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I'm trying to understand the following circuit, which I found in this document (on page 5). The purpose of the circuit is to go between a set of EEG electrodes and an EEG signal amplifier, and protect the EEG user from any accidental electrical discharge from the EEG amplifier.

I've already analyzed the circuit and I think I have most of it figured out. It looks like the left side connects to the actual scalp electrodes, and the right side feeds into the amplifier circuit. The 4xAA batteries creates a +6V voltage rail, so the DRL on the left is driven to +3V. The TLC071 is just a normal op-amp (powered by the batteries,) and there's an RC high-pass filter on the far left to eliminate any dc offset between electrode 1 and electrode 2. The op-amp and the optocoupler together form an isolation amplifier circuit that, in summary, sets the output differential signal (the difference between the two points on the right marked "signal") equal to the input differential signal (the difference between the two points on the left marked "signal.")

There are still 2 things I'm not sure about.

First, what do the "probe" and "shield" markers mean?

The "shield" seems to point to this strange structure that connects all 3 electrodes. I've highlighted this structure in red below. What exactly is it? I've never seen this in a circuit diagram before.

Second, what is the purpose of the section on the far right that connects to the "DRL" marker? Where is that rightmost "DRL" supposed to connect to? In the original document, the discussion of the actual amplifier circuit only mentions the two signal inputs, which makes me think that this rightmost DRL might be connected directly to the leftmost DRL, but I'm not sure if that's really the case. Below I've highlighted in red the part that I'm asking about.

One last thing: Is there any reason why the DRL pins are colored black and the signal pins are colored white, or is that just a stylistic choice?

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2 Answers 2

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Those circular symbols are meant to represent shielded connectors. The outer part connects to the shield of the electrode wire, the signal is the centre part. They are not electrically connected.

"DRL" is short for "Driven Right Leg". It is intended to provide a reference voltage for the body. The right leg is commonly used as it is a a distance from where the electrodes are placed. It is more commonly used for ECG (ie measuring heart voltages) than EEG.

Note that the "DRL" signal at the right must not be connected to the "DRL" signal at the left or the isolation will be compromised.

The resistor network at the right is to provide a suitable output voltage for driving the EEG signal amplifier. The output voltage will be centered on the reference voltage supplied by the DRL input on the right. The series 2k resistors are presumably to provide a suitable source resistance to imitate that of real electrodes.

The color of the pins is a stylistic choice meant to indicate the negative or signal common. It is not at all standard.

It is unusual to use that sort of photocoupler in photovoltaic mode, I'm surprised there is much signal output.

It would be useful if you put reference designators on the components so that responders can easily indicate which parts of the diagram are bing referred to.

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  • \$\begingroup\$ "Those circular symbols are meant to represent shielded connectors. The outer part connects to the shield of the electrode wire, the signal is the centre part. They are not electrically connected." Is there a reason why there is a connecting line drawn between the shielded connectors? Or are you saying they they are just 3 independent shielded cables? \$\endgroup\$
    – Svedberg
    Commented Sep 3, 2023 at 22:17
  • \$\begingroup\$ Also: "The resistor network at the right is to provide a suitable output voltage for driving the EEG signal amplifier." Would this mean that the 'DRL' on the right is connected to the reference voltage of the amplifier circuit (halfway between positive and negative supply rails) so that the dc offset of the input signal is centered at the reference voltage? \$\endgroup\$
    – Svedberg
    Commented Sep 3, 2023 at 22:22
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    \$\begingroup\$ @Svedberg - correct. I have updated the answer. \$\endgroup\$ Commented Sep 4, 2023 at 15:42
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  1. Think of the 4 NiMH cells as a split +3, 0, -3 VDC supply. The TL071 is a differential amplifier, requiring positive and negative rails. The shield is effectively connected to the middle, what might be thought of as "ground" -- though one does not want to probe into the human body with wires connected to an earth ground, since even body capacitance could couple dangerous voltages, to say nothing of a fault! The shield blocks extraneous voltage, while the signal is coupled to the IC through the 10 µF cap, across a 200 kΩ resistor (you might calculate the frequency response of that filter).
  2. The optocoupler provides negative feedback to stabilize the amplifier, as well as to couple output to the rest of the instrument.
  3. Again, if feeding a differential amp, the output must be balanced between rails, ergo DRL reference level (again, sort of "ground").
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