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Consider a simpler one-rung ladder with just L2/C4 and C3/L1/R3. Let the impedance of L2/C4 be: $$\small Z_1=\frac{a+bs}{c+ds}$$ and the impedance of C3/L1/R3: $$\small Z_2=\frac{e+fs}{g+hs}$$ where \$\small s\$ is the Laplace operator. The transfer function is $$\small G(s)=\frac{Z_2}{Z_1+Z_2} =\frac{(e+fs)(c+ds)}{(g+hs)(a+bs)+(e+fs)(c+ds)}$$ There will be ...


3

MY vna is 50 ohm source impedance, what values and which type of network would you use? First convert 4 - j22 to a parallel capacitor and resistor (8.258 pF and 125 ohms) Try this calculator. Assume an LC pad to will match 50 ohm to 125 ohm (may need a slight mod) Recognize that the LC pad needs too little capacitance compared to 8.258 pF Resonate out most ...


3

anhnha - perhaps you like my approach? From system theory (and from the classical feedback model) we know that - in case of feedback - the open-loop output impedance Zol without feedback must be divided by the expression (1-LG) with LG=loop gain. (We have a minus sign due to positive feedback) This opens the way to a relatively simple and straight-forward ...


3

With R_s = 0, an M1 impedance contribution and an C_gs impedance are connected in parallel; the output impedance is $$ Z_0 = {1\over{g_m}} || {1\over{s·C_{gs}}} $$ Neglecting C_ds, an R_s contribution is in series, only it should be scaled down. Common drain stage is a non-inverting current amplifier. Given a common drain stage's AC current gain $$ A_{cd} =...


2

Consider the ladder in the figure shown below. The ladder is shown in sub figure (A). At the pole frequency of the series branch \$Z1\$, the equivalent circuit looks like (B) due to infinite impedance of \$Z1\$. There is no connection from input to output at the particular frequency. Hence a zero. At the zero frequency of the shunt branch \$Z2\$, the ...


2

Conceptually, the starting point is the signal via and a ring of ground vias around it, treated as a short coax cable so that you can use the coax cable formula. The ring of ground vias cannot be smaller than the ground pads (since they must terminate there). The datasheet says edge-to-edge of the uFL groundpads is 1.9mm, so I imagine a ring of ground vias ...


2

Another aspect to keep in mind is the return current via. The path the return current takes also influences the impedance of the interconnect. It looks like the red stitching vias around this section will connect ground planes on the top and bottom layers. The return current for this signal will likely go through the nearest stitching via. Instead of ...


2

The MCU in Arduino has a successive approximation ADC. It works by briefly taking a voltage sample via a multiplexer into a small storage capacitor to handle multiple input channels with one ADC. With a high source impedance, the sample/hold capacitor may not have time to fully charge, and thus the sample of the voltage does not resemble the actual voltage. ...


2

Complex impedances that are purely series or parallel can be lumped similar to groups of resistors: $$ \begin{align} Z_{series} &= Z_1 + Z_2 + ... + Z_n\\ Z_{parallel} &= \biggl(Z_1^{-1} + Z_2^{-1} + ... + Z_n^{-1}\biggr)^{-1} \end{align} $$ The problem is that your circuit has 3 nets defining input and output, so it cannot be lumped ...


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What you need to measure depends on what you are doing and what is most useful for that task. For example, this paper deals with Retinol detection. It mentions the "Single Frequency Impedance technique." If you were trying to get results comparable to those in the paper, then you would have to implement that technique - nothing we tell you about ...


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The resulting asymmetrical Π will have C/2 and 2L as values: This is also one way to simulate it. Note that the source now has twice the impedance and twice the amplitude of the signal in order to match the outputs (so you should account for that when you're recalculating the filter). In response to the comments, I have re-uploaded a different image (the ...


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