# help with voltage reducing, then amplification

Discussion in 'General Electronics Discussion' started by David Stamper, Jan 5, 2016.

1. ### David Stamper

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Jan 5, 2016
This is probably going to sound weird but, this has been bugging me the past week. In my senior engineering class, we are designing a pressure sensor circuit that will measure the changes in air bladder pressure when more force is acting on it. The theory behind it is, when more pressure is applied, the higher the voltage output on the sensor, where less pressure applied yields a smaller voltage output.

I have a pressure sensor circuit (SM5420C) that we are using a maximum voltage supply of 5V. at its base pressure for the device, the voltage output of the sensor goes to 2.55V. When more pressure is applied to the device, the voltage changes to become 2.57V. This is a very small change in voltage (20mV), and I need to see that change from a 0V to 5V range (where no/smallest pressure applied is 0 volts, and highest pressure applied is 5 volts).

I went on the approach of using a non-inverting Op Amp circuit:https://upload.wikimedia.org/wikipe.../300px-Op-Amp_Non-Inverting_Amplifier.svg.png but am having trouble dealing with keeping constant gain.

if the voltage did not start off at 2.55 volts (maxing at 2.57V), and instead just started right at 0V (maxing at 0.02V) so we see the 0 to 20mV range only, then the gain for a 0 to 5V range of desired output would be approximately 250.

getting the resistors for that based on the picture shouldnt be the issue. The issue is because of the fact that the pressure sensors voltage output, always levels out at 2.55 volts when no pressure is applied.

Is there a way to reduce the voltage output of the pressure sensor by 2.55 volts so we focus only on that 20 mV change? Would resistors be the trick to do so? Or would that also affect that 20mV range we are looking at too? What would be the recommendation of approach?

Last edited: Jan 5, 2016
2. ### Gryd3

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Jun 25, 2014
You are using the sensor incorrectly

I looked at a datasheet, and I see 4 pins that must be used.
Which ones are you using?

(Look into how you need to use a wheatstone bridge. The datasheet, in addition to properly reading from a wheatstone bridge should fill in the missing puzzle piece for you)

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3. ### David Stamper

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Jan 5, 2016

I guess I could be wrong but it was my understanding that the +in and -in corresponded to the voltage supply, where the +out indicated that when pressure became applied, we would get a higher voltage, and for -out when pressure is applied, we get lower voltage.

4. ### Gryd3

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Jun 25, 2014
Take a look at the sample image here :

Then take a peek at the datasheet for your sensor here :
Page 3. Pinout section. It should look similar to the above

What ends up happening, is that sensor has a built-in wheatstone bridge.
What you are doing is measuring voltage at the + output, which will almost ALWAYS be very close to half the Voltage you apply because the 4 sides of the bridge will be closely balanced.
What ends up happening, is the + and - outputs need to be combined... the 'difference' between these two outputs is what will be useful.
eg... take a volt-meter or multi-meter and put the black probe on the sensor -, and the red probe on the sensor + and you will get a very small voltage on your meter!
You can do the same thing with an opamp
Just find the difference between the two, then amplify and you are home free.

Last edited by a moderator: Jan 7, 2016
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5. ### Gryd3

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Jun 25, 2014
Of course... you could always make your own 'reference' voltage and then do the same trick... take the difference between the reference and the + sensor and you can get similar results.

6. ### David Stamper

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Jan 5, 2016
wow, this really helps a lot, (and made me feel like an idiot for forgetting about this haha, but ill work on that later). Thank you so much! Ill follow up tomorrow on if it has worked. Thank you again!

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7. ### hevans1944Hop - AC8NS

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You didn't provide the complete part number for this Wheatstone-bridge piezo-resistive absolute pressure transducer. According to the Silicon Microsructures Incorporated datasheet this transducer is available in four full-scale ranges: 15, 30, 60, and 100 psi with 100 mV full-scale output.

The differential output is nominally zero with a ±35 mV tolerance using 5 V excitation when the applied pressure is zero. Note this is an absolute pressure transducer, so "zero" pressure is actually occurs when a good vacuum applied to the input pressure port. A full-scale range above atmospheric pressure (about 14.7 psi) is required to linearly measure your air bladder pressure, which starts at atmospheric pressure and increases from there. A full-scale range of 30 psi would seem to be appropriate. That would produce half of the full-scale output when the input pressure port is at atmospheric pressure.

