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I wanted to build a simulation of bypass capacitors in Matlab, especially to play around with pairs of different valued caps. So I thought I just take a model of a real capacitor as series connection of capacitor with a resistor (ESR) and an inductor (ESL) and derive the formula $$Z=\mathrm{ESR}+j2\pi f\cdot\mathrm{ESL}+\frac{1}{j2\pi fC}$$ separating into real and imaginary part: $$Z=\mathrm{ESR}+j\left(2\pi f\cdot\mathrm{ESL}-\frac{1}{2\pi fC}\right)$$ now take the magnitude $$|Z|=\sqrt{\mathrm{ESR}^2+\left(2\pi f\cdot\mathrm{ESL}-\frac{1}{2\pi fC}\right)^2}$$

Now I programed the Matlab. first the fuction that calculates impedance of the cap at given frequency/ies

function [ Z ] = calcCapImp( f,R, L,C )  
Z=sqrt(R^2+(2*pi*f*L-1./(2*pi*C*f)).^2);
end

The code to calculate and plot the impedance of a combination of two capacitors in Ohms and dBs:

clear all;
f=10000:1000:1e9;
Z=(zeros(size(f)));
for i=1:size(f,2)
Z1=calcCapImp(f(i),10e-3,700e-12,100e-9);%calculate first cap
Z2=calcCapImp(f(i),10e-3,1250e-12,4.7e-6);%calculate second  cap
Z(i)=(Z1*Z2)/(Z1+Z2);%Total impedance is parallel combination
end
loglog(f,Z);
ylabel('Resistance, Ohm');
xlabel('Frequency, Hz');
grid on;
figure;
Z=20*log10(Z);%convert to decibels
semilogx(f,Z);
xlabel('Frequency, Hz');
ylabel('Resistance, dB');
grid on;

While the results look plausible,

enter image description here

I fail to reproduce graph I have seen elsewhere.

In particular, the above code attempts to recreate Figure 2 from here (same range):

enter image description here

I get nowhere near as dramatic spike at ~10 MHz as on the above mentioned external figure (aprox 10 times/20 dB smaller)

Have I done the calculations correctly? Is there an error in my simulation?

EDIT: here the corrected code and graph per The Photon's anwser below

function [ Z ] = calcCapImp( f,R, L,C )

Z=R+1i*(2*pi*f*L-1./(2*pi*C*f));

end

clear all;
f=10000:1000:1e9;
Z=(zeros(size(f)));
for i=1:size(f,2)
Z1=calcCapImp(f(i),10e-3,700e-12,100e-9);%calculate first cap
Z2=calcCapImp(f(i),10e-3,1250e-12,4.7e-6);%calculate second  cap
Z(i)=abs((Z1*Z2)/(Z1+Z2));%Total impedance is parallel combination, take the magnitude
end

loglog(f,Z);
ylabel('Impedance, Ohm');
xlabel('Frequency, Hz');
title('Impedance in Ohms');
grid on;
figure;
Zdb=20*log10(Z);%convert to decibels
semilogx(f,Zdb);
xlabel('Frequency, Hz');
ylabel('Impedance, dB');
title('Impedance in decibels');
grid on;

corrected graph Now the result is practically identical to those used for reference

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  • \$\begingroup\$ The two graphs look pretty close to me. Or is the second one not your results? If not, you should include a graph of your results to make it clear what is the difference between your expected result and your actual result. \$\endgroup\$ – The Photon Sep 9 '16 at 1:49
  • \$\begingroup\$ I compare it to this one i.cmpnet.com/planetanalog/2007/10/C0238-Figure2.gif from planetanalog.com/document.asp?doc_id=527551 couldnt include it too because of limitation. this is one I attempt to recreate \$\endgroup\$ – Andrey Pro Sep 9 '16 at 1:50
  • \$\begingroup\$ The key to the difference is that the spike is greater than either of the impedance curves on their own, therefore you are getting constructive interference between the components, i.e. the impedances are opposing in phase so their imaginary components cancel. With that in mind, read the Photon's answer. \$\endgroup\$ – Brian Drummond Sep 9 '16 at 9:18
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function [ Z ] = calcCapImp( f,R, L,C )  
Z=sqrt(R^2+(2*pi*f*L-1./(2*pi*C*f)).^2);
end

This is calculating the magnitude of Z, not Z itself.

Z(i)=(Z1*Z2)/(Z1+Z2);%Total impedance is parallel combination

This is combining the two magnitudes as if they are two resistances.

You need to calculate the complex impedance, and combine the complex impedances with the parallel impedances formula

$$Z_{eq} = \frac{Z_1 Z_2}{Z_1 + Z_2}$$

What you calculated was

$$Z^*_{eq} = \frac{|Z_1| |Z_2|}{|Z_1| + |Z_2|} $$

(* superscript indicates a wrong formula)

which doesn't account for the interaction of phase changes in the two combined impedances, so it doesn't give the correct result.

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  • \$\begingroup\$ Oh, I see. I also somehow lost the sight of that I can directly manipulate complex numbers in Matlab and made things more complicated than neccessary.I corrected the code and the result is at the very least much better now. \$\endgroup\$ – Andrey Pro Sep 9 '16 at 10:11

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