Understanding Electronics Basics #1

[Muttley600]

On the issue of capacitors:
Essentially correct. These are the Polarized caps. They are marked as such by having a leg marked + or - .
These are the Non-Polarized caps. Their orientation in the circuit is NOT dependent + or -. They are generally used to pass an AC signal with a circuit and block any DC present with that AC signal. They are also essential to an oscillating circuit, which is where we're heading.

Going to have to try & remember that

VERY good with the inductor insight. You're starting to get the deeper correlations now. Although in one aspect they differ (from polarized caps) considerably: an inductor (at room temp) cannot hold a charge like an electrolytic cap. When power is removed from an inductor, the EMF collapses rapidly (but predictably), expending the energy it had momentarily stored. But then, that's what makes them useful in oscillating tank circuits.

Yes, I know what you mean, I was relating the oscillilation, I take it the current oscilates as well or have I misunderstood that bit, we haven't talked about current much

Sort of miss KISS, ya know. Might have to PM him.

Maybe he's on holiday **broken link removed**

Yeah, it's been 5 weeks, but given where we started and where we are now, I think that's pretty impressive.

I'm happy, I am starting to understand what the actual components do now, why that seems more exciting I don't know but it does **broken link removed**
I've never been one for adding something cause I'm told it works, it's great to understand why

Thank you **broken link removed**

just washed bike for mot tomorrow morning, went out last weekend & it was filthy, OCD makes me want to take it to garage clean **broken link removed**
The week I've been having this week, I'll be glad to get back to work for a rest **broken link removed**

 

[cowboybob]

I take it the current oscilates as well or have I misunderstood that bit, we haven't talked about current much

While there can be electrical potential (a battery not hooked to anything), there can never be work being performed (battery connected to a motor) without the flow of current: If you've got one, you got the other, however slight either of them might be.

To recap, electrical potential is the motive (moving) force that causes current (power to do work) to flow. AC or DC, makes no difference.

 

[Muttley600]

Ok, this makes my understanding a bit blurred, I was seeing waveforms as voltage as that's what we were measuring on sillyscope, but then we had ammeter readings as well, I get what your saying now, it is both volts & current, no current = no waveform, that kinda makes it all a bit simpler if that makes any sense at all.lol

I've been racking my brains on how this is actually all working, so if the caps are smoothing the voltage & the inductor is doing something with the current, that isnt making sense unless the caps have voltage & 'I' running through them but dont affect the 'I'flow & chokes/Inductors again have both but dont affect the voltage, i need to see them as one thing with two readings don't I to grasp this properly.
So we can affect a circuit to give us any desired 'V' or 'I' (now I know why they shortened words to letters, they got fed up of repeating words.lol) but basically off ONE, I repeat ONE thing called electricity!!

Do you know, I think it's finally sunk into this thick head of mine, so now you'll prob be disheartened to see me relate electric to a jam cake, one cake but with a pastry 'V' base & jam 'I' filling.....sorry but it will help me to remember it.lol
So basically you can't move the jam if you have '0' base, I've finally understood it haven't I, I know I have

 

[cowboybob]

I was seeing waveforms as voltage as that's what we were measuring on sillyscope... (and) it is both volts & current, no current = no waveform

Well, yes and no.

It IS the voltage potential we're seeing on the scope. One can think of it as "current-less" and that's fine.

But in truth, we are bleeding of a bit of current. EXTREMELY small amounts (pico-amps, if that much), but some. It's the nature of the beast that to test for a voltage level it's going to cost some current. But it is a vanishingly small amount.

So, strictly speaking, no current would mean no waveform but in the real world, there is some being used to create the waveform, it's just not obvious and we don't worry about it. It's a small price to pay in order to see tha shape of the waveform, which is everything.

And as to your last edit:

Do you know, I think it's finally sunk into this thick head of mine, so now you'll prob be disheartened to see me relate electric to a jam cake, one cake but with a pastry 'V' base & jam 'I' filling.....sorry but it will help me to remember it.lol
So basically you can't move the jam if you have '0' base, I've finally understood it haven't I, I know I have

If it's Jam Cake that allows you conceive of all this, that A-OK by me.

