Author Topic: Vocabulary for beginners  (Read 1633 times)

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Offline stmdudeTopic starter

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Vocabulary for beginners
« on: August 29, 2017, 08:37:04 pm »
I've been thinking about this for a while, and finally decided to write it down and get some input from other members.

What pushed me over the edge is Scottjd's excellent post here: https://www.eevblog.com/forum/beginners/probe-oscilloscope-ground-blow-up-issues-usb/msg1254104/#msg1254104

I think that many of us are so experienced with what we're doing that it's hard to explain things in a way that a complete beginner can reasonably understand. Many things are just so second-nature to us that we simply don't reflect on the fact that we're sometimes talking to people that are missing _massive_ parts of rudimentary knowledge.

I would love it if Dave would do a video for beginners about terminology more experienced people use and do some simple explanations/demonstrations of them.

It's been a while, so I've forgotten (repressed?) a lot of what I went through as a beginner, but one thing stands out in my mind.
I had a simple grounds-referenced circuit with two voltages. Lets say 5V and 10V. When I realized that putting a voltage-divider between the 5V and 10V rail, I would get 7.5V (referenced to ground) _as well as_ 2.5V (referenced to 5V), it was like a thousand light-bulbs turned on. "Ooohhhh. So _thats_ what they meant with potential, referenced voltages, etc!". I had heard the terms before, but obviously, I never understood what they actually meant.

I actually think it all boils down to that we need to give the beginners the vocabulary as well as the understanding of the words for them to be able to follow along in the more advanced topics.

So, what words do I think is needed?  In no particular order or relevance,
Referenced voltages (such as grounds-referenced), aka "voltage is relative, not absolute"
Differential signals
Floating circuits
Inrush
Ringing
Reflection
High-side / Low-side
Open-Drain / Push-Pull

I'm sure there's more though..

What do you guys think?
 

Offline rstofer

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Re: Vocabulary for beginners
« Reply #1 on: August 30, 2017, 12:06:52 am »
I guess I would wonder why.  All of those terms can be found via Google and there is always an empty tab on my browser.  If need be, I flip to a blank page and enter the term.
I expect those that are following along to go searching.

Besides, the list could get so long and cumbersome that nobody would use it.  Even if it was a sticky...
 

Offline rstofer

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Re: Vocabulary for beginners
« Reply #2 on: August 30, 2017, 12:09:13 am »
I guess I would wonder why.  All of those terms can be found via Google and there is always an empty tab on my browser.  If need be, I flip to a blank page and enter the term.
I expect those that are following along to go searching.

Besides, the list could get so long and cumbersome that nobody would use it.  Even if it was a sticky...
 

Offline buck converter

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Re: Vocabulary for beginners
« Reply #3 on: August 30, 2017, 12:42:37 am »
What do you guys think?

biasing
resonanc
transients

just my 3 cents
Just me and my scope.
 

Offline kalel

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Re: Vocabulary for beginners
« Reply #4 on: August 30, 2017, 12:48:09 am »
I might not be able to contribute much, but I would be interested in learning from it. :)
 

Offline Brumby

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Re: Vocabulary for beginners
« Reply #5 on: August 30, 2017, 02:19:16 am »
Here's something I wrote a while back:

What is the difference between positive, negative and common?
To put it as simply as possible .... it depends on where you put your multimeter probes.

The following diagram shows the same battery setup, repeated 3 times.  Each of these has been labelled with voltages.  All of them are correct in themselves (you cannot mix them).  The trick is - you pick one point as your reference point and call that zero volts.  Once you've done that, connect the black lead from your digital multimeter to that spot and leave it there.  Use the red lead for your voltage measurements.  (Note: This is safe with digital meters.  Analogue meters usually require special care.)


Quote

Example:

I always thought that if is do this I will simply double the voltage. 
So say the power supply's are 6V each, if I connect a bulb across the V+ and common or V- and common I would have 6V going through the bulb right?
Absolutely correct.

Quote
And if I connect the bulb across V+ and V- I would get 12V?
Again, absolutely correct.

Quote
So what is all this positive and negative voltage? :rant: :rant:
If I wanted to get pedantic, I would ask you - if you connect that bulb across the V+ and V-, have you put +12v across it or -12v across it?

The answer is, as I originally stated, it depends on where you put your multimeter probes.

So, if they are all the same thing, then you might ask - "Which is the best?"  The answer to that is "Whichever is the most convenient" - which leads us to an answer for this question:
Quote
Its especially confusing in Op-Amps. Why would I amplify a voltage to a negative?

There are several ways to try and explain this - but I will choose this one: Because the mathematics is easier.

If you take the case of a sine wave, mathematically, it cycles between + 2 and -2, so setting a reference point on a circuit where the signal follows the same movement makes it a lot easier.

For example, say we have an amplifier circuit that takes a sine wave input signal and produces an output that is twice the magnitude....

1. Using the middle example of mine, lets say the input signal varies from +7v and +11v  i.e. 4v peak to peak.
OK - let's multiply everything by 2!
This gives an output signal varying from +14v to +22v i.e. 8v peak to peak.  But that can't be right.  The high voltage of 22v is impossible.  The maximum is 18v.  If we were to put the signals on a scope, we would see that our maths is wrong.

Our maths needs some "tweaking".

The problem is that the amplifier does not amplify signals relative to our 0v reference - it amplifies them relative to somewhere between the two power rails.  With no input signal, we can see this is around the +9v mark and our scope observations confirm this.

We can now re-do the maths using this "offset" of 9v ... and it looks like this:
 - Input signal varies between +7v  and +11V   (4v p-p)
 - Subtract the offset and the "input signal" varies from -2v and +2v  (still 4v p-p)
 - We multiply the signal by the amplification factor of 2
 - The "output signal" varies between -4v and +4v  (8v p-p)
 - We add back the offset and the output signal varies from 5v to 13v (8v p-p) (confirmed by scope observation)



2.
Now let's try it with the first example.
 - Input signal varies between -2v and +2v   (4v p-p)
 - We multiply the signal by the amplification factor of 2
 - The output signal varies between -4v and +4v  (8v p-p)

Much easier.

Followed by this:

What I don't understand is how anything below 0V have any power?

Your confusion seems to stem from the notion that negative voltage equates to "negative energy".  This is not the case.  (Let's leave out the quantum mechanics for this ... OK?)

Let me offer you the following illustration: A person is standing, looking at a swing that is moving back and forth ... something like this:


Let's also say that that person considers where they are standing as the zero point - and any motion to the right is measured as positive.  Now, let's say when the swing is at it's furthest left position, it is 1m away from the observer and when it is at it's furthest right it is 5m away.  The swing moves through 4m and carries a corresponding amount of energy.

Now let's put the observer on the other side of the swing.  They are still facing the same way and still want to keep the same reference criteria: where they are standing as the zero point - and any motion to the right is measured as positive:



Now, when the swing is at it's furthest left position, it is 5m away from the observer, but in a negative direction and when it is at it's furthest right it is 1m away and, again in a negative direction.  The observer will write down that the swing as moving between -5m and -1m.  While these numbers are negative, the swing still moves through 4m and still carries the same energy that it did before.  Nothing has changed about the swing at all - just the observer's point of reference.



As a final example, let's say the observer stands in the middle (but out of the way enough that they don't get hit by the swing) with the same reference criteria: That they are still facing the same way, the zero point is where they are standing and any motion to the right is measured as positive.

They now report that the swing is moving between -2m and +2m ... yet it is still the same swing ... with the same amount of energy.
 


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