What is negative voltage? - Page 1
Forum section: electronics / beginners

author: electrolust
author_rank: Unknown
timestamp: July 25, 2017, 09:22:43 pm
content: For a split supply IC op amp, with V+ and V- terminals, what happens on the V- terminal? If voltage is negative, does it mean current is flowing towards the supply? Surely there are no self-powered elements within the op amp, so any current flowing to V- must be coming from V+?When amplifying an AC input signal, does the positive half of the output signal come from V+ and the negative half goes to V-?My dual power supply claims the 2 units can be connected in series (well of course they can), but also that the PSU can only source current and cannot sink it. But I am having no problem powering a pair of LM741's +/- 15V with output +/-4V offset at -10V.I have the op amps in non-inverting configuration, so I take it almost no current is flowing. But there must be some tiny amount.The manual goes on to say that each output has a diode to protect against reverse voltage, and reverse current must not exceed 3A continuous. Does that really mean that in fact the PSU can sink 3A?

author: WastelandTek
author_rank: Frequent Contributor
timestamp: July 25, 2017, 10:10:38 pm
content: voltages are entirely relativeour concept of "ground" is a completely arbitrary, if convenient, reference

author: jmelson
author_rank: Super Contributor
timestamp: July 25, 2017, 10:44:48 pm
content: Quote from: electrolust on July 25, 2017, 09:22:43 pmIt's all Ben Franklin's fault. He had to guess what the charge carrier was, with no way to figure it out. So, he could only have a 50% chance of guessing wrong, right?So, he labelled polarities positive and negative, and defined current as flowing from positive to negative. The standard charge carrier turns out to be the electron, and as Franklin defined things, it ends up having a negative charge. So, the plus terminal of a battery actually accepts electrons, and "current" flows in the opposite direction as the electrons are moving. Ahh, this is just to keep the neophytes away from all things electrical.By the time this was figured out (Crookes tubes and such) it was WAY too late to redefine everything.Jon

author: electrolust
author_rank: Unknown
timestamp: July 26, 2017, 01:18:39 am
content: mmm yeah I understand (I think) that voltage is a differential property. It is always in reference to some other potential. So does current always flow from higher potential to lower potential? Thus negative voltage on the op amp terminal means that any current flow is out of the op amp and towards the PSU. And then all the other questions I had.

author: WastelandTek
author_rank: Frequent Contributor
timestamp: July 26, 2017, 02:26:47 am
content: electrons spring from the pole we think of as lower and flow to the pole we think of as higher

author: TimNJ
author_rank: Super Contributor
timestamp: July 26, 2017, 03:04:13 am
content: Remember a voltage is a difference in electric potential between two points, with "difference" implying two quantities. When you define a node in your circuit as "ground" you are making one of those two quantities a constant. That is, when you make a voltage measurement, one of the two probes is always on the ground node. This simplifies your understanding of the circuit since all of your voltages are relative to that node.So, if you were to connect a resistor between a positive voltage and ground, the current would flow from the positive voltage and towards the ground node. If you were to connect a resistor between a negative voltage and ground, the current would flow away from the ground node and towards the negative voltage. At the same time, you could just as easily call the negative voltage "ground" and the "old ground" is now a positive voltage.

author: basinstreetdesign
author_rank: Frequent Contributor
timestamp: July 26, 2017, 03:24:13 am
content: Quote from: electrolust on July 25, 2017, 09:22:43 pmYES! (for conventional current flow NOT for electron current flow)Quote from: electrolust on July 25, 2017, 09:22:43 pmYES!Quote from: electrolust on July 25, 2017, 09:22:43 pmYES!Quote from: electrolust on July 25, 2017, 09:22:43 pmYes, there is a quiescent current flowing from V+ to V- pins through the op-amp. The maximum (and maybe the typical) values for this are quoted in the op-amps datasheet under Electrical characteristics/Supply current.Quote from: electrolust on July 25, 2017, 09:22:43 pmDont know without seeing the manual. If it really has protection diodes as you say then no reverse current should be possible beyond the diodes leakage current which would be a few microamps at most.Quote from: electrolust on July 26, 2017, 01:18:39 amYES! and YES! (at least if there is a path for it)

