking the integral of both sides (why not di since I is on the left side)?If the integral of a constant (say integral 1 dx) is 1x, then shouldn't the left side be I^2*t(t) since it's being integrated with respect to t?Why differentiate the u substitution?Finally, why are limit numbers used (or why infinity) when the it began as an indefinite integral?

author: dmills
author_rank: Super Contributor
timestamp: February 25, 2020, 09:53:35 pm
content: We square because instantaneous power is I2RWe use dt because we want the energy (We are evidently assuming the pulse is short compared to the thermal time constant of the fuse), and the total energy is the area under the power-time curve.Energy in joules = power in watts times duration in seconds. As power is clearly changing, we integrate wrt time to get the energy.