The internal Wheatstone bridge configuration is not particularly well-balanced, so you will have to add an external offset adjustment to produce zero output with no pressure applied. Once the offset is zeroed out with the pressure port exposed to atmospheric pressure, the differential output must be amplified by a factor of 50 to produced a full-scale output signal of 5 V. Note the full-scale output is only 100 mV no matter which pressure range your transducer happens to accommodate. Fortunately it is "proof tested" to three times the rated full-scale pressure, so there is some room for gain calibration error without destroying the piezo-resistive strain-gage elements inside the transducer.

There are many different circuits available for bridge conditioning, which is the appropriate term to use for the offset and gain circuitry. Your transducer has rather high-valued piezo-resistors (5 kΩ), and moderate sensitivity. The "quick and dirty" way to compensate for offset is to parallel one of the bridge arms with a variable resistance that is much larger the arm resistance. A "quick and dirty" way to obtain the x50 gain is with two op-amps to produce a differential amplified output followed by a single op-amp to convert the differential output to a single-ended output. See image below.

You can often get by without an offset adjustment at the bridge if the gain of the differential input stage is set to a low value, say x2 or x10, so that the op-amp outputs remain in their linear range even after the offset voltage is amplified. The output op-amp would then require a gain of x25 or x5 to produce an overall gain of x50. The offset is then removed by injecting a small voltage in series with the "grounded" resistor of the output op-amp.

Read up on strain-gauge bridge amplifiers here. Commercial products are available but tend to be a bit pricey. I've always "rolled my own" unless a customer insisted on a fancy commercial product.

Hop

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8. ### David Stamper

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Jan 5, 2016
This explains a lot, sorry forgot to mention one using. We are using a 15 psi (Absolute) sensor. I did some initial testing again this morning and found that there would be about 50mV change in voltage, depending on the pressure applied (at least for when i connected the + and -Sig to a voltmeter).

But as you mentioned there is still an offset of about 0.17 volts that would need to be corrected.

i looked up into those strain-gauge bridge amplifiers. I came across a signal conditioning report by texas instruments that mentions the three op amp amplifier you mentioned as well as a single op amp differential amplifier. The single one though probably doesnt correct for that small voltage offset.

http://www.ti.com/lit/an/sloa034/sloa034.pdf

The three op amp instrumentation amplifier would correct for the offset?

So say that I want the output voltage at 5 volts, so in the form under the three op amp instrumentation amplifier, it mentions Vo = (Sig+ - Sig-)(R4/R2)((2Rf/Rg)+1). Say in the case that sig+ - Sig- is about 50mV when Vo is at 5V. then theoretically i can use that to find what resistor values for this gain.

You mentioned that one way to correct for the offset is paralleling one of the bridge arms with a variable resistance, would it be possible to correct for the offset within the Op Amp itself, rather just in one of the op amps in any case?

Thank you also very much for explaining all this to me! This has been bugging me all week.

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9. ### David Stamper

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Jan 5, 2016
So If i am to make the instrumentation amplifier as shown by attachment, I have V1 (The +Sig of sensor) roughly 2.58 volts and V2 (-Sig of sensor) at 2.45 volts, If I want the desired Vout of 5 volts for a Vin ((+sig - (-Sig)) [mind typo in picture] then I could use that possible configuration? Or would that gain be incorrect for what i want as desired amplification?

On another note for a set up like this, does this require rail-to-rail I/O Op Amps?

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10. ### hevans1944Hop - AC8NS

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No. The only purpose of the three op-amp circuit is to convert a differential input signal into a single-ended (ground referenced) output signal. The input offset correction is a separate issue.

