Noise in an unconnected biopotential electrode lead wire

Question

My question stems from a particular statement on page 256 of the textbook Medical Instrumentation - Application and Design, Fourth Edition (by J. Webster):

Relatively high potentials can often be induced in then open wire as a result of electric fields emanating from the power lines or other sources in the vicinity of the machine

Such a situation also arises when an electrode is not making good contact with the patient

The "open wire" above refers to an ECG or EEG electrode lead wire that has become disconnected from the electrode element for some reason.

Why is it the case that "relatively high potentials" are induced on an open lead wire or one without a good connection, as opposed to a wire with a good connection to the electrode?

I've heard plenty of times that "wires act like antennas", but why is an open/crappy connection lead more of an antenna? It seems intuitive, but I'd like a first principles explanation here.

thanks!

Answer 1

Lets view the electric field coupling from a wall wire or a simple extension corde, into the biopotential wire.

We will use the parallel_plate model of capacitance, because of the presence of myriad other wires and PCBs that absorb energy also. This makes fields non_dipole, letting the parallel_plate equation become rather useful

Assume the wires are 1 meter long, 1 milliMeter diameter (now we have the AREA), and have a Distance between them of 1 meter.

The parallel_plate capacitance is

• C_parallel_plate = E0 * Er * Area/Distance

which becomes

• C_pp = 9e-12 farad/meter * (1meter * 1mm)/1meter

• C_pp = 9e-12 * 1milliMeter = 9e-15 = 1e-14

Now assume a clean sineusoid at 60Hz, 160 volts peak. (spikes from fluorescent lightes, light_dimmers, or motors will greatly worsen this)

The induced current will be

• I_displace = C * dV/dT == 1e-14 farad * (377 radian/second * 160volts)

• I_displace = 1e-14 * 60,000 volt/second (assuming CLEAN power line)

• I_displace = 6 * 1e-14 * 1e+4 = 6e-10 ~~ 1 nanoAmp

and the voltage across 100,000 ohms is 1e-9 * 1e+5 = 100 microVolts

And you need to avoid voltages of the 100 microvolt level.

============================================

In a floating wire, the voltage ---- which you CANNOT monitor with a scope probe -----will be defined by the TWO SERIES CAPACITORS.

If the biopotential_wire is halfway between the 60Hz power wiring and a ground system (a mess of wires, that are not moving very much in voltage), then your biopotential_wire will have 160 volts peak / 2, or 80 volts on it.

But you cannot monitor that voltage, because you greatly change the voltage division ratio.

More practically, suppose you DO monitor a long wire, one meter from the 60Hz power wiring.

We'd expect the scope (10:1 probe, 10 pF || 10MegOhms) to show?

Ignoring the 10,000,000 ohms, its just a capacitive divider

• 0.01pF / 10pF = 1,000:1 division, thus 160 milliVolts on your scope.

What is effect of the 10,000,000 ohms inside the probe?

It is a High Pass Filter, because 10,000,000 is constantly removing charge.

The 3dB frequency will be at 10pF || 10Meg ohm, or 100 uS time constant, which is 1,600 Hertz corner frequency.

The previous answer of 0.16 volts will be smaller by 60/1,600.

Answer 2

The impedance of an electrode-skin-body layer is on the low side (10k to 100k). The coupling into the wires is capacitive, modeled as a current from a very high impedance source.

When the wire is connected, this current flows through the low impedance of the electrode, resulting in a low voltage.

When the wire is disconnected, the current flows into the amplifier input impedance, resulting in a high voltage.

Answer 3

^{simulate this circuit – Schematic created using CircuitLab}

When you touch an unterminated 10:1 scope probe you raise signal level from <1mV to > 50V due to your body capacitance and E-field going thru you as a conductor.

This is a rough approximation of AC stray E fields coupled by femptofarads of line capacitance between bench power lines and your body into high impedance INA to generate a high impedance high E Field.

The immunity is created by balanced differential signals and RF grounding or Right Leg Drive from the common Mode.