calculus in electronics

[Ratchit]

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?

To answer your question, dv/dt depends on how you apply the voltage, doesn't it. Is it a ramp, square wave, sin, step, or what? Lets apply a 10 volt step to a 2.0 henry inductance with an internal resistance of 2 ohms. Applying a step voltage to a coil will cause it to produce an infinite backvoltage for an infinitesimal amount of time. This backvoltage will stop any current change in the coil at t=0. Any elementary textbook that delves into the physics of the coil will show you that the current with respect to time is

If the current is plotted, the result is

You can see the current is limited by the internal 2 ohm resistance to 5 amps. Now, what will the current plot be if the coil has no resistance? Can you figure it out? Give it a try.

Ratch

 

[Parth86]

You can see the current is limited by the internal 2 ohm resistance to 5 amps. Now, what will the current plot be if the coil has no resistance? Can you figure it out? Give it a try.

Ratch

Internal 2 ohm resistance limit current going through coil. I think if there is no resistance, inductor will behave as short circuit

Have you seen my post 14. There is example,

 

[William Brohinsky]

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?

I've updated my profile, but let's just leave it at "New England". It hardly matters with regard to electronics, though: I learned it as a tech in the Navy, worked in research labs and contracting to the Navy for about 32 years before being told I was unemployable _as an electronics technician_ without an associates degree, so I completed two (AS Electronics Technology and AS Lasers and Fiber Optics) at night and have all the credits (but not the paperwork) for a third in computer science technology. Very little of it changed my approach to electronics (and it included Calc 1 and 2, Linear Algebra, and Differential Equations). There was very little hands-on, even in the Controls courses and circuits and systems: it was mostly predicated on abstract maths. When it came to the circuits and systems labs, I ended up becoming a de facto teaching assistant, because the younger students had no idea which end of the oscilloscope to touch to the circuit, and multimeters were both too complicated and not complicated enough for the smartphone crowd. They all responded well to relating integration and differentiation to a simple RC circuit, though!

I will note, not incidentally, that having two AS degrees and a third waiting for paperwork (which I'm not really that interested to complete) has not made me one bit more interesting to employers.

 

[cowboybob]

What was your rate? (I was an ET - 1598)

I will note, not incidentally, that having two AS degrees and a third waiting for paperwork (which I'm not really that interested to complete) has not made me one bit more interesting to employers

Sadly, a life of practical experience, in today's work place, appears to be useless.

 

[Ratchit]

Internal 2 ohm resistance limit current going through coil. I think if there is no resistance, inductor will behave as short circuit

Have you seen my post 14. There is example,

Wrong! You are treating the inductor as a resistor. Your post #14 pertains to a capacitor, not an inductor. Everything you need to know is given in my last post. Use the math, and may the calculations be with you. Again, what will the current curve look like if a step voltage is applied to a inductor with no internal resistance?

Ratch

 

[tcmtech]

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.

You have it backwards. Calculus is a mathematical representation of that, not the otherway around. The ability to guestimate was here long before calculus existed and is used by creatures far less advance than us as well.

Mathematics is an approximation of reality. Reality is not a product of mathematics.

As in' "Theory says theory and reality are the same. Reality however shows otherwise."

 

[Parth86]

Wrong! You are treating the inductor as a resistor. Your post #14 pertains to a capacitor, not an inductor. Everything you need to know is given in my last post. Use the math, and may the calculations be with you. Again, what will the current curve look like if a step voltage is applied to a inductor with no internal resistance?

Ratch

Hello Ratch
I think step voltage means constant voltage source. Voltage will be same for all values of time. Example. 10 volt will be same for 1s, 2s, 3s....... Etc. When step voltage applied across ideal inductor (no internal resistance), current through inductor will increase while the voltage across it remain constant. Is it right way? I will do calculation and post soon.

 

[William Brohinsky]

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?

I think you have the wrong idea about dv/dt here.

dv/dt describes the change of voltage with time. The voltage in question is what you are applying to the capacitor: you have control over it. So you know how it changes over time. C is a scalar value, Capacitance, which is defined as the ratio of the charge impressed upon a capacitor to the change in potential caused by that change. This can be restated as the quantity of charge (usually expressed in a count of electrons) which must be added to raise the voltage by one unit of electrical potential.