The Texas Instruments paper is a really good introduction to signal conditioning, but there is so much more to learn, especially if you happen to have a low-sensitivity bridge circuit with, say, 10 mV full-scale output instead of your 100 mV. The three op-amp circuit is a classic design used by virtually all instrumentation amplifiers. It has the advantage of symmetrical inputs, high input impedance, and good common-mode rejection if the resistors in the output op-amp are carefully matched. By "matched" I mean the impedance "seen" by the inverting op-amp input is exactly the same as the impedance "seen" by the non-inverting input. That is why TI recommends using a resistor pack for these four resistors: the four resistors are more likely to all be closer to the same value than four randomly selected discrete resistors, even if the discrete resistors have low tolerance, such as 0.1% or 0.01%, for variations from their nominal resistance. TI also recommends that the output op-amp be configured for unity-gain because it is easier to obtain a good CMMR (Common Mode Rejection Ratio). However, with careful selection of resistor components, moderate gain can be accommodated as I mentioned in post #7.

The three op-amp "classic" design does not eliminate offset. Its purpose is to convert a differential signal to a single-ended (common referenced) signal, optionally with some overall gain. Whatever offset is present on the inputs will be amplified and appear in the output. If you don't want the offset to overwhelm the two input op-amps, the gain of that part of the circuit must be low enough so [ (+In) - (-In) + Voffset ] [Gain1] does not drive either op-amp to its positive or negative rail. That is why some transducers are nulled (offset removed) before applying their differential signal to the instrumentation amplifier.

With individual strain-gauges having all eight leads available for connection as a Wheatstone bridge, there are more options than just paralleling one of the bridge legs with a large-value adjustable resistor for removing offset. The simplest option just inserts a low-value trim potentiometer (typically a hundred ohms or so) in series with the two resistors on one side of the bridge. The output is then taken between the potentiometer wiper and the junction of the two resistors on the other side of the bridge. Unfortunately your transducer doesn't provide access to each of the piezo-resistive elements individually, so other methods are required.

Almost every offset null method has some disadvantages, the most serious being temperature compensation, or the lack thereof. The four piezo-resistors will track each other closely for changes in ambient temperature because they are micro-machined parts in the same thermal environment. That means temperature changes will have a minimal effect on bridge output. This goes out the door when you add external components to remove offset. These external components must have minimal effects as a function of ambient temperature change if their offset value is to remain constant. This isn't always easy to do if you are trying to work with microvolt signal changes in the presence of a few volts of common-mode signal. Fortunately you are working with millivolt signals (fairly high level for this type of instrumentation) so you can probably get by with a simple voltage offset from a stable reference voltage to remove the bridge offset. The question is: where and how do you insert the offset correction voltage?

I prefer to lift the "ground" end of the resistor (R3 in TI's Figure 5 Three Op Amp Instrumentation Amplifier) between the non-inverting input and common of the output op-amp and insert the offset voltage there. There are many ways to do that without messing up the CMRR of the output stage, but perhaps the easiest is to use the output of another op-amp connected as a unity-gain voltage follower. You apply the offset correction to the non-inverting input of that op-amp and connect the op-amp output to the "ground" end of the resistor.

BTW, all this is much easier if you use symmetrical bi-polar positive and negative power supplies to power everything, including the transducer. Because you are working with differential signals, the polarity can swing either positive or negative. It just makes life a lot easier if you don't have to worry about that during the signal conditioning stages. When you get to the point where you need to send the now high-level signal to a uni-polar input analog-to-digital converter an absolute-value circuit, or even just a simple small signal diode, can remove the output polarity you don't want or need.

If I were you, I would try TI's Figure 2 Single Op Amp Differential amplifier first to see how well it works. You could inject the offset voltage into the bottom of the resistor (R3 in TI's Figure 2 Single Op Amp Differential Amplifier) going to common. And you don't need a rail-to-rail op-amp or a bi-polar power supply. I don't much care for this circuit because half your excitation voltage appears as a common-mode signal at the two inputs. Unless everything is well-balanced resistor-wise, any changes in the common mode voltage will be converted to a differential signal that will appear as changes in the output, indistinguishable from changes in the pressure. The CMRR of the op-amp is important too, but most modern op-amps have very high CMRR and PSRR (Power Supply Rejection Ratio) specifications. The problem is in symmetry, or the lack thereof, of the resistances.

