School me. Capacitive Reactance.

[Kipper4]

A few days back looking at this

http://www.muzique.com/lab/sat2.htm

I noticed that the 6th and 7th examples in on is a 3.9nf cap and the next a 39k resistor.
So I did a little digging and found a capacitive reactance calculator.
I guessed at a 1khz signal and came up with a very near result 40.something k ohms. I assume Jack must have gone for the nearest value.

The calculator while handy does not really teach me much.
Please school me on capacitive resistance.

Xc =    1
       -------
       2xPi f C

An example would be nice with decimal point. I say this because I saw an example with this 10 -9
In it and don't understand it.  Link here

http://www.electronics-tutorials.ws/filter/filter_1.html


Thanks.

[DrAlx]

10³ = 1000  (i.e. kilo)
10^-3 = 0.001  (i.e. milli)
10^-6 = 0.000001  (i.e. micro)
10^-9 = 0.000000001 (i.e. nano)

[DrAlx]

So take this as an example with some numbers.  I assume you know how a voltage divider works.

Lets say you have a "voltage divider" consisting of a 22 nF capacitor and a 47k resistor like this:

IN----- 22nF-----OUT-----47k----GND.

The "resistance" or rather reactance of the capacitor is frequency dependent.

At a frequency of 100 Hz:
The capacitor has a reactance of  1/(2 * 3.14 * 100 * 22 * 0.000000001) = 72379. (i.e. 72k)
Therefore Vout/Vin at 100 Hz = 47000 / (72000 + 47000) = 0.4.
So the output voltage is 0.4 of the input.


At a frequency of 10000 Hz:
The capacitor has a reactance of  1/(2 * 3.14 * 10000 * 22 * 0.000000001) = 723 (i.e. 720 ohm)
Therefore Vout/Vin at 10000 Hz = 47000 / (720 + 47000) = 0.99.
So the output voltage is almost the same as the input.


So the voltage divider is a high pass filter.  It lets high frequencies through, and attenuates lower frequencies.

[Kipper4]

Ok Thanks Doc I get the power to thing now.

[R.G.]

/quote]
You are 100% spot on. People think I'm crazy for telling them not to use online calculators until they understand what they're calculating, but the fact is that unless you understand - which you are trying to do! - using the calculator actually impedes understanding because it lets you get an answer, maybe even a correct one, but it doesn't get you to think.

Good on you!

[PRR]

> example with this 10 -9 in it and don't understand it.

220 nano-Farads. nFd.

Young guy like you musta seen "nanos" around before. (I'm still resisting pFds.)

"nano" sounds like "nine" and is entered as 9*10^-9. That's easier than "0.000,000,009".

You REALLY should run your finger over a Reactance Chart. Pick some popular cap value and trace them over some favorite frequency range, you get a feel for what reactance that cap has in your work.

Or just work it out by octaves (well, decades are easier):

0.01uFd:
16Hz = 1 Meg
160Hz = 100K
1,600Hz = 10K
16KHz = 1K

[Transmogrifox]

Lets say you have a "voltage divider" consisting of a 22 nF capacitor and a 47k resistor like this:

IN----- 22nF-----OUT-----47k----GND.

...

At a frequency of 100 Hz:
The capacitor has a reactance of  1/(2 * 3.14 * 100 * 22 * 0.000000001) = 72379. (i.e. 72k)
Therefore Vout/Vin at 100 Hz = 47000 / (72000 + 47000) = 0.4.
So the output voltage is 0.4 of the input.

Even though this introduces more advanced mathematical concepts, I want to point out that the simple voltage divider equation is not correct when combining real and reactive elements.

In this example the capacitive reactance is -sqrt(-1)*72k ohms:
Physicists and mathematicians would write this:  -72k*i ohms
Engineers would write this as :  -j*72k ohms

i = j = sqrt(-1),  I use "j".

The resistor divider is now:
47k/(-j*72k + 47k)

The denominator is a complex number and it can be represented as a magnitude and phase.

The magnitude is the same as a formula for computing the hypotenuse of a right triangle:
|Z| = sqrt[ im^2 + re^2 ]
im --> means imaginary
re --> means real

the "im" part is the part with the "j" in front.  The "re" part is the part without the "j".

To find out the magnitude of the output (what amplitude you will actually see on an oscilloscope), the formula is this:
47k/sqrt[ (-72k)^2 + (47k)^2] = 0.545

So the output at 100 Hz is actually 0.545, and not 0.4 as the simple voltage divider equation would tell you.

Now what happens when Xc = R?
47k/sqrt[(-47k)^2 + (47k)^2] = 0.70712
20*log10(0.70712) = -3.01 dB

Look familiar?

This is called a "pole".  This happens where the magnitude of the imaginary part is equal to the magnitude of the real part.  In a 1-pole low-pass or high-pass filter, the magnitude of the filter response is -3dB at this frequency.  We often call this the "3 dB cutoff".

How do we find out at what frequency is the pole?

Xc = -j/(2*pi*f*C)
If pole (f3dB) is when  -Xc = R

R = 1/(2*pi*f*C)

Then, simple algebra tells us:
f = 1/(2*p*R*C)

Now maybe it is more apparent how we get the formula from the "-3dB cutoff" on RC filters from the resistor and capacitive reactance.