For our system of electronics, we use the Farad which is defined as one coulomb, which simultaneously is considered a dimensionless number (being the charge that passes any point in a circuit in 1 second when 1 amp of current is flowing) and a measure of electrons (6.24... x 10^18 e-), and is the charge to add to a capacitor of 1F capacitance to raise its potential by 1 Volt. (This is the fundamental basis of electronics units, incidentally: 1A for 1s moves 1 C of charges. 1C added to 1F raises potential 1V. 1A through 1Ω causes a 1V drop in potential, etc.)

So, if you have a capacitor with 1F capacitance and 1V potential, you know it has 1 coulomb of charge on it. If we add another coulomb, the charge rises by 1V to 2V. Similarly, if you apply 1A of current for 1s to the capacitor, charging the already-charged plate, and stop, the potential across the capacitor will rise by 1v, but if you apply the current to charge the other plate for one second, the potential will become 0V, regardless of the absolute number of electrons on either plate.

dv/dt doesn't describe how much voltage is applied, but rather the rate of change. In fact, this is a rather simple thing to determine, because it is saying that, if the change of applied voltage is constant (i.e., constantly rising or falling) the current created will be constant. This is where most people's heads explode. So let's look at it differently.

We know that a constant current will provide the same amount of charge per unit time. The voltage, according to I = C dv/dt (where I is average current, and i would be instantaneous current) that C is a constant and I is (because we're providing a constant current) a constant, and therefore the only thing that changes is voltage across the capacitor. Therefore, we know, if dv/dt applied to C is a constant, I will remain constant.

Similarly, we can tell that, if there is no change in applied voltage, dv/dt = 0 and I =0. Without changing the voltage on the capacitor, we won't cause current to flow. At all.

If dv/dt is a very high rate of change, a high current will be involved. If dv/dt is very small, a very low current will be involved. If dv/dt is varying, the current rate will vary as well as the first derivative of the voltage! This is important because if v=A sin(t), then current will vary as cos(t), and the amplitude will be dependent on A and C.

These are all qualitative features of i = C dv/dt. The fact of the identity means you can calculate values when you know values, but this is really secondary to understanding what the equation says.

The circuit that implements i = C dv/dt is simply a capacitor with a controllable current source, ideally without source resistance, etc. In the Real World, you won't see these things work out exactly, mostly because the charge rate will be controlled by the source internal resistance, the resistance of the wires, and the leakage of the capacitor. Because of these things, which we can theoretically lump into one resistance, we can picture it (as a realistic world view with ideal parts) as a controlled current source with no internal resistance, in series with all of the circuit resistances, in series with an ideal capacitor with all of the circuit's capacitance lumped together. The result, then, is to modify the instantaneous i by taking into account the time constants (which are R X C in seconds).

When the capacitor is in a series circuit with a resistance (which it _always_ is in the real world), the effect is the same as an RC circuit which has a signal source to the left, a series capacitor, a resistor from the capacitor's right lead to ground (which we declare to be the terminal of the source which is not connected to the capacitor, and our reference 0v point for measurements) and the output taken across the resistor. Why do we take the output across the resistor?

1. the resistor is a passive component whose voltage is independent of active-component effects, so we can be sure that its voltage is an accurate representation of the current flowing through it.
2. the current in a series circuit is the same at every point in the circuit
3. so measuring the voltage across the resistor gives us a scaled readout of the current, I=E/R. For a 1Ω resistor, I will be the same value as E, in amps (because E/R is V/Ω = A).

Would something bad happen if we measured across the capacitor? Possibly. We might provide an increase of leakage current effect by paralleling the measuring device (which requires some current to be able to make a reading). The capacitance of the capacitor might resonate with or otherwise affect the measuring device. So we go for the safe component.