Well, my understanding is you have chosen the wrong transducer range, 15 PSIA, so with nothing connected to the pressure port the transducer output is already at full scale (100 mV for 5 V excitation). That may account for the differential readings you get on your multimeter with no pressure applied. And some of that reading could also be inherent offset caused by manufacturing tolerances, which can be anywhere from -35 mV to +35 mV.
Either way, you are fast approaching the maximum input pressure for a linear response. If possible, I would replace the transducer with one rated for 30 PSIA full-scale. Other than that little nit-picky detail, your configuration is correct. I am not sure why you expect the output to vary by 200 mV thoughout the zero to 15 PSIA pressure range, unless you expect to excite the transducer with a +5 V and a -5V supply. If the full-scale output is indeed 200 mV then a gain of 25 is appropriate for a 0 to +5V output.

If you are using a single +5 V supply to excite the transducer bridge, then you will have about +2.5 V of common mode voltage on each of the inputs, which will cause the first two op-amp outputs to also be approximately +2.5 V with a differential voltage ideally equal to the differential bridge voltage multiplied by (1+2R2/R1). When this differential voltage is applied to the third op-amp to convert it to a single-ended voltage, voltage Va must be negative with respect to voltage Vb to produce a positive output at Vout. All that means is the V1 input must be slightly negative with respect to the V2 input when pressure is applied to the transducer. If you want Vout to go to zero with no external pressure applied to the transducer, you either need to use a bi-polar power supply for its op-amp or choose a rail-to-rail op-amp... and you need to add-in the offset correction, which could be either positive or negative depending on initial conditions at the transducer. Such considerations are why I always use bi-polar op-amp excitation: it makes life more simple.

The classic three op-amp circuit does offer a distinct advantage: it has a high input impedance that will not heavily load your transducer. An unavoidable disadvantage with single-supply transducer excitation is common mode voltage equal to half the excitation voltage at both transducer outputs. This means both input op-amps must accept and tolerate the common mode voltage, which appears un-amplified on their outputs, and it means the output swing of each input op-amp is thereby reduced by the common mode voltage, effectively limiting the maximum gain of that stage before the output swings into one of the power supply rails. This is a bigger "problem" with a single polarity op-amp supply and higher gains where the op-amp outputs cannot swing negative with respect to common.

I am not a big fan of simulation, but you should simulate your circuit, with LTSpice for example, and "play around" with the input voltages to see what happens. This is no substitute for actually bread-boarding the circuit, powering it up, and looking at various nodes with a real voltmeter or oscilloscope, but it can give you some idea of what to expect.

11. ### David Stamper

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Jan 5, 2016
In the lack of symmetry of resistances, do you mean via the wheatstone circuit within the sensor, or the amplifier circuit i had put up?

I think we will be switching over to the 30 PSIA one, it will probably receive a better linear response and not fast approach the maximum.

I will have to look into that as well, I have used LTSpice before so maybe that may help in at least the theory of the work.

I had just thought about something in that, the reason for seeing this offset voltage probably has to do with the fact that it is not a completely zero vacuum acting on the sensor, correct? Granted, the sensor is reading from within the tube we have set up, but I don't think we can get it to a zero vacuum. So I wonder, if it would just be easier to find that maximum voltage output on the sensor based on the maximum pressure that can be applied before the air bladder gets damaged or blows, then use a nominal pressures voltage output as say, our minimum rating value.

it wouldnt be the zero voltage we are looking for just yet, but maybe amplification of the signal would be easier when we have the maximum output as you mentioned. or maybe that might not be a good idea for the end measurement we are looking for at amplification?

Thank you again!

12. ### AnalogKid

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Jun 10, 2015
Since this is your first foray into differential signals, I strongly recommend that you buy an integrated instrumentation amplifier in a chip rather than try to grow one. Growing one from scratch is all about the resistors. Even with 0.1% parts, nothing you assemble will equal the performance of a \$5 or \$10 chip from Analog Devices or Burr Brown (now TI).

ak

13. ### hevans1944Hop - AC8NS

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It looks like you are getting a good handle on Wheatstone bridge signal conditioning and the use of instrumentation amplifiers. These are really old problems that go back to early in the last century when strain gauges were used to measure stress (change of length per unit length) in structural members... everything from bridge and building girders to automobile and aircraft parts. If it could be stressed by applying a load to it, someone wanted to measure how much could be safely applied before failure occurred. Strain gauges are still used today, pretty much for the same purposes, but the technology has changed somewhat.

