# Understanding Electronics Basics #1

Discussion in 'General Electronics Chat' started by cowboybob, Feb 16, 2012.

1. ### cowboybobWell-Known MemberMost Helpful Member

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Graham, just for the sake of clarity, this example derivative was calculated by taking the exponent of X (in the first equation), multiplying it times the multiplier of X (2, in this case), reducing X's current exponent by 1, eliminating the constant and putting it all back together as the first derivative Y=6X[SUP]2[/SUP].

The great thing about derivation is that I never could figure out how Newton came up with the concept. For my part I just accepted on faith that it was how it was done and it worked. I'm sure there's a proof of it somewhere, but in the end I didn't care...

Graham, don't miss post #78!!

2. ### KeepItSimpleStupidWell-Known MemberMost Helpful Member

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Look at the animated picture at the left margin where the line moves.

Also look at the very first picture of a sine wave on the right, a point and a line on the right hand side. They are illustrating the idea of a line having a slope which is the derivative at the given point.

3. ### cowboybobWell-Known MemberMost Helpful Member

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Also, Graham, in post #78, I forgot to include that The RC time constant is
where T = time in seconds..

For instance, in the example circuit, If C=1uF (0.000001F) and R=1MΩ (1,000,000 ohms), then T = 1 second.

That is the time for the capacitor to reach 63% of its full charge. As the link points out, 5 times the value of RC (in this case, 5 seconds) gives the time it takes for the cap to reach about 99% of the source voltage level.

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5. ### KeepItSimpleStupidWell-Known MemberMost Helpful Member

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Remember, I said that the capacitor initially acts as a short circuit when a voltage is initially applied? With the derivative concept, this could be more easily understood. In the limit wn power is applied the change in voltage over time is infinate (12-0)Volts/0 time. Na, you can't have 0 time and i(t) = C(dv/dt); and you can't have infinite voltage either. This is where the practical aspects of a capacitor come into play. i(0+) or just past turn-on would say that the current would be infinate and that can't be. The ESR or Effective Series Resistance would limit that.

The impact of this derivative thing is that it changes the phase angle of the voltage when we talk about sine waves. The derivative of the sin(x) is the cos(x), hence you can see a huge phase shift in the voltage.

In the real world, you can't have a pure capacitor. This phase shift makes inductors and capacitors very different from all other electrical components.

CBB: Chime in. Head can't do it right now. How I tried to explain it, is a bit different than
I learned, but it should suffice. I need to add the leading and lagging explanation. Explaining it using derivatives, makes a bit more sense in my opinion. The leading and lagging concept can be easily explained for inductors and capacitors just by using derivatives. The weather has to get better for me to attempt this.

6. ### cowboybobWell-Known MemberMost Helpful Member

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I have heard lame excuses before, but I must say...

(Chortle...)

90 degrees, to be exact. Of course what your saying is correct. It's just that when starting out in this arena, the ideal solution derived (like the "ideal" OpAmp) from the math solution butts heads with the "practical" application. You know how it is: the math says use a 1.3μF cap, but try to find one.

A lot fudge factoring goin' on...

Graham.

While I'm thinking of it, here's the oscillator I spoke of. Something to play with this weekend.

View attachment 61411

Enjoy!

Last edited: Feb 24, 2012
7. ### Muttley600New Member

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Ok, back again.....sigh
I was thinking...????what is he going on about animated picture

Today's lesson..never underestimate an iPhone, the only picture I could see was the red & black lines explaining a tangent.....good job KISS mentioned animated.lol switched to full site & there was an animated picture & the letters I was wondering where they came from

Give me time to catch up, don't panic about me missing anything CBB, I'm about two pages behind
Man, you two can yak.lol

WOW! Going back to post #71, I can see where we are going now, we are learning caps still, the whole shebang, this should be fun

So going backwards to go forwards: I had to read it, couldn't help myself, need to know what the letters are relating to, so we have a cartesian plane which is basically a way of measuring points on wave.
Then function, this where the 'f' came from.
The 'x' simply relates to the input measurement or if it is 't' this is simply a time input along along 'x'
& f(x) is the measurement of the output otherwise known as 'y'
these are known as an ordered pair
So f is the wave within a component or whatever your working on that basically has an input & output, then you could progress to a compisite (basically another funtion, which takes the output of 'f' & the input of 'g' to work out 'g(F(x))' which is basically working out the output value of a second component running through the first, that seems easy enough.

