# How to make single ended voltage measurements

I’m in the process of setting up a test rig to automatically measure and test induced PCB voltages. The device I’m using (we’ll call it the meter) makes single ended measurements. The PCB output drives a 24V volt fan via two pins (which I assume are analogous to live / neutral) when not connected to the test rig.

Firstly I’m confused over the concept of single ended measurements (I’ve been working in purely SW for several years now). If I wire the live and measure it relative to the meter’s internal ground surely I could get a zero voltage back as there is not necessarily a completed circuit if the meter’s ground doesn’t correspond to the PCB’s ground? A slight aside: I tried an experiment where I wired a single terminal of an AAA battery through a meter to the chassis of my PC and got a zero volt reading. I’m having difficult conceptualising this as I would expect that if V(batt) = 1.5V then V(+ve/earth) - V(-ve/earth) should also be 1.5V. I’ve asked a couple of people this question and been given the not very satisfying answer of “it’s not a completed circuit”. If that is the case would you not expect current to flow from a charged balloon into earth through a wire?

An alternative could be to simultaneously measure both the live and neutral pin relative to the device’s earth and perform a subtraction. However, I want to preserve as many screw terminals on the meter as possible as there are several other voltages that require testing with the rig.

The meter also has screw terminal labelled GRD and SGRD. Can I wire in the fan pin neutral to this terminal and get a single end measurement relative to the PCB’s earth?

Hope someone can give me some clarity here. Electronics is not my primary field so if some of these seem basic concepts good links and also terminology would be useful (although I'm in the process of trawling the net for answers).

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Basically there is no such thing as a single ended measurement. That term is sometimes used for convenience when the reference point is understood from context, usually ground. –  Olin Lathrop Aug 21 '12 at 1:28

Indeed, if you hold only one probe of your meter to the circuit, then the meter's circuit isn't closed, so there's no voltage difference to be measured.

Absolute voltage does not exist, it's always relative to some other point. So if you only use one probe of the meter it will say "compared to what?".

Ground is your reference for all voltages in your circuit. If a voltage is +12 V that will be relative to ground, unless otherwise noted. So you place the meter's reference probe, the black one, on the circuit's ground, and then that same +12 V will also seen by the meter as +12 V, because circuit and meter share the same reference.

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I appreciate the basic concept that voltage is a potential difference. However, my question was more subtle than this; why does it appear (at least in the example with the battery as described in my original post) that you cannnot measure, for example, the voltage difference between two points that are not otherwise connected by a closed circuit. –  user11623 Aug 19 '12 at 19:12

why does it appear (at least in the example with the battery as described in my original post) that you cannnot measure, for example, the voltage difference between two points that are not otherwise connected by a closed circuit.

(The above quote is not from the OP's question but from the OP's comment to stevenvh's answer).

If I understand your question correctly, you're trying to measure the voltage between two nodes - that are not connected in any other way - with a voltmeter.

To understand why this is not possible, you must realize that while there may indeed be a voltage across two nodes that are not connected in any way, the Thevenin equivalent of this is a voltage source with an (effectively) infinite series resistance.

Since any physical voltmeter has a finite input resistance, by voltage division, there will be no voltage across the voltmeter.

In other words, the open circuit cannot drive any current through the finite input resistance of the voltmeter. By Ohm's law: no current through the input resistance implies no voltage across the input resistance - the meter reads zero volts.

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Voltage as per wiki is "Voltage, otherwise known as electrical potential difference or electric tension (denoted ∆V and measured in units of electric potential: volts, or joules per coulomb), is the electric potential difference between two points — or the difference in electric potential energy of a unit test charge transported between two points." This is the meaning of " circuit is to be closed" to measure the voltage.

I hope this will clarify your confusion.

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When trying to explain electricity, it is common to use the water analogy. This can be seen here and in Wikipedia. There are some more good links, explanations, and pictures in this Q&A.

To explain your battery example, think of the battery as a reservoir of water that is capable of some pressure (pressure is analogous to voltage). Think of your meter as a very long pipe with a device to measure pressure at the end. If you just connect the pipe to the output of the reservoir, the water will go up the pipe to a certain level but never reach the pressure sensor.

In your experiment with the battery and the PC chassis, you had two reservoirs with pipes. One being the battery the other being the PC power supply. The only difference is that the power supply had the pipe connected to the input. Neither sensor has water flowing past it so neither sensor can measure any pressure.

The only way to measure the pressure of the battery reservoir is to loop the pipe back to the input. That way the water flows from the output back into the input continuously. This way the water continuously passes the pressure sensor and can be measured.

The reservoir always has a pressure. But that pressure cannot be measured unless it is allowed to flow. And in electricity, in order for there to be flow, the electrons must be allowed to return to the source.

