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.
Note how increasing capacitive value reduces the 3db frequency curve point.
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
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.
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.
Whats one of those?Methods of determining frequency, 1. Frequency counter (high accuracy)
understood2. A scope (low accuracy)
lets not go there just yet, step by step **broken link removed**3. Scope with math functions
how do you determine frequency by cursor?4. Scope with cursors (like the sim's)
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
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.
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.
B)
4. Set Xc = say 10K at 10 Khz
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).
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.
Capacitors do store a fair amount of energy and thus can delay the turn off of the soldering station.<snip>
They both work, but now we have to teach what a db and logs are? Darn math.
The RC product is tiny so the frequencies are higher where this occurs.
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.
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.
The point here is, that resistors can divide voltages at all frequencies and the voltage across a capacitor is frequency dependent.
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).
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.
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.
That's where the math comes in.
If you look closely at the way the LOG scale is divided, you may see the method to the madness.
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.
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
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.
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
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