# What is reducing the NOISE when probes are connected to a circuit?

When the oscilloscope probes aren't connected to anything, the following waveform is observed:

The oscillations are more then 1.5mV.

When the probes are connected to measure the voltage on a 27Ohm resistor with 30uA current flowing:

the big oscillations are gone and the new ones don't oscillate more than 200uV. Given that the probes are physically in the same location, and the circuit is battery powered (no earth grounding), what reduces the big oscillations?

Motivation

I came to this problem while trying to understand how much noise in the second graph is due to the probes. Wasn't expecting to get a larger error.

• As far as I can tell the frequency of the noise in the first graph looks like 50Hz. Guess where that comes from. Oct 7 '16 at 11:58
• @WesleyLee Ok, but what causes it to disappear once connected. The voltage on the resistor in the circuit are very low compared to the 50Hz noise. Oct 7 '16 at 12:05
• Take a look at this question and answers. Oct 7 '16 at 12:06

An ideal voltage measurement device will have infinite impedance. This is impossible as all voltage measurement devices need a small amount of power when making the voltage measurement. Common oscilloscope probes have an impedance of 10 million ohms. There are variations available.

In a normal room wired for main (120 volt AC service) there is a good amount of associated electrical noise. This noise can induce a small current in any conductive material. How much that effects a voltage measurement depends on the impedance of what is being measured. We can use ohms law to find out how much current is needed for your 6mV peak to peak oscilloscope measurement above (assuming you are using 10Mohm ohm probes):

V = I x R
6mV = I x 10Mohm
I = 6E-3 / 10E6
I = 6E-10
I = 60nA


As you can see, only a very small amount of current is needed to deflect the oscilloscope trace 6mV.

To find out how much this noise will effect your voltage reading while connected to your 27 ohm resistor we first need to find the over all resistance:

1/R(over-all) = 1/10E6 + 1/27
R = 26.9999271


Then we can find the expected additional voltage due to the noise we measured:

V = I x R
V = 60nA x 26.9999271
V = 1.62uV


The second scope trace is set to a little more than 1/10th volt per division. So it is unlikely a deflection of 1.62uV will be observable.

The electric field intensity many metres from ac household or office wiring can be up to hundreds of millivolts as measured on a high input impedance device like an oscilloscope.

Put a 27 ohm resistor across the scope probes and what might be hundreds of millivolts into a 1 Mohm impedance turns into about 1 micro volt and what you might see is more likely the input noise self generated by the oscilloscope.

The probe impedance is measuring the stray AC Line E field in mV/mm due to the length of the probe ground lead acting as an antenna.

• If your finger touches the probe tip without touching ground , your body becomes the antenna and may pickup as much as 100Vpp of the AC voltage depending on proximity to AC line voltages and wiring loops.
• Then when you shunt another finger to earth ground or probe ground, this attenuation of 100pF or more at line frequency becomes a voltage divider from the stray impedance coupling to the stray capacitance (xx pF) to become a capacitance voltage divider.
• The resistor does the same thing as a Zc:R voltage divider so you can compute the stray capacitance to the source at 50/60 Hz
• since Zc(f) drops with rising f, this ratio increases resulting in more high frequency noise, which can also be used to compute Zc:R ratio except now there is also a current loop with magnetic coupling of current to the loop antenna.
• some noise may be thermal, while other may be from other sources (logic, SMPS etc.)