So this example:
For example, the rule f(x) = 2x : simply means that the output is twice the input whatever numbers you are working on

For example, ƒ(x) = 2x+1 : simply means the output equals two times the input plus one again whatever numbers your working on

ok, that bit makes sense, I will understand at some point what this derivitive is, however I feel I've suddenly been landed half way along a path in terms of understanding, I know I need to know this stuff, but to those who are explaining things to me please remember that unless we've covered it, I won't know what the actual letters mean. Yes, I can see these are calculations to show us what something is doing, but I haven't learnt yet how to use the caculations.

& I'm willing to eat humble pie if I'm wrong, but I'll bet my last £ that when I work it out, a derivitive means a point on a graph.lol

guess I did ask, it may as well be now that I jump in feet first

I will continue editing this post until I understand before proceeding, you might like to go out visiting for the weekend or something, I may be some time....hahahaha

please keep an eye on this post & correct me where I'm misunderstanding, but as always, I really appreicate the help you guys are giving me, think I owe you both a xmas present

Ok, first question of the day, I'm still looking at the derivitive, all of a sudden while looking at animated diagram I noticed the -&+ signal is different to the Y-T graph we have been working with on sillyscope, so have I understood correctly the sillyscope graph is:
Y-T graph:
Y= V/I
X=T

But the Y-X graph is:
X= input - f
Y= output - f(x)

Is that correct? So instead of having having a signal above (+) or below (-) gnd we have now changed to measuring the .........
I'm at a loss, the -/+ signal moving has messed with my head to the point I don't even know what we are measuring now let alone understand it?

Ok, here's where I'm at, I have worked backwards until I could understand something, I got as far as algebra & realised that even that looks complicated at the moment, so hopefully you know me by now & I did say I wouldn't quit, I think my goal of understanding this subject is way past what I'm capable of at present, so I shall go off & learn algebra, calculus then come back in the future when I understand what your actually talking about......there is no point learning if I don't understand.

Well, I have just worked 24hrs in two days, I have had a long time to think about the way forward with this (& a nice long soak in the bath to sooth my shattered soul last night) & I can see that I do need to learn & more so that I need to learn to give myself time when I'm struggling, a sure sign of this is I go very quite while I'm considering options of what to do, I still don't understand as yet & I don't know what a realistic timescale is to learn algebra & calculus but I'll get there if it takes two years or longer if needed. You probably won't understand this but I struggle to let things go when I haven't fully understood them.

I don't know what else to say at the moment
signed:.......a confused friend

Last edited: Feb 26, 2012
8. ### RatchitWell-Known Member

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cowboybob,

A resistor does not generate voltage across itself. That would take energy, and a resistor is not a energy storage element. It instead dissipates energy. Since voltage is the energy density of the charge (joules/coulomb), the energy loss across the resistor causes the energy density of the charge to lessen, and thereby always causes the voltage to decrease, not generate across the resistor.

You really mean flow of charge. Current means charge flow, so current flow literally means "charge flow flow", which is redundant and ridiculous. So charge is what flows, and current is what exists.

Muttley600,

The electrical energy is what ties RLC components together.

cowboybob,

Both Newton and Leibniz simultaneously and independently formulated the fundamental concepts of calculus as an effective discipline. The basic concept is not that complex. Archimedes used integral calculus in his geometrical computations. Any good calculus book can show you the derivations of the calculus formulas.

Ratch

9. ### cowboybobWell-Known MemberMost Helpful Member

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Ratch,
That gave me a laugh. I have been accused of the same thing...

And you gotta love semantics:

Granted. I should have said, "a voltage is presented across the resistor".

Redundancy is an oft used teaching tool. Having worked on the Ohio, pushing coal one way and empties the other, I often heard of the term "river flowing". Course, if the river weren't flowing, it would be a lake, or a pond, or a puddle. Current flowing, while redundant, is an acceptable verbal trick to reinforce a particular concept. No change.

Which is repetitive. I say again, which is repetitive.

Thought that was solder.

This was FUN!

CBB

Last edited: Feb 25, 2012
10. ### KeepItSimpleStupidWell-Known MemberMost Helpful Member

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Not really. The derivative is the slope of a line tangent to the curve. The derivative is a function. You can think of the function as Δy/Δx = the derivative. Thus if the function is f(x) the derivative could be called f'(x); Note the single quote read as f prime of x.

