Understanding Electronics Basics #1

[cowboybob]

so you work out frequency on a circuit by simply dividing V = R by cap value

Going to have to burst a bubble here.

V (voltage, or amplitude) has nothing whatsoever to do with (or affect) frequency. Nothing.

There are, to be sure, circuits (called voltage controlled oscillators) where a voltage level (amplitude) is used to alter the frequency produced by the circuit.

But in our case (a simple RC circuit) the voltage level (amplitude) of the VG (sine wave voltage generator), or the SG (signal generator supplying the VG) has no bearing on the frequency produced.

None.

In an RC oscillator circuit, the ONLY components affecting the center oscillating frequency are the resistor and capacitor values (the RC,or "time constant").

Period.

In your three examples of the RC circuit (Post # 347) where you varied the VG's output ammplitude, the ONLY variation in the signal on the scope was in the PP value shown on the scope.

As to changing the resolution of the Signal Analyzer, just type in the number. Best to use a number like 128, or 256. Generally, though, 100 is adequate. Higher resolutions slow up the graph production and don't really add anything to the understanding of the effect demonstarted.

 

[Muttley600]

Morning CBB
I have had many bubbles burst it is all part of the learning process

I think I'm getting muddled between circuit resistance & RC, it sounds like they are two different things, so RC is purely to alter the cap resistance over time altering the frequency, but surely this resistance must be allowed for in the circuit? Or is it so small that it is negligible ?

 

[cowboybob]

but surely this resistance must be allowed for in the circuit? Or is it so small that it is negligible ?

Yes, overall circuit resistance IS a factor.

But, generally speaking, it is of little consequence in determining the resonant (or "natural" frequency) of an RC, or RL, or RCL circuit, especially in the lower (say less that 1MHz range). This is due to the FAR greater R of the RC circuit that the R of the surrounding and associated circuitry.

And, in our case, there IS NO surrounding circuitry, just the VG, the R and the C (and, of course, the VF output terminal).

For the most most part, the RC value is the only significant controlling factor.

And, with respect to what we saw in the SG examples, We altered the frequency being applied to a fixed RC circuit (their values did not change) to observe how that circuit did, or did not and to what degree, allow the voltage level to make it through to the output.

As KISS was explaining, there's "resistor" resistance (CONSTANT and linear) and capacitive "resistance", more accurately called "reactance" that VARIES and is NOT linear because of its varying response to frequency, again, as demonstrated by the FG examples.

Does that help?

We'll keep at this concept as long as necessary...

 

[Muttley600]

As KISS was explaining, there's "resistor" resistance (CONSTANT and linear) and capacitive "resistance", more accurately called "reactance" that VARIES and is NOT linear because of its varying response to frequency, again, as demonstrated by the FG examples.

Does that help?

We'll keep at this concept as long as necessary...

Sorry, been playing bikes again, have been helping friend with his this time

I think I understand now, I need to think of 'Reactance' as more of a control of a component rather than resistance even though they sort of mean the same thing, this is just a miniscule amount in real terms hence measuring on graphs so it enables to see what we are on about & set the required frequency

So now I moved my understanding towards the phase (which is the start of the cycle (Hz))
but am I now right in thinking that the reactance is the ability to control the time of the complete cycle by using different value caps giving us required frequency

Just had another look at LOG graph, so what your showing me is the fact that the lower the V the higher the frequency, the higher the V the lower the frequency, it is simply a graph showing what gain (V) (or my words-charge) you would have at a set frequency in the cap over one Hz

So that would make KISS's example of -3db always 707m the charge capacity at a given frequency, does that make Xc a simple way of finding a charge rate for cap over one Hz, then somehow it is going to have to be tied in with the phase - start time of the cycle

This would mean you cannot choose any old R as it affects the frequency, it is part of the circuit in a bigger way than I ever imagined, they actually a crucial role to play, so if the cap is the cycle monitor, R must be the control room setting

I know Im close but I could see R as setting the gain (which is the cycle) I'm struggling to break them down into their own jobs, maybe I have it backwards & the cap alters

I don't honestly know, going for a soak in bath cause my heads spinning, I just can't see this at all, I know you've explained it but I don't know how to make sense of it

Will try again later hopefully with a clear head

 

[KeepItSimpleStupid]

You want to make it simple: Frequency was the input (control) and a resistance was the output.