This means that the capacitor voltage can be deduced (calculated when we have numbers) from the resistance voltage. Now, we can control the circuit current by controlling the voltage across the entire string, we know the resistor voltage, and we can subtract that from the capacitor to know the capacitor voltage. And what we find is:

1. with an applied pulse of voltage V for a given time t, the resistor's voltage instantly at start is 100% of the applied voltage. This is because the capacitor acts like a short, i.e. just a wire.
   
2. as the capacitor starts to charge, it develops a voltage, tiny at first, but increasing quickly, and this voltage is subtracted from the resistor (because the entire circuit is provided V for the period t, and the capacitor and resistor act as a simple voltage divider at any instantaneous time.)
3. After one time constant, if t is longer than one time constant, C will have acquired 63% of the charge it is going to be able to hold (based on its capacity) and that time constant's length is R x C.
4. After 4 times constants, it will be up to about 98.12%, and after five, it will have risen to 99.3%. At this time, the resistor will be dropping a very low voltage, about .7% V. This will decrease at a decreasing rate until, some time far in the future, the resistor will drop 0v, the capacitor will drop 100% of the voltage (because of the charge stored in it) and no current will flow.

If, then, the source is disconnected, no current will continue to flow because of the open circuit, but if it is left connected and we reach time t, it will drop to 0v. At this point, we have (through the source) the equivalent of a short between the input end of the capacitor and ground. Now, the capacitor will discharge. The Capacitor will become, effectively, the source, and its entire output will be dropped across the Resistor, which will allow maximum current. Not infinite current, of course, because the resistor is not a short. Instead, the current that flows is the Capacitor charged voltage divided by the Resistor's resistance: i=e/R. (Remember, small i and e are instantaneous values.) As the capacitor discharges through the resistor, its voltage falls, according to the i = C dv/dt equation, and that reduces the voltage across the resistor. As e -> 0, I -> 0. The discharge of the capacitor is governed by the same physics as the charging, so after one time constant, 63% of the charge will have left, and the capacitor's voltage will have dropped to 63%.

What if the input from the controlled source is sin(t)? -cos(t)? A triangle-wave? A square wave? Work out each of these qualitatively first, then play with the math and see if your results match your qualitative expectations.

It is this working out of qualitative effects that is most useful in electronics: the numbers can always be done afterwards to determine if what ought to work really will. But it is unlikely (enough to be called serendipitous) that sitting around manipulating math is going to result in a working, useful circuit.

 

[William Brohinsky]

Hello Ratch
I think step voltage means constant voltage source. Voltage will be same for all values of time. Example. 10 volt will be same for 1s, 2s, 3s....... Etc. When step voltage applied across ideal inductor (no internal resistance), current through inductor will increase while the voltage across it remain constant. Is it right way? I will do calculation and post soon.

This is quite inside out.

The equations for both capacitors and inductors contain differentials/integrals (depending on what you're calculating). That's a hint.

The steady state of the pulse is only of interest for the duration that it is active, and the duration (probably the same amount of time) after it stops.

When a step voltage is applied across an ideal inductor, there is an effect caused by the initial rise of voltage. No matter how instantaneous, this rise of applied voltage tries to move current through the inductor. The instant that current changes, the inductor either grows or reduces its magnetic field as a result. But as the field grows, it cuts through neighboring coils of the inductor, and because it is moving the opposite way from the field growing out of each of those turns, it creates a Counter Electromotive Force (CEMF) which fights to reduce the current in those turns. And, the fields growing out of those turns do the same thing to their neighbors. The overall effect is that, at initiation, extremely little current flows, and very little magnetic field grows. Over time, the overall effect is reduction in the speed with which the current flow increases and the magnetic field grows. In the ideal case, that time is negligible and the magnetic field springs into existence. In the real world, the resistance in the circuit (which may be just the resistance in the wires with an 'ideal' source, but probably includes the source internal resistance) causes the growth of the magnetic field and thereby the current to be slowed, and it increases exactly the same way that the capacitor does. Likewise, the voltage across the inductor starts at 0 (with 0 current) and grows... but you can only see it if there is a resistance to drop a voltage from the current and make a divider. If the inductor is connected across a constant current source, the current changes, but you don't see the voltage develp: it's just pegged at the source value.

