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DIY Muscle Sensor / EMG Circuit for a Microcontroller

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Step 6: SIGNAL CONDITIONING - Smoothing + Amplification

In this last phase of circuit assembly, we will be using an active low-pass filter to filter out the humps of our signal to produce a smooth signal for our microcontroller.

You will need the last TL072 chip (chip D), the two 80.8 kOhm resistors, the 100 kOhm trimmer, the 1 kOhm resistor and the 1.0 uF ceramic disc capacitor.

First, plug in chip D and connect +9V to pin 8, -9V to pin 4, and GND to pins 3 & 5. (image #1).

Now, grab one of the 80.6 kOhm resistors and connect one end to chip C’s pin 7. Connect the other end to chip D’s pin 6. Next grab the other 80.6 kOhm resistor use it to connect chip D’s pin 6 and 7. Do the same thing for the 1.0 uF capacitor. (image #2)

That’s the end of the filter circuit. However, since this is an active filter, there is a side effect of inverting the signal. We will need to invert the signal one more time (and have the ability to amplify it more if desired) using another inverting amplifier circuit with a trimmer configured as a variable resistor.

Use a jumper wire, connected to chip D’s pin 7, and the 1 kOhm resistor to bridge the board’s center gap. Use another jumper wire and connect the 1 kOhm resistor to chip D’s pin 2. Next, place the trimmer one row over with the pins laid out and a jumper wire connecting two of the pins as pictured. Finally, place the last two jumper wires as indicated. (image #3)

By using a screw driver and turning the trimmer, you will be able to adjust the gain of your signal to account for different signal strengths from different muscle groups. Start out with it set pretty low and go up from there (~20 kOhms).

 
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joshsh1 year ago
This may be a silly question, but how important is the value of the 80.6 kOhm resistors? Was this a convenient choice or a precise requirement? I have substituted 82 kOhm resistors in my muscle sensor, and it works nicely. Will it work that much better if I get the resistors just right (putting several resistors in series to get 80.6 or giving Digi-Key some more business), or is 82 close enough? Thanks!
Gundanium (author)  joshsh1 year ago
The 80.6k resistors set the cut off frequency of the smoothing low pass filter to 2Hz. If you change them to 82k, then the cut off frequency will decrease to ~1.9Hz.

So yes, 82 kOhm should work fine. =)
Got it! For the benefit of anyone else reading these comments, the cutoff frequency of an active low-pass filter follows the formula f = 1 / 2πRC (see Wikipedia).  In this case, C is 1.0 µF and R is 80.6 kΩ (or 82 kΩ in my case), so f comes out to just under 2Hz.

Many thanks for this Instructable, btw.  This is a pretty complex circuit to understand if you have never dealt with anything similar (I haven't), and this format makes it about as straightforward as possible.
Gundanium (author)  joshsh1 year ago
The 80.6k resistors set the cut off frequency of the smoothing low pass filter to 2Hz. If you change them to 86k, then the cut off frequency will decrease to ~1.9Hz.

So yes, 82 kOhm should work fine. =)

elzurdo861 year ago
Hello I have a question about the circuit, does it match with the schematic at the end? I'm having problems trying to understand it . I have assembled the circuit and I have an osciloscope with me, but from the third phase to the fourth phase my isgnal is lost, how is the connection from those two phases?

Gundanium (author)  elzurdo861 year ago
Hi, yes these instructions match the schematic at the end. Are you talking about the transition to the rectification phase? The 10k Ohm resistor connected to pin 1 of chip B is the connection between that phase and the previous phase.
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