# calculus in electronics

Discussion in 'General Electronics Chat' started by Parth86, Aug 29, 2016.

1. ### Parth86Member

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Hello
Integration and differential are most important concept in electronics Calculating voltage or current through capacitor require both integral and differential calculus. Differential calculus is procedure of finding derivatives. Integral calculus calculate area under the curve. Dv/dt means derivatives rate of change of voltage over time. As same di/dt means rate of change of current over time.
I can solve differential integral calculus but I am having problem to understand in electronics. I can put formula and can find the but don't understand how the are related in electronics Can someone explain concept behind integral and differential calculus in electronics?

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2. ### MikeMlWell-Known MemberMost Helpful Member

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How do you find the voltage across a capacitor as a function of the current through the capacitor?
How do you find the current through an inductor as a function of the voltage applied to the inductor?

3. ### Parth86Member

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Can we start with example? Suppose I have 1 f capacitor, connected with 10 volt DC battery. Now I want to find out voltage and current through capacitor. I know voltage and current represent over time
Formula i(t) =C dv/dt
Known value C= 1f, V =10v DC
dv/dt is rate of change of voltage. We need derivatives of voltage over time. How to find out dv/dt?

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5. ### specWell-Known MemberMost Helpful Member

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It is not possible to connect a capacitor to a battery, because with a perfect battery and perfect capacitor an infinite current will flow and the universe will be destroyed.

spec

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6. ### Parth86Member

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OK battery is not connected, can we assume 10v DC applied to to capacitor. My main purpose is how to find out derivative of voltage over time. What should be know to find out derivative of voltage over time. It's better for me if someone can explain with example

7. ### Les JonesWell-Known Member

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If you Google " derivation of capacitor charging equation" you will get many hits. This is the first hit which should answer your question. The way you asked the question shows that you do not understand the way capacitors and inductors behave. Asking the question with an example of a capacitor connected to a constant current source would be valid provided you also specified a reasonable time that it was connected as the voltage would eventually reach infinity. Also you can connect an inductor to a constant voltage source (A perfect battery.) but again you must specify a time as the current will evenually reach infinity. If you connected a perfect constant current source to an inductor the inital voltage would be infinity. All REAL situations will have a component of resistance in the circuit so you can connect a perfect battery to a capacitor or inductor in series via a resistor. You can also connect a perfect constant current source to a capacitor or inductor as long as there is a resistor in parallel with the constant current source.

Les.

8. ### specWell-Known MemberMost Helpful Member

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I was being flippant vead because it is one of two related impossibilities in electronics. The other is that you cannot force a current through an inductor because an infinite voltage will be generated. A third, and more basic tenant, is that you cannot place a short circuit across a voltage source, because an infinite current will flow.

(1) If you apply a voltage to a capacitor an infinite current will flow.
But, in practice, if you force a constant current into a capacitor the voltage across the capacitor will increase linearly for ever.
Thus, if you force a constant current of one Amp into a capacitance of 1 Farad, the voltage across the capacitor will increase by 1V per second. Thus dv/dt =1

(2) If you force a current through an inductor an infinite voltage will be generated.
But, in practice, if you apply a constant voltage to an inductor the current through the inductor will increase linearly for ever.
Thus, if you apply 1V across an inductor of 1 Henry the current will increase at a rate of 1 Amp per second. Thus di/dt= 1

As you imply with your question, applying mathematics to electronics is quite difficult and is one of the great failings in electronics in general.

You need to understand electronics and understand mathematics to relate the two effectively.

Unfortunately, much of the literature is confusing but it is much better now than in my early learning days.

spec

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9. ### Parth86Member

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OK thank you for your detail explanations I want to understand concept behind the formula. We have formula
I(t) =C dv/dt
I have some points related to that formula
Does it valid for DC circuit?
What we should know to find out derivative of voltage?

10. ### specWell-Known MemberMost Helpful Member

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Afraid I don't really understand your question. Perhaps one of the ETO maths experts can help out.

spec

11. ### Les JonesWell-Known Member

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spec has answered this question in answer 1 post #7 dv/dt (Rate of change of voltage) = I/C The units are volts per second. capaciance in Farads and current in amps.

Les.

12. ### Parth86Member

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OK I try to make simple
Formula
I(t) = C dv /dt
C is capcitance, dv/dt rate of change of voltage over time
Question : what should I know to find out dv/dt?

13. ### specWell-Known MemberMost Helpful Member

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For a capacitor:
I/C = dv/dt

Where
I is in Amps (DC)
v is in Volts
s is in seconds

For an inductor:
V/L = di/dt

Where:
V is in Volts (DC)
L is in Henrys
i is in Amps
t is in seconds

spec

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14. ### specWell-Known MemberMost Helpful Member

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A capacitor is analogous to a bucket with water flowing in to it.
The cubic capacity of the bucket represents Farads
The flowing water represents amps (flowing electrons)
The height of the water in the bucket represents the volts
Seconds, as usual, is seconds.

spec

15. ### Parth86Member

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I think it should be
I(t) /C = dv/dt
I think i(t) and I are two different things.
I is current
I(t) I is function of time t
Is it right?