When the step voltage ends, the falling edge essentially provides a short to the inductor, and it discharges. Again, the _change_ in current causes the field to start to collapse... but as it impeded current from building up due to CEMF, the field now moves in the opposing direction, and it reinforces the current which is already flowing. Thus the discharge of the coil is the reverse of the charge, and resistance in the circuit causes time constants to be non-negligible.

The fun part, in the real world, with inductors is that, if they are sufficiently powerful, and the current sufficiently strong, disconnecting a coil with full current running through it may cause that current to pile charges up at the break, enough to create a potential field between the broken contacts which is sufficient to ionize the air between: in this case, a spark will form along the ionized path, and with sufficiently large flowing currents and sufficently large inductance, this can create some interesting mayhem. Vide:
https://video.search.yahoo.com/vide...simp=yhs-prodege_001&hspart=prodege&vm=p&tt=b

 

[Ratchit]

Hello Ratch
I think step voltage means constant voltage source. Voltage will be same for all values of time. Example. 10 volt will be same for 1s, 2s, 3s....... Etc. When step voltage applied across ideal inductor (no internal resistance), current through inductor will increase while the voltage across it remain constant. Is it right way? I will do calculation and post soon.

Certainly the current will increase when voltage is applied across an inductor. Current will not stay at zero or reverse, will it? The question is what will the current curve look like when the voltage is first applied and afterwards. Yes, a step is discontinuous at a finite point in time, usually zero. Something like this is what I had in mind.

Don't be discouraged, because the answer is not that hard to find.

Ratch

 

[Ratchit]

"William Brohinsky, post: 1273117, member: 262933e b

I will ask you a question similar to what I asked Vead. What will the current curve of the inductor, with the same specifications as I gave Vead, look like if a 1 amp step current is applied to the inductor. By a curve, I mean the beginning and end of the step current.

Ratch

 

[Parth86]

Don't be discouraged, because the answer is not that hard to find.

Ratch

OK

 

[Ratchit]

OK
View attachment 100988

You gave an answer for R=2 . I asked for the current curve for R=0. Back to the books for you.

Ratch

 

[William Brohinsky]

I will ask you a question similar to what I asked Vead. What will the current curve of the inductor, with the same specifications as I gave Vead, look like if a 1 amp step current is applied to the inductor. By a curve, I mean the beginning and end of the step current.

Ratch

I'm sorry, but I don't think you're asking a sensible question.

If you apply a 1-Amp current to an inductor, that says you are going to do anything possible in the world to force 1 amp through that inductor. This will not happen, unless your ideal inductor (with R=0) also happens to lack the ability to build a magnetic field when a current is applied. In which case, at the beginning of the 1Amp current pulse, you will go from 0A to 1A, and at the end from 1A to 0A. However, by simplifying the inductor to an idealized short, you have eliminated any kind of inductor action, and by applying an idealized 1Amp source as a pulse, you've eliminated the possibility that the "current curve" will be anything else but that input current 'curve.'

If you allow the inductor to build a magnetic field, and you force 1A to flow in the first instant of application to the inductor, you create an illogical situation. I'm not sure there is a lot of benefit in pursuing that, myself.

 

[Ratchit]

I'm sorry, but I don't think you're asking a sensible question.

If you apply a 1-Amp current to an inductor, that says you are going to do anything possible in the world to force 1 amp through that inductor. This will not happen, unless your ideal inductor (with R=0) also happens to lack the ability to build a magnetic field when a current is applied. In which case, at the beginning of the 1Amp current pulse, you will go from 0A to 1A, and at the end from 1A to 0A. However, by simplifying the inductor to an idealized short, you have eliminated any kind of inductor action, and by applying an idealized 1Amp source as a pulse, you've eliminated the possibility that the "current curve" will be anything else but that input current 'curve.'

If you allow the inductor to build a magnetic field, and you force 1A to flow in the first instant of application to the inductor, you create an illogical situation. I'm not sure there is a lot of benefit in pursuing that, myself.