Absolute pressure gauges require a de-formable membrane or diaphragm, to which strain gauges can be affixed to measure the amount of deformation. Usually one side of the diaphragm is exposed to the pressure port and the other side is exposed to ambient air pressure. These are differential pressure or gauge-pressure gauges. Obviously their output will change with atmospheric pressure if the measuring port is maintained at a fixed (absolute) pressure. OTOH, if the reference side is sealed and evacuated (doesn't have to be a perfect vacuum) then the diaphragm will only respond to absolute pressure at the measuring port and will be insensitive to pressure variations (there are none!) on the reference side. It doesn't have to be a perfect vacuum on the reference side because there isn't enough gas left there to cause significant pressure variations with changes in temperature. The diaphragm is another matter. It needs to be carefully constructed with matched strain gauges to be insensitive to temperature changes. MEMS (Micro Electro Mechanical Systems) solves this problem nicely. You can even integrate the electronics signal conditioning into the MEMS device, although this does increase the cost.

For your instrumentation, it appears there is a minimum gauge pressure (pressure above atmospheric pressure) you need to measure and a maximum gauge pressure you also need to measure. Ideally, the conditioned output will vary linearly from zero volts to +5 V between those two limits. Getting that to happen is what (strain gauge) bridge conditioning is all about.

Since the transducer you have chosen happens to respond to absolute pressure instead of the more common gauge-pressure transducer, you will have an output that is something more than zero but less than full scale at your minimum gauge pressure. At your maximum gauge pressure you want the output to be nearly full-scale, but not greater, in order to obtain best linearity. Hence the need for a 30 PSIA full-scale range. Your transducer has 3X over-pressure capability, without damage to the strain gauges or the diaphragm, but it is not necessarily linear for input pressures between full-scale and 3X full-scale. Actually, no transducer is perfectly linear. Most of them also exhibit hysteresis, where the pressure readings for increasing pressure are not the same for decreasing pressure. Nothing can be done about that AFAIK, but piezo-resistive transducers are better than most at minimizing hysteresis.

The lack of symmetry applies to both the bridge and the subsequent signal processing. You can't do anything about the bridge, other than to try to null out the imbalance with an external resistor network. The same applies to the differential-to-single-ended conversion circuitry. As long as there is a common-mode signal present on the differential inputs, any asymmetry in the signal paths will convert some of the common-mode voltage to a single-ended voltage in the output that is indistinguishable from the bridge output signal. This can be reduced (or even eliminated) by using bi-polar bridge excitation and carefully balancing the bridge so there is zero common-mode voltage. Imagine you build the one op-amp differential amplifier, and instead of applying a differential signal to the two inputs, you tie the two inputs together and apply a signal from there to common. Ideally, with perfect symmetry and infinite CMRR at the op-amp input terminals, there would zero output no matter what signal you applied. This never occurs. There is always some lack of symmetry that leads to conversion of the common-mode input to a differential signal at the op-amp input terminals and its subsequent appearance in the output.

If you want a real headache, imagine how this works at AC frequencies instead of at DC. One of my first jobs as a technician, after leaving the Air Force and way before I had earned my college degree in electrical engineering, was to build an instrumentation amplifier with state-of-the-art CMRR from DC to at least 120 Hz (to suppress power-line effects). This took about a month of research and "playing around" on the bench to find out what would actually work. Commercial instrumentation amplifiers were readily available back then (late 1960s) but my boss wanted to find out how good I was. Ended up using a (now obsolete) integrated-circuit pair of matched input transistors on an integral temperature-controlled substrate. Along with some picofarad-range trim capacitors to null out the stray capacitance the high CMRR objective was achieved. High CMRR wasn't the only objective: low offset voltage, low offset bias current, and high input impedance were also design goals. At the time, I think we could have purchased state-of-the-art instrumentation amplifiers every bit as good, maybe even better, than the one I built for a few hundred dollars. Today you can purchase the same or better performance for just a few dollars. And they will have a footprint about the size of a pencil eraser instead of requiring a small Bud box.