I am just taking another example to understand things
V is function of t
V=f(t)
Derivatives of function v=3t(may be anything)
Apply differential calculus
dv/dt=d(3t)/dt=3
Does it make any sense? I am just showing my best efforts to learn things.

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16. ### William BrohinskyNew Member

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Of course, this is not at all true in the Real World: Only in the theoretical world can we lay claim to perfect batteries, capacitors, inductors, wires, etc.

In order to understand how the math models the action of the active and passive components, it really helps to understand the real basic tenets of electricity and electronics:
1) There are no ideal components: Wires have resistance, batteries and capacitors have leakage currents, inductors' effects are weakened by eddy currents (induced into any nearby metal including their own windings) and even at initial conditions (i.e., immediately upon connection of voltage across an inductor) have resistance which limits currents from becoming infinite.
2) Only when you simplify equations to make calculations easier, do you run into impossible situations, which are dissolved by adding back the things removed to make things simpler.

An example: A battery connected to a capacitor.
If we are working in a simplified 'ideal' system, we presume that the voltage which is marked on the battery is delivered to the terminals of the capacitor, which immediately begins to draw infinite current. There is _nothing wrong with this_, because we've already discounted that any heat/work is created by ignoring lead resistance, and by the same simplification, we've ignored the capacity of the plates (which would fill instantly without resistance.) Which is also OK, because then we can add an in-line resistance in the form of a perfect, ideal resistance, whose value is exactly what we say it is in ohms, and which doesn't have _any_ capacitive or inductive characteristic.
In the lab, even if you connect a battery to a capacitor, reality keeps you from destroying the universe. The resistance of the wires, small though it may be, prevents infinite current. The internal resistance of the battery also prevents infinite current. That alone is sufficient to cause a rise-time in the charge of the capacitor. If the capacitor value (Farads) is gigantic, a great big current will be drawn, and the internal resistance of the battery will drop nearly all of the voltage generated by the battery, causing the voltage differential between the battery terminals to drop to a very very small value. As the capacitor charges, the capacitor's voltage rises, and the current that flows is reduced as the capacitor charges. As the current decreases, the battery's internal resistance also sees less current and the voltage dropped inside the battery decreases, and the voltage at the terminals is able to rise. It is a complicated dance, which only matches reality when the mathematical model takes into account every single _real_ characteristic of the components.

So, to make it easier to learn, we reduce the variables we pay attention to, and we don't design laser ray guns to shoot down ICBMs... because we know we've simplified to allow learning. When we've learned more, we add more variables and pay attention to more physical characteristics. When we've learned a whole lot, the aspects we are ignoring have such small effects that we can once again ignore them without too much worry... as long as their values aren't large in comparison to the electrical characteristics of the circuit.

And here's an example of that: When we play with semiconductors like diodes and bipolar transistors, resistances under 1Mohm are fairly common. A 100V supply which is connected across a 1Mohm resistor draws only 1ma of current, which 'isn't all that much'. 2-3Amps can make problems (and lots of heat.) In transistors, you control a large current with a small one: it's magic! Except if your small current is too large, the large current is gigantic, and the real-world heat buildup can damage the transistor. What if we had a transistor with a control element that didn't draw current? Enter the IGFET: instead of controlling current with current, you control current with a voltage, and current through the control terminal is very very small (pico-amps and smaller). So that's even more wonderful! Except...

Except the insulated gate is very thin. And that means the voltage which causes breakdown is very high. That's not all that bad, though, because there aren't too many sources of pesky megavolt sources, right? Wrong. Static electricity can build up gigantic voltages, and when the breakdown potential is exceeded, the current through the insulation gate is enough to destroy it. This is why ESD (Electrostatic Discharge) is such a worry in the microelectronics field.

So we don't sweat impossibilities created by oversimplifying the model and then smacking it with real-world problems. We start simple, learn the fundamental relations and explore the fundamental equations, and if we really want to succeed, we experiment. Learn to solder. Learn to operate an oscilloscope. Learn to fix TVs or Radios or (heavens forfend) pocket calculators. And keep track of what you learn.

17. ### William BrohinskyNew Member

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Now, that said, it is possible for technicians who have never encountered calculus to design, analyze and understand electronic circuitry. Why is that?

Largely, it is a byproduct of engineers who _do_ understand the calculus, finding the closed equations and physical descriptions of what is going on, and providing them. For instance, the first thing I learned about capacitors (when I was training to be a technician) is that feeding a constant current to a capacitor would cause the voltage across it to rise in a linear relation to the supplied current. So if you had a constant current source, and it had a capacity to continue producing a constant current into a capacitor even as the capacitor's voltage is rising (which requires that the constant current source starts with a gigantic voltage capability), you can charge a capacitor to a gigantic voltage. Then, if you were to disconnect the capacitor from the battery and form a circuit with the charged capacitor and a constant current element, the capacitor would discharge, and the voltage across the capacitor would decrease linearly. (There is, in fact, a constant current device, but for now let's just assume that we have a black box that provides that function and not worry about it too much.)

