# How can I accurately and precisely measure current in a 110V AC circuit, with a 1kohm-50kohm load, using an oscilloscope?

I need to measure the electrical resistance of a volume of ground.

To do this I need to cause a current to flow between two metal electrodes pushed into the soil (let's call them the "current" electrodes) and measure the potential difference created by this flow using a second set of electrodes connected to a voltmeter (lets call them the "voltage" electrodes).

Since the charge between the current electrodes causes ions in the soil-water to accumulate around the electrodes I need to use an alternating square-wave current source (in the 50-150Hz range - low enough so we don't get problems with complex impedance). And since I'm interested in the way in which the voltage between my measurement electrodes changes over time I will use an oscilloscope to make my measurements - which must therefore be voltages in a reasonable range (1mV to 10V).

The ratio of the current flowing between the injecting electrodes to the voltage between the measurement electrodes gives me a "resistance" value using Ohms law and a simple geometric modification. And that solves my original problem.

I can use an off-the-shelf 12vDC to 110V AC square-wave "mains" converter to supply the voltage to drive the square-wave alternating current powered by a simple 12v battery. 110v is high, I know, but the contact resistance of the electrodes with the soil is in the 1kohm to 50kohm range and to get a current flow high enough to measure with precision (at least 1mA) I need to drive the current with a voltage of tens to hundreds of volts - one commercial system I own uses voltages as high as 800v to measure ground resistance, though 100v maximum is more common.

The current will still be too low to measure precisely using a Hall-effect current sensor so I need an alternative.

So here's my problem: The off the shelf circuits I can buy to measure small current accurately and precisely, by conversion to a voltage of reasonable range, are DC only but the current flowing into the ground must be square-wave AC. So what simple circuit, perhaps a version of a shunt-plus-amplifier, can I use to convert the AC current to an equivalent (zero-crossing) AC voltage measurement in the <+/-10v range my oscilloscope can accept, without significantly affecting the current itself?

• How accurate, and how precise, do you want? – Simon B Oct 27 '20 at 8:32

Edited because of new information in a later comment)

As your after a 4 wire measurement, lets begin by lowering the voltage to be within the safe range of your scope

For a 4 wire measurment you can either know the current your sourcing, and measure a voltage, or know the voltage and measure a current, due to the application, measuring the voltage will be much easier

Lets start with say a 6.3V transformer, its peak voltage will be within your scopes ability, (about 9V peak), we only need to know the amount of current so we will use a current shunt resistor to measure it, as the current will be low due to the resistance of the soil we will keep it rather high, lets say about 10K,

1 scope probe attaches across this current shunt and gets the current via measuring a voltage, now the scope probe will shift this value slightly due to acting like a 1M or 10M resistor in parrellel, but the effect can be calculated out.

So 1 probe to 1 side of the transformer, 1 probe via the current shunt to the other side, and we have out "Source" probes

Next up the measure, due to the small amount of current, we will use amplification to make up for it. using a "differential amplifier", these have a fun side bonus that its easy to choose what voltage is outputs that difference against, you will need to provide power to this device, so may need to build a small rectifier and capacitor to feed it DC voltage off the same transformer

The relationship for the gap between the probes is important so you may wish to mount these 2 measure probes on something to keep them a fixed distance apart.

the second scope channel measures the differential output, and you have a voltage, and a current, leaving you with a resistance,

The input voltage minus the measurement voltage is then the soil voltage, you know the current based on the measurement voltage leaving a simple calculation for its resistance