You are on the right track. First establish the minimum pressure you want in your air bladder. This may indeed be a partial vacuum if you intend to evacuate and collapse the bladder (remember, we have no idea what you are trying to accomplish), but it will certainly produce a non-zero output from your transducer bridge. This will be your "zero" and you will need to "null" the bridge output at this pressure (whatever it is). Next you need to establish your maximum working pressure, which will likely be some value less than 30 PSIA but for sure less than 90 PSIA. Whatever the bridge output is at this pressure, you set the bridge conditioner gain to produce +5 V output. There may be some back and forth adjustments between the zero adjust and the gain adjust before everything is responding the way you want, but it's just that simple. Zero first, then gain adjust. Wash, rinse, and repeat.

As I said earlier, it is desirable to null the transducer output at the transducer and leave the gain adjustment to the conditioning amplifier. Normally, you vary the resistor that is common to the two input op-amps (R1 in your circuit diagram) to vary the overall gain. You can also vary the bridge excitation voltage to vary the gain, but I don't recommend that. Try to keep the bridge excitation voltage constant and instead vary the amplifier gain.

If you want to try to null the bridge output using a high-value external resistance across one of the bridge resistors, start with a 1 kΩ trim pot (connected as a variable resistor) in series with a 100 kΩ fixed resistor, the series-connected pair is then connected across one of the bridge resistors. Do this at the minimum pressure and measure the differential output before and after connecting the zero-trimming network. If the differential output increases, move the network to the resistor on the opposite leg. If the differential output decreases, you are on the right path.

Vary the pot from minimum to maximum to see if the differential output can be set to zero. If there is not enough adjustment range, decrease the ohmic value of the fixed resistor until the differential output is zero with the pot adjusted near the middle of its adjustment range. If the zero adjustment is too sensitive, use a smaller valued pot down from 1 kΩ to 100 Ω or so. If it isn't sensitive enough, use a larger value trim pot.

You are paralleling a nominal value of 5 kΩ, and you want to change that value by perhaps one to ten percent by varying the parallel resistance. This requires some experimentation to find the "sweet spot" of fixed resistor in series with variable resistor, but you only have to do it once. After that, only an occasional tweek might be necessary to restore the zero, along with a tweek of the gain potentiometer to set the full-scale output.

If you have to change out the transducer because you broke it (good thing they are cheap!), you might get lucky and be able to use the same zero-trim configuration with the new transducer. However, there is no guarantee the replacement transducer will be offset the same amount and in the same direction as the old transducer, so you may have to repeat the "finding" procedure again to determine where to connect the external zero-adjust network. The values you end up with for the fixed resistor in series with the trim pot will remain the same.

The 1 kΩ for trim pot and 100 kΩ for the fixed resistor are just starting values to get you into the ball park. You could include these in your SPICE simulation to see their effect on bridge output, and perhaps narrow down the "sweet spot" combination quicker.

Hop

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14. ### hevans1944Hop - AC8NS

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Aww! How's he gonna learn anything from that?

15. ### AnalogKid

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"A man's got to know his limitations."

ak

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16. ### David Stamper

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Jan 5, 2016
Our group has been looking into that actually yes, in terms of the end product we are going for anyway this will probably be useful. Some of the chips we have been using right now are TI-LM301AP Op Amps and UA741CN Op Amps, which we weren't really sure were the best for this set up or not, so our group is looking into some better ones. Thank you for letting me know what companies to look at!

I can see now why the need for the higher range. Our group has actually been using three different sensors from si-micro, them being the 15, 30, and a 60 PSIA ranges. Attached is a small pcb weve been using to try them all (minus the chips actually being on them). We could tell that the pressure capability isnt going to be linear, hence it will require some more calibration on our end but it is a start! Some kits have been developed that explains its calibration but we havent gotten chance to order them yet so, we are working on that.

