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Art of Electronics, 3rd Ed, pg. 179, gives a model for the capacitive coupling between independent channels of an AD7510DI analog switch. The authors ask to calculate the coupling between channels given a 1 MHz signal applied to the input for the four combinations of either channel being on/off. The following figure shows a SPICE simulation of the circuit, where the signal is applied to Channel 1 and the output is taken across R2 on Channel 2, with Channel 2 input grounded (but retaining the 10k implied source impedance). In the figure, the Channel 1 is 'on' (75 Ohms) and Channel 2 is 'off' (1 Meg).

enter image description here

Channel 1, Channel 2 : Coupling

  • OFF, OFF: -43 dB
  • OFF, ON : -34 dB
  • ON, OFF : -29 dB
  • ON, ON : -31 dB

It's obvious why there is least coupling when both channels are off. What I don't understand is why there is significantly more coupling when Channel 1 is on and Channel 2 is off (-29 dB) compared to the opposite situation (-34 dB). I would have thought the situations would be quite similar. I'm also confused about why both channels on gives less coupling (-31 dB) than having Channel 2 off (-29 dB).

So, qualitatively speaking, why does the greatest amount of coupling occur when Channel 1 is on and Channel 2 is off rather than both channels being on? And why is the coupling so much greater than the opposite situation when Channel 1 is off and Channel 2 is on?

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  • \$\begingroup\$ You're probably measuring output of ch2 with input from ch1, and output ch1 with input from ch1. If so, you should switch the input to ch2 to measure crosstalk output from ch1. \$\endgroup\$ – a concerned citizen May 4 '18 at 13:43
  • \$\begingroup\$ @aconcernedcitizen I don't think I understand. The point is to measure crosstalk between channels. Therefore, the output is measured at whatever channel the input is NOT applied to. I don't see what I've done wrong. \$\endgroup\$ – pr871 May 4 '18 at 13:50
  • \$\begingroup\$ And you're right to think so, but do you also move the input source to ch2 when measuring to ch1? Or are you just leaving it where it is, at ch1 in, and then emasuring ch1 out? In the beginning of your question, you say it right: ch1 in, measure at ch2 out, with source at ch1 in. Good. But when you measure ch1 out, do you also move the source to ch2 in, or do you leave it at ch1 in? \$\endgroup\$ – a concerned citizen May 4 '18 at 13:54
  • \$\begingroup\$ I just noticed you have the parallel resistors mismatched. Is it supposed to be so? One of them is 75 (R3), the other is 1Meg (R1). \$\endgroup\$ – a concerned citizen May 4 '18 at 14:02
  • \$\begingroup\$ @aconcernedcitizen I'm not measuring the output of Ch1 at all. The input is always at Ch1 and the output is always at Ch2. What's different is which channel is 'on' and which is 'off'. There are four measurements corresponding to the four combinations of on/off for the two channels. The mismatched resistors show one of the four combinations: Ch1 on (75 Ohms) and Ch2 off (1 Meg). \$\endgroup\$ – pr871 May 4 '18 at 14:47
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It's normal to be so. Here is the equivalent resistance circuit @1MHz:

equivalent

When ch1 is ON and ch2 OFF, the 75\$\Omega\$ is shorting C8, thus the signal comes unaltered through ch1, while through ch2 it first goes through C4, whose reactance @1MHz is ~319k\$\Omega\$, thus forms a divider with the 10k\$\Omega\$ grounded at ch2 input.

When ch1 is OFF and ch2 ON, C8 will conduct, mostly, with a reactance of ~159k\$\Omega\$ (in parallel with 1M\$\Omega\$), and the signal is attenuated first through C8 and C7||R4, then attenuated even more through the divisor formed by C5 and C2||R2,while on the other branch, it\'s only attenuated through C4 and R5, then going clean through the 75\$\Omega\$ of ch2 ON.

As you can see in the picture, when ch1 is ON it's ~21.5mV, while the other is ~10mV.

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For trustable modeling, you need to include the inductance (start with 10nH) in the GND pins and the various (+10, -10) volt VDD supplies. These inductances mean the common rails ARE NOT QUIET.

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