Buffer Input OR Instrumentation Amp Input Floating Voltage Level

I have implemented an Instrumentation Amp circuit as shown in the picture below:


The inputs ‘SIG_POS’ and ‘SIG_NEG’ are coming from a remote location (not on-board)
Problems:
Case1:
When both inputs ‘SIG_POS’ and ‘SIG_NEG’ are floating, the actual voltage levels measured at input ports (SIG_POS) & (SIG_NEG) are apprx 4.3V with respect to 0V-GND. Why are they sitting at level nearer to the Rail? (Would this happen to any OPAMP I use OR just specific to this type of OPAMP?)
Output for above case (Signal Out) was apprx 40uV
Case2:
Now ‘SIG_NEG’ was connected to 0V-GND
SIG_POS was left floating. The output was now measured as 4.29V. I am thinking this because the floating positive input is sitting at 4.31V whereas Negative input is at 0V-GND so the output is the difference between the two. But the question here is same as above why is floating input sitting at 4.31V?
I am using the output to measure using an ADC but was hoping for output to go LOW when there is no sensor connected at the remote location (i.e., SIG_POS floating but SIG_Neg connected to GND -- GND pin of sensor shares other GND on the remote board). I didnot use any pull down resistor between SIG_POS and SIG_NEG as the remote sensor has series output resistance (opamp with several series output resistances and I would like to measure the actual output signal from sensor output pins without loading it
Help please

Hi,

a measurement with (one or two) floating input gives no meaningful output.

If possible pull them with high ohmic resistors to V_suplly/2.

***
In your case I don´t see why it should be pulled close to VCC, therefore I assume it is caused by stray currents on your PCB (or other wiring).
Did you use ultrasonic cleaning of your PC?

Do you really need that GigaOhms of input impedance?

For more help we need to know the type of diode, resistors, capacitors and the PCB layout.

Klaus

Apart from inappropriately floating input signal, I don't understand the purpose of the "amp-rail" diode circuit. Why are you reducing the amplifier supply voltage?

To understand the input circuit, what's the signal source, how is it referenced to 0V-GND? Is one of the circuits/are both circuits grounded to PE or connected to other grounded circuit? Is one of the circuits truely floating?

@KlausST & FvM Thanks for the valuable comments

Under normal operating conditions, the remote board has sensor connected so the inputs are Not floating. But if ADC sees 4.3ish V at its input, it would NOT know if the input is actually coming from the sensor OR is it a Fault case where SIG_POS is floating
What would be the benefit of pulling both inputs HIGH to VSUpply/2 using High ohmic resistors? (Bit more help on this comment please)

I have actually tested & simulated the circuit and it measures 4.3ishV at its inputs when inputs are floating. I just tried simulating by disconnecting the protection diodes (D112 & D113) and found that inputs are now nearer to negative rail. This means that 4.3V is coming from protection diodes. I will try it on hardware tomorrow to check if it is the case on Real circuit aswell.
All resistors are 0402 & 0603 packages (1%), Caps are ceramics.
Diodes D112 & D113 are RB751V40T1G from ON Semi - Farnell 1459171 **broken link removed**

Diodes D116 & D117 are small signal diodes (BAS16) **broken link removed**

The purpose of Amp Rail Diodes is to keep the Total Supply to Opamp within 5V --- from -0.7ishV to +4.3ishV . This is to ensure that OPAMP output goes all the way down to 0V for very small input signals 10mV input signal

Quite simple, there's no chance to detect an undefined (e.g. floating) condition. If it's possible to detect a particularly defined condition instead, depends. Without knowing all possible regular circuit conditions, it can't be decided.

Since there is no current, the Input Floating Voltage will obviously be internal voltage of the OPAMP because input circuitry of these OPAMPS supplies a biasing voltage to either BJT or MOS.
It's normal to read-out this biasing voltage.

Hi,

Leaving the inputs floatin in unconnected state is in your case unknown voltage.
Pulling the signals to a known voltage give predictable outputs.

For sure you may use any other voltage.

Klaus