This is close enough for now.

-3db is nothing more than 70.7% or 0.707 or √2 divided by 2 of the reference input. The frequency this occurs at for a resistor and capacitor in series is f=1/(2∏RC)

 

[cowboybob]

I think I understand now, I need to think of 'Reactance' as more of a control of a component rather than resistance even though they sort of mean the same thing, this is just a miniscule amount in real terms hence measuring on graphs so it enables to see what we are on about & set the required frequency

Resistance and reactance BOTH are that quality of a component that lowers the charge flow (current) in a circuit.

One resistor, all by its lonesome (at a constant temperature) will resist current very predictably and always the same way, whether it is an AC or DC current. The resistance is totally predictable (and "linear": a straight line, no curves) based on its resistive value ONLY. A 1k resistor will provide the same current reduction in a circuit for DC or for AC (regardless of the frequency). Always.

If you were to put another resistor in series (or parallel) with the above resistor, its resistive function in this new circuit would be, once again, the same, linear and utterly predictable, no matter the signal type nor frequency.

In the case of a capacitor (say, 1uf), however, the amount of its resistance to current is NOT always the same. In particular, it varies considerably with a variation in the frequency of the signal being applied to it.

And if you add another component (in series or parallel), such as a resistor or an inductor, that component will radically alter how the capacitor "reacts" to current. It is predictable, but NOT linear.

So now I moved my understanding towards the phase (which is the start of the cycle (Hz) but am I now right in thinking that the reactance is the ability to control the time of the complete cycle by using different value caps giving us required frequency

Two issues here. 1. Yes, the phase value is the start of the cycle, generally in relation to zero degrees (up to the 360° in one complete cycle). And 2. by controlling the timing (which is to say, frequency) with capacitors (and other components as well) one can alter the frequency of the signal.

Just had another look at LOG graph, so what your showing me is the fact that the lower the V the higher the frequency, the higher the V the lower the frequency,

Well, if you're looking at the graph from right to left, then, yes. But keep in mind that the V is nothing more than a representation of the amount of signal getting through the cap. As the frequency goes up, eventually NONE of the signal gets through.

it is simply a graph showing what gain (V) (or my words-charge) you would have at a set frequency in the cap over one Hz

This part is correct. But, please see the last sentence in this post.

I think you may be confusing the term Cycle with Hz. One cycle is one cycle, that all. Hz is the number of repetitive cycles over a given period of time (generally one second), i.e., frequency.

So that would make KISS's example of -3db always 707m the charge capacity at a given frequency, does that make Xc a simple way of finding a charge rate for cap over one Hz, then somehow it is going to have to be tied in with the phase - start time of the cycle

You've mixed and matched several differing concepts here.

1. dB. A method of considering the "gain" (think volume) of a circuit. It can be positive (+ going up) or negative (- going down) but ALWAYS relative to the initial input level. It is a value that is a result of a circuit. It is not something that affects the circuit (it's an output, not an input).
2. Xc is something that has to derived. That is, whereas a resistor has clearly stated on it what its resistance is, a capacitor NEVER has its reactance stated because its reactance varies all over the place depending on many differing situations.
3. And neither of these is affected by phase. We should probably skip a more thorough discussion of phase fro a later time. It's important, but not all that relevant at the moment.

This would mean you cannot choose any old R as it affects the frequency, it is part of the circuit in a bigger way than I ever imagined, they actually a crucial role to play, so if the cap is the cycle monitor, R must be the control room setting

(My emphasis).

Yes to the first part. I'm not sure of your analogy in the underlined part.

Maybe it would help to realize that the SA (Signal Analyzer) example graphs only show a voltage level (you could almost think of it a DC level) as a sinewave of increasing frequency is presented to the RC circuit. Granted, the graph looks like a wave, but it isn't really a wave.

