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Just a general electronics question: What is negative voltage, like -5 Volt?

From my basic knowledge, power is generated by electrons wandering from the minus to the plus side of the power source (assuming DC power here). Is negative voltage when electrons wander from + to -?

Why do some devices even need it, what is so special about it?

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    \$\begingroup\$ You will do better to think about electronics as a kind of applied math, and forget about the electrons entirely. \$\endgroup\$
    – markrages
    Feb 17, 2011 at 3:34
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    \$\begingroup\$ Yeah, don't think about electrons. Engineers use conventional current instead ("just a convention"), since charge can flow in both directions at the same time. amasci.com/amateur/elecdir.html And power is not "generated" by electrons. Energy is carried in electric and magnetic fields. amasci.com/elect/poynt/poynt.html \$\endgroup\$
    – endolith
    Feb 17, 2011 at 15:13
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    \$\begingroup\$ xkcd.com/567 \$\endgroup\$
    – user
    Apr 23, 2013 at 9:54
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    \$\begingroup\$ Do the electrons wander, or do the holes they want to fall into wander? Might be a question for the Electrical Philosophy SE site? \$\endgroup\$
    – user65586
    Oct 12, 2015 at 1:37
  • \$\begingroup\$ @endolith thinking about the electrons is exactly where I'm stuck right now, trying to think about how high-voltage power supplies for e-beam guns are specified. If conventional current + means the "electron rich" side of a circuit, then to power an e-beam gun I would need a some-kilovolts + supply? But for an ion-gun (i.e. gallium) I'd need the gallium connected to a - supply (even though we know Ga ions are positive in charge)? Also how do I know what such power supply mfgs specify with, conventional current,or particle charge (these supplies are generally marketed for this purpose) \$\endgroup\$
    – nmz787
    Nov 30, 2020 at 14:23

11 Answers 11

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Someone may have better words to explain this than me, but the big thing you have to remember is voltage is a potential difference. In most cases the "difference" part is a difference between some potential and ground potential. When someone says -5v, they're saying that you are below ground.

You also need to keep in mind that voltage is relative. So like I mentioned before, most people reference to "ground"; but what is ground? You can say ground is earth ground, but what about the case when you have a battery powered device that has no contact to ground. In this situation we have to treat some arbitrary point as "ground". Usually the negative terminal on the battery is what we consider from this reference.

Now consider the case that you have 2 batteries in series. If both were 5 volts, then you would say you would have 10 volts total.

But the assumption that you get 0/+10 is based off of "ground" as being the negative terminal on the battery that isn't touching the other battery and then 10V as being the location of the positive terminal that isn't touching the other battery. In this situation we can make the decision that we want to make the connection between the 2 batteries be our "ground" reference. This would then result in +5v on one end and -5v on the other end.

Here is what I was trying to explain:

+10v   +++   +5v
       | |
       | | < Battery
       | |
+5v    ---   0v
       +++
       | |
       | | < Another Battery
       | |
0v     ---   -5v
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    \$\begingroup\$ Thanks. Just wondering though: Why would devices "choose" to require -5 Volt instead of +5 Volt? If I have a device that's powered by a "normal" Power Supply, why wouldn't the manufacturer just require +5 Volt? (Some LC Screens requite negative voltage) \$\endgroup\$ Feb 17, 2011 at 5:17
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    \$\begingroup\$ It's specific to the device. Read up on how LCD's work at a technical level and I bet it'll explain it. \$\endgroup\$
    – davr
    Feb 17, 2011 at 5:43
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    \$\begingroup\$ @michael, LCD screen contrast provides the reference for each of the cells. It may be rated for common 3.3V or 5V for easy integration, but the cells may need more of a drop in order to read them (molecules need to be twisted). Often 5V LCDs can use 0V contrast, even without an integrated (-ve) charge pump. \$\endgroup\$
    – tyblu
    Feb 17, 2011 at 5:51
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    \$\begingroup\$ @MichaelStum: Many LCD controllers (especially those patterned after the Hitachi 44780) require a 3.3V-5V supply to operate the logic, and 3.3V-12V to power the glass. In such situations, it's generally easier to have one rail in common between the two supplies than to have four independent rails. Hitachi found it more convenient to have the two supplies share a positive rail than a negative one. Since the magnitude of the LCD drive voltage may be higher or lower than the magnitude of the logic voltage... \$\endgroup\$
    – supercat
    Dec 28, 2011 at 18:27
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    \$\begingroup\$ ...the voltage on the negative LCD supply pin may be above or below the voltage on the negative logic supply pin. Note that from a logic-design perspective, the negative logic supply is "ground", but from the standpoint of wiring nomenclature, designating the positive rail as "ground" makes clearer the fact that the two negative rails are independent, and either one may be higher or lower than the other. \$\endgroup\$
    – supercat
    Dec 28, 2011 at 18:30
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Imagine you're measuring the height of a car. You could take a tape measure and measure the distance from the ground to the roof of the car. "The roof of this car is 4 feet above the ground."

