3
\$\begingroup\$

I'm building a 64:1 multiplexer, using 9 ADG408s (64:8:1). Last night, I had one of the chips on a bread board and was taking some measurements.

Everything worked as expected, except I observed a very large crosstalk signal (many orders of magnitude higher than the datasheet suggests). When I add a 1 kΩ pull-down resistor to the output, the crosstalk goes away; details below.

My setup looks basically like this:

enter image description here

enter image description here

Pins A0-A2 set the channel. (0, 0, 0) sets channel S1 and (1, 0, 0) sets channel 2. The top figure shows channel 1 selected, and the second figure shows channel 2 selected.

I started by applying a 5 Vpp, 1 kHz sine wave to channel 1 and a 5 Vpp, 1 kHz square wave to channel 2. I switched pin A0 between GND and 3.3 V and observed the change in signal on my oscilloscope. Everything looked as I would expect.

However, if I select Channel two (1, 0, 0) and disconnect the 5 Vpp square-wave signal from pin 5 (channel 2), I observe a 50 mVpp sine wave (crosstalk from channel 1). This is much larger than the -85 dB quoted value in the datasheet. See figure below:

enter image description here

In the datasheet, figure 27, the crosstalk measurement shows a pull-down resistor between the output and ground. It also shows a similar resistor across the input. See figure below, which shows how the crosstalk was measured:

enter image description here

Note that channels 1 and 2 are switched in my setup (I'm applying the signal to channel 1 and measuring the crosstalk on channel 2. Sorry for any confusion).

When I add 1 kΩ resistors to channel 2, connecting it to ground, and across the output to ground, the crosstalk goes away completely (at least I can't see it on my oscilloscope). See figure below:

enter image description here

If I remove the resistor on the input, the signal stays flat (i.e., no crosstalk). So it seems the resistor at the output is doing most of the work. I did some research and it seems that this might be charge injection and the resistor is allowing the charge to bleed off.

But I still have some confusion as to what exactly is going on and how to generalize these results. Since my final design will use 9 ADG408s (64 signals going into 8 ADG408s going into a single ADG408), I'm wondering how many resistors I need to add to my PCB design. Should I add these to the inputs just to be safe (that would be a lot of resistors)? Also, these resistors will affect the singal amplitude as they will form voltage dividers with the internal resistance of the ADG408 (which is about 70 Ω, but is not constant). If anyone can shed some light on this, I would greatly appreciate it. The AD website has a few tidbits here and there. It seems that crosstalk and multiplexers comes up a lot, but I couldn't find anything comprehensive. I also checked Horowitz, but there was no mention of crosstalk withing the context of multiplexers.

\$\endgroup\$
1
  • 1
    \$\begingroup\$ +1: Nice question \$\endgroup\$
    – RussellH
    Sep 17, 2022 at 17:47

2 Answers 2

3
\$\begingroup\$

You have a capacitive divider between the source, a row of pins, and the probe. Each pin is ~11pF to GND, and there is approx. 4pF between pins due to the solderless breadboard (assuming this is the PDIP chip version and you're using it that way, that is). The scope probe itself is on the order of 10 or 20pF to GND, and quite high resistance (~10M), assuming a typical 10x probe anyway. Evidently, all these together, give you a ~1/100 ratio. Which seems reasonable: if the ratio is about 1/3 per pin pair, then 1/4 on the last, that's 1/108 total -- these are certainly very rough numbers, but the order of magnitude is correct.

Grounding an unused input should solve the issue. Going to a PCB will also help (less capacitance between pins; guard traces can even be added if necessary).

If you're selecting one of 64, but one can be allocated as never-used, then tie it to GND and it will short out the whole chain while selected. If you need to use the EN functionality, and a load resistor on D is not practical for other reasons, then consider adding one more switch to short D to GND in that case.

The circumstances are not charge injection, well -- it's through capacitance, so charge is involved, but not in the usual meaning of that term, which refers to switching edges being coupled through internal capacitances.

\$\endgroup\$
1
  • \$\begingroup\$ Thanks for the response. I played around with the circuit some more and it seems that putting a grounded resistor across any other pin (other than the output pin) has little effect. \$\endgroup\$
    – njs
    Sep 18, 2022 at 14:26
2
\$\begingroup\$

The cross talk model shown below is from an Analog Devices Tutorial MT-088.

enter image description here

Crosstalk here is "input crosstalk" formed by the voltage divider \$C_{SS}\$ and \$C_{S}\$. Off isolation can appear as crosstalk. It's pathway is the voltage divider \$C_{DS}\$ and \$C_{D}\$.

For your setup with S2 on and S1 off with a signal applied to S1, the load resistor will reduce both the isolation crosstalk and the input crosstalk since it will appear in parallel with \$C_{S}\$ through \$R_{on}\$. So the load resistor will have a great effect on crosstalk.

With the load resistance open, then you may be able to see the effect of the 1k source resistor on S2 on crosstalk reduction.

I couldn't find anything comprehensive.

The tutorial has a lot of great information on using this kind of device.

\$\endgroup\$
1
  • \$\begingroup\$ Thanks for the paper. It seems the crosstalk is only observable when there is no signal at the selected channel (e.g., apply a signal to ch1 only and then select ch2). If there is a signal on the selected channel, I cannot see the crosstalk. For example, I tried putting an 8 kHz square wave on ch1 (±2 V) and a 1 kHz square wave on ch2 (0-2V), selecting ch2 and then zooming into the 0-V level on my oscilloscope to look for the higher-frequency signal from ch2. Only the faintest lines could be observed, and if I changed the ch1 signal to any other waveform, they disappeared completely. \$\endgroup\$
    – njs
    Sep 19, 2022 at 14:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.