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I'm attempting to create my own hot wire liquid water content sensor, the control circuitry of which is similar to many hot wire anemometers:

schematic

simulate this circuit – Schematic created using CircuitLab

RS, the sensor wire, is a 5 cm long, 125 micron diameter platinum wire with a resistance of 0.5 ohms. RT, the temperature compensation wire, is an 8 cm long, 25 micron diameter platinum wire with a resistance of 17 ohms. My understanding of how hot wire anemometers are supposed to work is this: as air flows over the electrically heated wire RS, its resistance goes down, making the Wheatstone bridge unbalanced. This increases the difference between the two inputs of the op amp, which then outputs a higher voltage, increasing the temperature and resistance of RS until the bridge is balanced again.

The issue I'm running into is that the op amp is unable to provide enough power to heat the sensor wire, which needs to be about 50°C. Im using a LM358P op amp. I think that part of the issue may be that because the resistances in the bridge are so low, the op amp needs to deliver more current than it is capable of to supply the output voltage expected. I've been supplying the op amp with between 3-32v single supply, and the output voltage and current have been around 30 mV, 30 mA respectively.

I study mechanical engineering so the workings of op amps is something I've been trying to teach myself, any suggestions for increasing the power through the sensor wire to heat it properly would be greatly appreciated!

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The issue I'm running into is that the op amp is unable to provide enough power to heat the sensor wire

Based on your circuit (and values) you're going to have a more basic issue. That circuit has only one stable point and that is when the op-amp is delivering 0 volts at the output. In other words you need to build in what is known as an "over-temperature" so that Rs “runs” warmer than ambient temperature by a known margin.

This entails adjustment of typically R2. And, you need to know the "over temperature" because, to calculate the thermal conductance (or resistance) of the water (or air or other medium), you need to know how hot Rs is running compared to ambient temperature.

Then, you'll have three stable points of operation - the first one is when the op-amp is delivering 0 volts (unwanted), the 2nd when delivering a positive voltage and the third when delivering a negative voltage.

So, what you need to do now is force the op-amp so it can't deliver stability at 0 volts or negative voltages and, to do this you simply run the op-amp from a single positive supply because the output can never get to 0 volts or go negative.

Then you have something like a working system and adding an emitter follower current booster is almost trivial in comparison. I've done it using emitter followers and inverting common emitter amplifiers but, with CE you have to invert the polarities of the op-amp inputs. The circuit works well and delivers good performance if you are looking for accuracy: -

enter image description here

The inverting input above is not wired to the emitter on the circuit you need. You still wire it to the junction of R1 and Rs. All the bridge resistors connect in the above circuit replacing \$R_{LOAD}\$. The non-inverting input connects to the junction of R1 and R2 as per your original circuit.

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    \$\begingroup\$ Your "more basic issue" is simply false. Given that Rs is less than 0.5Ω at any temperature less than 50°C, the opamp output will be driven high until the bridge is (nearly) balanced. If something sucks heat out of Rs, the opamp output will rise in order to compensate, just as the OP desires. The only issue here is the current requirement. \$\endgroup\$
    – Dave Tweed
    Commented Feb 6, 2020 at 17:01
  • \$\begingroup\$ @DaveTweed that’s not how I read the question. The electrical wire won’t have heat taken away unless it is at a different temperature to the reference resistor Rt. And, as shown there is no impetus to drive the RS wire to get hotter than ambient. \$\endgroup\$
    – Andy aka
    Commented Feb 6, 2020 at 17:03
  • \$\begingroup\$ Hi Andy, thank you for your answer. My understanding of the Wheatstone bridge is that it will be balanced when R2*RS=R1*(R3+RT). Therefore, while keeping the ratio R2/R1 constant, to keep the ratio (R3+RT)/RS constant, increasing the net resistance (R3+RT) will heat up RS. You are correct in that at ambient temperature, RS is about 0.5-0.6 ohms. I've tried hooking up a BJT (BC547 B 331) as you have suggested, with R3 changed to a 20 ohm resistor instead of a 3 ohm, but the sensor wire isnt heating up, and the BJT is getting very hot when both it and the op amp are being given 2 volts \$\endgroup\$
    – mattsons
    Commented Feb 6, 2020 at 17:19
  • \$\begingroup\$ You might need a beefier transistor and a power supply that can produce more current. Ask yourself what current do you need to Rs 50 degrees above Rt? Let me know also. Being given 2 volts??? What does that mean.. \$\endgroup\$
    – Andy aka
    Commented Feb 6, 2020 at 17:29
  • \$\begingroup\$ @Andyaka by given 2 volts I mean that I am using a linear power supply, the positive terminal of which is connected to the collector of the BJT and the positive power terminal of the op amp. The negative power terminal of the op amp is connected to ground. \$\endgroup\$
    – mattsons
    Commented Feb 6, 2020 at 18:34

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