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I would like create an ECG circuit based on this schematic (from the AD620AN datasheet):

ECG circuit

I don't know this part of the circuit and how it works. I know this is called a right leg driven circuit which is reducing the effect of the noise. But I don't know exactly how negative feedback works in this case. Can someone help me?

Right leg driven

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  • \$\begingroup\$ isnt it a buffer for the voltage divider R2/R3? \$\endgroup\$ – PlasmaHH Sep 15 '17 at 11:10
  • \$\begingroup\$ @PlasmaHHp, essentially. The voltage divider is sort of a cheat, that lets you recover the common mode signal from the input stage of the U.S.. The bigger deal is that you're feeding this back! \$\endgroup\$ – Scott Seidman Sep 15 '17 at 11:25
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The right leg driver tries to drive the average voltage of the body to cancel out noise. The right leg is chosen because it is far from the heart, so any signal injected on there will be common mode to two electrodes near the heart.

The right leg drive is much more tightly coupled to the body than ambient noise it picks up from capacitive coupling to things like the AC power in the room.

The network in the feedback path of the right leg driver opamp provides some low pass filtering of the signal.

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  • \$\begingroup\$ A better way to look at this is that you're effectively reducing the resistance between the body and the reference electrode. \$\endgroup\$ – Scott Seidman Sep 15 '17 at 11:26
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This circuit, and the need for it, make much more sense when you consider some things that aren't depicted. First, remember that is is necessary to establish some sort of reference voltage on the body, so that the voltage at the measurement electrodes has some reference with respect to the circuit.

Picture this reference being established by a Right Leg electrode directly connected to circuit ground. If a zero-impedance connection to the body could be made like this, we'd be done, and there would be no need for a driven leg connection.

In fact, the connection between the reference electrode and the circuit can be kiloohms, or tens of kiloohms. Now, because of the common-mode voltages riding on the body, and the fact that the reference electrode is connected through highish impedances to ground, there are stray currents. (This is less of a problem on the signal electrodes, which go into very high input impedances, as opposed to ground).

What the Driven Leg circuit does is use feedback techniques to measure the common mode voltage, and feed it back through the reference electrode. This effectively reduces the impedance of the connection at the reference electrode by a factor of the gain of the feedback. enter image description here

I'm attaching Fig 1 from Winter, Bruce B., and John G. Webster. "Driven-right-leg circuit design." IEEE Transactions on Biomedical Engineering 1 (1983): 62-66., which show electrode impedances drawn in, but I highly recommend reading the paper if you can get it, as it shows very clear derivation of the effective reduction of impedance.

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    \$\begingroup\$ This is also an excellent example of why we don't ask engineers to draw people in art class ;-) \$\endgroup\$ – Cort Ammon - Reinstate Monica Sep 15 '17 at 19:21
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    \$\begingroup\$ I do not want to know where you're supposed to stick Cb. \$\endgroup\$ – Jules Sep 15 '17 at 19:35
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    \$\begingroup\$ @CortAmmon agreed. Takes me back though. I imagine in 1983, this was Rapidograph pen and stencil work, maybe with a pantograph for lettering, and then off to the photographer. \$\endgroup\$ – Scott Seidman Sep 15 '17 at 19:37
  • \$\begingroup\$ @Jules -- you just figured out why stickman is smiling. \$\endgroup\$ – Scott Seidman Sep 15 '17 at 19:38
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I saw this weird circuit solution for the first time yesterday and it immediately caught my attention. Obviously there was some clever idea of suppressing common-mode signals. What was it like?

In order to grasp the fundamental idea, I first removed all the minor details that hindered understanding and started trying to see familiar circuit building blocks and principles. I simplified and sketched out the circuit diagram and focused around the part with AD705 op-amp (А3):

DRL circuit - idea

Structure. I saw two single-ended input voltages (VIN- and VIN+) between the signal electrodes and reference electrode. Surprisingly their input "sources" were not grounded... but connected to the op-amp output. What the hell was that?!? Aha... they were connected to a "moving" ground, which probably allowed their simultaneous (common-mode) variations to be suppressed.

The input voltages were buffered by amplifying stages (A1 and A2) with high input impedance. In regard to the common-mode signals, these stages acted as voltage followers. That is why I did not draw the network of three resistors between the outputs of the input amplifiers because it was important only for the differential mode.

But what did the 2-resistor circuit between the follower outputs serve? I realized that an op-amp inverting summer was built with the help of Rf and A3.

Operation. Imagine that initially both input voltages are zero. So the op-amp output voltage VREF (of the right leg) is zero as well.

If both input voltages tries to increase (due to some common-mode noise voltage above the real "immovable" ground), the op-amp output voltage decreases (approximately) with the noise voltage below the real ground. And since the input voltage "sources" are connected not to real but to "movable" ground, their voltages move down with the noise voltage. Figuratively speaking, the op-amp output "pulls down" the input voltages with the magnitude of the common-mode voltage (the op-amp output subtracts equivalent voltage from the common-mode voltage). As a result, in respect to the real ground, the common-mode signal will be (almost) zero.

So, in respect to the common mode, the weird RDL circuit can be thought as of an op-amp inverting summer with input sources "grounded" to its output instead of the true ground. Because of this "movable ground", the common-mode signals are suppressed.

If we combine both input voltages and resistors in one, we can think of this arrangement as an inverting amplifier with gain of 200, which output is fed back by VIN... i.e., there are two negative feedbacks - local (implemented by Rf, R1 and R2) and global (by VCM).

DRL circuit as inverting amplifier

I have attached the genuine circuit diagrams sketched with pencil and rubber yesterday to illustrate more realistically the course of my thoughts that led me to this explanation. Of course, I can outline them beautifully... but so they will become less informative...

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