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A driven right leg (DRL) circuit is often added to a biopotential amplifier to reduce the common mode voltage (i.e. to increase the common mode rejection ratio, CMRR).

A basic form of biopotential amplifier is just an instrumentation amplifier which has two stages: a buffer stage and a single output amplification stage.

The DRL is added after the first stage, as illustrated in the picture below.

Driven right leg circuit

The big picture regarding why the DRL improves the CMRR is that the human body acts as an antenna to pick up electromagnetic interference (EMI) from nearby power lines or other electrical devices and the driven right leg somehow siphons part of the common mode voltage away after the first stage but before the second one.

Unfortunately, I'm having trouble understanding how precisely...

Also, passive grounding is when the right foot is actually connected to ground, if I understand this correctly. Why is it still called grounding if the right foot (or any part of the body) is not actually connected to ground? Why is the right leg (and not another part of the body, maybe even the left leg) used to ground the circuit?

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  • \$\begingroup\$ The choice of grounding point is arbitrary, either leg would work. As for the right leg circuit, it's an inverting amplifier that takes the common voltage, inverts it and the add it back to the body, hopefully canceling it out. You didn't really ask a question, but maybe that answers whatever you wanted to know. \$\endgroup\$ Nov 15, 2021 at 0:40
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    \$\begingroup\$ It kind of does answer the question, but I really would like a more detailed explanation of why the driven right leg circuit manages to extract precisely the common voltage. Also, it makes sense that the choice of leg is arbitrary, but where does the name "right leg" come from? Seems a bit arbitrary, why not just call it active (or driven) grounding and be done with it then? \$\endgroup\$
    – David Cian
    Nov 15, 2021 at 1:02
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    \$\begingroup\$ The average of the two outputs (centre point of the two R3 resistors) is taken as the common mode that is then buffered to the right leg. Don't forget that the DC component is not important in an ECG. This is mainly to get rid of 50/60Hz pickup. \$\endgroup\$ Nov 15, 2021 at 1:07
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    \$\begingroup\$ electronics.stackexchange.com/a/329491/11684 \$\endgroup\$ Nov 15, 2021 at 2:48
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    \$\begingroup\$ Does this answer your question? What does this op-amp circuit do? (part of an ECG) \$\endgroup\$ Nov 15, 2021 at 2:48

2 Answers 2

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The classic instrumentation amplifier circuit consisting of opamps 1 and 2 in your diagram have a common mode rejection ratio of 0db, otherwise called "no rejection at all". In other words, the outputs are still differential, both containing common mode noise, and still requiring a third differential stage to remove that noise. This can be seen here:

schematic

simulate this circuit – Schematic created using CircuitLab

enter image description here

V1 produces a 1V low frequency input offset common to both inputs. This represents common-mode input "noise", from a source with some impedance Rnoise. This noise is added to the signal from V2, a smaller and higher frequency 100mV signal, representing the ECG potentials to be measured. It can be clearly seen that the two differential inputs IN1 and IN2 contain this "common mode" noise offset.

Take a look at the outputs:

enter image description here

Here we see than OUT1 and OUT2 (blue and orange) are complements of each other, which both still contain the 1Hz common mode noise. OUT3 is the result of subtracting OUT2 from OUT1. Obviously, what you want is OUT3, a proper single-ended single signal representing the potential difference between IN1 and IN2.

Usually instrumentation amplifier ICs include this final difference stage, to provide a single-ended output, with the common-mode element eliminated.

However, your diagram performs common mode rejection without that third stage, by offsetting the body's potential to half-way between the two outputs. This is bootstrapping the body to have the same potential as the output's mid-point, thereby shifting the potentials of all three participants (body, and opamps) into the same regime. Common mode noise (V1 in my example) is eliminated because the bi-potential amplifier itself is measuring potentials relative to its own, imposed, idea of what the body's potential should be.

Here is that scenario simulated:

schematic

simulate this circuit

enter image description here

Without some proper analysis, I can't tell you the significance of different amplitudes of the signals at OUT1 and OUT2, but it is clear that the common-mode noise is gone. And you still benefit from the balanced inputs afforded to you by the instrumentaion amplifier setup.

I don't know offhand what advantage this bootstrapping offers over a simple difference stage following the instrumentaion amplifier stage, but it does work. I imagine that by employing this bootstrapping technique and a difference amplifier for OUT1 and OUT2, you can achieve really good common-mode rejection.

Perhaps there are other benefits that your book can point out. I would be interested to know.

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  • \$\begingroup\$ your last schematic misses the feedback resistor on OA3 :-) \$\endgroup\$
    – tobalt
    Nov 15, 2021 at 7:05
  • \$\begingroup\$ The point I was making doesn't need it, really. It's just a voltage follower. \$\endgroup\$ Nov 15, 2021 at 10:11
  • \$\begingroup\$ I'm not sure what the advantage is in using a DRL instead of a simple difference stage, but I do know that almost all EEG headsets have a DRL, often connected to one of the mastoids (bone behind ear). \$\endgroup\$
    – David Cian
    Nov 15, 2021 at 12:33
  • \$\begingroup\$ This doesn't answer the DRL question at all. That's what the OP asked about. \$\endgroup\$ Nov 15, 2021 at 12:54
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The case for the driven leg can be seen in Winter, Bruce B., and John G. Webster. "Driven-right-leg circuit design." IEEE Transactions on Biomedical Engineering 1 (1983): 62-66..

It becomes pretty apparent when you actually draw in the resistances between the electrode and the signal you're trying to amplify, which are largely at the skin interface. The DRL circuit uses feedback to effectively reduce the impedance between the body and the "ground" electrode.

To quote the article:

Connecting the electrode directly to the common is undesirable for two reasons. 1) If the circuit is not isolated, dangerous currents could flow through the third electrode. 2) A poor electrode contact may present up to 100 k\$\Omega\$ of resistance between the patient and the common. The most common and effective use of the third electrode is to connect it to a driven-right-leg circuit [2], [3]. This circuit overcomes both of the problems listed above. It reduces the effective electrode resistance by several orders of magnitude, and it allows only a safe amount of current to flow through the third electrode.

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  • \$\begingroup\$ Yap! "The DRL circuit uses feedback to effectively reduce the impedance", that is one part. \$\endgroup\$
    – jay
    Nov 15, 2021 at 15:20
  • \$\begingroup\$ @jay -- actually, with respect to CMRR, it's the only part. \$\endgroup\$ Nov 15, 2021 at 15:24
  • \$\begingroup\$ I am thinking, yet, trying to think about AC impedance created in the feedback loop,... while wondering why filtering effects are not considered in the discussion. I really like the discussion in the link you suggested. I was thinking of the "active impedance reduction" and your link exactly explained it. Meantime, I thought there is another. In the CircuitFantasist's second picture, the AC impedance effectively (?) slides (biases) the reference point to the center of Vin, thus actively creating the common mode voltage from the feedback. \$\endgroup\$
    – jay
    Nov 15, 2021 at 15:36
  • \$\begingroup\$ @jay -- yes, the circuit obviously feeds back the common mode voltage. That doesn't really explain why CMRR is improved. It's effectively improved because it minimizes impedance differences at each electrode. \$\endgroup\$ Nov 15, 2021 at 15:43
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    \$\begingroup\$ Thanks Scott, that is where my thoughts are circling around, about "is it the same thing?". I will probably come back there, if i come across a chance. \$\endgroup\$
    – jay
    Nov 15, 2021 at 15:48

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