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Flow sensor circuitry,help needed to understand

  1. Jun 20, 2013 #1
    I am mechanical engineer , and I have presentation somewhere next week about sensor of my choosing.

    The thing is:I picked sensor which is a research project ,and the sensor is flow sensor,using hot wire technology .

    Not only this entire subject confuses me (measuring flow using heat exchange ? Sounds not the most accurate thing ),which confuses me the most is the circuitry designed to keep steady temperature of the "hot wire" resistor.And in that research,they have the circuit that confuses the hell out of me.

    Anyway,here it is :
    http://img401.imageshack.us/img401/5583/ld52.jpg [Broken]

    It has two wire sensors,to measure upstream an downstream.It measures nasal respiration.
    Now we learned in our class the basic circuit for keeping temperature of the hot wire constant,but this goes way beyond.
    Now upper and down circuits are the same,to reduce the response time.Anyway,beyond the fact the both are supposed to keep the temperature of the hot wire constant,and the final comperator is to detect the direction of the flow,I don't understand the choosing of circuitry.Why integral amplifier is used?Why the transistor afterwards?

    Thanks in advance.
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Jun 20, 2013 #2


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    The transistors are there to act as 'drivers' for the resistor bridge. The chip cannot supply enough current on its own and the emitters of the transistors will give as much as is needed.
  4. Jun 22, 2013 #3

    jim hardy

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    well i want to apologize for this being so long.
    But i have a dear friend who is a mechanical and we used to spend hours on this type stuff - it's where i picked up my meager ME knowledge.
    It just takes time to get a mechanism working in one's head.

    So i hope you'll dwell on this and work the steps in your head one at a time. it'll take a few minutes. I think it'll help you with your talk..


    Let's take the bottom amplifier, heater 1.

    Do you understand that: the bridge is balanced at only one temperature of the wire?
    And that: when the bridge is balanced, the voltage at left corner is same as voltage at right corner?

    If so - good , we can address that 'integral amplifier".

    If not say so, so we can get you to that point.

    Accept for a moment that the amplifier attempts to hold the bridge balanced by controlling power applied to it. That's all it can do, it can't tweak R3 for there's no motor in that schematic. It can only control power applied.
    Presumably R1, R2 and R3 have zero or very low temperature coefficient;
    but heater wire Rh1 has a positive one.
    So as the whole bridge gets heated by the power applied to it, only Rh1 changes resistance.
    So there is some design temperature where Rh1 equals R3 and the bridge is balanced. R3 sets that temperature.
    And power applied to bridge will be whatever is necessary to make power into Rh1 equal power lost to air flowing over it at that design temperature!.

    Aha ! More airflow = more power removed = more power must be applied.

    So - how does the little amplifier go about that?

    Fundamental rule of opamps is this - they will hold their two inputs equal. It is incumbent on designer to provide circuitry that allows the opamp to hold them equal.
    The other important trait of opamps is their input is electrically quite 'stiff' - it won't accept any current.

    Okay now back to the bridge

    This is important - voltage at left corner is always fixed at some fraction of voltage applied by opamp. Let us assume half, for simplicity's sake. If R3 isn't = R1 it'll be some other fraction but let's stay simple with half.
    The little opamp's + input is tied directly to left corner of bridge.
    So the opamp will do whatever is within its power to hold its - input at same fraction of applied bridge voltage, half.

    ...........Okay now it's thought experiment time....

    Assume we are at equilibrium - steady airflow over wire and it's holding steady at design temperature so voltage on both corners is half applied bridge voltage.
    Power is whatever is necessary to keep right corner at that voltage, and steady because we're at equilibrium...
    There's no current through R4 so no voltage across it, so opamp's -input is also at half applied bridge voltage.. .

    Now in your mind increase airflow ever so slightly.
    Wire cools a teeny bit so Rh1 decreases, voltage at right corner of bridge drops ever so slightly.
    R4 is in between that right corner and the opamp's - input. Opamp will do something to keep his - input at half applied bridge voltage - what can he do?
    This is where the capacitor comes in.
    There's only one thing opamp can do - move his output higher to draw some current through capacitor C1 and R4 , raising voltage at his - input back up to half applied bridge voltage.

    Aha there's the mechanism: cause some current through R4.
    How about putting a number on it?
    Allow me license to assume a number for purpose of demonstrating a principle....
    (Edit - I tried to clean up the arithmetic, edits in this color)
    The voltage to be made up by current through R4 is, let us just guess- one millivolt, .001 volt.
    Ohm's law says we need 0.001/R4 amps through R4. That's 1/R4 milliamps..
    Well, that current flows also through C1.
    Current through a capacitor is in proportion to rate of change of voltage across it, I = C X dv/dt.
    ( Usually we deal with sinewaves which are a mathematical curiosity. Let us stick with DC here.)

