# Why is there a voltage difference between these two grounds?

I just started studying robotics and we are taking a introductory class in electrics. We where supposed to perform the following task:

1. Wire up the breadboard in the following way: 1. Use a multimeter to measure the voltage between each of the wires. Under is a table the different combinations. The top row with the bold letters means the red wire (+) from the multimeter was used while non-bold means black wire (-) was used. So AB means the red wire from the multimeter was placed on wire A, the black wire plased on B and the measured voltage was 5.1V.

A B C D
A 0 -5.1 -5.1 -10.15
B 5.1 0 0 -5
C 5.1 0 0 -5
D 10.15 5 5 0

The measured voltage on combinations like AA, AB, DC was expected. But what I did not expect was the measured voltage on combinations like AC, AD and BC. I was expecting both the power sources to be +5V and both grounds to be 0V relative to each other. Meaning there would not be a voltage difference between A and C for example. But this is clearly not the case.

From searching the web it seems the ground voltage is only relative to the source voltage, meaning the power rails on each side have different voltages. I could not find the manufacturers name or model off the usb power source and could therefore not find documentation about it. But from reading the text on the usb power source saying +5V and -5V and going over the data in the table, I thought this might be a explanation:

The left side has a power source voltage off +5V and the ground is 0V (relative to each other) while the right also has a +5V source and a 0V ground when measured relative to each other. But when the right side is meassured relative to the left side, it has a source voltage of 0V and a ground voltage of -5V. Using this logic, all the measurements like AD, AC and BC would make sense.

But I don't really understand why the circuit is working like this. Here are my questions:

1. Is my explanation about the data above correct?

2. It does not make sense to me that the ground on the right side can be -5V when measured relative to the left side and 0V when measured to the right side (my head hurts trying to visualize it). Does this mean creating a circuit with the +5V as the power source and the -5V as the ground would give the circuit a voltage off 10V? Would that not require twice the energy as a 5V circuit?

3. Why does it say +5V on the right side and -5V on the left side? If you compare the voltages to each other, the right side should say 0V. But if you don't compare the voltages to each other, the right side should say +5V since its relative right?

4. Why would you not design the usb power source in such a way that both the power sources where +5V relative to each other?

5. Can someone give me a good site to read more about voltage, grounds and how they are relative to each other? (like the physics behind it).

• The left side clearly states +5V while the right states -5V. These are the output voltages relative to 0V. The 0V point will therefore be halfway between the two and you'll measure 10V across the whole supply. This is not really any different to putting two batteries in series and using the midpoint as ground; one end will be positive and the other negative. Aug 27 at 11:12
• Take a look here electronics.stackexchange.com/questions/392010/…
– G36
Aug 27 at 11:19
• There's one simple but important thing you missed, which doesn't explain everything but should at least make some things less confusing: B and C are grounds. This is true despite the red and black markings on the breadboard, which suggest that D is a ground; that's misleading. (Those markings are just paint. They are suggestions for how to use the breadboard. The suggestion is not being followed on the right side here.) Aug 27 at 12:06
• With that clarified, you should be able to see that the labels on the circuit board (+5V, GND, GND, -5V) are lined up with the corresponding pins. Although voltage is always a relative measurement taken between two points, if you want to think in terms of "absolute voltage", it's traditional to consider ground to be "0V", and then you should be able to see that the pins labeled +5 are at 5V, and the pins labeled -5 are at -5V (yes, negative voltage; that just means "voltage below the point we're calling ground.") Aug 27 at 12:11
• you chose gray to be +5 V and green to be gnd, so don't get fixated on red and black to mean something specific Aug 27 at 18:59

Here's an equivalent circuit which behaves exactly like your USB supply board. The green box represents the module: simulate this circuit – Schematic created using CircuitLab

Potential (voltage) is rather like altitude; when you state an altitude, it's not clear how "high" that actually is, unless everybody agrees about where "zero altitude" is. That could be sea level, or the airport, or the centre of planet Earth, or any arbitrarily chosen reference point.

In the same way, we must declare some node (for example, A, B, C or D here) to be our "zero volt" reference point, and every other potential will be stated relative to that. That reference point is called "ground" (in keeping with the altitude analogy), and will be our "zero volt" point. Every other node will have some potential (voltage) relative to that.

In the above circuit, I have declared that node B shall be our "ground", and I tell other readers this with my use of the ground symbol.

Every voltmeter is measuring a potential difference, which is the difference between potentials at two points in the circuit. The reading on a voltmeter is not telling you the absolute potential at some point (with the exception I'll explain at the end). The voltmeter only tells you how different the potential at its red (+) terminal is from its black (−) terminal.

VM1 reads +5.0V. All this means is that node A is 5V higher in potential than node B. It is not saying that A is +5V, or that B is 0V or anything about the absolute potentials anywhere. I repeat, all this tells you is that node A is 5V higher in potential than node B, and without any more information, we can't know the absolute potential at either node.

