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Physiological amplifiers seem to often (always?) have both ground (G) and reference (R) electrodes along with one or more active (A) electrodes. Typical voltage at an active site is determined AFAIK as [(A-G) - (R-G)] = A-R. I haven't managed to find out why the ground electrode is needed at all and why the amplifier cannot simply do the A-R calculation directly without the need for the ground electrode? Is it because when there are multiple active electrodes that A1-R and A2-R couldn't be done without them interfering with each other?

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    \$\begingroup\$ Touch the tip of a 10MOhm scope probe, and you'll likely see 100 or 200 volts at 50 or 60Hz; you are the antenna, the 2nd plate of the capacitor gathering charge from all the power line wiring around you. A physiological amplifier has that same input --- 100 or 200 volts--- plus the tiny signal of interest. That GND (to your ankle) greatly reduces the undesired voltage. \$\endgroup\$ – analogsystemsrf Nov 23 '17 at 20:33
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Differential amplifiers such as instrumentation amplifiers, with or without unity gain buffers, have a certain working range for the common-mode voltage.

enter image description here

In general they will stop working if either input voltage gets too close to the power supply rails. For example, in the case of the INA117, both input voltages must be within the range of V+-4V to V-+4V, so if you have +/-10V supplies, each input must be within the range of +/-6V. There are also internal nodes in instrumentation amplifiers which can saturate. Here are some curves for the INA118 which show the permissible input voltages:

enter image description here

The reference ground makes sure that the inputs remain within the common mode range for proper operation and minimizes the common mode AC voltage. The latter will show up in the output even if the inputs remain within the common mode range, attenuated by the common mode rejection ratio (CMRR).

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After looking at some schematics: (Google: "physiological amplifier circuit" and select Images tab), this is one example:

enter image description here

Many amplifiers do use A-R directly similar to instrumentation amplifiers. The ground connection might be needed for:

  • shielding around the A and R signals to prevent coupling to external noise like other equipment.

  • safety: since the amplifier must be mains isolated (to prevent electrocuting the patient) any charge build-up cannot escape without a ground connection. If a charge were present it could escape via the patient, not harmful but still unpleasant.

  • safety: if the mains isolation fails on the amplifier then the ground connection prevents the mains voltage from reaching the patient.

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  • \$\begingroup\$ Thanks. Interesting to know that it doesn't have to be. I'm guessing that #1 - shielding is the reason in this case (signal levels are typically 20-50uv) then as the ground electrode is always attached to the patient too! (Often very close to the reference electrode for EEG.) \$\endgroup\$ – jacanterbury Nov 22 '17 at 16:17
  • \$\begingroup\$ -1: You need a ground connection or your recordings will be mostly useless. @Spehro Pefhany's answer seems to be the most correct. In short, without a ground you have two very high impedance lines between two electrically isolated systems. Without establishing a common voltage point between the two systems (the patient and the amplifier), any voltage readings will be mostly useless. It would be similar to attempting to measure a voltage with a single lead from a voltmeter. \$\endgroup\$ – sbell Nov 22 '17 at 20:46
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    \$\begingroup\$ @sbell Although your common mode argument is valid for the amplifier schematic included in my answer, it is not for the amplifier schematic in Trevor's answer. It is possible to define the common mode voltage through the inputs. It would be similar to attempting to measure a voltage with a single lead from a voltmeter. That's a nonsense argument because all these circuits use differential inputs which means 2 wires just like any voltmeter. This even disproves your first point (about needing a common mode level) as a voltmeter can measure between 2 points without a ground reference. \$\endgroup\$ – Bimpelrekkie Nov 22 '17 at 20:55
  • \$\begingroup\$ @Bimpelrekkie Physiological amplifiers mostly follow the type of circuit that you posted. Although the one from Trevor's answer is sometimes seen, it would perform poorly due to the inherent source impedance (in the range of 3k - 1M) in almost all physiological connections. It's true that the voltmeter has two leads, but the ground is a very low impedance connection that joins the two isolated systems (meter and DUT) to a common voltage range. I suppose a better analogy would be two voltmeters with a common ground, and trying to record a difference between the two leads on a floating system. \$\endgroup\$ – sbell Nov 22 '17 at 21:10
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The reason is actually because of the buffers.

A simple differential amplifier does not care about the reference point of the input. Without any other common reference, all it cares about is the difference between the positive and negative inputs. The difference itself of course has to be within the tolerable range of the inputs.

Notice in the circuit below, the output is correctly -1V.

schematic

simulate this circuit – Schematic created using CircuitLab

Why is that?

Generically, the inputs stage of all op-amps basically looks something like this...

enter image description here

That is the input is a balanced circuit. That balance is disturbed by any applied voltage/current difference between the two inputs. The reference point for those inputs is, in reality, the junction between those two emitters. As such the applied voltage need not be referenced to ground as long as the two inputs are referenced to each other. You should be able to see, how attaching a battery between the plus and minus pins in this circuit should be acceptable.

The buffer on the other hand needs a reference.

Obviously the circuit below, attaching a voltage with no reference, to a unity buffer makes no sense.

schematic

simulate this circuit

But how about this..

schematic

simulate this circuit

In fact what you end up with here will depend on the op-amp. Some may get you what you expect +0.5V on one -0.5V on the other, some may rail out to either or both sides. All will be very sensitive to any noise.

So what does all that mean.

schematic

simulate this circuit

It means a standard differential amplifier measures the difference voltage \$Vr\$. An instrumentation amplifier measures \$V_a - V_b\$. Mathematically that may be the same, electrically, it is not.

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  • \$\begingroup\$ There are still common mode input range requirements on a purely differential amp, so, yes, it does care about absolute voltage on the inputs. \$\endgroup\$ – Scott Seidman Nov 23 '17 at 1:40
  • \$\begingroup\$ @Scott yes indeed. good eye, I missed the without a ground part. Corrected. That's wot happens when you get interrupted mid answer. \$\endgroup\$ – Trevor_G Nov 23 '17 at 2:03
  • \$\begingroup\$ No. Put the 10k resistor from GATE to SOURCE (GROUND), not from GATE to DRAIN. \$\endgroup\$ – Jim Fischer Nov 23 '17 at 3:51
  • \$\begingroup\$ @JimFischer ????? \$\endgroup\$ – Trevor_G Nov 23 '17 at 4:13
  • \$\begingroup\$ @Trevor: Whoops. Ignore that comment. That was meant for a different question. I remember typing that reply on my cellphone, and somehow the app I'm using swapped message threads on me while I was typing. (Perhaps I unknowingly selected a different message as I was holding the phone and typing? I dunno...). \$\endgroup\$ – Jim Fischer Nov 24 '17 at 0:16
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Touch the tip of a 10MOhm scope probe, and you'll likely see 100 or 200 volts at 50 or 60Hz; you are the antenna, the 2nd plate of the capacitor gathering charge from all the power line wiring around you. A physiological amplifier has that same input --- 100 or 200 volts--- plus the tiny signal of interest. That GND (to your ankle) greatly reduces the undesired voltage.

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