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

  • \$\begingroup\$ isnt it a buffer for the voltage divider R2/R3? \$\endgroup\$
    – PlasmaHH
    Commented Sep 15, 2017 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\$ Commented Sep 15, 2017 at 11:25

3 Answers 3


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.

  • \$\begingroup\$ A better way to look at this is that you're effectively reducing the resistance between the body and the reference electrode. \$\endgroup\$ Commented Sep 15, 2017 at 11:26

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.

  • 6
    \$\begingroup\$ This is also an excellent example of why we don't ask engineers to draw people in art class ;-) \$\endgroup\$
    – Cort Ammon
    Commented Sep 15, 2017 at 19:21
  • 3
    \$\begingroup\$ I do not want to know where you're supposed to stick Cb. \$\endgroup\$
    – Jules
    Commented Sep 15, 2017 at 19:35
  • 2
    \$\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\$ Commented Sep 15, 2017 at 19:37
  • 1
    \$\begingroup\$ @Jules -- you just figured out why stickman is smiling. \$\endgroup\$ Commented Sep 15, 2017 at 19:38
  • \$\begingroup\$ Is that why I need to touch GND on my Arduino for a "clean" signal? \$\endgroup\$ Commented Dec 30, 2022 at 16:41

EDIT: I saw this weird circuit solution for the first time a year ago and it immediately caught my attention. Then I managed to reveal and roughly describe the idea behind the circuit. Now I want to expand these explanations with some new insights. Here is another more detailed circuit diagram - Fig. 1:

Detailed right leg driven circuit

Fig. 1. Detailed right leg driven circuit (chegg.com)

Obviously there was some clever idea of suppressing common-mode signals. What was it like?

Basically, the next amplifier stage with differential input suppresses the common-mode input voltage. However, since ECG amplifiers operate at extremely low input voltages, it is desirable that there is no common-mode signal at all (only the differential signal).

In order to grasp the fundamental idea, I first removed all the minor details that hindered understanding - Fig. 2, 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:

DRL circuit - idea

Fig. 2. A simplified circuit diagram of the right leg driven circuit

Structure. First I saw two single-ended input voltages (VIN- and VIN+) between the signal electrodes and reference electrode. Surprisingly their input "sources" (the human body) were not grounded; instead, they were 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. Usually, they were implemented as two paired non-inverting amplifiers (Fig. 1) but, in regard to the common-mode signals, they acted as voltage followers. That is why I did not draw the network of three resistors between the outputs of the input amplifiers (the input part a classic instrumentation amplifier) 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 with the purpose to produce the common-mode voltage.

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 try to increase (due to some common-mode noise voltage above the real "immovable" ground), the op-amp senses this change at its inverting input and begins decreasing its output voltage below zero until makes its value equal to the common-mode noise voltage. 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 even when the input voltages change in a differential manner.

So, in respect to the common mode, the weird RDL circuit can be thought of as 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 - Fig. 3:

DRL circuit as inverting amplifier

Fig. 3. DRL circuit presented as an inverting amplifier

We can see here two negative feedbacks - local (implemented by Rf, R1 and R2) that is used to make the op-amp summer and global (by VCM) that compensates the common-mode voltage. The latter has another important function - to compensate for the voltage drop across the contact resistance between the human body and the reference electrode. How does this magic work?

This undesired resistance is put into the negative feedback loop and the op-amp observes the voltage after this disturbance (at its inverting input)... and keeps it almost zero. For this purpose, it lowers its output voltage even more thus compensating for the undesired contact resistance... like in any op-amp circuit with voltage type negative resistance.

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


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