Question in short:

If I have a multiplexer, and my selected input is X volts, I suppose that the multiplexer's output will be X+-ΔX volts. How can I calculate ΔX ?

Question with more details:

I have four Wheatstone bridges composed by loadcells liks this one, its output is a differential signal S- and S+ that I need to amplify with a differential amplifier.

I was thinking about using just one amplifier and multiplex its input signals with a multiplexer like the CD4052B.

The excitation voltage of the bridge is 5V thus the difference between S+ and S- is in the range of 0mV to 5mV. The amplification gain is 1000.

If the ΔX introduced by the multiplexer is bigger than 0.005mV, it is not good for me, than I would like to know how can I calculate the ΔX so I choose the best multiplexer or go to another solution.


1 Answer 1


ΔX assumes that as the signal gets bigger or smaller there will be a non-linearity. Let's look at that first: -

enter image description here

Let's say you are powering the 4052 from 5 volts (left graph). The change in resistance between -1 volt and +1 volt on your input is about 470 ohms - 350 ohms = 120 ohms. If you have a signal this large AND you are feeding the output of the multiplexer into a low impedance input (say 1 kohm) then yes, you will have a substantial ΔX misshaping of the output.

But, in all likelihood you are using an instrumentation amplifier than has input impedances approaching 1 giga ohm (or at least several Mohms). Let's say worst case 1 Mohm and let's say the nominal resistance of the analogue switch is 400 ohms.

What will the basic potential divider effect be: -

\$\dfrac{V_{OUT}}{V_{IN}}=\dfrac{1,000,000}{1,000,400}\$ = 0.9996 or an error of 0.04%

On top of this is the ΔX error of maybe +/-60 ohms - this will be a worst case error of about: -

\$\dfrac{V_{OUT}}{V_{IN}}=+/-\dfrac{1,000,000}{1,000,060}\$ = +/- 0.006%

However, your input aint +/-1V it's much, much smaller so I'd say forget it.

The error you DO need to be concerned about is leakage currents in and out of the input pins to the load cell - it has a 1 kohm resistance and the leakage quoted on page 5 is +/- 100nA (worst case).

This gives rise to an error voltage across the load-cell of 1 kohm * 100nA = 100 \$\mu\$V.

Can you live with this? It doesn't sound like you can.


I use DG309 multiplexers for miniscule signals from thermocouples and have no problem with them - however, a thermocouple has a very low source impedance of maybe 1 ohm (or maybe 100 ohms if you include the cable). At ambient temperatures the DG309 exhibit a leakage current of +/-5nA max and this gives rise to a dc error of about 0.5 \$\mu\$V. These will work better than the CD4052 I reckon.

  • \$\begingroup\$ Andy's graphs show the temperature dependency of switch resistance.. though it's not distortion as you asked about, it will have an effect on the signal- the 5V curve near 0V changes about +1 ohm/°C, so a 1M ohm load will make the output change by -1ppm/°C (typical). It's not trivial to make ppm-level instrumentation. \$\endgroup\$ Apr 13, 2014 at 15:49
  • \$\begingroup\$ @SpehroPefhany - the graphs shown not only indicate temperature dependency but the channel resistance variation with input signal level. Do you have in mind a different distortion Spehro? In my answer I am talking about the change in resistance between a signal level of -1V and +1V. Is this not clear enough? \$\endgroup\$
    – Andy aka
    Apr 13, 2014 at 16:12
  • \$\begingroup\$ He's been asking questions about overall accuracy and there is some overlap between distortion (this question) and accuracy effects, so I was taking advantage of your graphs. \$\endgroup\$ Apr 13, 2014 at 16:34
  • \$\begingroup\$ @SpehroPefhany you cold-blooded exploiter you!! \$\endgroup\$
    – Andy aka
    Apr 13, 2014 at 16:51
  • \$\begingroup\$ @Andyaka Thanks for your answer, it was was very educational :) \$\endgroup\$
    – koike
    Apr 14, 2014 at 20:34

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