Electronic – How do these current pull-up sources work

The issue is likely to be insufficient voltage for the desired operation.

Based on the description, this is the schematic for the circuit being used:

^{simulate this circuit – Schematic created using CircuitLab}

For the current source to function as designed with a 5 Volt supply, the maximum resistance across the terminals of the sensor would have to be under 25 kOhms.

Explanation:

• If the sensor has a resistance of, say, 100 kOhms, then it would require V = I x R = 10 Volts across the sensor alone, to allow 100 uA to pass. Add a minimum of 2.5 Volts for operation of the REF200, and the minimum required supply voltage is 12.5 Volts. For stable operation, take 15 Volts as a safe margin.
• If the sensor has a resistance of, say, 22 kOhms, then it will develop just 2.2 Volts across its terminals. This leaves 5 - 2.2 = 2.8 Volts across the REF200 current source, which is within its recommended operating range of 2.5 - 40 Volts.
• In the schematic shown above, add a supply voltage via a battery, replace the sensor with a potentiometer, and vary the potentiometer to simulate this behavior for yourself.

Thus, depending on what the sensor's effective resistance is, at the desired operating point, the supply voltage for the current source to provide a stable 100 uA may need to be higher than the 5 Volts provided.

Most buck converters (including this one) have an internal voltage reference and a feedback loop, which regulates the output voltage. The output voltage is set by the resistors R₁ and R₂. The output voltage is independent of the input voltage fluctuations. The output voltage formula is not related the optional filter.

Switch mode converters can be layout sensitive, so it's usually a good idea to follow a reference design. On the other hand, this one has a top frequency of only 250kHz, so it may be more forgiving than switchers with higher frequencies.

There's more details about the principles of operation of buck converter here and here.

edit: A somewhat odd thing, however, is that the resistor values in the drawing don't quite check with the output voltage

\$V_{out}=1.25\left( 1 + \cfrac{R_2}{R_1} \right)=1.25\left( 1 + \cfrac{3.8}{1.2} \right) = 5.2 V\$

should be 5.0V. I wonder if there's a reason for the extra 0.2V?