Understanding Electronics Basics #1

[Muttley600]

Still on post #317

Play with all you like. Also, in particular, note that we are using a logarithmic scale for the frequency. This is because it gives us a much better graph of the capacitor's charging curve. You're welcome to change that as well.

ok, I've had a play with log/lin/db curves & heres what I've found

Changing the amplitude range on SA or VG has no affect on whatever it is we are measuring in any curve which is strange because changing voltage was affecting caps in my post #302 by changing the waveform on previous sim (which should therorectically change the frequency as I'm understanding things so far)
So if this waveform isn't changing, we are measuring something totally different, is that correct **broken link removed**

Lowering the cap value makes the wave a higher frequency, lifting the value of the cap lowers the frequency

Lowering R gives a higher frequency but lifting value of R gives a lower frequency

onto phase & playing with values of caps, a small cap gives me a curve that drops at a low frequency while lifting value has a higher frequency before dropping

of course, I can see what is happening in front of me & how values of components are affecting the curves, now I'm hoping to read on & hopefully understand what the actual curves mean **broken link removed**

So a log curve is the charging curve of cap, one down two more db & lin

Feel free to correct me

Note how increasing capacitive value reduces the 3db frequency curve point.

I must be missing something, I can't see what your relating to **broken link removed**

 

[Muttley600]

If you make a circuit out of a VS, a load and gnd, the multimeter will read the frequency. I don't know the trick, yet, for doing that with any other circuit.

View attachment 62145

**broken link removed** That is SOOOOOO cheating, if you set the VG with frequency we don't need multimeter

Thats the best none answer so far,
sorry, but it seems funny **broken link removed**

but what you have indirectly shown me is that this frequency isn't something that is as obvious as V or A or even ohms, so this rates up there with phase in measuring terms so far

see, I always look deeper, so your none answer was a good answer **broken link removed**

 

[Muttley600]

Oh, btw I am a a self-taught (with some help) machinist. I had full machine shop privs at one time and that's nothing more than a mechanical drawing.

I used to do technical drawing at school, 3 lessons into new year, I got bored & started doing my own thing, got took to one side to ask what I was doing, I simply replied I'm bored so went through workbook until I found something interesting, teacher relented & marked my work, propbabley the only thing I did well at school, I asked what the problem was, he replied, that was the end of year exam piece **broken link removed**

One lead is longer to also denote polarity and to make it easier to insert. LED's have a long and a short lead too.

ahah, the anode & cathode, I was going to relate each one to + or - but I've just read that depends on component as to which way around they sre, that stopped me in my tracks, duly noted

 

[KeepItSimpleStupid]

My version of your story was "If you have something better to do, don't bother coming to class". I arranged to be a ghost in two courses. The one above was the second time.

Teach gave us time to do the homework in class. I kept finishing every assignment he gave me before anybody got finished with the first one. Then he gave me a problem not in the book "How about doing a loan amortization table". I asked "How do I do that?" Well, I finished that too.

 

[KeepItSimpleStupid]

The anode is (+) and the cathode is (-). That doesn't change. Making that relationship would have been right.

Lead length is a way of identifying polarity without reading the part. For larger LED's, you might notice a flat on one side. That is the cathode or the straight line symbol part of a diode.

 

[Muttley600]

Methods of determining frequency, 1. Frequency counter (high accuracy)

Whats one of those?

2. A scope (low accuracy)

understood

3. Scope with math functions

lets not go there just yet, step by step **broken link removed**

4. Scope with cursors (like the sim's)

how do you determine frequency by cursor?

 

[Muttley600]

CBB: I was thinking more along the lines of:
A
1) voltage source or SG and 2 resistors in series to ground, say 1K, 9K, 1K
b) The 1K to ground and the voltmeter or scope across the 1K
c) Vary the amplitude

I knew V altered the frequency 1V/5V/12V

View attachment 62241 View attachment 62242 View attachment 62243

 

[Muttley600]

My version of your story was "If you have something better to do, don't bother coming to class". I arranged to be a ghost in two courses. The one above was the second time.

Teach gave us time to do the homework in class. I kept finishing every assignment he gave me before anybody got finished with the first one. Then he gave me a problem not in the book "How about doing a loan amortization table". I asked "How do I do that?" Well, I finished that too.

A man after my own heart, art & drawings I loved, math & english **broken link removed**

 

[Muttley600]

The anode is (+) and the cathode is (-). That doesn't change. Making that relationship would have been right.

Lead length is a way of identifying polarity without reading the part. For larger LED's, you might notice a flat on one side. That is the cathode or the straight line symbol part of a diode.

oh, I read this **broken link removed** second paragraph

 

[Muttley600]

wow, I forgot I left sim in Tina running, then computer fan kicked in, so I checked the CPU, look what happened when I turned sim off

**broken link removed**

 

[Muttley600]

Post #324

B)
4. Set Xc = say 10K at 10 Khz

Is the 10Khz for the sinewave

Whats the 10K for, the cap?

Or am I doing this all wrong

View attachment 62245

 

[Muttley600]

missed this reply

Nope.. Remember that RMS is a measuring technique. You don't do it after the fact. e.g. if you had a unit sq ft, you wouldn't have a unit (sq ft sq meters).

So 175 is max, so needed to work backwards, so how do you work it out backwards?