[Fast Pistoleros]

a good way to think is of it as impedance, and not resistance which is a fixed value and adheres to ohms law- you are now mixing ac with dc in a circuit and ac has a frequency

I was just given this book and trying to wrap my head around a few concepts lol  - I really dont have time to read it just yet as I am still working n rectifiers and power supply ..I may have gotten in over my head enrolling in EE 

[PRR]

> the simple voltage divider equation is not correct when combining real and reactive elements.

Quite true.

The cave-man hack is to assume that "flat is flat" and "slope is slope", but at the "corner", where flat meets slope, R=R network gives half-voltage (-6dB), the R=X network gives 0.707 (-3dB) at the corner, -1dB off flat/slope an octave either side.

See Bode Plots.

[~arph]

Ah memories, pole plots. Takes me back to highschool/uni years. Sadly all my analog electronics math skills have nearly evaporated.

Side track:

One thing I have always wanted to do is to make an online javascript based version of the duncan tonestack calculator. For a couple of reasons.  One of them being a non-windows person. But the other to dust off my analog systems math like Norton and thevenin equivalents, reactance etc.

Anyone with a fresh skill in analog circuit math interested in collaborating? like taking a certain tonestack (say james) and show us how to calculate the frequency response at a certain frequency. Then I will convert this to a interactive web based tool. I know there is tons of simple filter calculators online, but nothing interactive focused on tonestacks. Which is why I still see a lot of referrals to Duncan's app.

[Kipper4]

Bode plot?
Time for a Google.
Thanks guys

[DrAlx]

... I want to point out that the simple voltage divider equation is not correct when combining real and reactive elements.

I know but given that the OP wasn't familiar with the 10^-9 notation I assumed a non-technical background and figured trying to introduce complex variables into the explanation would have been too much. I should have mentioned that the division doesn't actually work as I  said and that to get the correct result you have to do a messier  calculation.  Not sure what the best way to "school" someone on reactance without introducing complex variables or pictures with phasors.  For anything other than the simplest circuit fragments like HPFs or LPFs the calculations will quickly become messy.

If I was going to "school" the OP on reactance with a hand-wavey explanation I would summarise by saying it is NOT the same thing as resistance (since there is no energy loss) but it shares some similarities with it.  It is similar to resistance in the sense that something with low reactance  placed in series with a signal will tend to let it through (as would a low resistance).  Conversely a high reactance will tend to block signal (as would a high resistance).  The other thing I would tell them is that reactance is frequency dependent, and will cause phase changes on the signals.  So for example a high frequency signal may see a circuit component (e.g. a capacitor) as having a low reactance, but a low frequency signal will see that same component as having a high reactance.
This frequency dependent behaviour makes it become useful (in fact necessary) to describe a circuit's in terms of how different frequencies see the circuit (i.e. the circuits frequency response).  A Bode plot is an example of that.  It shows how some frequencies get through while others are blocked.

Because reactance and resistance are not the same thing, the maths for combining them and calculating with them is complicated. So people tend to let computers calculate things for them and show them results as nice graphs for how  frequencies are affected.

Now I am not sure how much better an explanation this is than saying that "capacitors tend to block low frequencies and pass high frequencies".   There's an inkling that there may be similarities to simpler concepts like resistance, but when it comes to calculations, I can't see a way of explaining why the calculations work as they do without making things abstract, whether that's introducing complex variables or diagrams with phasors, or god forbid... Smith charts..

It's all well and good to say that you represent a capacitors reactance as an imaginary number but the person wanting to be schooled may ask why on Earth would you do that and who came up with that anyway?  So you might say that it's related to Fourier transforming the equation that descibes how the voltage and current behave in the capacitor.  They say what's a Fourier transform?  It goes on...

So my own take is that for someone with a non-tech background, its probably best to give an intuitive understanding of what reactance is, but as far as trying to calculate with it, use a tool to do the job.

[Kipper4]

I'm still at the basic stage you are correct Dr Alx
i intend to do some calc and get familiar with it before i move on to the later stuff
this is all appreciated

[Transmogrifox]

I think it's the pedant in me can't let it slip by without at least mentioning there's more to it  

Kipper -- I think DrAlx did a good job of keeping it simple.  Even if you don't understand the concepts I presented, just keep it in the back of your mind that there is more going on than a frequency dependent resistor.

As PRR pointed out the actual error between the simple approach and the "correct" approach is not much once you go an octave out.  Bode plots are a nice graphical way to understand what's going on, so it's worth looking them up.  Anybody who seems to hint that Bode plots and Smith charts are from the devil probably hasn't worked through the pages of calculations needed to get the same answer you can pull off the chart at a glance.

Ultimately this all relates to the time dependence of the current-voltage relationship of a capacitor (as Alx hinted when he talked about Fourier Transform):

It takes a large current to change the voltage quickly on a capacitor.
If it takes a large current to change voltage quickly, then that results in a larger voltage drop on a resistor feeding a capacitor. 

Slow changes then result in very little voltage drop across the resistor, fast changes result in a lot of drop on the resistor.