So in to finding this sweet spot we are looking for, mainly it requires first finding the max output of sensor at max pressure first, then designing amplifier based on those values which will later be used for the amplification. I imagine this set up will probably have to be different for the 15, 30, and 60 PSIA chips, so that will be in part the steps for today!

Thank you for your help and explaining this! And also Thank you Hop for your Service in the Air Force, I myself will be commissioning into the Air Force in May, its going to be a challenge, but a great one!

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17. ### AnalogKid

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Jun 10, 2015
Not.

Mentioning the 741 usually brings on a flurry of childish hyperbolic borscht about how old and terrible it is. But for what you are trying to do there are in fact many better individual opamps and way better integrated instrumentation amps, parts that would have been black magic back in the 60's. So unless you are in Russia or central Africa and can't get contemporary parts, leave the grandchildren behind on this one.

ak

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18. ### David Stamper

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Jan 5, 2016
We will probably still be working with instrumentation amplifiers, in terms of some we found at TI this came to mind: http://www.ti.com/lit/ds/symlink/ina114.pdf with a gain setting of G = 1 + 50k/R1, R1 obviously being the bridge resistance, but in terms of other ones we could use, such as just a single op amp, do you have any specific recommendations? if you don't mind.

19. ### AnalogKid

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Jun 10, 2015
A single precision opamp configured as a difference amplifier with 0.1% resistors will have a CMRR of less than 60 dB, while the INA114 has a higher input impedance and a minimum CMRR of 96 dB, 63 times better. Both the lower input impedance and lower CMRR contribute to what appears as a gain non-linearity error, overall apparent gain varying with signal amplitude.

ak

20. ### hevans1944Hop - AC8NS

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@David Stamper Ah! You are conditioning at least three of those little puppies, and it looks like you want to insert the PCBs into a tube, perhaps along with some other electronics, and eventually produce a commercial product. Good job! In which case, the commercial instrumentation amplifiers are definitely the way to go as @AnalogKid suggested. I thought you were perhaps in a university environment and learning how all this works rather than shooting for a product you can market.

I have been a fan of Burr-Brown and Texas Instruments since I first started using their products in the 1960s. When Burr-Brown was purchased by Texas Instruments for \$7.6 billion in September 2000 things just got better. With the TI mother-ship support they have continued to excel in design and production of analog devices, which some pundits thought were a dying breed. They are still my "go to first" source, although there are others such as Analog Devices and Linear Technology. Analog design is far from dead. In fact it is thriving and now integrated with digital, as can be seen in TI's MSP430 and Microchip's PIC line of processors, just to give two examples.

As for op-amp selection (if you still want to go there), that's a whole 'nother ball game. I like the generic 741 in all its many variations (single, dual, and quad packages) because it is "bulletproof" and cheap, but it is very old technology, being the successor to the even more venerable Fairchild μA709 op-amp, which required external compensation to keep it from oscillating, and external trimming because its offset voltage which was terrible. Since the 741 was introduced, technology has gone on to offer FET inputs, CMOS, and chopper-stabilized op-amps, laser-trimmed offset compensation and power op-amps with dozens of amperes and hundreds of volts output capability. And that's just the tip of the iceberg of technology available today.

If you are new to op-amps, TI offers an application note that will get you started. It's about a two megabyte download, so a bit large to upload here. A "pretty good" instrumentation amplifier is the Burr-Brown/Texas Instruments INA121, available in an SO-8 surface-mount package. You might want to start there. It does require a bi-polar power supply (as does the INA114 you linked to) but ±2.5 or ±5 V will work fine. You may need to use ±15 V supply to get a ±10 V output. See discussion below. Be sure you null the bridge output before applying the bridge signal to the amplifier.

I don't understand this. The gain-setting resistor has nothing to do with the bridge resistance. The bridge output is applied to the two high-impedance inputs and the bridge resistance has no effect on the instrumentation amplifier gain.

No, it requires you to know the change is bridge output between the minimum and maximum pressures. It is this change in signal you want to amplify, preferably after removing whatever offset is present at the minimum pressure. Thus, you start with zero input to the instrumentation amplifier at minimum pressure (by nulling the output at that pressure) so the output of the amplifier will be zero for any gain (ideally) at that minimum pressure. Then you set the instrumentation amplifier gain to produce +5 V at your maximum pressure.

