What's the difference between Negative and Ground? Like a negative terminal and a copper ground? I was reading that they are both the same; one person stated that negative is just more negative than ground. What's the difference? Do I even need ground if I have a negative terminal?


Strictly speaking, ground is a connection to the Earth. When we speak about a "safety ground", it is this kind truly earthed connection.

In circuits, the "common return" path to the power supply is informally called "ground", even though it is not actually earthed. Battery-powered devices, and the electronics inside airplanes, still have grounds.

All circuits have to have return paths to the power supply. Ground is only a convention. The designer decides that a particular network connected to the power supply is at a 0V potential. Other voltages are measured with respect to this potential. But, of course electrons do not read schematics and do not understand what ground is, and what point is 0V. Moreover, conventional current is opposite to the flow of electrons: Benny Franklin got it wrong. So when the negative terminal is grounded, electrons are not actually returning to ground, but emanating from there.

If the power supply is a battery, and the devices in the circuit are designed for a positive supply, then the negative terminal serves as the ground. The positive terminal could also be ground, if all the circuits are designed for a negative supply.

"Designed for a positive supply" means that the key voltages in the circuit are positive, measured from the negative return. For instance, many NPN transistor circuits, such as an emitter follower, have a positive sense. The collector is connected to a positive supply, and the emitter branch of the circuit is on a network which is goes to a negative ground. The input signal is understood to be between ground and the transistor base.

However, the same circuit could be built using a PNP transistor instead, using a negative power supply, whose positive terminal serves as ground.

There also exists, in wide use, circuits that run on dual voltage power supplies. The power supply provides a positive voltage, a negative voltage, and a ground. The two voltages are usually approximately equal (but opposite) so the ground is nearly exactly in between them.

  • \$\begingroup\$ An example of NPN circuits with a positive ground is ECL logic, where the collectors were connected to the positive terminal as you say, but that terminal was labelled 0V and was the ground reference. The emitters carried the logic levels, with resistors pulling them down to -5.5V. Emitter followers are extremely fast (no Miller capacitance) so ECL was the way supercomputers were built for at least a decade... \$\endgroup\$ – Brian Drummond Dec 17 '12 at 11:02

Ground is a reference and as such can both source and sink current while maintaining it's reference value. Power supplies often can only either source (for a positive rail) or sink (for a negative rail). You can build power supplies that act as virtual grounds. Additionally your circuit may behave correctly using a sinking negative rail. So the correct answer is - it depends, more information is needed.


Ground is simply referring to a common reference charge level across the circuit. It is often most negative than the other charge levels and therefore often connected to the negative terminal of a battery, but it doesn't have to be this way. You can easily have circuits that have ground (i.e. 0V), positive and negative voltages.

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    \$\begingroup\$ Ground doesn't refer to a reference charge, but to a reference potential. Otherwise, nice answer! \$\endgroup\$ – The Photon Dec 17 '12 at 4:04

In any circuit there can only ever be one point which is truly at 0V ( the - terminal). Since the ground plane has a finite non-zero resistance, any current flowing through it will develop some voltage across it.

For example, consider a ground plane that has a resistance of 10mOhm/inch and points A, B, C, -terminal D in line, with currents Ia (20mA), Ib (10mA), & Ic (10mA).

Vc = (Ia + Ib + Ic) * 10mOhm = 400uV

Vb = Vc + (Ia + Ib) * 10mOhm = 700uV

Va = Vb + (Ia) * 10mOhm = 900uV

So it looks like the further you get away from the negative terminal in terms of current paths crossed, the greater the voltage drop at that point on the plane.

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    \$\begingroup\$ 0 V is not always the "-" terminal. \$\endgroup\$ – The Photon Dec 17 '12 at 4:03
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    \$\begingroup\$ ah, yes... touché \$\endgroup\$ – alxcpa01101 Dec 17 '12 at 4:44

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