# How does low impedance measurement reveals ghost voltage?

First, I'm not sure about the terminology and whether it is correct to call this "ghost voltage".

When I was seaching info related to my previous question, I just came across the following video about ghost voltage: https://www.youtube.com/watch?v=mRwsxmn2P-s

I guess large 50Hz AC floating voltages between the SMPS output terminals and earth(as in my previous question) goes into this category "ghost voltage"(?)

The person in the tutorial video uses low impedance setting in a multi-meter and reveals whether the voltage is ghost or not.

What is the logic behind using low impedance and voltage being disappeared?

What is the logic behind using low impedance and voltage being disappeared?

The "ghost voltage" you refer to is a real voltage and although high in amplitude it is potentially very weak due to it being sourced via a high impedance. A high input impedance measurement device (o-scope, multimeter etc.) can measure this voltage with little or no obvious reduction in the amplitude of the voltage but, a low impedance device will try and take current through the inherent high impedance in series with the voltage, and the voltage will collapse in amplitude, sometimes just to a few milli volt.

It's not a ghost voltage in that someone or something hasn't created it following a death; it's just a voltage that is in-series with a very high impedance.

For instance you can capacitively couple to AC voltages in house wiring and this coupling might only be 50 pF. An oscilloscope will display the voltage on your body and tell you that the voltage is a few volts peak to peak yet, 50 pF has an impedance at 50 Hz of 63.7 Mohm. This will form a potential divider with your o-scope input (usually 10 Mohms) and the 120 VAC might drop to around 10 or 20 volts. Prior to connecting the multimeter the voltage would be higher.

• I see so in this case we should think of it as a voltage source with a very big output impedance and if we connect a low impedance almost zero voltage will drop across our low impedance resistor. But does this model with Fluke low impedance setting also apply to floating AC voltages at SMPS outputs wrt earth? Im trying to prove this AC voltages at SMPS is harmless to some people. Jan 5, 2017 at 18:04
• The devil is in the detail but if you establish that the source of voltage does come from a high impedance you can then use an ammeter to measure the current. An ammeter has virtually zero input impedance of course. Anything around 1 mA is pretty harmless but there are tables of current versus "shock" that are googleable. I'll try and leave a link. Jan 5, 2017 at 18:06
• Jan 5, 2017 at 18:08
• I would use a fused measurement device. If properly designed, the meter would blow the fuse to protect the circuit but, read the manual first. Jan 5, 2017 at 18:12
• If it blows it doesn't tell you much. If it survives then the current will be below (say) 2 mA. I'd use a 10mA fuse I reckon. Jan 5, 2017 at 18:26

Wow, the term "ghost voltage" sends a chill up my spine. That's like saying a current source can't produce real voltage across a load.
On to your question...consider the following circuit with a high-voltage V1 of 220v rms and a 5v DC source where two externally accessible terminals are available to take measurements (this is a "black-box" measurement scenario):

simulate this circuit – Schematic created using CircuitLab

On the left, if you connect an ohmmeter from either terminal to ground, you measure infinite resistance, because C1 and C2 block the DC currents applied by the ohmmeter. You can now say that battery BAT1 is floating (at least for static DC situation).
Now you take a single AC voltmeter VM1 and connect it from one of the terminals to ground. This voltmeter has 10 MEG ohm internal resistance. What does it measure? You have a classic RC circuit comprising C3 in parallel with C4 whose combination is in series with 10^7 ohms. Capacitive reactance is 15.9155 MEGohms. You measure 117v rms.
Now you move your meter, and try to measure AC voltage from the other terminal to ground. Again, you measure 117v rms.
Now you grab another identical AC voltmeter, and try measuring both terminals to ground at the same time. Now you read 66 v rms on both meters, because you now have two voltmeters in parallel (5 MEGohms).
These voltages are real, not "ghosts". AC current is flowing, but it is small.
If you substitute voltmeters having lower internal resistance, you'll measure lower AC voltages. This is what the video person has done, and has concluded that those high voltages were "ghosts".

• Im trying to understand this amazing explanation. I would be so glad if you can edit this question and model the similar thing similar way for an SMPS leakage capacitor and how I measure 90V AC but when loaded the current is so small. Can we model that as a simple circuit with caps and battery as you did? Jan 6, 2017 at 1:03
• What I do not understand is also why current is very tiny if a human is connected between BAT2 terminal instead of VM2. In that case human is 100Meg Ohm lets say and he will be exposed to around 200V AC and he will die. But they say in the ghost case current will be small. Im confused. (I hope I could pose my question well) Jan 6, 2017 at 1:11
• @doncarlos Capacitors in this model may differ from your particular SMPS, and V1, V2 may be less. If you grab either terminal of BAT1, and you are grounded, a small current flows through C1+C2 through your body. Your skin resistance might be 100 k ohm. But even if skin resistance is zero, current is still small: limited by the very high reactance of C1+C2. Andy's current-measuring test should ease your mind. Jan 6, 2017 at 2:06
• Can I measure that reactance? Imagine 10Meg voltmeter measures 90V floating AC from one DC terminal of the SMPS to the earth in a 50Hz 230V system. So it means voltage drop across the multimeter is 90V through 10Meg so I guess the reactance in that case is 10 * 90 = 25Meg? Jan 6, 2017 at 2:24
• So basically you are saying if human is connected the the voltage drop across him will be too small due to this 25Meg Ohm capacitive divider? Jan 6, 2017 at 2:26

We use this effect ----- low impedance ------ to manage Efield interference into Signal Chains. We view the signal traces between OpAmp stages as being the MIDDLE plate of a 3 plate capacitor. The top plate is the interferer, possible a SwitchReg an inch away or a foot away. The bottom plate is (sum of)the Ground Plane under the signal trace + the output capacitance of the prior stage + the input capacitance of the next stage.

This 3-plate capacitor forms a voltage divider, just like a scope probe does high frequency division with its 2 capacitors.

The various Resistances on the signal trace: Rout of prior stage, Rin of next stage, Rdiscretes from trace to GND or to VDD, combine in parallel to create a HIGH PASS FILTER with the top 2 plates.

All this is modeled, with editable PCB trace (or silicon aluminum layers) dimensions, or even COAX between stages, in the free tool "Signal Chain Explorer" from robustcircuitdesign.com

To enable the Efield interference (or the Hfield interference, the Power Supply trash and the GND plane trash), simply click the "Gargoyles" button at top center, then click "Update" near the topright.

To insert OpAmps into the Signal Chain, look at left margin and click "Amps"; then DoubleClick on your preferred stage; you can edit the gain, the resistors, the Johnson Noise, the lead parasitics, the chip ESD, the Power Supply filtering.

• Are you writing about SMPS supplies only? How about if one replaces SMPS with a linear power supply, would that minimize this ghost voltage? I cannot find any document about quantitative comparison of these two type of supplies' leakage. Feb 22, 2017 at 9:49
• Yes----a linear power supply would not have "ghost" voltage. The rate-of-change of voltages is the key to ghosting. An unregulated 120Hz DC voltage with 2 volts of ripple, perhaps 1milliSecond rise time, is much less coupling than 200 volts being chopped by MOSFETS with 100nanosecond edges. The linear voltage risetime is 2 volts in 1mS; the second is 200 volts in 0.0001milliSeconds (100nanoSec); thus the second has 200/2 * 1mS/0.0001mS = 1MillionX more coupling, thru the formula I = C*dV/dT Feb 27, 2017 at 3:59