    So, a current of .001/R4 amps through capacitance C1 requires what rate of change of voltage?

    i = C1 X dv/dt
    .001/ R4 amps = C1 X dv/dt
    dv/dt = .001/ (R4C1) volts per seond

    so opamp will begin raising his output by 1/R4C1 milli volts per second to hold his inputs equal...

    and that's a constant rate of change in positive direction.., and more positive increases power to bridge....

    AHA ! So an increase of airflow causes the little opamp to begin steadily raising his output voltage.
    That will heat the wire more, making up for increased heat loss to airflow, thereby re-establishing bridge balance , thereby reducing need for current through R4,
    so you should see a nice asymptotic approach to equilibrium at somewhat higher power.

    That's what you'd expect for an increase in airflow, isn't it?

    You should work that thought experiment in your head until it's intuitive both directions - both airflow increase and airflow decrease.

    ...........end though experiment......

    Now observe that if dv/dt = ki, v = k∫it.

    That's why they call it an integrating amplifier.
    In this circuit it's an integral controller.

    Most electronic folks will understand if you just point to it and call it an integrator. Point out it's wired to hold the bridge balanced.

    If your audience is mechanicals, tell them it's a 4 bar mechanism held in place by an integrator that tweaks the length of leg Rh1..

    There are some genuine educators in the forum, i'm not one so if i'm off topic it wasn't by intent.

    Sophie already pointed out the transistor lets them use an inexpensive low power (but precision if you're going to integrate) opamp.

    old jim
    Last edited by a moderator: May 6, 2017
  5. Jun 23, 2013 #4
    Hollllllllly Mother....

    I couldn't hope for anything better.
    Thank you ,thank you,at this moment I fill my PP presentation with your explanation.Thousand thanks to you !!
    Thank you for the simpleton explanation (to my opinion ,farty words and dry explanation of lecturers means they do not "feel" the material themself),and you passion in explanation.
  6. Jun 23, 2013 #5

    jim hardy

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    Let us hear how your talk goes, will you ?

    In presentations it's always a nice touch to have a real life exhibit.
    Perhaps you'd want to get a junkyard automobile Mass Airflow Sensor. The heated wire is on a thin film inside, your audience could pass it around and see it. (run through the dishwasher first to degrease)

    ...thanks for the kind words.

    old jim
  7. Jun 23, 2013 #6
    Well,the presentation is about flow meter which is tiny as hell and micro-machined (produced by the same method as PCB's),and its fit is inside nasal passage .
    I doubt automobile sensor would do the trick,it's kinda hard to stick it inside the nose :). Also,what is much more important was to show the control system of circuit,and establish its static and dynamic characteristics,which I couldn't do cause I didn't understood how the circuit works .

    Also,in bibliography,special thanks to you :) .

    P.S:Final question -Why does it have to be integrator?Wouldn't regular op.amp with a feedback do the same job ?
    Last edited: Jun 23, 2013
  8. Jun 23, 2013 #7


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    The answer is in Jim's post. The circuit measures changes in velocity, you need to total all of those changes.

  9. Jun 23, 2013 #8

    jim hardy

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    Fair enough. I had no idea what it looked like.

    Reason I suggested the automobile part is:
    1. it's big enough to see - the sensor is on a thin plastic film in middle of airflow;
    2. the first thing you notice about it is how they made it preposterously low mass and plenty of area for heat transfer - that makes it fast responding. But it looks too fragile for a car part.

    Integral doubtless could improve on following answer with some better math than I can do -
    but at intuitive level here's how I think of it:

    Now you're into control theory basics.
    An opamp with just gain would control, but with error equal to 1/ gain.
    When you increase the gain to reduce the error you'll reach a point where it breaks into oscillation, and likely before you have error as small as you'd like it..
    That is because of the time response of the heater.
    Its response lags because of its thermal inertia.
    If you have substantial gain the system will oscillate at the frequency where that lag time is one half a cycle, and it will probably make a nice sinewave.
    This is explained beautifully in your Mark's Mechanical Engineer's Handbook chapter on automatic controls. The lag makes the feedback that you intended to be negative , become positive at that frequency where lag = a half cycle.* There's a term "Phase Margin" that is related , it tells you how close you are to oscillation.
    You've heard this feedback oscillation in school PA systems - the whistle when microphone is too close to speaker.
    *(remember - for a sinewave, a half cycle phase shift is same as multiplying by -1.)

    So control system wise,
    1. that integrator is not so fast as an opamp with DC feedback so it's more tolerant of delay in system response ; and
    2. it will keep on integrating until error is zero.
    Those are the two reasons for using the integrator.
    Again Marks Handbook has a good chapter on this.
    I think of the integrating controller as 'eventual' infinite gain.
    It is widely used in industry where you want fine control. It'll drive error to zero for you.

    If you guys are prototyping this, try making R4 adjustable and see if you can make it go unstable. Typical oscillation from too much integral is at 2/3 the integral time constant R4C1.
    If you're simulating it have fun with thermal inertia and heat transfer coefficient.
    Last edited: Jun 23, 2013
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