VM3 also reads +5.0V. This means only is that node C is 5V higher in potential than node D. It is not telling you anything about the abosulte potentials at either node, only that nodes C and D are 5V different, and that node C has the higher potential.

As you know, the straight lines joining component terminals are wires, the purpose of which (besides carrying current) is to bring all interconnected terminals to the same potential. For instance, all points along any of the blues wires in my schematic have the same potential. In fact, they must therefore be the same node, and the labelling of nodes B and C is somewhat misleading here.

Nodes B and C are actually the same node, because they are physically connected together, and should really only have one name. Since they have the same potential, it should come as no surprise that voltmeter VM2 between them is measuring zero potential difference.

Since B and C have been designated "ground", this is my way of saying "here is 0V":

\begin{aligned} V_B &= 0V \\ \\ V_C &= V_B = 0V \\ \\ \end{aligned}

Knowing this we are now able to state absolute potentials for A and D. VM1 tells us that A is 5V higher than B:

\begin{aligned} V_A &= V_B + 5V \\ \\ &= 0V + 5V \\ \\ & = +5V \end{aligned}

Similarly for node D, VM3 says C is higher than D by 5V:

\begin{aligned} V_C &= V_D + 5V \\ \\ V_D &= V_C - 5V \\ \\ &= 0V - 5V \\ \\ &= -5V \end{aligned}

The module behaves as if it contains two 5V batteries. All a battery does (or any other voltage source for that matter) is impose a potential difference between its two terminals. It doesn't say what the absolute potential at either terminal is, only what the difference is. By connecting two such batteries in series, you are ensuring that the joined terminals have the same potential, but still none of the terminals in this arrangement have an quotable absolute potential until you designate some arbitrary potential to one of them: simulate this circuit

Here I have declared that the top of BAT2 shall be called +100V. That battery ensures that node X shall be 10V more positive (check the polarity and orientation of BAT2):

$$V_X = +100V + 10V = +110V$$

BAT1 causes node Y to be 20V lower in potential than X:

$$V_Y = V_X - 20V = +90V$$

Understanding this will enable you to better see what's happening in the module. BAT1 and BAT2 in the module schematic are connected in series in such a way that BAT1 ensures node A is 5V higher in potential than our arbitrarily designated ground (0V) B/C, and BAT2 sets node D 5V lower than ground, at -5V.

Hopefully, now it should be clear why voltmeter VM4 shows +10V. That is the difference in potentials between nodes A and D:

\begin{aligned} V_A - V_D &= (+5V) - (-5V) \\ \\ &= +10V \\ \\ \end{aligned}

A voltmeter never shows an absolute potential, unless its black negative terminal is connected to ground. This is not the case for voltmeter VM4, and therefore you should not expect it to show any kind of absolute potential. VM1 and VM5, however, do have their negative terminals connected to 0V. Notice how they both show the absolute potentials $$\V_A\$$ and $$\V_D\$$ respectively:

VM1 is showing $$\V_A - V_B = V_A - 0V = V_A = +5V\$$.

VM5 is showing $$\V_D - V_C = V_D - 0V = V_D = -5V\$$.

In case this hasn't sunk in yet, there's no such thing as 0V, only potential differences.

The physics of it, briefly is like this; "potential" is named after potential energy. Like a mass having gravitational potential energy, charges have electrical potential energy. A mass's potential energy is increased by moving it further away from Earth (physically lifting it higher), but the amount of potential energy it possesses is always stated relative to some fixed altitude. The ground, or if you're at the top of the cliff, the sea, or if it can fall into a hole towards the Earth's core, the centre of the Earth. We measure gravitational potential energy by how far it is permitted to fall.

If an electric charge is permitted to fall in an electric field, from node to node, it can't fall further than the node with lowest potential in the circuit, and that's what defines the maximum amount of energy it can impart to whatever it travels through on its way there. That's why voltage is always relative. If ever it reaches an "altitude" of 0V, but there exists still lower potential that it can fall to (say, -5V), then it will continue to fall (if permitted), in the same way that if a rock falls from 1000m to sea level (0m), it can still sink to the sea bed at, say -100m.

There the rock must stop, and all the potential energy it had at +1000m altitude has been spent heating the air and the sea, and making water waves and sound waves. Even though we defined sea level as a point where everything has zero potential, relative to that arbitrary level, the rock still has 100m further to travel (to the sea bed), and so the total energy it had to "spend" was actually related to the difference $$\(+1000m)-(-100m) = 1100m\$$.

