0
\$\begingroup\$

I cannot for the life of me understand how mV/g or LSB/g can be related to the g sensitivity. For example, how can I find 0.0025 G sensitivity accelerometer given the sensitivity in mV/g? I haven't found resources online to explain this clearly to me. Can someone please enlighten me? Thanks!

\$\endgroup\$
11
  • 1
    \$\begingroup\$ You're talking about neither the gravitational constant nor grams. g is a symbol used for the acceleration of gravity, which is roughly 9.8m/s^2 on the earth's surface (varies by location). I'm not sure what you mean by "how can I find 0.0025 G sensitivity accelerometer given the sensitivity in mV/g" .. are you asking how to define the noise level of the accelerometer? \$\endgroup\$ Commented Mar 14, 2020 at 2:31
  • 1
    \$\begingroup\$ @SethShill I think by writing 0.0025 G sensitivity you really mean detectability. (See this.) Tell me if I'm wrong about that. \$\endgroup\$
    – jonk
    Commented Mar 14, 2020 at 3:36
  • 1
    \$\begingroup\$ @SethShill Strange thing: I was, myself, thinking of using the ADXL335 for an example when considering writing an answer. But as I started to consider it, I realized you have a LOT to learn. And this means I'd need to use a slow progression, building up block by block, to reach an answer. So I kind of decided, after seeing Neil's (too short in my mind for what I think you need) answer, that maybe his answer is enough and to just back off. \$\endgroup\$
    – jonk
    Commented Mar 14, 2020 at 18:16
  • 1
    \$\begingroup\$ @SethShill To use these IC accelerometers well, you need not only solid embedded hardware and software backgrounds, but also solid mathematical grounding and solid signal processing background (especially on a class of filters called "optimal" (Weiner, Kalman [digital], and Kalman-Bucy [analog.]) (And there's more, but of secondary need.) Your detectability will depend upon all these skills used, themselves, in an optimal combination. \$\endgroup\$
    – jonk
    Commented Mar 14, 2020 at 18:25
  • 1
    \$\begingroup\$ @SethShill It's kind of like LEDs, in a way. They seem simple enough. And they are. If you just use them to turn on and off as indicators. But when you try and create a full color display using them, one that deals with human perceptions of color and intensity and ability to notice flicker, etc., the whole project requires vastly greater skillsets to be brought together. If you want to use the ADXL335 as a simple "upside down" or "right side up" mode (like an LED on/off), then it's not so hard. But if you want to really use it to the max its capable of, it's a whole different story. \$\endgroup\$
    – jonk
    Commented Mar 14, 2020 at 18:31

3 Answers 3

0
\$\begingroup\$

You have been reading literature that uses the word 'sensitivity' carelessly to mean two slightly different things. The usual sense is 'what is the smallest change in g that can be detected'. However many people will also say that a higher gain accelerometer is 'more sensitive'.

Unfortunately, while sensitivity and mV/g are related, you need other values specified to relate them.

The mV/g is simply the gain of the accelerometer, the ratio of output voltage to input acceleration. That's quite easy to measure. If the acceleromter is sensitive to less than 1g, you measure the output, turn it over, and measure again. That's the change of output you get for a 2g change in input signal.

The sensitivity depends on the noise of the total system, including the accelerometer, any analogue signal processing, the ADC, and any digital signal processing, and crucially, how you define sensitivity. Usually, one of these components will dominate the noise level. If it's the ADC resolution, then a simple scaling of the ADC LSB with the gain will give you a g figure. However, the ADC might be more noisy than one LSB, or you might do some post-processing to reduce the noise below one LSB (it can be done, for some ADCs).

So you need to know the noise. An active accelerometer will generate some of its own noise. Unless that's specified in the data sheet, you won't know that until you buy it. It's likely that you'll be able to build a lower noise signal AQ system and the accelerometer noise will be the limiting factor. The noise from a passive strain gauge should be possible to estimate if you're told the resistance in the data sheet, and your amplifier will probably be the limiting factor there.

\$\endgroup\$
4
  • \$\begingroup\$ The noise is not my concern. I think @jonk pointed out correctly, I am looking at the minimum detectability and trying to find that value. For example [url=sparkfun.com/datasheets/Components/SMD/adxl335.pdf] ADXL335 [/url] lists the sensitivity. But nothing w/ respect to resolution. \$\endgroup\$
    – Seth Shill
    Commented Mar 14, 2020 at 15:47
  • \$\begingroup\$ @SethShill you can't detect something that's below the noise. Perhaps I'm using noise in a slightly more general sense, to include the quantisation noise, the step size, of the ADC that's setting your minimum step size. In a low-noise system, the ADC step determines detectability. In a high noise system, even with a 24 bit ADC, the noise determines it. My general noise means 'anything that isn't signal', and 'that which determines how small an effect you can detect'. \$\endgroup\$
    – Neil_UK
    Commented Mar 14, 2020 at 16:27
  • \$\begingroup\$ @Neil_UK I know you are talking about this device, but making broad statements about "you can't detect something that's below the noise" might mislead in other circumstances. For example, spread-spectrum techniques routinely violate your statement (well, in the US only the military are permitted to use such techniques -- commercial systems are required to be more "detectable.") And you can easily detect a blinking LED from across a bright fluorescent-lit room if you first "lock on" with a PLL when the room is dark. No bright lines, it sometimes seems. \$\endgroup\$
    – jonk
    Commented Mar 14, 2020 at 19:30
  • \$\begingroup\$ @jonk but in such application the noise after the demodulator will be low. If you always account for global noise in all bands.. well, you see where that would be going. And Neil is perfectly right that post-demodulator noise is the only important figure for base sensitivity. In fact, only the 1/f of that noise to be precise. The white noise level does not matter at all, as longer integration can always overcome it. \$\endgroup\$
    – tobalt
    Commented Oct 27, 2021 at 4:21
0
\$\begingroup\$

It seems that if I assume an ideal (no noise) system the minimum detection can be found as the range/scale divided by all possible discrete levels of output, so for example 32/2^14 for a +/- 16g scale and 14 bits of output.

\$\endgroup\$
1
  • 2
    \$\begingroup\$ It's not clear if you're posting this as an answer to your own question (which is OK and you can even "accept" it) or an addition to your question, in which case this should have been edited into your original question. Can you clarify? \$\endgroup\$
    – Transistor
    Commented Mar 14, 2020 at 23:06
0
\$\begingroup\$

As Neil_UK points out, what you are actually interested in is the noise performance. That will determine what kind of minimum acceleration you can reasonably expect to measure.

It's all explained on page 11 of the ADXL335 datasheet you link.

The question you need to ask yourself is what integration time you can allow for your application. Say you want a bandwidth of 20Hz, then according to formula in the datasheet, you get 849ug rms noise.

Now we refer to the peak to peak noise table in the datasheet: If you assume your minimum sensitivity to be 2rms, or 1.7mg, then you will be wrong 32% of the time. That probably isn't acceptable. So we can set the threshold higher, depending on how much we can tolerate errors.

At 6rms, you will be wrong only 0.27% of the time, which is probably ok for most applications. 6rms is 5.1mg.

This of course assumes that all of your analog readout circuits work flawlessly and have a lower noise level than the accelerometer. It might be more reliable to go for a sensor with a digital readout.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.