I've never seen an electrical analogy to software. But since you say you've worked purely in software, I'm going to attempt one. Let me know if this helps your understanding (or if it only confuses you more) because I'm always in need of a better way to explain electronics to software engineers.

As has been explained in other answers, voltage is a potential difference. The key word is difference.

Take the following code:

cout << "abc";

What is the difference between the location of 'a' and the location of 'c' in the stream?

Now take this code:

cout << "abc";

cerr << "xyz";

What is the difference between the location of 'a' and the location of 'z'?

Answer: Undefined because "abc" and "xyz" are in separate streams. You can't measure the location of a character in one stream in reference to the location of a character in a different stream.

Location of a character in a string isn't a great analogy for voltage/pressure. But hopefully the concept of measuring between two separate streams translates to measuring the pressure of one reservoir with respect to another or the voltage of a battery with respect to a power supply.

Edit:

In your question, you state, "wired a single terminal of an AAA battery through a meter to the chassis of my PC". This leads me to believe you connected one lead of the meter to the + of the battery and the other lead of the meter to the chassis of the PC.

If there was a complete circuit and if the meter leads were in teh correct connectors on the meter, this would be how you measure current, not voltage. Have a look at this page for a general discussion on the difference between measuring voltage and measuring current. And have a look at this tutorial for some basics on using a meter.

If you connected the + of your battery to your chassis through a wire and then measured from chassis to earth ground, you would get 0V. This is because chassis and earth ground would be a the same potential and therefore there would be no difference.

Electrons won't go on a trip unless they know they can come back home. When you connect the battery to the chassis using only one wire, the electrons won't leave the battery because they know that earth is not their home. The - of the battery is their home and they won't budge unless they know they can return there.

If you connect the - of the battery to the chassis and measure from the + terminal to earth ground, then you should see a voltage because you are connecting earth ground to the electrons home base at the - terminal. And as long as there is a path back home, they will take the trip, current will flow, and voltage can be measured.

Edit 2:

First off, DO NOT connect an ammeter directly to a battery without a load. The ammeter is a very low resistance and you will essentially short out the battery. The battery will then get very hot in short order. Always connect a resistive load in series when using an ammeter.

You are correct in stating that an ammeter is just a galvanometer. A voltmeter is just an ammeter with a series resistor. So when you are using the voltmeter directly across the battery, you have a complete circuit with a load. That is why you measure a voltage when you measure directly across the battery.

If you use an external resistor in series with your ammeter and complete the circuit, current will flow and you will read some amount of milliamps depending on the value of the resistor.

I'm an engineer and you're a physicist. So when you ask "is this due to the chemical nature of the battery?", we're both getting out of our areas of expertise. Although you're probably better off because I can tie this back to physics principles that I probably don't understand as fully as you do. So while I can't give you a good, complete answer, I can point you towards some search terms to help you understand it a bit more.

The field of study is known as electrochemistry.

The flow of electrons is always from anode to cathode outside of the cell or device, regardless of the cell or device type and operating mode, with the exception of diodes

And the reason for this is the physical principle of charge conservation:

Charge conservation is a physical law that states that the change in the amount of electric charge in any volume of space is exactly equal to the amount of charge flowing into the volume minus the amount of charge flowing out of the volume.

Sorry about all the Wikipedia links but it's as good a place as any to start from. Once you have the terminology, you can ferret out more complete and definitive sources of information.

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Hi there thanks for your reply. I should clarify a couple of points: when I said chassis I meant not the power supply but just that I had connected the terminal of battery to earth (the chassis). Because the terminal of the battery was at a potential I was expecting some kind of current to flow between it and earth. Either this current flowed over an unmeasureablely short period of time or no current could flow because there was no where for the electrons to go or perhaps the nature of the battery is such that voltage is only maintained when hte circuit is complete. –  user11623 Aug 20 '12 at 20:43
Also, my main degree was in theoretical physics (a long time ago) so I do have some knowledge of the what a potential and a field is. If I think of this electrostatically (in terms of individual electrons) then I would expect some flow of charge whenever there is a potential difference and this flow of charge does not necessarily have to complete a circuit; we will have charge moving from a low to a higher potential. If there is a potential diff across a battery of V(batt) then I would expect V(batt) = V(+ve/Earth) - V(-ve/Earth). –  user11623 Aug 20 '12 at 20:51
You state that "that pressure cannot be measured unless it is allowed to flow." If there is a potential difference does current not flow until equilibrium is reached and if two PDs are connected then there is a path for this current to flow. All I can think is as I stated above; that equilibrium is reached over an unmeasurable short period of time. –  user11623 Aug 20 '12 at 20:56
See my edit. Especially this link. –  embedded.kyle Aug 20 '12 at 21:20
HI Kyle. Thanks again and sorry to draw this out. –  user11623 Aug 20 '12 at 21:27
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