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I want to use an ADXl326 accelerometer to measure vibration in one plane only. The accelerometer is fed with 3.3 V and returns 1.65 V at 0 g +/- 57 mV/g

What I want to do is remove the 1.65 V bias to get an AC signal, then amplify that signal so it is close to -1.65 / +1.65 V so that I can add the bias back and have a DC signal again, which is close to the full ADC range of 0-3.3 V.

I considered using an AD8221 instrumentation amplifier with no gain resistance on pins 2 & 3 i.e. unity gain, followed by a PGA112 programmable gain amplifier with SPI control and 1.65 V on the reference pin.

However, I was wondering if anyone had any suggestions which would not need a -5 & +5 V supply?

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  • \$\begingroup\$ Very well asked. I would have edited your question for proper punctuation and voltage V units, but am on mobile now. What is the maximum G you are expecting? Input peak AC voltage... \$\endgroup\$
    – User323693
    Mar 6, 2017 at 14:45
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    \$\begingroup\$ There's no need to remove the bias and then put it back in. Just amplify the signal with the bias built into the amplifier. There are many examples of such "signal conditioning" (Google that phrase) circuits on the Internet. \$\endgroup\$
    – Dave Tweed
    Mar 6, 2017 at 14:58
  • \$\begingroup\$ Add an analog high pass filter. \$\endgroup\$ Jun 4, 2019 at 10:31
  • \$\begingroup\$ Why wouldn't you just run the signal from the ADX1326 directly into the ADC? \$\endgroup\$
    – MOSFET
    Mar 31 at 20:28

2 Answers 2

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schematic

simulate this circuit – Schematic created using CircuitLab

The zero G null adjust pot only has to correct source null error away from 1.65/3.3=50% so ratio of 1k/20K is 5% can be reduced to 2% for example.

Higher Gain is possible but choice of low input bias current, rail to rail type, unity gain stable Op Amp is necessary.

This is just one of many examples, not final version... after you specify full scale g and frequency limit.

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  • \$\begingroup\$ The accelerometer measures -16/+16 but I wouldnt expect anywhere near that. The application is bearing condition monitoring so a healthy bearing with a very low vibration, say 0.1g's will need more amplification to use max adc range than a bad one with 5 or 6 g's hence the pga. Frequency limit would be 1600hz. \$\endgroup\$
    – baart_cm
    Mar 6, 2017 at 18:03
  • \$\begingroup\$ You have not accepted an answer yet. You don't need a bipolar supply or signal. Only need to convert 57mV/G for +/-0.1G to say 10V/G = +/1V but still centred at Vdd/2 to avoid offset error with a matched resistor pair. (0.2%) or use an offset fine error null. Then your gain becomes 10V/57mv = 17.5 x for R Ratio or boost to full scale with 16.5V/g . OK? \$\endgroup\$ Oct 7, 2019 at 1:11
  • \$\begingroup\$ This will have a lot of noise at the output, since C1 will increase gain with frequency. Better remove C1 altogether, and ensure that \$V_{REG}\$ is clean. \$\endgroup\$ Mar 31 at 13:23
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To avoid the need for a negative supply rail, you can bias the op-amp to operate around +1.65V, like this:

schematic

simulate this circuit – Schematic created using CircuitLab

This produces the following relationship between input \$V_{IN}\$ and output \$V_{OUT}\$:

$$ V_{OUT} = (V_{IN} - 1.65)\left(1+\frac{R_2}{R_1}\right) + 1.65 $$

From that expression it can be seen that the first operation is to subtract the offset of 1.65V from the input, then multiply by some gain, and then re-add the 1.65V, which is exactly what you require.

Gain is \$1+\frac{R_2}{R_1}\$, for which it's easy to find values R1 and R2 that achieve whatever gain (within reason) you require. The only problem is obtaining the 1.65V reference for the bottom of R1, node X, required for this to work.

You could derive 1.65V using a resistor potential divider between 0V and any positive supply potential, as follows:

schematic

simulate this circuit

By drawing the Thevenin equivalent for those resistor dividers, you can see how these might be incorporated into the above amplifier design. Thevenin resistance \$R_{TH}\$ is the parallel combination of R3 and R4:

$$ R_{TH} = \frac{R_1R_2}{R_1 + R_2} $$

For the above dividers, these are their Thevenin equivalent circuits:

schematic

simulate this circuit

Taking the left example (using the +5V supply), we can see how the following two circuits are functionally identical:

schematic

simulate this circuit

Having substituted R1 with the divider formed by R3 and R4, we simply assume that R1 is now the equivalent Rth, with value 3.3kΩ. Now we can choose R2 to obtain the required gain. For instance, if we require gain 4:

$$ \begin{aligned} 1 + \frac{R_2}{R_{TH}} &= 4 \\ \\ R_2 &= 3R_{TH} \\ \\ &= 3 \times 3.3k\Omega \\ \\ &= 9.9k\Omega \end{aligned} $$

So, using a 5V supply, this circuit would center the signals about +1.65V, and apply a gain of 4:

schematic

simulate this circuit

enter image description here

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