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Project Help and Ideas » Need Help with Strain Gage Project
March 05, 2010 by scaubrey84 |
So I finished the nerdkits guide and am attempting a variation on the weigh scale project. I am using the same code as the tempsensor project in an effort to simply verify the output of the number of steps in the ADC (no polarity noise cancellation and sending the reading to the LCD). The output of the amplifier (AD620ANZ) has simply replaced the output of the temp sensor on the board. I have replaced one of my resistors in the wheatstone bridge with a potentiometer to try to zero out my reading while the gage has no strain. Gain is set to 150. I understand that when I turn the potentiometer one direction the steps will increase to a maximum and go no more (usually a little past 800). However, when I turn it the other way it never gets to zero steps. Once it reaches about 150 steps or so, it stops and won't decrease any more, though the potentiometer can continue to change the resistance. If I stop the potentiometer right when it hits 150, I can get a nice linear reading from strain gage (though with quite a bit of noise). But if I go too much farther, I get no reading but a constant 150. Can anyone tell me where that signal is coming from giving the steady 150 steps? Shouldn't I be able to get a reading over the bridge with the potentiometer that shows up as 0 steps in the ADC? It doesn't seem to be dynamic noise (which I do get in the "sweet spot" of the strain reading as described above). I'm relatively new to developing my own electronics, although I do have a degree in Mechanical Engineering. Any help would be appreciated. I may also post in the forums. Thanks! Cameron |
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March 06, 2010 by BobaMosfet |
scaubrey84- The calculations in the tempsensor project are based on the mV that the LM34 inputs into the ADC. In order to do the same thing with the component you are using instead, you need to obtain the datasheet for that component and find out what it's mV output range is, adjust your calculations accordingly. Remember, output from the LM34 is LINEAR (about 10mV per degree as I recall). Your inability to reach zero is because too much current is allowed to flow. Soak this up and it will bring it inline with what you need. Do you know what a 'voltage divider' is? BM |
March 06, 2010 by scaubrey84 |
Hi BM, The 'component' that I am using is a wheatstone bridge (4 resistors, which act as a pair of voltage dividers) where the reading across the voltage divider is sent to an amplifier, which is sent to the pin on the ATmega. I am changing the resistance in one of the arms using a potentiometer to try to balance the bridge. This is linear, is it not (V=IR)? The strain reading is also linear. So I should be able to adjust the potentiometer to get close to zero output from the amplifier before straining the gauge (at least be able to pass the zero point), and then the strain reading should unbalance the bridge and give me the signal I want. Maybe there is too much current flow, but it shouldn't be from the bridge. Any other ideas? |
March 06, 2010 by scaubrey84 |
Or maybe you're referring to the amplifier, BM? Also, if I implement this to switch the polarity of the bridge excitation like shown in the video, can I use any of the free pins on the MC to do that? It's just a matter of setting the pins high or low, correct? Anyone? |
March 06, 2010 by scaubrey84 |
I just found this website (http://www.scitec.uk.com/lockin_amplifier/notes/wheatstonebridge.php), which explains my first question I think. "The potentiometer Rg can be used to correct for this mismatch and could, in an ideal world, bring the output from the lock-in to zero. Unfortunately, it will be found that although either the X or the Y outputs of the lock-in can be made to reach zero, it is not possible to bring both of them to zero at the same time. This can most easily be seen by looking at the R output which calculates the amplitude (or modulus) of the input signal. Adjusting Rg will decrease the output R to a minimum but will not bring it to zero. Due to the very small parasitic capacitance associated with the circuit a small phase shift is seen down either branch of the Wheatstone bridge. This small phase shift stops the two signals generated down either branch from cancelling out each other completely." |
March 16, 2010 by mrobbins (NerdKits Staff) |
Hi Cameron, I just replied to your e-mail that asked me about measuring the alternating excitation voltage with a multimeter. Just for future reference and other NerdKits users who may come across this: A multimeter showing you a DC voltage is showing you a time-averaged measurement. So if you look between your two alternating excitation pins, you'll measure zero, because half of the time it's +5, and half of the time it's -5V. But if you look between ground and either pin, you'll read 2.5V. Although your first reaction may be that a digital voltage pin should be either 0 or 5, the multimeter is seeing the time-average, which is 0 for 50% of the time, and 5 for 50% of the time. The same concept is used for creating a DAC using PWM. As far as your initial question, where the AD620 output only ranges between 150 and 800 on the ADC (which means a voltage of about 0.7 volts to 3.9 volts): The first thing to do is do a "sanity check" with a multimeter, and see whether the ADC's reading matches the true voltage. This tells you whether your issue is on the analog/amplifier side, or on the ADC side. However, the actual answer as to why the results never go past 150 or 800 is that the AD620 is not a "rail to rail output" amplifier. If you look at the AD620 datasheet, on page 4 they list "Output Swing" as being "-Vs+1.1v to +Vs-1.2v". That means that for a "single-sided supply" like we're using, they only guarantee that the part will work over an output range of 1.1V to 3.8V! Notice that this matches the 0.7-3.9V range you measured pretty well -- the part actually performs just slightly better then the datasheet promises. Hope that helps! Mike |
March 17, 2010 by scaubrey84 |
Great feedback. Thanks for the help, Mike! |
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