What does the circuit shown in the picture do?

Question

What does the circuit shown in the picture do? Does it act as a current source? If so, how does this happen?

Answer 1

Searching for something familiar

If we tidy up this mess of resistors and op-amps a bit, we can recognize the so-called "improved Howland current source". The idea behind this famous circuit is exceptionally simple and intuitive.

Basic idea

In this circuit solution, the voltage V producing the current is not constant but follows ("lifted" with 1 V) the voltage VL across the load.

^{simulate this circuit – Schematic created using CircuitLab}

As a result, the resulting voltage VRI across the current-setting resistor RI...

... and consequently the current IRI = VRI/RI through it (and the load), is constant.

Implementation

Classic circuit

The following voltage source V from the conceptual circuit above is implemented by an op-amp summing circuit that sums Vin and VL. As a result, there is a constant transfer ratio of unity between the points A and B of the current-setting resistor RI, i.e., R3/(R3 + R4).(R1 + R2)/R1 = 1.

^{simulate this circuit}

This means that when the voltage across load VL varies, the voltage at point A follows the voltage at point B (the so-called "bootstrapping"), or the difference between them (the voltage drop across RI)...

... respectively, the current through it IRI = IL = VRI/RI is almost constant.

As you can see, the IL curve is not absolutely horizontal. There is a small error because the voltage divider diverts a small current from RL.

OP's circuit

This problem is solved in the OP's circuit by inserting an op-amp follower U2A between the load and divider.

^{simulate this circuit}

Now the IL curve is absolutely horizontal.

See more in my question and answer What is the basic idea behind the so-called "improved Howland current source"?

Answer 2

This is a current source for a Platinum Resistance Thermometer known to be more linear and stable than thermocouples.

The interface indicates a PT100 is used which is 100 ohms at 0°C. The US std. is 0.392 ohms/°C and the EU std. is 0.385 ohms/°C for commercial grades. In order to choose your temperature range you would have to attenuate the gain ratio and adjust 5V for offset. You might have 0°C at some offset voltage to measure negative temps and reduce gain to reach 500°C. Different grades exist from 0~100°C up to -70~850°C

Tiny constant currents are used to limit self-heating.

The platinum FPC strip should provide minimal error at 100 μW at 0°C using 1 mA.

Since this circuit does not provide much voltage gain a third amp. is needed.

ADI has a low offset, low noise op-amp for this based on a U of Bulgaria 3-wire RTD invention. (See link and image below.)

https://www.analog.com/media/en/reference-design-documentation/reference-designs/CN0337.pdf

Ref: https://peaksensors.com/resources/resistance-thermometer-information/

This design appears to work well with a 12-bit ADC and a single supply 3.3V.

Answer 3

You already have a description of how the circuit could function.

"Pt100" suggests that it is used with a Platinum (Pt) RTD (Resistance Temperature detector) sensor. In particular, one with a nominal resistance of exactly 100.0Ω at 0°C.

For that, we would like a constant current (infinite impedance) or a high negative source resistance (much, much higher than 100Ω).

By passing a current through the sensor (typically of the order of 1mA), a voltage can be derived that is related to the resistance (or more closely to the temperature in the case of the slightly non-constant current).

Probably it's more common nowadays to measure resistance and do the curvature correction digitally.

Answer 4

For simplicity, assume R4=R6 and R7=R10.

Pin5 of U2B is then (A1 + Temperef)/2.

Pin7 of U2B is then 2x this, or A1+tempref. Thus R3 has Tempref across it, no matter what value A1 is, and so acts as a current source.

The same analysis will hold as long as R7/R10 == R4/R6.