EMC CAPACITIVE COUPLING BETWEEN CONDUCTORS
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Solution Summary
EMC calculation of crosstalk voltage induced in victim conductor due to capacitive coupling between conductors are examined.
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The above can be redrawn more formally as two parallel networks connected together by the mutual capacitance as shown below
As no source resistance is specified we set R_S=0Ω so that all the source voltage appears across the input network. Also since no secondary circuit source voltage is specified we set V_S2=0 as shown by the modified circuit below
We now find the parallel combination of R=R_2 ||R_L2 in the secondary network from
R=R_2 ||R_L2=(1/R_2 +1/R_L2 )^(-1)
R=R_2 ||R_L2=(1/1000+1/2000)^(-1)=667Ω
Then we can replace both R_2 and R_L2 by a single equivalent resistor Ras shown below
The above circuit is simply a voltage divider circuit with two complex impedances denoted by Z_12 and Z_2 as shown
We use circuit theory to derive the magnitude of the impedances at a angular frequency ω as shown below
Z_12=1/(jωC_12 )=-j 1/(ωRC_12 )
Which has magnitude
|Z_12 |=1/(ωRC_12 )
Also
Z_2=(1/R+1/(1/jωC_22 ))^(-1)=((1+jωRC_22)/R)^(-1)
Z_2=R/(1+jωRC_22 )=(R(1-jωRC_22))/(1+(ωRC_22 )^2 )
Thus
Z_2+Z_12=(R(1-jωRC_22))/(1+(ωRC_22 )^2 )-j 1/(ωRC_12 )
Z_2+Z_12=R/(1+(ωRC_22 )^2 )-j{(ωR^2 C_22.ωRC_12+1+(ωRC_22 )^2)/({1+(ωRC_22 )^2}ωRC_12 )}
Z_2+Z_12=(ωR^2 C_12-j{ω^2 R^3 C_22.C_12+1+(ωRC_22 )^2})/({1+(ωRC_22 )^2}ωRC_12 )
Which has ...
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