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My question is about wiring a "floating differential input" signal to an instrumentation amplifier.

I've read that in the case of a floating signal two high value resistors are used to provide DC bias for the amplifier inputs as shown in Figure 1 below(the 10k resistors in this case):

enter image description here

As you see in Figure 1, the two 10k resistors are used for such purpose and my questions are:

1-) If the 10k resistors in Figure 1 were not equal what would change at the output?

2-) Looking at Figure 2 above: if we don't use such resistor pair but instead if we just ground one end of the input signal, what would be the consequences?

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  • \$\begingroup\$ Because it matters to this conversation, is a thermocouple part of your design (as implied by your diagrams)? Because the details with the construction of the thermocouple can have a large impact in these types of connections. \$\endgroup\$
    – Chris Knudsen
    Commented Jan 20, 2023 at 21:01
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    \$\begingroup\$ Where did the pictures come from i.e. there should be context to these faulty looking pictures. Please link the document that contains them. \$\endgroup\$
    – Andy aka
    Commented Jan 20, 2023 at 21:09
  • \$\begingroup\$ @ChrisKnudsen Thermocouple is an example to represents a signal with "floating differential inputs". \$\endgroup\$
    – ty_1917
    Commented Jan 20, 2023 at 21:10

2 Answers 2

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If the 10k resistors in Figure 1 were not equal what would change at the output?

Think of the thermocouple as a piece of wire (3 ohms is its typical resistance). This means that the two inputs of the instrumentation amplifier are joined and connected through the two 10 k resistors in parallel to ground.

In this situation, the two input bias currents of the amplifier are combined. They pass through a common resistance and create a common voltage drop... and there is no problem.

So it does not matter if the resistances are the same or different. In fact, only one resistor is sufficient to provide a path for the input bias currents to ground.

Two resistors (and of equal resistance) would be needed if the sensor had a high resistance. Each of the input bias currents would then flow through its own resistor and would create its own voltage drop.

Looking at Figure 2 above: if we don't use such resistor pair but instead if we just ground one end of the input signal, what would be the consequences?

It follows from our reasoning above that there will be no negative consequences.

So the general conclusion is that it does not matter if the resistances are the same, different, only one or zero.

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If the 10k resistors in Figure 1 were not equal what would change at the output?

The DC offset would change also called the bias point. Since the temperature is proportional to the voltage, this will create error. At some point you need calibration anyway or you need the total error to be low enough to avoid calibration. It will also depend on why type of thermocopule you have.

Looking at Figure 2 above: if we don't use such resistor pair but instead if we just ground one end of the input signal, what would be the consequences?

There would still be a bias, but it would be relative to ground. If there was any noise coming through the ground you would see it in the output of the instrumentation amplifier whereas the other circuit the noise would be common to both terminals and would be subtracted out.

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  • \$\begingroup\$ Regarding the second answer for Figure 2, does that mean we are making the inAmp totally useless for common mode interference because we corrupt the balanced impedances on the input lines? So the issue in Figure 2 is we convert the CM interference to differential noise? But if the 10k resistors would be non-equal lets say 10k and 10 Ohm, wouldn't that also create the same imbalance issue besides DC bias point at the output? \$\endgroup\$
    – ty_1917
    Commented Jan 20, 2023 at 21:23
  • \$\begingroup\$ "...other circuit the noise would be common to both terminals", that is if the construction of the thermocouple is not of a "grounded" or "bonded" type. This is why my inquiry about the actual use of a thermocouple or not. \$\endgroup\$
    – Chris Knudsen
    Commented Jan 20, 2023 at 21:38
  • \$\begingroup\$ Is it correct to say that the resistor divider in this case is the functional DC equivalent of a decoupling capacitor in AC circuits, i.e. it balances the signal about ground, removing any offset? \$\endgroup\$
    – Thagomizer
    Commented Feb 5 at 18:55
  • \$\begingroup\$ It balances the common mode voltage, its not really comparable to a decoupling cap. Decoupling caps are to form an LRC filter to smooth out local power fluctuations. Instrumentation amps use a bias current, if it isn't balanced it can create error. \$\endgroup\$
    – Voltage Spike ♦
    Commented Feb 5 at 19:22

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