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Understanding Electronics Basics #1

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Now your both going to be relieved I understood if you alter the phase, it starts the cycle of the wave at a different level on Y
Ok,ok, I'm going.lol
 
PHASE

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

It does not have to be relative to another waveform. It can be that point in time that a single wave form starts or arrives at.

For instance, if you wish a process to start at the 50% point of its most positive swing from the beginning of inception, as in, for example, an AC amplitude rise used in a comparator circuit.
 
Now your both going to be relieved I understood if you alter the phase, it starts the cycle of the wave at a different level on Y

Yes. Close enough. Very good.

YEEHA. We're rollin' now!...
 
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So what your actually saying is if you start the phase at different degrees, not only is it at a different level on the graph but also delayed in time? Along X by the phase amount

But what has any of this got to do with a slope, I'm going to regret asking that aren't I :-/

I have to deduct smoke breaks from my working day to make sure I do my hours I'm paid for, at least I'm making use of breaks.lol
 
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It does not have to be relative to another waveform. It can be that point in time that a single wave form starts or arrives at.

For instance, if you wish a process to start at the 50% point of its most positive swing from the beginning of inception, as in, for example, an AC amplitude rise used in a comparator circuit.

We must be on a roll, that makes perfect sense (except the comparator circuit bit, what's that?) so we can simply decide where to start the process :)

Now reading posts backward as I have a mental block on 75, so the slope is used to measure a distance in time along the wave, don't get excited, this is a different graph.lol
So trying to get my head around the input/output graph, the slope is used to measure a distance in time relating to the output, so if you change the input, it s a way of measuring the different curve? Can you tell I'm starting to guess now :-/

So you saying a input/output graph still has a phase, of course it does if it is still power your measuring, now I'm starting to see it.......
All you had to say to get me to understand that was: if you were measuring power you can calculate the phase of the signal on a curve at any point by using the slope which gives you the - or +
That's if I've understood it right?
 
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The sin(0) = 0 and has values of -1 to +1. If you time shift it on the x-axis by altering the phase, it starts at a different spot. It still has the same shape and it still repeats.

Slope has little to do with distance. Think of a hill on a road. when you are going up the hill the slope of the hill is positive. Going down, it's negative. At the top, it's zero. The slope is how steep the hill is. If it's a flat road the slope is 0. If there was a wall in the middle of the road, it would have infinate slope.
 
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You better not be telling me we went through all that to explain that we are simply measuring a sinewave further along phase cycles.......you wouldn't do to that to me would you???????

If so, IM getting drunk tonight
 
If you were measuring rms voltage, rms current or power you have to include a complete cycle. It doesn't matter where you start.
 
One tiny little addition.

at any given moment as you cross it.

Like this:

View attachment 61512

They are lines tangent to the slope of the curve at precise points (times) on the curve.

I totally get that bit as well

Have one (or a bunch) for me, bro...

haha, I'm too old to do drunk any more, but I will have one for you **broken link removed**

Ok. so I take it this was all just about measuring the phase, but we covered that already didn't we when we were learning waveforms

So which bit of all this mess is the derivitive, the slope?
& are dy/dx the ends of the slope or the point where it meets the wave?

you do realise I'm going to have to go back and try and make sense of all the gobbledegook don't you **broken link removed**

you'll be pleased to know I have moved onto post #78, although I still don't understand what the bike example was, but if the derivitive is the slope, that will do me, gonna knock sim up
 
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ok, looking at post #78 I have played with cap values first, so the lower the cap value, the shorter the time before reaching charge according to attached, but once you hit the linear line it doesn't go upside down, so do they do positive & negative caps **broken link removed** but if you keep rising it loses voltage, I take it this is because your past its working capacity, i.e. it is taking too long to charge
 
