I'm working on a circuit involving an AC-excited sensor bridge, and am trying to figure out ways to detect a short between the two bridge legs, as that causes the bridge to fail to the "wrong side" (i.e. indicate all is well, when the sensor says all is not well). One idea I had for this was to bias the in-amp that measures the bridge into its operating range from the input side, so that shorted inputs would cause the output common-mode voltage to shift dramatically, rendering the failure detectable by downstream circuitry.

The problem

So, I set to design a simple bias network to do this task:


simulate this circuit – Schematic created using CircuitLab

From this design, I went to solve the circuit for R by treating it as two voltage dividers with equations as follows:

$$V_1=5\frac{R+1M\Omega}{R+2.23M\Omega}$$ $$V_2=5\frac{1M\Omega}{R+2.23M\Omega}$$

However, when I plugged in a V1 of 2.2625V and a V2 of 2.2375V (2.25V +/- 12.5mV) and went to find my unknown R by treating the equations as a system, I was unable to come up with a solution, either by hand calculation or using Wolfram Alpha. Initially, this was because I had R1 set to 1MOhm, which was correct for an original center voltage of 2.5V, but wrong for the current center voltage of 2.25V. So, I solved a simple voltage divider for that center voltage and transposed the R1 I found (1.23MOhm, in E192) into my working circuit, but that didn't work either -- I'm getting far closer now, but it seems like I may have run out of resistor precision before I can get the small differential voltage drop I want?


1 Answer 1


You have 2 equations and 1 unknown variable R. Your system of equations is overdetermined. Just use the second equation, solving it is easier than solving the first one.


For the record, it is sometimes useful to have an overdetermined system. One can use least-squares method, averaging, median to "improve" the measurement.

Edit : The original data does not make sense, V1 should be greater than 2.5V since R1 = R3 = 1Mohms and R2 should be greater than 0...

  • \$\begingroup\$ I tried solving the second equation, got an answer, but if I try plugging it into the first equation I get a voltage that's way off the mark I need it to be at... \$\endgroup\$ Mar 20, 2020 at 3:10
  • \$\begingroup\$ Perhaps you made a mistake in V1 or V2. Are you sure you have the right values? \$\endgroup\$
    – Ben
    Mar 20, 2020 at 3:12
  • \$\begingroup\$ I am sure I have the right values \$\endgroup\$ Mar 20, 2020 at 3:15
  • \$\begingroup\$ R2 is greater than 0, right? R1 = R3 right? Then the minimum value for V1 is 2.5V, it cannot be lower than 2.5V \$\endgroup\$
    – Ben
    Mar 20, 2020 at 3:16
  • \$\begingroup\$ You made a mistake, V1 = 2.7625V \$\endgroup\$
    – Ben
    Mar 20, 2020 at 3:17

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