It's just showing us how much of the sinewave, at any given frequency, that is going into the RC circuit is getting through it to the output of the circuit.

 

[KeepItSimpleStupid]

Graham: Stay out of this
CBB:

In no particular order.

1) Should we wait until most of the current stuff gets understood at whatever level we are at?

2) Do you want to at this point to TRY to illustrate phase? e.g. in your RC SIM, just measure the voltage across the capacitor and the voltage across the resistor on a dual trace. Remembering that the voltage across the resistor is directly proportional to current.

3) Take a timeout and introduce Pythagoras and the right triangle, vectors and then on to R, Xsub L and Xc?

4) Show how to use cursors in the SIM?

1,4,2,3 ?

 

[cowboybob]

KISS,

Sounds like a plan to me. So, yes.

And if that's a suggested order at the bottom, then also yes, as you've arranged it.

We can see how Graham gets on with the above.

Meanwhile, I'll work on sim examples of phase and cursor usage. As a side note, isolating the current phase from the voltage phase will be easy enough to demo, but a bear to explain, at this stage.

OK, Graham, you can look now...

 

[KeepItSimpleStupid]

As a side note, isolating the current phase from the voltage phase will be easy enough to demo, but a bear to explain, at this stage.

Hopefully it won;t be as bad as you think. Instead of using the voltage across the resistor, how about putting an ammeter in the circuit. It might help the understanding? In that case you do have V and I and not something "directly related" to I.

[Crossing fingers and toes]

 

[cowboybob]

Good idea. Will do.

 

[Muttley600]

ok, bath was nice nice & phoned mate to drown my sorrows about being a numpty, drowning sorrows didnt work cause I dont drink lots of beer (flaw 1**broken link removed**) but had nice social chat about normal stuff which has helped clear my head, just got back to (flaw 2, cuppa tea tastes nicer than beer **broken link removed**)

You want to make it simple:

oh go on then **broken link removed**

Frequency was the input (control) and a resistance was the output.

but you said you were going to make it simple, this is whats confusing me, your saying frequency was the INPUT, I thought a cap altered (what we are working towards) Frequency, my misunderstanding might be greater than you could imagine **broken link removed**

This is close enough for now.

but you know thats not true really don't you, I won't let this go, you know I won't, yes I might struggle to understand but once I can see what we are trying to achieve peace will be restored

Before I read on:
I was seeing frequency as the length of the sinewave (which I think still stands) & I thought we were working on how to set frequency, but it seems not, have we jumped a step straight to phase?

I'll continue reading now

 

[KeepItSimpleStupid]

The length of one cycle is called the period and it's measured in time units, like 20 mS. 1/(period) is the frequency. So 1/20 mS is 50 Hz. See if you can get that answer on your calculator?

Capacitors change their behavior based on frequency. When capacitors are part of a timing circuit, the quantity R*C is usually involved. That quantity is called the TIME CONSTANT. Here is a reference to a popular IC used for generating a square wave. **broken link removed** You'll find the actual datasheet under documents.

 

[Muttley600]

Sorry guys, please dont hate me but I need sleep, had 1.5hrs in last two days which could possibly not be helping me to concentrate
I did look quickly at other posts but can't take it in & I don't want to make same mistake as last night to still misunderstand

Back tomorrow, I can't believe I caught up to end up back at square one

Promise I'll be able to keep my eyes open tomorrow

 

[KeepItSimpleStupid]

Yep, you were up too late last night. The temperatures are starting to get warmer on this side of the pond.

 

[KeepItSimpleStupid]

Oh, your looking at caps as doing one thing, but the wrong thing. The property of a cap is that "they will not let the voltage across them change instantaneously". That's what they do.

They are used to start motors, they are used to change the phase (we'll get to that) on say a synchronous motor, they are used to reduce noise (RF suppression), they are used to filter (the power supply example), they are use to remove DC, they are used to couple (pass) AC signals (amplifiers), they are used to perform mathematical differentiation, they are used to perform mathematical integration, They are used in oscillators and they are used as brief backup power sources to name a few applications.

