I have only one input to measure 16 channels, my intention is to use this circuit below based on the LTC1380 MULTIPLEXER:

enter image description here

The multiplexers have different addresses (AO A1). The output of both is linked to the same output buffer. Therein lies my confusion how to deal with both outputs connected together

My questions is

  1. Will I have to close top multiplexer off, while I take a reading from bottom multiplexer and vice versa. (I would assume that this is the case, which leads to the next question...
  2. What would happen if I have both S0 lines open on each multiplexer
  • \$\begingroup\$ "Therein lies my confusion how to deal with both outputs connected together." Your diagram is taken straight from the LTC1380 datasheet which says that is the way to make a 16 channel multiplexer. Why are you doubting the manufacturer? \$\endgroup\$ – Oldfart Jan 11 '18 at 18:17
  • \$\begingroup\$ Table 2 in the data sheet tells you that you can switch one output off when wanting the other enabled. \$\endgroup\$ – Andy aka Jan 11 '18 at 18:30
  • \$\begingroup\$ @oldfart, OP was questioning how to deal with the effects of connecting them together, which is a reasonable question, not questioning whether to connect them together or the manufacturer's assurances \$\endgroup\$ – TonyM Jan 11 '18 at 19:34
  • \$\begingroup\$ @oldfart as tonyM pointed out I dont doubt, Im trying to understand it. \$\endgroup\$ – Pop24 Jan 12 '18 at 8:48

You are correct in your statement about having to toggle between the two multiplexers to take a proper reading of any single input. You do this by setting or resetting the Enable bit which is included in the SMBus-delivered configuration word ( Page 8 of the Linear Tech data sheet). You must select one or the other of the two multiplexers. So, there are a few precautions.

A. You must avoid enabling both multiplexers at the same time as this will effectively connect two of your inputs together. Since these inputs are sourced by some entitiy, you would effectively be shorting these two sources together. Maybe OK, maybe not, depending on their source impedances and electrical composition. So, best to avoid this situation entirely.

B. When switching between two inputs on opposite multiplexers, always go to the "both off" state first, then select the appropriate multiplexer in a subsequent SMB transaction. (See "C" below, as this is a transient floating condition.) This is essentially a "break-before-make" strategy. Not to worry when you are switching between inputs on a single multiplexer because the LTC1380 has "Guaranteed Break-Before-Make" (Page 1 of data sheet).

C. If you disable both 1380's simultaneously, the NI input of your unity-gain buffer will be left floating - maybe OK, maybe not, depends on a number of factors. Unless you leave both MUX's disabled for an extended period of time ( say more than 100 milli-secs ), I wouldn't worry about it. You could connect a high value resistor ( 1 meg or greater) from the NI input to DC Common to eliminate this floating situation, but that could create accuracy issues, depending on the source impedances of the input signals and a few other factors. As a practical precaution you should probably assign one of the inputs as a "parking" position. Always select this input when you are not actively utilizing any of the other inputs.

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  • \$\begingroup\$ Thanks for taking the time to answer! Some useful information it helps alot, particularily point C, which i was completely unaware of. \$\endgroup\$ – Pop24 Jan 12 '18 at 8:53
  1. Yes, as you suspect.

  2. The voltages from S0a and S0b would be connected together by the low impedance paths through each enabled mux. Unless these connected voltages are equal, they will cause a current to flow between them. The current value depends on each voltage source's internal resistance and the sum resistance of the two enabled mux channels, which you can get from the mux datasheet. If this current flow is high enough, it may cause destructive power dissipation in your mux or voltage source components. It will also cause the voltage seen by the op-amp to be somewhere between the two voltage source levels.

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  • \$\begingroup\$ thanks for answering, I marked another answer as "the" answer just because of additional information. This is valid and useful to me thank you very much! \$\endgroup\$ – Pop24 Jan 12 '18 at 8:54

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