You could also stand on the roof of the car and dangle the same tape measure down to the ground. "The ground is 4 feet below the roof of this car."

Voltage works the same way. The negative sign is just a convention, in the same way that the car has the same height, regardless of which way you measure it. Flip your multimeter leads, and the negative sign will disappear.

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    \$\begingroup\$ Ground is just a convention, too. You can pick any single point in a circuit and label it ground, but we choose specific points out of convention. \$\endgroup\$
    – endolith
    Feb 17, 2011 at 15:16
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    \$\begingroup\$ @michaelStum, @pingswept, I always like to remind people that there is potential energy from height if you take physics. (mgh) is potential energy, if I pick the atmosphere as my reference we all suddenly have negative potential energy. \$\endgroup\$
    – Kortuk
    Feb 17, 2011 at 23:11
  • \$\begingroup\$ Good points, the both of you. \$\endgroup\$
    – pingswept
    Feb 17, 2011 at 23:30
  • \$\begingroup\$ Gravitational potential energy is referenced negative by setting it to zero at infinity. Electrical potential energy is negative between two charges of opposite sign and positive between charges of the same sign. Say there's an electrical potential drop from 5V to 3V. A positive test charge q would add -2q to its potential energy (becoming less positive), while a negative test charge -q would accelerate in the other direction (+2V, from 3V to 5V), but also adding -2q to its potential energy (becoming more negative). In both cases potential energy decreases. \$\endgroup\$
    – Eryk Sun
    Feb 27, 2011 at 13:06
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    \$\begingroup\$ @eryksun: While it's true that potential energy is measured by taking zero at infinity, if one thinks in terms of height (a more familiar concept, and the original one above), one could recognize that while one could define absolute altitude (distance to the Earth's center of mass), it is in most contexts more helpful to define elevation relative to something that's a little closer [e.g. the "natural" height of the terrain]. If e.g. a house is built in a flat area, one may describe the floor of the basement as having an elevation of -2 meters relative to the surrounding terrain. \$\endgroup\$
    – supercat
    Apr 17, 2013 at 15:57
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Voltage is a difference in potential. If I connect terminal A of a device to potential of 30 volts and terminal B to a potential 20 volts. The potential from A to B is 10 volts, but the potential from B to A is - 10 volts.

Think of it a tall building To go from floor 30 to floor 20 you go down 10 floors.

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oh why not ..

To say you have a voltage of +5V at point A means that point A is 5 volts more positive than your chosen ground.

To say you have a voltage of -5V at point B means that 'ground' is 5 volts more positive than point B.

The sign just tells you the polarity of the voltage, with respect to the ground node.

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The first part of this question has already been answered very well.

As for the second part, you could take the lowest supply terminal from the power supply and call that 0V, then every other voltage would be positive. However, that would be very inconvenient for many circuits. For example, common power supplies for an op-amp circuit are +12V and -12V. You could relabel the -12V output as "ground", then the old ground would be +12V and the old +12V would be +24V. Also, all your signals would be referenced to +12V and you would have to take that into account any time you measure things. Also, a lot of the power is used between the highest and middle output (in effect the charge in the top output originally came from the middle output and wants to go back there), same with the lowest output. All in all it is easier to label the middle output 0V (ground) and work with positive and negative voltages.

All this is ignoring earth grounding issues. In real life often the ground output on the supply is literally grounded, and thinking of the whole Earth as being at +12V would just be weird.

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Some OP-Amps, for example the veteran 741, require their supply as two voltages, one positive and the other negative with respect to the ground or zero level of the signal input and output.In this context it is natural to speak of the supplies being +15v and -15v (these are the values commonly quoted for the 741)

An alternating voltage, such as the a.c mains supply, swings positive and negative with respect to the neutral line, which is very close to the earth potential, so "neutral" is regarded as at zero voltage.