Now, if you provide a resistance in series with the capacitor and connect them across the terminals of a constant voltage source, what happens in the moment the circuit is completed? The Capacitor has no voltage across it, and acts very much like a short: it allows maximum current to flow, limited by the resistor. The resistor starts with all of the source's voltage dropped across it. This is a Law: all of the voltage raised in a source is dropped in the circuit. (And for now, we're going to be simplistic about it, and ignore the effects of internal resistance in the source, which is the equivalent of presuming that our constant voltage source is capable of providing infinite, or at least very great current, at need. That's the definition of an ideal constant voltage source!) Now. The source raises a voltage differential which we've connected across the capacitor and resistor. The capacitor is discharged (because I discharged it before I gave it to you to put in the circuit!) and the resistor has a resistance, but no voltage across it because no current is yet flowing. We close the circuit (finish connecting everything) and suddenly, there's a voltage across the resistor and capacitor! Current immediately flows. But even at that instant of initial current flow, the plates of the capacitor haven't charged at all, so the capacitor voltage remains zero. Therefore, all of the circuit voltage must be dropped across the resistor. So for the first instant of current flow, the current in the circuit is the applied voltage divided by the resistance of the resistor (by Ohm's Law). If the resistor is 10 ohms and the voltage is 10 volts, the current is 1 amp (which is a lot). If the resistor is 1000 ohms (1kΩ) and the voltage is 10V, the current is 1V/1kΩ = 1ma, which is pretty small. (And as a side note, the power is E x I (in volts x Amps) so we're producing 10V x .001A = 10mW, which isn't a lot!). Now, as you can probably calculate, a smaller resistor will let the capacitor charge faster, a larger one will make it charge slower. But by how much?

At this point, the technician student is presented with a graph of e^x, marked with five equal divisions of time. A curve of this sort is at http://munro.humber.ca/~lloyd/tech101/CapacitorChargeCurve.gif. The first division of time shows 63% charge, meaning that the capacitor will charge to 63% of its capacity. Each successive time division accounts for a little more charge, in decreasing amounts, until the cap has charged to about 99.3% after five divisions. At this point, it's going to take a long time to get to 100%, but for most situations, that's close enough to call it done. The time divisions are called time constants, and contain the same amount of time. To find out how much time that is, multiply the resistance and capacitance, with the units being Ω x F = s.

Now engineers do the same thing, but instead of memorizing the perecentage of change over each of the five time constants, they get the whole equation. And so I did, when I got around to going back to college and taking engineering courses. But knowing the circuits was immensely helpful: knowing that, if the current goes through a resistor and across one end of a capacitor, the other end being grounded, that the output voltage will be integrated, but passing through a capacitor with a resistor going to ground (load-resistor, as it were) will differentiate the output voltage made it a lot easier to see what was going on.

So I can't recommend tinkering enough. Then, since you can choose the components and voltages to use and measure them over time (with a meter and a watch, or an oscilloscope) you can see what the voltages do (and if you put a very small resistance inline with the current, measure across that and scale it up, what the currents are doing). This is how Electrical Engineering got where it is, and while it is possible to understand it with nothing but abstract math, as long as it is abstract, it leaves you open to making mistakes that would be obvious if you had a concrete experience with it.

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18. ### cowboybobWell-Known MemberMost Helpful Member

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Welcome to ETO, William!
Well said. I learned the more complex concepts (beyond E=I/R) by practical experience and only later learned the math behind them (differential and integral relationships). The real-world applications made the math much, much easier to grasp.

If I may ask, where do you call home?

19. ### tcmtechWell-Known MemberMost Helpful Member

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I doubt that being I have never used calculus for any electronic's or electrical or any other practical real life application ever and I have been designing building and fixing things electrical and mechanical for ~25+ years now.
To be honest I have found that calculus is one of the most over hyped least practical forms of mathematical drivel created when comes to solving day to day problems.

Sure it can give you an answer but it's way overly complicated method to find what practical knowledge and knowhow would tell you can be solved with basic applied math.

20. ### nsaspookWell-Known Member

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Every time you adjust the heat while cooking a roast or a thick steak on a grill you use calculus to compute the cooking time to perfection (to cook all the way through while not burning them). The mental process of adjusting cooking times and applied heat is a calculus rate of change equation.

Most of us use calculus daily if you drive a car, the few that don't get tickets after the wreck if they're lucky.

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21. ### RatchitWell-Known Member

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I and D calculus are a tool, not a concept.

It requires a knowledge of how to solve differential equations. The kind of differential equations determine whether you need I or D calculus.

Tell us something we don't already know.

I think you need to proofread your posts a little better. The first and second sentences above are missing words. Understanding what? Find what? You need to study the physics of components like R,C, and L. Then you will understand how integrations and differentials are related to the voltage and currents of those components.

Ratch