That's like declaring the "ground" point in our circuit, but then also providing an even lower potential node, at -5V, towards which those charges could continue their journey, if permitted. Alternatively, we could have called node D ground. Then $$\V_D=0V\$$, and $$\V_A=+10V\$$. The two scenarios are equivalent. Whatever way you look at it, charges found at node A have 10 units of potential energy per charge more than charges found at node D, as indicated by VM4. That's the amount of energy that charges have to "spend" on their environment as they journey from A to D, and this will be the case irrespective of where we call "zero".

A charge starts its journey with a certain amount of potential energy, and "falls" towards a point in the circuit with lower potential, losing that potential energy on the way, donating it to whatever it encounters. If ever it reaches the point of lowest potential in the circuit (-5V in your module), its journey ends, unless it can somehow be "lifted" to the top again (+5V).

Potential (voltage) is a measure of how much potential energy is possessed by charges found at some location, some node, relative to the energy possessed by charges at another location. It's therefore also a measure of how much energy it has to impart during its journey between those locations, as it "falls" in the electric field, to cause heating, or turn a motor, or cause an LED to emit electromagnetic waves of light.

Oh, and by the way, due to the similarity between the scenarios of falling masses and falling charges, it really, really helps to draw schematics as you would draw a picture of a rock rolling down a cliff; with higher potential nodes at the top.

As humans we are all very familiar with principles of gravity, masses, and falling, and we all share a common idea of "up" and "down". It makes sense to help your readers understand the operation of a circuit by drawing schematics that conform visually with those same principles. It also makes circuits easier to troubleshoot.

It's not always possible, of course, but do it when you can.

If you turn your photo clockwise by 90°, I bet you'll suddenly see how obvious and simple all this becomes. Try it.

Also, I recommend that you use your multimeter to measure the electrical resistance between B and C (while it's un-powered, of course). It should be 0Ω, proving that they are physically connected together inside the module, somehow.

Notice how every voltmeter measurement between B and C in your table is zero, telling you that there's no difference in potential between those two nodes:

$$V_B - V_C = V_C - V_B = 0V$$

You may interpret this to mean that whatever absolute potential $$\V_B\$$ exists at node B, it must be equal to the absolute potential $$\V_C\$$ present at node C.

Therefore, in response to your titular question "why is there a voltage difference between these two grounds?": Grounds B and C are not different. Your own measurements have shown this.

• I think this is the best answer I have heard to any question, ever. Aug 28 at 5:23
• @volticus That's very kind of you, cheers! Aug 28 at 5:24

But there isn't a voltage between your two grounds. You have two GND outputs from the PCB, and there is 0V between them, they are your wires B and C, as they are the exact same wire on the PCB.

As ground is the level defined as 0V, your PCB has +5V output which is your wire A and the PCB also has a -5V output which is your wire D, which is why you have 10V between wires A and D.

If you are confused how the grounds connect to red and black on your breadboard, then forget the colors. The point seems to be there is 5V between red and black, but on the other side there is +5V red and 0V black, and on the other side there is 0V red and -5V black.

It basically just can be simplified as you having two 5V batteries connected in series to make a 10V battery, but since the midpoint of the two batteries is defined as 0V, you have both +5V output and a -5V output.

I think the important thing to remember is that voltage is always relative. You should imagine that every wire has a number associated with it called its "potential." (Potential can change over time.) It's impossible to measure the potential of a wire, but a voltmeter will tell you the difference between two potentials—specifically, a voltmeter will tell you the potential of the red lead minus the potential of the black lead. The difference between one wire's potential and another wire's potential is called "voltage."

By measuring the output of this board, you've found out the following information:

• The potential of B is the same as the potential of C.
• The potential of A is about 5.1 V higher than the potential of B (or C).
• The potential of D is about 5 V lower than the potential of B (or C).

If you want to visualize this, it's easy! Take out a sheet of lined paper and write the letters B and C both on the same line. Write the letter A 5 lines above B and C, and write the letter D 5 lines below B and C. To "measure the voltage" between one letter and another one, point a red pen at one letter and a black pen at another letter, and then count how high up the red pen is compared to the black pen.

Why is the circuit behaving like this? Why are B and D different? Well, let me turn the question around: why would B and D be the same?

I suspect that you're used to power supplies that have only two output wires. If we have such a power supply, we can call the higher wire "source" and the lower wire "ground," and if you measure the voltage between two things that are connected to the ground wire, you'll get 0 V. Now you see this power supply with four outputs, and your prior experience has probably led you to think that there must be a source and a ground on the left, and a source and a ground on the right.

However, that's not the way that it works. When we're designing a power supply, we're not limited to giving it just source pins and ground pins; we can make any pin have any potential. (Well... any potential within the limits of what's physically possible.) The designers of this power supply decided that they wanted to make B and C have the same potential, make A be 5 volts higher than those, and make D be 5 volts lower than those. So that's what the power supply does.

Note the labels on your circuit board (from left to right: 5V, GND, GND, -5V). Compare the breadboard: red, blue, red, blue.