 

[Muttley600]

Yes, but this is a major breakthrough, that makes all caps/semiconductors 'V' related doesn't it?

 

[cowboybob]

Yes. The whole enchilada.

 

[Muttley600]

So know you see how I'm looking at this, go to 'maplin' electronics, click on components in blue band across screen under main header, we have a few familys:
Resistors
Opt- something, basically LEDs
Semiconductors- basically in voltage family
Then others yet to be worked out, hopefully once I understand family's, indivual components should be easier to work out.....I did say hopefully :-/

Night, back tomorrow

 

[KeepItSimpleStupid]

yes, but this is a major breakthrough, that makes all caps/semiconductors 'v' related doesn't it?

no !!!!

 

[Muttley600]

no !!!!

Nice to see you on here KISS, go on then, what have I missed now?

 

[cowboybob]

Hey KISS! Good to see ya back.

no !!!!

What did I miss??

 

[KeepItSimpleStupid]

CBB:

You did a good job with everything. Sleeping on the "no", I'm not sure how to back-up what I said. Muttley's conclusion is wrong. The teaching has been on an intuitive level. Capacitors, Inductors have not yet been introduced with a bit of calculus or complex numbers. These relationships for an inductor or capacitor haven't been introduced, neither has complex numbers or circuit analysis the hard way (non-sim). Neither has the RC time constant been introduced.

[latex]v(t)=L\frac{di(t)}{dt}; i(t)=C\frac{dv(t)}{dt}[/latex]

We're not even sure we know that Graham knows what they are.

So, going back to an intuitive approach.

We have devices that exhibit resistive behavior V = IR
But, a LED, for instance invalidates Graham's conclusion. It operates with current and has a voltage drop. A transistor also operates with current with voltage drop. A FET operates as a voltage device. A capacitor's voltage depends on the derivative of i(t).

You did good so far.

Graham:

Do you know what a derivative is? Have you had Calculus or Differential equations? How about rectangular and polar co-ordinates? Some of the concepts are necessary at this point, but not the full power of Calculus.

CBB:
Where do we go from here? First, let's see where Graham's at. The capacitor's job of smoothing AC was good, but that's not all it can do, right?

 

[Muttley600]

KISS, I havent done anything with values yet, I was trying to build up a general idea of family's with components, not really understand what there full potential is.......yet.lol

I know it's a long road ahead, step by step

Just popping out for lunch, back in a bit

 

[cowboybob]

KISS,

or circuit analysis the hard way (non-sim)

Believe me. At times a sim is actually more difficult to work out a circuit than bread boarding. Wish it weren't so, but that's how it is. One problem is TOO much information. And you can change EVERYTHING about a component, which can lead you down very dangerous component alterations that have no bearing on the actual behavior of a real circuit.

Neither has the RC time constant been introduced.

That was next, as we head towards AC generation (an oscillator).

Sorta trying to keep the original PCB schematic in mind as we travel the info highway on how it all works. keeping in mind that some stuff, in particular the math behind it all, can be pretty daunting. I've had the Calculus and even so the proofs for a great deal of electronic theory just leaves me in the dust.

As they say, "Ignorance is bliss" and I'm a happy guy...

You did good so far.

Thank you for that. It's always reassuring to get positive (and negative, for that matter) feedback.

CBB

 

[Muttley600]

We're not even sure we know that Graham knows what they are.

Not yet **broken link removed**

But, a LED, for instance invalidates Graham's conclusion. It operates with current and has a voltage drop.

But if you see MUTT21 attached, LEDS are in optoelectronics, I was just trying to figure out which family of components altered V & which with I

A transistor also operates with current with voltage drop.

So a transistor, although it operates with both V & I as I would hazard a guess all do if there isn't any jam without base, what is it's job, to alter V or I, or something entirely different**broken link removed**

A FET operates as a voltage device.

