# Improving equation for selective amplifier's LC tank impedance

#### MrAl

Joined Jun 17, 2014
11,725
I would like to incorporate resistor - I name it R6 in attached screenshot (which is in series with inductor and both parallel with capacitor)
into equation of LC tank's impedance. Why? Because real inductor has resistor connected in series with it.
I am including only LC impedance equation, not entire script, which calculates AC voltage gain.

// 1ST EQUATION
// produces result of AC voltage gain equation: -103.09889, which is far from peak inside bode plotter inside simulator
var ZC = 1 / (2*Math.PI*f*C);
var ZL = 2*Math.PI*f*L;
var Zlc = (ZL * ZC) / (ZL + ZC);

// 2ND EQUATION
// this one produces result of AC voltage gain equation: -1373.9779, which corresponds to measured value inside bode plotter inside multisim
var Zlc = (2*Math.PI*f*L) / (1 - ((2*Math.PI*f*L)*2*Math.PI*f*C));

I don't understand why second equation works, because it is derived from first equation (I took both equations from
https://www.geeksforgeeks.org/lc-circuits/ ).

Please see attached second screenshot, to see where they are originally on website.

Increasing R6 resistor value, the AC voltage gain at resonant frequency drops down inside simulator.
Where in my second equation to put it's resistance variable, so it would reflect the simulated one?

Hi there,

As you know by now, you have to use complex math to get the right result. There is no way around this unless you just need the magnitude of the impedance, but it's better to know the full story using complex numbers.

There is a way to make it slightly easier though, by using what is sometimes called the Laplace variable, lower case 's'.
Using this notation, the impedance of a pure inductance is:
zL=s*L
and for a capacitor it is:
zC=1/(s*C)
and for a resistors it is just:
zR=R.

Since you want R in series with L, the total impedance is the sum of the two. That changes the definition of L a little:
zL=R+s*L

Now if we want them in parallel, we use the parallel impedance formula which I see you know about. That could be:
zP=zL*zC/(zL+zC)

and since L has built in resistance R as above this would come to:
zP=(R+s*L)*(1/(s*C))/(R+s*L+1/(s*C))

Note in the above we have no imaginary operator yet.
Computing the result, we get:
zP=(R+s*L)/(s*C*R+s^2*C*L+1)

It is now that we have to include the imaginary operator 'j'. To do that, we substitute j*w in for ''s'. We end up with:
Zjw=(R+j*w*L)/(j*w*C*R+j^2*w^2*C*L+1)

Next, if we look up what the square of 'j' is, we find that it is equal to -1, so we can change this expression to:
Zjw=(R+j*w*L)/(j*w*C*R-w^2*C*L+1)

That's a bit simpler, but note that we have a complex number in both the top and bottom of this expression. To convert this into just a single complex number, we can multiply the numerator and denominator by it's complex conjugate. For the complex conjugate, we just change the sign of the imaginary part of the denominator to the opposite sign. Since the sign is positive, we change it to negative to get the complex conjugate. That gives us:
cc=-j*w*C*R-w^2*C*L+1

Now multiply the numerator and denominator of Zjw by cc and we get, written out:
Zjw=((R+j*w*L)*(-j*w*C*R-w^2*C*L+1))/((-j*w*C*R-w^2*C*L+1)*(j*w*C*R-w^2*C*L+1))

and then we simplify that and get:
Zjw=(R-j*w*(C*R^2+w^2*C*L^2-L))/(w^2*C^2*R^2+w^4*C^2*L^2-2*w^2*C*L+1)

and note there is no longer an imaginary operator in the denominator, only in the numerator.
We then expand the numerator so we can separate the real from the imaginary, and divide both parts by the denominator:
Zjw=R/(w^2*C^2*R^2+w^4*C^2*L^2-2*w^2*C*L+1) - j*w*(C*R^2+w^2*C*L^2-L)/(w^2*C^2*R^2+w^4*C^2*L^2-2*w^2*C*L+1)

and here the left side term of the numerator is the real part, and the right side term of the numerator is the imaginary part.
That's the complex impedance.

I should note that if you insert the values first it is simpler, but in a program that has to deal with different values for all the variables you have to keep the original variables intact.

Since this wasn't really homework I was able to provide more help. For homework problems though we don't do that.
Also, since this was a little long you should go over all the calculations carefully to make sure they are correct.