# What does the circuit shown in the picture do?

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

• The main advantage of this circuit solution is that the load is grounded. Otherwise, an inverting or non-inverting amplifier also does a good job and is simpler and more accurate. The disadvantage of this solution is that it requires exact matching of the resistance values. Commented Aug 7 at 15:40

# 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"?

• Of course, this is for the PT1000 not the 100 ohm version and the actual PT low Max/Min ratio needs a 3rd stage for gain. :) Commented Aug 7 at 15:54
• D.A.S., Admittedly, I'm not familiar with the specific implementation, and my answer only covers the general principles of designing a grounded current source. By the way, since I'm from Bulgaria, I was wondering who the inventor U from Bulgaria is :-) Commented Aug 7 at 17:04
• Page 1 from my 1st link, = The novel circuit for 3-wire RTD lead wire compensation was developed by Hristo Ivanov Gigov, Associate Professor and PhD, and Stanimir Krasimirov Stankov, Engineer and PhD Student, Department of Electronic Engineering and Microelectronics, Technical University of Varna, Varna, Bulgaria. BTW I'll be in Armenia Sept 4 from Toronto Commented Aug 7 at 17:22
• @D.A.S., it's incredibly detailed, but it's a real pleasure to see fellow inventors featured in such a prestigious publication. Commented Aug 7 at 19:20

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

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

• Tip: HTML entities &deg;, &Omega;, &omega;, &mu;, &pm;, etc. all work in posts (but not in the comments). Commented Aug 7 at 13:30
• TY Mr Q. °>) ... Commented Aug 7 at 15:58

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.

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.

• "temper" appears to be a typo Commented Aug 7 at 15:10