When you learned y=mx+b; m was the slope and b was the intercept. For this sort of equation m is also the derivative for an equation in this form. This is the slope intercept form of the equation of a straight line. It has a constant slope of m. Remember a Δy/Δx = m

The difficulty is defining "tangent". We don't really want to go there except saying that the slope of the line will be the derivative at that (x,y) point.

The line in the animation is the line with the slope of the derivative. Polynomial derivatives is one of the first ones you learn. Then you learn some of the simple trancendental functions like sin(x) and cos(x).

We only need to address math to the extent that it helps understanding. If there is a concept glitch, we'll work on it.

and yes; y=x+1 and f(x) = x+1 is just different notation. It even relates to programming say you could have a function Life_Expectancy(age, sex, mode); mode could be single, joint with right of surviveorship

Are you still having issues with the concept of phase? The derivative?

Last edited: Feb 25, 2012
11. ### KeepItSimpleStupidWell-Known MemberMost Helpful Member

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Although this looks pretty nice: http://tutorial.math.lamar.edu/Classes/CalcI/DerivativeIntro.aspx It's also the reason I don't want to go there. We can do a LOT with an intuitive understanding. Math is a symbolic language. It helps us understand the phenomenon around us.

In an EE program, one typically learns math, physics (electricity and magnetism) and circuit analysis before one even attempts any sort of design. Unfortunately the engineers are clueless whether you can jump a car with a household extension cord or which end of the soldering iron is hot. We want to to concentrate on the latter.

12. ### KeepItSimpleStupidWell-Known MemberMost Helpful Member

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I noticed earlier, that you were having trouble with the y = mx+b concept. So go here: http://mathforum.org/cgraph/cslope/mxplusb.html and then go back to the contents on the left.

One thing that the above doesn't address is a slope of 0 which is a horizontal line and an infinite slope for a vertical line where you end up dividing by zero.

The ONLY reason why I wanted to introduce the concept of a derivative is to help understand the capacitor.

In electronics the x-axis being time or frequency is common. I-V cures of components are common as well. Also, some of the functions available with your function generator are not a pure math function. A ramp doesn't really have an easy mathematical counterpart.

Last edited: Feb 26, 2012
13. ### KeepItSimpleStupidWell-Known MemberMost Helpful Member

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Another concept you had trouble with is phase: Take a look here: http://www.wolframalpha.com/input/?i=y=sin(t);+y=sin(t-PI/4) where I have shifted the phase of the sin curve by ∏/4.

The sin function is periodic, which means it holds true for all values. In this case, the phase is relative to t=0.

Let me know if we have to visit the y=sin(x) function more. There is a lot more to visit, for instance, what would a 50 Hz, 240 VRMS wave look like mathematically.

Last edited: Feb 26, 2012
14. ### Muttley600New Member

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ok, you've told me so many times it's a slope.....I believe you, it's a slope, why do we need a slope to measure a curve on a graph, I cannot see the logic in having a slope, can you tell me why we need a slope in the first place

man, you think it's a mess with non-polynomials I'll save it for the later appropriate post

that's me scuppered then

I know your itching to tell me what this: with respect to something is, heres me trying to figure out what we were measuring but we wasn't measuring anything, I had nothing to relate to

I figured that bit out & looking at CBB's next sim (I am so looking forward to that now) this plane runs at 90degrees to the Y-T graph, but I'm struggling as that would mean your measuring width of something & we aren't are we?

Is dy/dx simply the ends of the tangent line?

that bit I get, change the input & the output changes, hence the curve changes

but why, what is the difference of having a slope & a point on the curve?

We'll have to leave that until I understand what we are talking about first

bet you know the answer to that now

Don't forget this is on about post #75

Thinking about this (dangerous I know) lets go to the actual curve a minute, you are on about a curve that measures differently to the Y-T graph (so this graph is restricted to the same input/output values whereas the Y-X graph can measure varying input/output values a bit like a dyno power graph for engines speed-output in BHP. So your not trying to tell me the signal (coiled spring/ my image of wave) has a roundness as this is already covered by the ampitude p-p measurement

Last edited: Feb 26, 2012
15. ### KeepItSimpleStupidWell-Known MemberMost Helpful Member

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It goes back to this. The current through a capacitor:

The capacitor ONLY responds to the change in voltage. A steady voltage across a capacitor does nothing - It's like the capacitor isn't even there. If the voltage attempts to move, the capacitor RESISTS the change.

