At page 13 of this Analog Devices document there is a circuit which consists of a servo amplifier and a bridge. The operation principle is explained in the document. Here is the the schematics given:

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I wanted to simulate this circuit in LTspice with following same circuit:

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According to document node A is the output. But this doesn't work at all. So basically:

How come the suggested circuit function at all? Initially there is no voltage at node A.

If it can be modified and work in LTspice, now my challenge will be simulation. Since this circuit tries to make Rf constant with respect to cooling effect, how could this be simulated to see the transient analysis to mimic its operation?

Edit: The second part needs to be a separate question


You choose the wire resistance at the hot temperature then the voltage divider on both inputs are matched. (Virtual gnd. or null)

Since feedback is open at Vout=0, very slight cooling will force Conduction as the inverted input lowers and output rises. Since when Vout=0, it has open loop gain, the changes are negligible for the set point.

R0 & R2 are reversed. Pls fix that.

  • \$\begingroup\$ Thanks for the first part of the question. The second part maybe I should open as a sepetate question. I want to somehow simulate how this regulates the Rf with cooling. \$\endgroup\$ – cm64 May 31 at 16:27
  • \$\begingroup\$ A HWA will have a wind speed , v vs ΔT/ ΔW vs ΔR vs bridge ΔV transfer function that needs a datasheet to work out. i.e. thermal conductance vs electrical resistance \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 May 31 at 17:28
  • \$\begingroup\$ But you can test and measure this to calibrate it with a suitable anemometer and fan to measure steady state and step response time to try to achieve the 1.5k to 3kHz BW or rise time =0.35/BW for 10 to 90%tr, higher W yields more BW \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 May 31 at 17:38

Sorry, but you've badly misread the article.

Most grossly, from the curve presented in the article, the resistance of Rf will be in the vicinity of 1.3 ohm, not 6. Dynamic resistance can be higher, as much as about 3 ohms. Note that the 1.3 ohms is stated at the top of the second column of the first page of the article, and it represents a straight-line approximation of the resistor behavior.

In order to get the circuit to "work" (although it won't work for fixed resistor values), you need to choose a resistance for Rf, calculate the voltage produced by the voltage divider R1/Rf, then match that voltage by choosing the ratio of R0/R2. However, the best you can get with this approach is to get the differential voltage at the input of the op amp to zero.

More generally, you'll need to replace Rf with a rather complex model which produces an apparent resistance which models two factors:

First, it must compute the power being dissipated and convert that to an effective temperature, then apply the temperature to model the apparent resistance.

Second, it must allow for a term in 1) which compensates for heat lost due to airflow. More airflow, more heat loss, so lower internal temperature and lower resistance.

Also note that the circuit shown will only be accurate for particular air temperature. It's not intended as a precision general-purpose anemometer.


Think it through. The op-amp output will go high when the inverting input is lower than the non-inverting input. You've selected values that will keep the inverting input higher. Try \$\frac{R_0}{R_2} > \frac{R_1}{R_f}\$ for a start.

I suspect that this also depends on the op-amp output riding a bit above ground at rest, at least enough to start powering the circuit. The ideal op-amp in LTSpice may not do that without a nudge to the initial conditions.

Your second part is about advanced LTSpice usage, and should be its own question, possibly with the title "Model self-heated temperature dependent resistance in LTSpice".

  • \$\begingroup\$ The cold temp is lower than the balanced servo, so initial conditions will be enough to force conduction. Rf is chosen for hot temp. Therefore your analysis needs reconsideration . Until then,-1 \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 May 31 at 15:11
  • \$\begingroup\$ @SunnyskyguyEE75 The resistor values, as shown, guarantee that for any positive output, the op-amp output is forced low, thus turning off the output. What more is necessary to show "doesn't work"? \$\endgroup\$ – TimWescott May 31 at 18:50
  • \$\begingroup\$ The solution , as I saw it, was reversed values for R0,R2, then it is balanced in regulation for a hot wire resistance of 6. When cooled < 6 Q1 turns on. SO what you suggest trying should be the equal equation for the servo at setpoint. \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 May 31 at 20:44

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