author: ejeffrey
author_rank: Super Contributor
timestamp: July 26, 2017, 03:38:42 am
content: Try not to think about electron flow. It is unnecessary and confusing at best, and downright misleading at worst. Electrons have their place in the description of the physics of electrochemical cells and semiconductors, but they are mostly irrelevant for circuit behavior. In the few instances where you actually care about the "charge carrier" such as the hall effect -- the answer is not what you expect. P-type semiconductors behave for all the world as if current is carried by positive charge carriers (aka holes), even though the only mobile "fundamental particles" in them are electrons.Likewise, fundamental rules of electrical circuits such as Kirchoff's laws don't actually work if you try to think of current as "an electron flow but reversed because ben franklin got the sign wrong". For instance, Kirchoff's Circuit Law is the conservation of current -- the current flowing into a node equals the current flowing out of a node. What if my node is the plate of a (air/vacuum gap) capacitor that is being charged? Charge flows into the plate but not out. So is KCL wrong? If you define current as charge movement, the answer is yes. But if you define current in terms of maxwells equation: charge movement plus dE/dt (aka displacement current), then the electric field in the capacitor gap is also counted as a current, and KCL works. So while you could of course use either definition, one is obviously more useful. This total current = charge motion + displacement current is also the thing that causes magnetic fields.What you actually care about are fields. This is why it is OK that the electron drift velocity is measured in mm/s, but signals propagate at the speed of light: more specifically, at the speed of light in the dielectric surrounding the conductors. That is because the fields are almost entirely in the dielectric, not the conductor.Voltage, or electric potential is just a convenient way of keeping track of the electric field. The electric field is a 3-D vector. That gets cumbersome to try to design circuits with. However, it has an interesting property. If you start at point A, and travel to point B, at each point adding up your "electric field headwind", the answer you get is the same no matter what route you take from A to B.. We called that summed electric headwind the potential difference between A and B. You can show that the potential difference also has another nice property: The potential between A and B can be described in terms of the potential to an arbitrary third point, "O" (for origin). The relationship is: V_AB = V_AO + V_OB = V_AO - V_BO. That means if I pick an arbitrary origin, and calculate all the potential differences between every possible point and the origin, I can calculate the potential difference between any two points. Since the origin is arbitrary and doesn't affect any of my potential differences, I can just ignore it, and pretend I have something called "the voltage at a point". If I change my origin point, I add or subtract a constant to all of my "voltages" -- but remember: the voltage is just a convenient way of keeping track of the electric field. Changing my origin or changing the zero offset of the my potential doesn't change the electric field at all, and that is what I actually care about. The mathematical way of explaining this is that the electric field is the gradient (a type of derivative) of the potential, so adding an arbitrary constant doesn't change the result.Tl;dr: circuit design is applied maxwell's equations, and maxwell's equations don't care about the sign of the charge carrier. Even when the charge carrier matters, the useful thing is not what you might expect, such as in P type semiconductors.

author: Richard Crowley
author_rank: Super Contributor
timestamp: July 26, 2017, 03:44:42 am
content: Quote from: WastelandTek on July 25, 2017, 10:10:38 pmRather like "0" elevation which we arbitrarily associate with Mean Sea Level.

author: electrolust
author_rank: Unknown
timestamp: July 26, 2017, 08:26:35 am
content: Quote from: ejeffrey on July 26, 2017, 03:38:42 amAh, so that is why a coax cable (or any transmission line I guess?) has a velocity factor?QuoteAh-ha, I think I get it now. V+/V- have a specific meaning to the op amp simply due to convention. And to get the max headroom for a symmetrical (amplitude-wise) AC signal, the origin point or ground reference should be mid-way between V+ and V-.As far as the PSU is concerned, current flows from positive terminal to negative terminal and the circuit consumes power in the process. The fact that the negative terminal of the PSU is "V-" and not "GND" makes no difference to the PSU. The reason my PSU that "cannot sink current" works to power an op amp is because in fact it isn't sinking current; my circuit isn't generating any "new" current on its own that the PSU has to sink.

author: JacobPilsen
author_rank: Regular Contributor
timestamp: July 26, 2017, 08:22:30 pm
content: https://www.eevblog.com/forum/beginners/what-the-heck-is-negative-voltages/

author: bson
author_rank: Unknown
timestamp: July 26, 2017, 09:00:00 pm
content: Quote from: electrolust on July 25, 2017, 09:22:43 pmTheIn your particular case the V+ terminal can only source. The common or ground terminal can both source and sink. A V- terminal would only be able to sink and hence is not a ground.MostIn general, supplies commonly have V+ and COM outputs. COM can both source and sink while V+ can sometimes only source, while in other supplies (SMUs in particular) can both source and sink. This means if you set one output to 5V and the other channel's output to 12V, and connect them together with a 1k resistor (with the two COM outputs closed), if the 5V terminal can sink a current of (12-5)/1k = 7mA will flowintoit. These are generally referred to asfour quadrant supplies. If it can't sink, say if it's a simple 7805, it will face 12V on its output and possibly be damaged, or it could have a protective circuit to shut it off if this happens.