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Voltage between two physical points \$A\$ and \$B\$ is defined as

\$ V_{AB} = - \displaystyle \int\limits_{A}^{B} \vec{E}.d\vec{\ell}, \$

where \$\vec{E}\$ is the electric field on the path of the integration.

\$V_{AB}\$ becomes negative or positive (or just zero) according to this integration. For example, if you swap the points \$A\$ and \$B\$, the sign changes.

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If for example your power supply shows ground, 5v, -5v then your ground terminal is positive to the -5v, you get 5v from the ground that way. If you use the 5v and -5v terminals, then you will be using the -5v as the ground and you will get 10v from the 5v terminal If say there was also a 3v terminal, you would get 8v from the 3v terminal using the -5v as ground. Simple question, simple answer people. I probably know less about this than everyone else who commented here.

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Well, just to chuck in my two cents, let's say you have an ungrounded device. With +10 volts, you're expecting the current to come of the battery, through the widget, and then... where? It's only 10 volts, so arcing through the air to ground isn't really possible, but it could go through the case into the hand of the user, or the charge could simply stay on the far end of the widget. So now you have +10 volts on one side and +8 volts or something on the other, relative to ground. The wiget only sees a 2 volt difference though.

With a a +5V and a -5V the current is both pushed into the widget and pulled out of the widget.

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  • \$\begingroup\$ I like your approach, but if you only connect one side of a circuit to 10V and leave the other floating(both relative ground) you will find that the entire device should measure 10V, as you have just charged it to 10V, if it only went to 8V on the fare side you have a 2V drop and current is flowing. This would only happen if your measurement device was loading it. \$\endgroup\$
    – Kortuk
    Feb 17, 2011 at 23:10
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I'll try to explain it as I understand it...there is no +ve and -ve, its only a denotation of two directions. Voltage means Potential "difference". i.e, it is not a quantity, but a comparative value, in a direction. If you like a direction, then its +ve, if you don't like it then its -ve. An apple falls from the tree to the ground, both the ground and the branch of the tree exert a certain pull on that apple, but the ground exerts "X" amount more than the branch, the ground wins. Then Jupiter decides to take a cosmic stroll and climbs 2 feet above the tree, now the apple flies up. Actually there is no up and down, there never was. If the sight of an apple flying up distresses you, you say down is where earth is so towards the earth is the +ve "direction", and up is where Jupiter is so flying to Jupiter is -ve "direction".

In the same way there is no +ve and -ve in electrical engineering, its just a denotation of two directions (scalar not vector in the case of electronics), one we prefer, one we don't so much but still use it to our benefit.

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Power is not generated by electrons wandering around. In fact, an electron might wander all over the place with no voltage at all applied. Energy cannot be created or destroyed. It comes from someplace, and it goes someplace. For example, when you turn the crank of a generator, you are not really generating power; you are merely transferring the power that plants absorbed from the sun before you ate and digested them. You should market such a generator as "powered by thermonuclear fusion".

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    \$\begingroup\$ The question is "What is negative voltage?" This answer does not address the question. \$\endgroup\$
    – Transistor
    Jul 6, 2018 at 7:46
  • \$\begingroup\$ @Transistor Yes, and electrons don't wander all over, in fact they hardly even move when voltage is supplied. Its a common misconception that voltage application causes migration of electrons, but that is besides the point as you say. \$\endgroup\$ Apr 7 at 8:33
  • \$\begingroup\$ @DribbleNibble, the mobile charge carriers (electrons or positive ions) do move at a rate given by \$ i = \frac {dQ}{dt} \$. See Drift Velocity and the sample calculation for 1 ampere, and a wire of 2 mm diameter resulting in a drift velocity of \$ 2.3 \times 10^{-5} \ \text{m/s} \$. \$\endgroup\$
    – Transistor
    Apr 7 at 12:40
  • \$\begingroup\$ @Transistor as I said, "Hardly". At that velocity an electron would cover the distance of a tenth of milimeter at over four seconds. That is hardly the behavior of a sure headed migrant. Yes it is a current, but the problem with focusing on such a insignificant migration is that we start supposing this is the source of power. \$\endgroup\$ Apr 8 at 1:32

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