A doubt on interfacing EMG circuit to Arduino?

Question

So I have done an EMG detection circuit using TI's instrumentation amplifier chip INA128 as below (sorry if the schematic looks bad):

And the circuit is working OK for my requirements (although slight hum and background noise is there). I was trying to interface the V_out to the ADC of Arduino. Which interfacing circuit would be better for this purpose? I was trying the instructions given here using op-amp summing amplifier solution. I used LM741 op-amp in my circuit. But it was a failure. My output signal V_out varies between -100mV to 100mV.

Answer 1

Because you seem to have an LM741, the most simple solution is just to use 2 inverter configuration:

^{simulate this circuit – Schematic created using CircuitLab}

For the ADC of your arduino, you should check but the maximum input could be 5V, if it is the case you can raise R2 to 20K and you will get a 2.5 V offset.

Answer 2

The circuit suggested by you should work in principle. The OP you used (LM741) needs at least \$\pm10V\$ so if you used the same \$\pm9V\$ supply as for the INA128 that may be the error.

WARNING: MATH For just the solution see circuit below.

If not, here is a circuit I would use to convert the +-100mV to 0-5V using the formula from you link

$$V_0 = \left(V1\frac{R2}{R1+R2} + V2\frac{R1}{R1+R2}\right) \left(1 + \frac{R4}{R3}\right)$$

Lets say \$V1\$ is our input signal and \$V2\$ the Offset Voltage. As you can see in the Formula above we first add then multiply with this circuit just like in math. So to get a positive signal we add \$100mV\$ to the signal. To achieve this we can create a voltage divider with the 5V supply of the Arduino.

$$V2 = 5V \cdot \frac{R5}{R5+R6} = 100mV$$

Using a resistor calculator like this one we get the values for R5 and R6 as:

$$ R5 = 1k\Omega$$ $$ R6 = 22k\Omega + 27k\Omega$$

As you can see R6 is actually created from two resistors in series to match the ratio. Lets call the second Resistor R7 $$ R6 = 22k\Omega$$ $$ R7 = 27k\Omega$$ $$V2 = 5V \cdot \frac{R5}{R5+R6+R7} = 100mV$$

Now to add V1 and V2 without any weighting of the values \$\frac{R1}{R1+R2}\$ and \$\frac{R2}{R1+R2}\$ have to be equal. So in fact R1 has to be equal to R2

Lets chose a value for R1 and R2, say $$R1=R2=10k\Omega$$

Lets see what we have now:

$$V_0 = \left(V1\frac{10k\Omega}{10k\Omega+10k\Omega} + V2\frac{10k\Omega}{10k\Omega+10k\Omega}\right) \left(1 + \frac{R4}{R3}\right) = \left(V1\frac{1}{2} + V2\frac{1}{2}\right) \left(1 + \frac{R4}{R3}\right) = \left(V1 + V2\right) \frac{1}{2} \left(1 + \frac{R4}{R3}\right)$$

So now that the \$\pm100mV\$ Signal is offset by \$100mV\$ to a range of \$0-200mV\$ we need to boost it to \$0-5V\$

$$\frac{1}{2} \left(1 + \frac{R4}{R3}\right) = \frac{5V}{200mV} = 25$$ $$\frac{R4}{R3} = 49$$

By using the calculator above we get $$R4 = 330k\Omega$$ $$R3 = 6.3k\Omega$$

This is actually not a perfect fit (3% off) but as resistors can have large margin of error (5%) is is close enough. (You can go better with three resistors: R4=290k R3=10k||39k)

Now we have finally all we need. Final circuit:

^{simulate this circuit – Schematic created using CircuitLab}