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Now this is more interesting, changing the R value, should change the voltage as higher R goes it should be losing/gaining V (I've forgot which way around it is now) hold on it is only a 10V battery, so it can't go higher than 10V, that means R is only affecting the time?
but I take it the cap is trying to maintain V on the discharge cycle, although it is taking longer to achieve full value, didn't we discuss the energy was always split between the components that = the full V potiential, this circuit is not working like that, theres no V on cap side **broken link removed**

ok, this has confused me

Next experiment to change them both but only slightly now I know what each one is doing

& then a little sim for KISS **broken link removed**
 
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ok, last test threw me a bit, not sure I've got this properly but I carried on, almost predictable results
MUTT29 = higher R equals slower charging time even with lower cap value but keeps V
MUTT30 = Keeping higher R & adding higher cap blew V
MUTT31 = so KISS can tell me numbers on this sim so I can see dy/dx, what do you mean, course I've let it go **broken link removed**

That was nice doing something easy, Thanks CBB **broken link removed**

interesting, changing the batt between 5V & 15V kept the charge time at 4s on original setting, so the cap is keeping charge steady at that time (yes, I've just edited V out **broken link removed** as that did change)

Time to walk bobby, feel free to correct me where I'm misunderstanding
 
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Well, as usual, while I was preparing this post, you snuck one in on me (chortle)...

So, first things first.

Example 1. Yes. the RC constant (rise time) to full charge is so short (100msec) that it looks like a straight line when your time base (5) is that long,
Try the transient analysis again with the "End Display" time set to "100m" (100 milliseconds) and you'll see the classic rise curve.

Example 2. Fine.

Example 3. Now your rise time is too short for the time base.
Try the transient analysis again with the "End Display" time set to "10" (10 seconds)and you'll see the entire classic rise curve towards the 10 VDC rail.

Example 4. It looks your capacitor is set to "1k" (or, 1,000 μF). That being the case, your time RC constant is 1MΩ X 0.001F which gives an RC time constant of 1,000 seconds, or 5,000 seconds for 98% charge (or 1 hour and 13 minutes). As a result, your trace was only 5 nanovolts (5 millionths of a volt) above zero after 5 seconds. And too, it looks like a straight line because for that brief a period ( 1/1000ths of the RC constant) it should.

Good job, Graham.

Some of these comments should address some some of the voltage values you saw. If not all of them, give them to me again. On that note, the capacitor will eventually always charge up to the battery (or voltage source) level (unless stopped short by some other component or circuit). In these simple circuits, it will always get to VDC max (although mathematically it never does. It is an asymptotic function [A curve and a line that get closer and closer to each other but never intersect]).

Don't you just love it??

CBB
 
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Mutt31:
It's like 0.5 sec in time and 1.25 units in volts so Δy/Δx = 1.25 V/0.5 s. Just picking convienient numbers off the graticule (Axes). Too lazy to do the divide. So, you basically have the idea.

(The 1.25 isn't correct - fixed in a later post)
 
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Yea, I said that capacitors didn't appear to be there for DC. Applying power is a pulse. The power goes from 0 to some voltage nearly instantaneously. After about 5 time constants, it's 99% there. The problem is mathematically the capacitor will never reach the power supply voltage. Practically, 99% and climbing is probably close enough. Practical power supplies have ripple, a small amount of AC is present. Batteries don't have ripple.
 
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So, first things first.

Example 1. Yes. the RC constant (rise time) to full charge is so short (100msec) that it looks like a straight line when your time base (5) is that long,
Try the transient analysis again with the "End Display" time set to "100m" (100 milliseconds) and you'll see the classic rise curve.

ok, See differences in MUTT22 & MUTT32, these are same R & cap values just different 'X' values, so your saying if I knew what I'm looking at & adjusted time value (X) accordingly, we will always end up with curve that shows a charge time for cap, you will notice at the bottom of graph, it took me five goes to get to 30ms but that's half of learning to understand how to adjust graphs to get the proper readings isn't it, I'm guessing learning math gives you the ability to know whay time value your looking at without playing?

on a different note **broken link removed** don't laugh 'what do you think to me starting here to learn algebra yes, I know it's for schoolkids but I've never done anything like this, I gotta start somewhere to get the basics
 
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