 

[cowboybob]

Graham,

I think it would be useful to put AC aside for a bit and revisit DC. We need to nail down the relationship between to current, resistance and voltage.

Below I have laid out some basic DC circuits with nothing more that a simple resistive load.

Note the circuit(s) construction, with resistance, voltage, and current values for each example. I'll include one of the math solutions.

You'll have to do the rest.

Ohm's definition, V=IR, or I=V/R, or R=V/I.

Note ALL circuit component values!

One 1kΩ resistor circuit:

View attachment 62334View attachment 62335

The math: V=10mA * 1000Ω, Therefore V=10VDC. Now you're on your own.

One 500Ω resistor circuit:

View attachment 62336View attachment 62337

Two 500Ω resistors circuit:

View attachment 62323View attachment 62324View attachment 62325

Two 500Ω resistors circuit (again, with a change):

View attachment 62326View attachment 62327View attachment 62328

Do the math so you can see the why of the current.

Now here's one that shows only the voltage output of V1 and the current through R1.

View attachment 62329

What is the value of R1?

And, finally, at what voltage is V1 set to?

View attachment 62330

Have fun...

CBB

 

[Muttley600]

Sorry guys, woke up this morning around 10am, had a coffee & smoke & went back to bed til 12:30pm.lol
Been round parents all day as mothers day, just had tea, back in ten to start again

 

[Muttley600]

Yep, you were up too late last night. The temperatures are starting to get warmer on this side of the pond.

I was, I went to bed at 7.30am & was up again at 9.00am **broken link removed**

We get typical British weather here, we have winter, then it rains & rains **broken link removed**

 

[Muttley600]

I think it would be useful to put AC aside for a bit

You noticed I've hit a brick wall then, can't get my head back in gear **broken link removed**

I'll include one of the math solutions.

One 1kΩ resistor circuit:

The math: V=10mA * 1000Ω, Therefore V=10VDC. Now you're on your own.

Shall I look at that as I=10mA

One 500Ω resistor circuit:

20mA * 500ohms = 10V

Two 500Ω resistors circuit:

1= 10V / 1000ohms (2*500) = 10mA
2= showing 20V because nothing between battery & VF1 to change it but I'll give you the other reading anyway
VF2 - 20mA*500ohms = 10V

Mind block again, what is up with me, gonna have a smoke, don't know whats up with me, can't get back into this at all
3= already worked that out on previous but half due to battery change

Two 500Ω resistors circuit (again, with a change):

1= As above again
2= it's coming back, starting not to have to think now, thank god for that
3= As above

Do the math so you can see the why of the current.

Never mind the math, somethings clearing inside my head, I had a total mental block

Now here's one that shows only the voltage output of V1 and the current through R1.

What is the value of R1?

1.5V / 2.5mA = 600ohms

And, finally, at what voltage is V1 set to?

3000ohms*0.004mA = 12V

look at that, even decimal point in the right place

Have fun...CBB

I needed that, seems to be clearing my head

V can be altered around circuit but I is spread even throughout circuit, so I is only ever a product of V & R, I see where your going with this, duly noted in folder

Told you I'd woke back up, thanks CBB

 

[cowboybob]

V can be altered around circuit but I is spread even throughout circuit

Exactly. Very good. AND, that is true for AC as well. Remember this.

Now, to be sure, this is a simple series circuit. Parallel circuits get a little more tricky (we'll do them next).

But your quote is still true.

The math: V=10mA * 1000Ω, Therefore V=10VDC.

Good catch.

3000ohms*0.004mA = 12V

look at that, even decimal point in the right place

Also very good. The decimal thing has been bugging you. Glad to see you're getting past that snag.

What is the value of R1?
1.5V / 2.5mA = 600ohms

No. Look a little more closely at the displayed circuit values. Don't forget what the battery is called.

I'll post the parallel stuff tomorrow.

Proud of ya...