The only reason this frequency-domain stuff is useful is because some guy named Jean-Baptiste Joseph Fourier came up with a way to represent most any mathematical function as a sum of sinusoids of varying phase and frequency.

The key is that both signals and systems can be represented this way and so a pure sine wave applied to a system has a practical meaning with respect to complicated signals. Our ears confirm these mathematical gymnastics are valid when we use it to manipulate those complicated signals when they are coming from musical instruments.

[Kipper4]

It's all good Transmogrifox.
No such thing as too much information.
It may take me a while to get my head around some of it but get it i will.

Plus side it's an excellant resource for other members and guest to learn from.

[PRR]

> Bode plots and Smith charts are from the devil

Bode is heavenly.

Smith is hellish.

Reactance Chart forces you to think on a slant, but really is the fast no-calculator-error way to rough-out any R-C(-L) problem. You can't read it better than 10% but that's usually way-ample for us (especially if struggling to get "any" answer).

Problem is Reactance Charts have gone out of style. There's a decent one in the back of Radiotron _3rd_; that's where I learned and is still my go-to. There are a few on-line but they tend to go to "modern" frequencies, many MC (MHz). Try this one: http://www.vibrationworld.com/AppNotes/Image8.gif  Note that it calls "pFd" "uuF", no nanos.

Take AMZ's example with 39k against 0.01uFd.

39K is a horizontal line somewhat above "10,000 Ohms".

0.01uFd is the slanty line which starts from "0.01uF" on the right side.

These intersect at 400 Hz (somewhat left of "100~").

0.01ufd is more than 39K reactance from zero to 400Hz, less than 39K from 400Hz on up.

By inspection, we figure that 0.01uFd won't suck against 39K from bass to near 400Hz, and above 400Hz the 0.01uFd will suck the 39K increasingly hard. High-cut.

Whether you do-math with 2-part numbers (not imaginary, that's just a name), or cave-man it with a "-3dB at corner" Bode curve, you come to essentially the same result.

And note that we did not need a "nFd" or "10^-9" key on our calculator, nor slip zeroes or decimal points with fumble-fingers. And if you do math the right way, with a slide-rule, the Reactance Chart gives you the zeros and decimal points directly, the slide-rule gives you "408" where the chart was just "about 400". (e-Calculator gives 408.0895977, which is totally bogus accuracy and IMO detracts from the solution.)

[Kipper4]

I did some calculus in my downtime at work today.
I think I'm getting it. I'm certain this will help me design better in the future.
Great stuff

[Kipper4]

Here's my homework

Using Dead Astornauts. Ridiculously simple cab sim
Ref here
http://www.diystompboxes.com/smfforum/index.php?topic=113287.0

Signal freqauncy 100hz

Filter 1
Hpf 220n-4k7 to gnd

Xc = 1/2pi F C
Xc = 1/ 2*3.14*220*100*0.0000 00001
= 7238hz
Therefore Vout =Vin @100hz *4700 / 7238+4700 = 0.393 of Vin


Filter 2

Lpf in-2.2kohms out-100nf-gnd

Xc = 15923 say for argument sake 16kOhms
Vout = Vin * 2200 / 15923 *2200
Roughly 2200 / 16000 * 2200 = 2200 / 35000 = 0.063 of Vin


Filter 3
LPf   In - 2k2- out- 22nf - gnd

Xc = 72463 ohms
72kohms
Vout = Vin* 2200 / 72500 +2200 =
2200 / 74700 = 0.03 of Vin

Did I get it ?

[ashcat_lt]

2-part numbers (not imaginary, that's just a name)

I prefer to think of it as vectors, because it's a bit more visual, is more like basic trig, and really boils down to Pythagoras.

[PRR]

> Filter 1
> Hpf 220n-4k7 to gnd
> Xc = 1/2pi F C
> Xc = 1/ 2*3.14*220*100*0.0000 00001
> = 7238hz

How do you get "X" in Hz??

Keep your units straight or you will get all confused.

Yes, 7,238 OHMS.

The "4700 / 7238+4700" is dubious because 4700 and 7238 are not very different. Easily a couple dB error here, because R-X dividers are not R-R dividers.

>         220n
> in.......//........2.2k......2.2k.......out
>                /              /            /
>                /              _           _
>                / 4.7k       _ 100n  _  22n
>                /               /           /

> Signal freqauncy 100hz

So are you going to plot the response at every possible frequency? Tedious.

Plot the corners! 220nFd and 4.7K is 154Hz. Flat above, slope-down below. 100nFd and 2.2K is 724Hz, roll-off above. 22nFd and 2.2K is 3,290Hz, faster roll-off above. Already you can sketch the general shape, and read-off the approximate loss at ANY frequency. Remember that "sharp" corners are not possible, sketch them 3dB down at the corner.

When you have more than one R-C network, no buffers, no large impedance difference, everything interacts. The LPF load is not just 4.7K, it also pulls on the 2.2K+100nFd. In this case the 220 and the 100 tend to pull everything down. Plotting the exact response by hand will be very tedious. Simulators are labor-savers (still your job to know when they are lying to you).