You would need to do this for each transducer because there are three different full-scale pressure ranges of 15, 30, and 60 PSIA, but all have 100 mV full-scale output at their rated pressure. In other words, each transducer will produce a 100 mV change in output (with 5 V excitation) from 0 PSIA to its rated pressure range of 15 or 30 or 60 PSIA. The "zero" for each transducer will be different for the same pressure applied to all three transducers because your "zero" pressure isn't 0 PSIA unless your are pulling a vacuum on your air bladder. It will instead be whatever minimum pressure you apply to the air bladder, and this minimum pressure will cause a different bridge output depending on the full-scale range of the transducer.

You could hope that the bridge output at minimum applied pressure is small enough that, when amplified by the instrumentation amplifier, there is still enough "head room" in the amplifier output to accommodate the bridge output at maximum applied pressure. I don't know what your minimum applied pressure is, but it is likely to be a significant fraction of the transducer's rated full-scale pressure, perhaps even atmospheric pressure if you don't pull a vacuum.

Let's say you are using the 15 PSIA transducer and are at atmospheric pressure for the minimum pressure. Since atmospheric pressure is about 14.7 PSIA, you are nearly at full-scale for this transducer, half of full-scale for the 30 PSIA transducer, and one third of full-scale for the 60 PSIA transducer. Without nulling any of the transducers the instrumentation amplifiers could be set at a gain of 50 to produce +5 V full-scale output for each transducer, but the 15 PSIA transducer would almost immediately reach its linear full-scale limit (but not its maximum safe pressure limit of 45 PSIA) when the gauge pressure increases from atmospheric pressure, 0 PSIG, to 0.3 PSIG. That's probably a useless measuring range. The 30 PSIA transducer will be at half of full-scale under the same conditions so the pressure can increase from atmospheric pressure, 0 PSIG, to 15 PSIG before it reaches full-scale, but then you are only using half of the linear range of the transducer. The 60 PSIA transducer will be at one-third of full-scale under the same conditions, so the pressure can increase from atmospheric pressure, 0 PSIG, to 45 PSIG before it reaches full-scale, and now you are using two-thirds of the transducer range.

Unless your applied pressure is less than atmospheric pressure (you pull a partial vacuum on the air bladder), you will always be limited to less than the full-scale range of the transducer without external offset nulling of the bridge. That's just the way an absolute pressure transducer rolls. Some designers will just accept this and compensate for it in the analog-to-digital conversion, subtracting out the offset in software. What you sacrifice is dynamic range in the measurement since the instrumentation amplifier gain must be small enough to accommodate the initial offset and not saturate (become non-linear) when full-scale output, plus the offset, from the bridge is applied.

Your product has an inherent and unpredictable maximum offset of ±35 mV at 0 PSIA input pressure, which is one third of the linear output range of 100 mV. It is impossible to know the magnitude and polarity of this offset without measuring it for each transducer, but you can set the amplifier gain to accommodate whatever it is without external nulling. For example, choose worst case of +35 mV offset that is not nulled externally. With a gain of 50, this produces an output of +1.75 volts from the instrumentation amplifier, leaving 3.75 volts of headroom for signal if the maximum output is limited to 5 volts. However, increasing the output capability to, say, ±10 V restores the headroom and allows the transducer's 100 mV full-scale change in output to produce 6.5 V at the output of the instrumentation amplifier. Voila! No external nulling required. You can now remove the high-level +1.75 V offset output by subtracting it from the instrumentation amplifier output using a simple op-amp summing circuit, or scale and digitize the output with an A/D converter, removing the offset in software. The latter is the approach I would take to minimize external components associated with the pressure transducer.

It would help if we knew what you are trying to do rather than how you plan to do it. For example, one transducer with 60 PSIA range may be sufficient to use without external nulling if a high-resolution A/D converter (22 or 23 bits delta-sigma converter comes to mind) is used at the instrumentation amplifier output. Non-linearity and offset can be easily removed with software.

Hop

BTW: @AnalogKid knows what he's talking about. He's just less windy than I am.

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