First of all, I specified a step function, not a pulse. That means that that current source will be 1 amp from t=0 until the end of time. In this problem, whether R = 0 or R=2, is insignificant in comparison to the large resistance of the current source. Let me modify the problem a little. Suppose the current source is one million volts in series with one million ohms of resistance. Now , what will the current curve look like? Assume that the inductor will act like an inductor. Don't get hung up on the magnetic field surrounding the inductor when current is present.

Ratch

 

[Parth86]

You gave an answer for R=2 . I asked for the current curve for R=0. Back to the books for you.

Ratch

After Google search and reading book, I don't see the formula where I can find out time constant for inductor. There is formula for RC, RL combination And if I use same formula and put the value of r=0 than time constant t=L/R=2/0 which is not possible

 

[William Brohinsky]

First of all, I specified a step function, not a pulse. That means that that current source will be 1 amp from t=0 until the end of time. In this problem, whether R = 0 or R=2, is insignificant in comparison to the large resistance of the current source. Let me modify the problem a little. Suppose the current source is one million volts in series with one million ohms of resistance. Now , what will the current curve look like? Assume that the inductor will act like an inductor. Don't get hung up on the magnetic field surrounding the inductor when current is present.

Ratch

Uhhh...yeah. Don't get hung up on the magnetic field surrounding the inductor when current is present, indeed.

" Lets apply a 10 volt step to a 2.0 henry inductance with an internal resistance of 2 ohms. Applying a step voltage to a coil will cause it to produce an infinite backvoltage for an infinitesimal amount of time."..."You can see the current is limited by the internal 2 ohm resistance to 5 amps. Now, what will the current plot be if the coil has no resistance? Can you figure it out? Give it a try." <- your original attempt at this.
"Wrong! You are treating the inductor as a resistor. Your post #14 pertains to a capacitor, not an inductor. Everything you need to know is given in my last post." <- your response to Vead's first try
"I will ask you a question similar to what I asked Vead. What will the current curve of the inductor, with the same specifications as I gave Vead, look like if a 1 amp step current is applied to the inductor. By a curve, I mean the beginning and end of the step current." <- your posit to me
"First of all, I specified a step function, not a pulse. That means that that current source will be 1 amp from t=0 until the end of time." <- the beginning of your latest riposte. Note that your "By a curve, I mean the beginning and end of the step current". Compare that to "the end of time".
"In this problem, whether R = 0 or R=2, is insignificant in comparison to the large resistance of the current source." <- Nice of you to mention this, seeing as how you haven't mentioned it before.

I am doing my best to help, but you are doing a great job of trying to embarrass people, both those who are asking for help and trying to help. I don't know why. But if you want me to be your performing monkey, it ain't going to happen.

If you still feel a need to challenge me, you can try stating a problem with all the hidden gotchas on the outside at the outset. I might take the challenge. I might decide there's no point and ignore you. After all, all you've been doing, at least with me, is demonstrating how an engineer can start from stupid assumptions based on insufficient information and arrive at stupid conclusions. That's a bad way to operate, and a bad way to try to make other people appear to be operating.

I have to admit to feeling quite ill-used, and I am surprised that Vead is putting up with it, as well.

 

[Ratchit]

After Google search and reading book, I don't see the formula where I can find out time constant for inductor. There is formula for RC, RL combination And if I use same formula and put the value of r=0 than time constant t=L/R=2/0 which is not possible

As I said in post #25, everything you need to know is in post #21. Finding the limit of I as R approaches zero, we get

Finding the above limit if facilitated by using L'Hospital's Method, as described in any good calculus book.
If V = 10 volts and L = 2 henrys, the current will be 5*t when R=0. Plotting the current curve when R=0.000001, we get.

As you can see, the value of the current is about 25 amps at the end of 5 seconds when R = 0 instead of 5 amps when R=2 ohms.

Ratch

 

[Ratchit]

Uhhh...yeah. Don't get hung up on the magnetic field surrounding the inductor when current is present, indeed.