Just as CBB showed the capacitor, resistor and a DC source. Initially when power is applied, the capacitor RESISTS the change. It takes > 5*R*C to be at 99% of the DC source's value.

If that voltage source were to be changed to a current source, the voltage across the capacitor would be a straight line.

The major effect of all of this is that the voltage and the current are out of phase. The AC voltage lags the current by 90 degrees for a PURE capacitor. The actual amount of phase lag depends on the amount of resistance and capacitive reactance in the circuit and the frequency.

i(t) is the current as a function of time. v(t) is the voltage as a function of time.

Last edited: Feb 26, 2012
16. ### KeepItSimpleStupidWell-Known MemberMost Helpful Member

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SLOPE

Slope is the basis of triggering on your sillyscope. You have positive slope, negative slope and the waveform level, minus and plus including 0. That's at least an intuitive reason you need the understanding of slope.

17. ### Muttley600New Member

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By now you understand we have a lauguage problem, I don't understand it, but I'm determined it will not stop me understanding this concept

So your saying the slope is actually indicating the positive or negative symbol i.e. which way the oscillation is going

I'm wrong yet again, This plane is not running at 90degrees to the Y-T plane at all, this Y-X plane is just measuring input/output

Still unsure on this but I'll get there

I get this bit now

& this was wrong as well

Man, thats a lot of crosses in my understanding.lol

As you've prob guessed by now, I'm STILL working on post #75, once I understand the basic concept we should be able to move forward, I'm in a totally different place regards understanding what we are actually measuring now so I can actually see what the curve means, what the slope is for, just need to crack the dy/dx bit & we are away again
So dy/dx is used to measure minute detail that normally would not register........don't get your hopes up I'm still working on it

Ok, need to go to work, going to have a rest on this bit & have a look at CBB's sim tomorrow morning for a break cause I know I'll get my head around that, don't panic I won't give up trying to understand this concept, I know your trying to help me understand, I just don't understand the language, it may as well be Chinese, let's start with the easy stuff & go from there

Last edited: Feb 27, 2012
18. ### cowboybobWell-Known MemberMost Helpful Member

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This might help you visualize the concept of phase as it relates to a sine wave.

View attachment 61505

First, it is important to remember that phase is always expressed as a phase difference. Different from what?, you might ask.

Well, In other words, for our simulated examples, a single sine wave definition allows for a phase selection. What that is allowing you to do is select the delayed point in time that the starting point of the sine wave begins with respect to a default (or normal) starting point of"0" seconds on the "X" axis.

Major points in the sequence are marked with the Phase at that moment in time.

Keep in mind that during a single cycle of a sine wave (what is displayed in the graphic), the trace traverses 0° to 360° (like the degrees in a circle, or a compass, but laid out in a straight line). Then it restarts at "0" degrees all over again for each subsequent cycle.

The voltage (amplitude) of the sine wave is displayed on the Y (or vertical) axis. The X (or horizontal) axis represents the moment in time that that amplitude occurs.

Now, let's examine two sine waves, one of which (VF1) is started with a phase setting of "0", and the second (VF2) is started with a phase lag of 90°, as below.

View attachment 61507

Last edited: Feb 27, 2012
19. ### KeepItSimpleStupidWell-Known MemberMost Helpful Member

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That's wrong. It does start over, but 720 (2*360) deg is a valid input and that would be further down the time scale, but it would have an amplitude of sin(360 deg). We can also use radians for x which is a dimensionless number for the sin(x). 360 degrees = 2*∏ radians, thus x in the sin(x) can be anything. Typically you will see the sin(ωt+θ) for the sin function in electronics). ω is known as the radian frequency and is 2∏f where f is the frequency in cycles/second and θ is the phase angle.

SLOPE

Yep, slope positive means the signal is rising. A slope of zero means it's at a maximum or a minimum. A negative slope means it's falling.

Last edited: Feb 27, 2012
20. ### cowboybobWell-Known MemberMost Helpful Member

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Granted. Although it starts over at 361°, not 720.

But for the sake of brevity and simplicity, I'm sticking with my definition until Graham has a chance to understand what phase is.

For myself, not in 50 years of the practical application of phase shifting, have I ever used a phase shift greater than 180° (such as in IF circuits, or radar propagation manipulation).

Accuracy does not necessarily translate into clarity.

Last edited: Feb 27, 2012
21. ### KeepItSimpleStupidWell-Known MemberMost Helpful Member

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PHASE

Phase is a shift in the time axis relative to another waveform measured in degrees.