Finding Capacitance Change

[ChristianCMcKay]

Okay, so I've been checking a lot of capacitors lately. I've been trying to find out exactly what the capacitance was. So here was the method I was going about to do this.

First I solve for Xc

I know that Xc = 1/(2*3.14*f*C)

So then I set up a low pass filter. I then use the forumula

Vpp = (Xc/[R + Xc])*Vin

So I measure to see if I'm relatively close. In one case, I wasn't. Way off. I wanted to calculate to see what the capacitance changed to. I'm also really not so great at changing formulas to solve for something else. So what I was trying to do was...


Solve Vpp = (Xc/[R + Xc])*Vin for Xc

When I got Xc, I would then try to...

Solve Xc = 1/(2*3.14*f*C) for C

Can anyone help me with the last part of this?

 

[JimB]

This is incorrect:

Vpp = (Xc/[R + Xc])*Vin

You cannot just add resistance and reactance to get impedance, you must use:

z = √(R² + X²)

Which leads to Vc = (Xc/√(R² + X²)) * Vin
(I am using Vc rather than Vpp for the voltage across the capacitor.)

If this is then manipulated about a bit you get:

Xc =V.R/√(1 - Vc²)

You can then find C using C = 1/(2*Pi*F*Xc)

JimB

 

[ChristianCMcKay]

Oh, okay. I can see now that I was confused. So when I get back home, I'll calculate like this. I was assuming that Z = √(R²+Xc²) was the formula for the impedance of a whole circuit.

Forgive me on this. I've only gotten 2 of my 4 years for Electronics Engineering out of the way.

 

[MrAl]

Hi there Christian,

If you are studying EE, then you probably already have had complex numbers. In this case you can do this much more simply if you use a complex variable to help solve your networks with C and L in them.

The trick is that instead of thinking about xC, which is the capacitive reactance, we think of zC, which is instead the impedance of the capacitor. And this zC is equal to:
zC=1/(s*C)

Now for the low pass filter instead of doing this:
Vout=Vin*(xC)/(xC+R)

We do this:
Vout=Vin*(zC)/(zC+R)

which as you can see is exactly the same expression except now we are using zC instead of xC.

Next, we replace zC with the definition above and we get:
Vout=Vin*(1/(s*C))/(1/(s*C)+R)

Looks a little strange maybe, but then we simplify by multiplying top and bottom by s*C, and we get:
Vout=Vin*1/(s*C*R+1)

and here we have a nice neat little form for the low pass filter. We can divide top and bottom by R*C to get it into a form where the variable 's' is alone:
Vout=Vin*(1/RC)/(s+1/RC)

and there i have simplified R*C to just RC which is just R times C also.

We also sometimes simplify this by replacing 1/RC with just a constant like 'a' so we get:
Vout=Vin*a/(s+a)


Ok so now we need to think about what this variable 's' is.

s (lower case) is a complex variable, sometimes called the "complex frequency". To calculate the response for a low pass filter we only have to replace 's' with the imaginary term of the complex number which will be "j*w". 'j' is the imaginary operator, and 'w' is the angular frequency and 'w' is equal to 2*pi*f where 'f' is just the plain old frequency.
Ok, so replacing s with j*w we get:

Vout=Vin*a/(j*w+a)

Here we have a real number 'a' in the numerator and in the denominator, and we also have an imaginary number 'w' in the denominator and that is indicated by having that 'j' multiplied times that w.

Now here we often want to calculate the amplitude of Vout, and to do that we use this general rule:
Amplitude of Vout=sqrt(RN^2+IN^2)/sqrt(RD^2+ID^2)

where RN is the real part in the numerator, IN is the imaginary part in the numerator, RD is the real part in the denominator, and ID is the imaginary part in the denominator.

In the numerator we have:
RN=a, and IN=0 (note no imag part in the numerator)

and in the denominator we have:
RD=a, and ID=w

and note the imaginary part is NOT "j*w", it's just 'w'.

Now that we know all four parts, we can calculate the amplitude for any frequency f from:
Amplitude=sqrt(a^2+0^2)/sqrt(w^2+a^2)

and for this problem this simplifies to:
Ampl=a/sqrt(w^2+a^2)

or in terms of the frequency f we get:
Ampl=a/sqrt((2*pi*f)^2+a^2)

Using this last expression we can quickly calculate the amplitude for any frequency.

Using a similar method, we can also calculate the phase, which is the phase shift between input and output. If you are interested i'll post the equation for that too.