Can't even see one of those on list **broken link removed**

A capacitor's voltage depends on the derivative of i(t).

havent covered that yet

Do you know what a derivative is?

That'll be a no then **broken link removed**

Have you had Calculus or Differential equations? How about rectangular and polar co-ordinates? Some of the concepts are necessary at this point

Don't let us stop you now, your on a roll **broken link removed**

Where do we go from here?

Forwards my friend, forwards

First, let's see where Graham's at. The capacitor's job of smoothing AC was good, but that's not all it can do, right?

pray tell, what else does it do then **broken link removed**

but on a brighter note, watching that video off this site about lumped abstraction means those sums don't look scary anymore **broken link removed** only another 19 to go **broken link removed** just tell me when we get to each part & I'll watch it......

 

[KeepItSimpleStupid]

Simply, the derivative is the slope at every point on a curve. Mathematically it's a mess for non-polynomials and requires a lot of memorization and rules just like algebra. The derivative is usually taken with respect to something. If you have an x-y graph then dy/dx is the derivative of y with respect to x. The dy can be somewhat thought of as delta y or the change in y over the change in x. This goes back to the slope at any point definition.

Now with something that Graham is familiar with, his bike.
We have two variables; distance and time.
so if s=DISTANCE, VELOCITY (speed) = [latex]v=\frac{ds}{dt}[/latex] and the second derivative of speed with respect to t is ACCELERATION and the third derivative with respect to speed is the more unknown quantity, the JERK.

ACCELERATION is how fast the speed is changing per unit time. The JERK is how fast the acceleration is changing with time.

So, does this give you an intuitive "feel" for what a derivative is?

 

[Muttley600]

I must have read that ten times, speed= jerk bit relates to some of the riders I see on the roads.lol but going back to this, are you simply saying any given
point at which you would reference a curve on a graph?

So if the graph had a curve up to 60mph over one minute, how you work out how long it took you to get to 30mph

 

[KeepItSimpleStupid]

That's a simple algebra problem and one would assume linear acceleration and solve with ratios, thus: 60 mph/1 min = 30 mph/x min --> 60x = 30; x= 0.5 min


Don't read this article, it's too messy for you right now, but look at the first figure animation https://en.wikipedia.org/wiki/Derivative

 

[cowboybob]

OK. Here we go. Finally...

Below is a simple RC (Resistance - Capacitance) timing circuit to demonstrate what's called an RC "time constant". What that amounts to is that a capacitor in series with a resistance takes a very predictable amount of time to charge up to the level of the source supplying the charge (a battery, in this case).

View attachment 61408

This phenomenon is extremely useful, for instance, in controlling the frequency of an oscillator (by controlling the timing of the oscillations).

Follow the instructions and note the basic characteristics of the circuit.

It would also be very helpful to look at this site.

 

[KeepItSimpleStupid]

Sometime time ago you learned that in y=mx+b, that m is the slope. IF you were taking calculus, m would be known as the first derivative. The derivative of any constant such as b in the equation is zero.

In calculus you could be given [latex]y=2x^{3}+3[/latex] and be asked to find the first derivative. It would be [latex]y=6x^{2}[/latex]

So, given any x, you could find the SLOPE at any point. Again. don't worry how we got to this definition except that the derivative is the equation of the slope of a mathematical function at any point. Let's not even worry about how to compute them for now.

The derivative of a constant is zero. It doesn't has have a slope. It's value never changes.

 

[Muttley600]

but look at the first figure animation https://en.wikipedia.org/wiki/Derivative

Sorry, back with you for a bit, so you are on about the slope or the point at which the two lines converge that is called the derivative?

Ok, let's go back a few steps......what is the derivative used to find?
& without getting caught in math, why do we use it?

My brain needs to know what we are looking at & what it's for, you've prob told me, I just haven't understood why we need a slope, it must be telling us something?

Y=mx+b means nothing to me as yet????? Where did the m the x or the b come from, if I can relate them to something we have covered it would be easier, do you know if it was a named sim we have done?