Strain Gauge Oscilloscope Help

I am using Vishay 120 ohm Strain Gauges, (specifically EA-06-062AP-120)
to try and generalize certain readings taken from aircraft engines. We
have set up a small device to vibrate to simulate an engine on a small
scale. The strain gauge is mounted to this device and the leads are
then attached from the strain gauge terminals to a low noise signal
amplifier and then to a computer based oscilloscope.
Here is a link of a sample output image taken at 16.7 Hz. The small
wave is coming from a signal generator while the larger wave is from
the strain gauge and amplified by low noise amplifier.
formatting link

I understand that a strain gauge is in essence a resistor. However this
setup is just going through a low noise amplifier, and not a wheatstone
bridge. I understand what I would probably expect if I went through a
wheatstone bridge followed by the amplifier, but this is different. So
is what I am looking at in that graph the value of the resistance of
the the strain gauge with respect to the signal from the signal
generator? Can anyone help explain to me in english what exactly I am
looking at on this oscilloscope screen. What is the significance of the
Vrms value if it is a resistance? And I know the -24.8dBm is in
relation to the gain, but what does it mean?
The signal generator is at 16.7 Hz with a setting of 1.5V per division,
while the signal from the strain gauge is a setting of 50 mV per
division.
On a different note, the amplifier has a gain based on the resistance
value of a particular potentiometer mounted to a circuit board. Is
there a way to measure the resistance of that potentiometer while it is
mounted to the circuit board? Is it acceptable just to use an ohmmeter
across available terminals?
Any help is much appreciated.
Reply to
Whatever I Fear
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An 'amplifier' by itself can't measure a resistance. There must be some aspect of the circuit passing some current through the strain gauge. It may be built in to the 'control amplifier' you mention. The scope is showing you *voltage* output of the control amplifier. To work back and get the resistance, you need to know something about that exact 'control amplifier'. Sometimes referred to as 'excitation' current.
I would also guess that you have the scope on AC coupling. So any fixed DC signal from the control amplifer is filtered out and you're only seeing the changes in voltage (50 mV per div). So once you have the control amplifier's particulars, this can tell you the variation in resistance (but not the total resistance). Resistance variation may be enough if you have data about the particular strain gauge.
So with a tiny amount of vibration, the strain gauge has a small amount of resistance variation. Which in turn results in a tiny variation in 'control amplifier' output.
Generally, no. Your meter will be reading the combination of the potentiometer and any/all circuitry that is in parallel with it. So you will get a lower reading than what the potentiometer is actually set for. To accurately read the potentiometer (although, I can't see what good that would do), you'll have to somehow isolate it from the remaining circuitry.
daestrom
Reply to
daestrom
You need to find out
1) The wiring diagram of how the strain gage is wired. There must be some bridge completion resistors in the circuit. It's probably a wheatstone bridge circuit, but how many legs are active strain gages? You must know this.
2) Strain gages have a gage factor which relates strain to change in resistance. You should know what this is.
3) You need to calibrate your instrumentation by
a) replacing the gage with a voltage source or b) attaching a calibration resistor to the circuit or
c) applying a known physical load to the specimen
4) Lastly (actually firstly) you need to get a book on how to do all this and read it.
dave y.
Reply to
dave y.
I can't tell what is really going on without a circuit diagram. I can give you some pointers however.
The resistance change is proportional to strain. Typical gauge factor is about 2. So for your 120 ohm resistor a 1000 microinch/inch strain should cause the resistance to change by 0.12 ohm.
You should try a calibration run by changing the resistance of the gauge and seeing what the output of the amp does. For example, placing a 120 Kohm resistor in parallel with the 120 ohm gauge will produce a total (parallel) resistance of 119.88 ohms. Measure the amp output with the 120 ohm gauge without the 120 Kohm shunt, then with the 120 Kohm shunt: the change in amplifier output is the change due to 1000 microinch/inch of strain.
M Walter
Reply to
Mark Walter
the -24.8 dbm is a measure of the power usually assuming a 600 ohm load and is really just a voltage measurement.
when you use a strain gauge out of a bridge circuit there are loading considerations that will affect the reading. the balanced nature of the bridge gives greater sensitivity. if you are using the strain gauge in a different configuration it will be a voltage divider including the input impedance of the meter.
it depends on the external circuitry but the resistance across the pot in circuit with the power off will usually be close. most pots have the value printed on them.
Reply to
bob mcree
hI-
1. Try getting a free copy of PMV2000 image viewer. Then convert any images to be shared to jpg if you expect people to download them. BMP are large files and take too long.
2. To measure a pot on the fly, first measure the total resistance of the pot. Then when the pot is installed, connect the ohmmeter to the unused portion of the pot and this should not interfere so long as the ohmmeter leads are to the op amp output and to the unused section of the pot.
3. Analog Devices has a little monthly rag called Analog Dialog. They have discussed just about everything at some time.
4. Try looking for the info you need at Tinaja.com. This fellow has an encyclopedic site of just about everything electronic.
5. I couldn't understand your problem from a quick read of your message. Try writing more clearly so you can benefit from these contacts. I couldn't understand my own first writings and I knew the problem. Look at some patents and there find a good example of the most clear, unequivocal technical writing on the planet.
Charles Gilbert Consultant NonDigital.Netfirms.com
Reply to
NonDigital

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