I am new to this, but I wanted to ask how it would be possible to convert a dB value into resistance. I've been looking at an isolated analog switch from http://www.ti.com/lit/ds/symlink/ts5a3359.pdf that has an OFF isolation value of -64dB at 2.5v (more on pg 10), and I need help converting this. I'd appreciate any help.


You've got a voltage divider, the series element is a capacitor, and the output (shunt) element is a 50 ohm resistor.

Chances are you will not be loading your analog multiplexor with 50 ohms, but this is how the TI measurements are taken.

The attenuation level of -64dB (lets call it 60dB, for now, to ease the fears of using dB) means 1,000:1 voltage or 1,000,000:1 power.

Thus the crosstalk, thus the capacitances of FET junctions and FET gates and the non-ideal VDD bypassing and the non-ideal real-world package inductances, is 1/1,000 of the input level at 1MHz.

The shunt reactance is that 50 ohm resistor; the series reactance is 1,000X larger at 50,000 ohms at 1MHz ---- and is a capacitor.

Given 1pF at 1GHz is -j159 ohms, or -j159,000 ohms at 1MHz, we easily conclude there is an effective 3pF (to produce -j53,000 ohms at 1MHz).

Thus your crosstalk circuit, or your isolation circuit, looks like this:


simulate this circuit – Schematic created using CircuitLab

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    \$\begingroup\$ So my understanding is that basically the "Off iso" is 50k Ω at 1Mhz, which comes from reactance of the 3pF cap and shunt 50Ω. Is that right? And this measurement is taken at the intersection between the 3pF & 50Ω resistor correct? \$\endgroup\$ – Alex Feng Jun 6 '18 at 4:17
  • \$\begingroup\$ Yes. That is correct. \$\endgroup\$ – analogsystemsrf Jun 9 '18 at 3:59

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