Wrong again. It just means that the capacitor is suitable for removing ripple at higher frequency such as 100 kHz or more which is where switching power supplies operate. The filter caps that used at 50 or 60 Hz sometimes aren't suitable for use at higher frequencies.

fair enough, but we need math for that & my learning on that front has taken a backwards step due to time constraints, have a few days to try & catch up a bit now

Capacitors do store a fair amount of energy and thus can delay the turn off of the soldering station.<snip>

1 out of 3, better than none out of 3 **broken link removed** always the optimist

 

[Muttley600]

They both work, but now we have to teach what a db and logs are? Darn math.

don't forget lin, I'm hoping she's pretty **broken link removed** let me guess, wrong again, it has something to do with a graph......sigh, an theres me getting myhopes up **broken link removed**

The RC product is tiny so the frequencies are higher where this occurs.

What is RC **broken link removed**

In any event, we can predict the frequency where the output is 70.7%, 0.707 (really the sqrt(2) decimal or -3db down from the beginning of the sweep.

**broken link removed** We've covered that number but I haven't got round to updating my folder yet (slowly going back & making notes so I can refer to it easier when brought up again) this was some kind of rule wasn't it, but can't remeber for what at the moment

I did something totally absurd here, by using way too many decimal places on purpose. The capacitance in a real circuit could have a +-5% tolerance and the resistor a 5%, 1% or lower tolerance, thus the number of digits I calculated to doesn't make any sense.

Phew.....I thought it was just me **broken link removed**

The point here is, that resistors can divide voltages at all frequencies and the voltage across a capacitor is frequency dependent.

Is that due to the R being instant & the cap having a time delay

The graphs look funny because they are on a log scale. Earthquake magnitudes are on a log scale. e.g. a 4 on the Richter scale is 10 X more powerful than a 3 on the Richter scale and a 5 is 100 times more powerful than a 3.
Logs have bases. The two common ones are Log base 10 and log base e, usually denoted by LOG10(x) and LOG(x).

What sort of graph was you hoping for **broken link removed**

 

[Muttley600]

Look below and tell us which graph is better at depicting the effect of frequency on the reactance of a capacitor?
In my mind, the LOG frequency scale does a better job (than the Linear scale) of showing how a capacitor "reacts" (opposes voltage) to an increasing frequency. Just my opinion. Either scale shows the exact same thing, just differently.

I concur, it seems to show the bigger picture of how it is working due to givng fuller curve

Also notice how the LOG Scale example emulates the shape of a sine wave.
ALL this stuff, eventually, ties together in a nice, neat package.

I know it will all come together, quite exciting really, if you keep piling small rocks you could end up with a big mountain........unless someone pinched one from the bottom & it collasped **broken link removed**

That's where the math comes in.

Yes, but the image of works will make that so much easier

If you look closely at the way the LOG scale is divided, you may see the method to the madness.

that's something to do with that 707 number again isn't it, I'm going to have to go back & find that, it's starting to bug me **broken link removed**

Whatever the case, You're still going to have to learn what a LOG scale is.
dB is another issue. But that, too, will eventually have to be learned.

I'm happy to learn, although time isn't always on my side I feel like I'm starting to see a slowly emerging picture, I'll get there given time **broken link removed**

 

[KeepItSimpleStupid]

Ripple in RMS: You can only work backwards if the ripple is a sine wave. The best you can do is find the p-p or 0-peak value.

No math. The manufacturer of the capacitor has control over it. It might, for instance be the choice of the dielectric (insulator), the foil, and the electrolyte choice. High ripple caps are more expensive.

 

[Muttley600]

Here ya go:

[latex]Vrms=\frac{Vpeak}{\sqrt{2}}[/latex]

where [latex]{Vpeak}[/latex] is the 0 (zero) to peak VAC value. In this case 1.92VAC (1/2 of 3.84)

The [latex]{\sqrt{2}}[/latex] is a constant, in the case, 1.414. This value never changes, no matter what happens in the rest of the equation.

so,

[latex]Vrms=\frac{Vpeak}{1.414}[/latex]

Thus, [latex]\frac{1}{1.414}=0.707[/latex].

Finally, in this case [latex]Vrms=0.707 * 1.92=1.357[/latex] RMS VAC

I've found that annoying number stuck inside my head, so we are learning how to tie in V with frequency, I knew it was related, but now we are delving deeper into function of it rather than skimming the surface, layer by layer **broken link removed**

 

[KeepItSimpleStupid]

I was off by a power of 10 for the Richter scale. A 5 is 10 times a magnitude 4. I should have checked. The rain got to my head again.

https://en.wikipedia.org/wiki/Richter_scale

 

[Muttley600]

Ripple in RMS: You can only work backwards if the ripple is a sine wave. The best you can do is find the p-p or 0-peak value.

No math. The manufacturer of the capacitor has control over it. It might, for instance be the choice of the dielectric (insulator), the foil, and the electrolyte choice. High ripple caps are more expensive.

So with a mArms number like that, you would simply make sure you don't exceed it, would it be wise to allow a clearance?

 

[Muttley600]

I was off by a power of 10 for the Richter scale. A 5 is 10 times a magnitude 4. I should have checked. The rain got to my head again.

https://en.wikipedia.org/wiki/Richter_scale

Breath easy my friend, we forgive you **broken link removed**

of course we knew you were wrong, just didn't like to mention it









CBB, whats he on about **broken link removed**

 

[KeepItSimpleStupid]

RC is just R * C or the product of R & C.

i.e. R = 10 M ohms, C = 1 uf
RC = 10,000,000 * 1E-6