" Lets apply a 10 volt step to a 2.0 henry inductance with an internal resistance of 2 ohms. Applying a step voltage to a coil will cause it to produce an infinite backvoltage for an infinitesimal amount of time."..."You can see the current is limited by the internal 2 ohm resistance to 5 amps. Now, what will the current plot be if the coil has no resistance? Can you figure it out? Give it a try." <- your original attempt at this.
"Wrong! You are treating the inductor as a resistor. Your post #14 pertains to a capacitor, not an inductor. Everything you need to know is given in my last post." <- your response to Vead's first try
"I will ask you a question similar to what I asked Vead. What will the current curve of the inductor, with the same specifications as I gave Vead, look like if a 1 amp step current is applied to the inductor. By a curve, I mean the beginning and end of the step current." <- your posit to me
"First of all, I specified a step function, not a pulse. That means that that current source will be 1 amp from t=0 until the end of time." <- the beginning of your latest riposte. Note that your "By a curve, I mean the beginning and end of the step current". Compare that to "the end of time".
"In this problem, whether R = 0 or R=2, is insignificant in comparison to the large resistance of the current source." <- Nice of you to mention this, seeing as how you haven't mentioned it before.

I am doing my best to help, but you are doing a great job of trying to embarrass people, both those who are asking for help and trying to help. I don't know why. But if you want me to be your performing monkey, it ain't going to happen.

If you still feel a need to challenge me, you can try stating a problem with all the hidden gotchas on the outside at the outset. I might take the challenge. I might decide there's no point and ignore you. After all, all you've been doing, at least with me, is demonstrating how an engineer can start from stupid assumptions based on insufficient information and arrive at stupid conclusions. That's a bad way to operate, and a bad way to try to make other people appear to be operating.

I have to admit to feeling quite ill-used, and I am surprised that Vead is putting up with it, as well.

There are no hidden gotchas, and the problem has a succinct, reasonable answer. Just as there was a solution for the problem I gave Vead, there is a rational solution for the problem I gave you. If you don't know how to find the solution, just say so, and I will give you the solution which I am sure you will agree is correct. You can get a hint how to solve the problem by reading the solution I gave Vead.

Ratch

 

[MrAl]

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?

Hi there,

I see you are doing pretty good so far, that's nice to see in someone just starting out with this stuff.

May i suggest that you start with the simplest version and then work from there?

i(t)=C*dv/dt

If we rearrange a little we get:
dv=i(t)*dt/C

and for CONSTANT current we get;
dv=I*dt/C

This is really all algebraic.

So given I=2 and dt=6 and C=3, what is dv?

dv=2*6/3=4 volts

So the voltage changes by +4 volts every second, and that means that if it starts out at 0v then we have:
4 volts after 6 seconds,
8 volts after 12 seconds,
12 volts after 18 seconds,
etc.

You should fully understand what is happening here before moving on.

Next, we can look at a series RC circuit where the cap charges through the resistor. That creates an exponential response, but if you look at it little by little you'll clearly see the real mechanism behind this: it's the derivative again but because the current keeps changing we have to think a little deeper about how it works. After each time period the current is a different value so we have to change I in teh above formula.
For example, in the incremental model if we start with a 10v power supply and 10 ohm resistor and a capacitor of value 1F,
The cap charges up to 1v in 1 second because 10v/10=1amp and in the above formula that means in one second we get dv=1.
But now what happens is the votlage across the resistor is only 9v, so we now have 9/10=0.9v, so the next dv is:
dv=I*dt/C
dv=0.9*1/1=0.9

so now the voltage only changes by 0.9v, so we now are up to:
V=1+0.9=1.9v.

We're up to 1.9v now, so the voltage across the resistor is now only 10-1.9=8.1 volts, so now the current is only 8.1/10=0.81 amps.
Now after another second, dv=I*dt/C=0.81*1/C=0.81v, so now it only went up by 0.81 volts and the total voltage now is:
V=1.9+0.81=2.71 volts.

See how that works? Little by little the current decreases so dv decreases. Eventually the cap gets charged up to VERY NEARLY the full voltage but we just call it fully charged after that.

The next step would be to see what happens when we decrease dt. We used 1 second above, but we can lower that to 0.1 seconds and get finer results.

This is one of the simplest ways to visualize what is happening there. It's actually a type of numerical method, a simplified version of what modern simulators use. BTW simulators let you see results for any circuit you draw almost.