transistors: so confusing!!

Here is the not exactly right approximate run through.

There are two PN junctions in a transistor, one is the emitter to base junction and one is the base to collector junction. Normally, the base to collector junction is reverse biased, to produce an insulating layer between the base and collector with no movable charges.

Lets pick a polarity... NPN.

So the collector has a positive voltage with respect to the base, so the doped in electrons in the collector N material are attracted away from the base and the holes in the base are attracted away from the collector, leaving just insulating silicon between them.

When the base emitter junction is slightly forward biased (emitter relatively more negative and base relative more positive), the doped in electrons in the emitter are repelled toward the base, and the holes doped into the base are repelled toward the emitter. At about a half volt forward bias, the holes and electrons begin to find each other and the electrons tend to jump into the holes and both effectively dissappear. However, a well made transistor has the emitter much more highly doped than the base, so more electrons get pushed into the base than holes get pushed into the emitter.

So the holes that get pushed into the emitter are anihilated very quickly, but the electrons that get pushed into the base have to hunt around a while beforo they dissappear.

The small positive base voltage causes these electrons to wander toward the base lead (the most positive voltage around them). But the base layer is very thin, and the electrons drift rather slowly in that direction. If the temperature was very low, this is about all that would happen, and the forward biased base emitter junction would have almost no effect on the collector curret.

But at normal ambient temperatures (well above absolute zero) the movement of the electrons is randomized by the thermal energy in the silicon, so they stagger quite randomly, with only a little progress toward the base lead. And since the base layer is so thin, most of them will never make it to the base lead. They will fall off the cliff into the highly stressed charge-empty reverse biased base collactor junction. There, instead of wandering in a drunken stagger through a very small electric field (volts per meter) they will whoosh out of the reverse biased junction, because it is much more highy stressed with e-field. They become collector current.

The more strongly you forward bias the base emitter junction, the more electrons are pushed into the base layer, and the more stagger over the cliff into the collector, though there will also be more that make it out the base lead. Over a wide range of collector current, the collector be a fairly fixed multiple of the base current. This is the transistor's current gain or beta.

So the electrons are drunks being encouraged with a slightly tilted sidewalk onto a slightly down hill, vibrating curb, next to street that tilts away from the curb, very steeply. Most never make it to the end of the curb, but fall onto the street where they slide into traffic.

This is the drunken bum on a crazy street transistor analogy.

I have the same problem to some degree.

I have two suggestions that might help though.

Learn the 3 common mode configurations and play around with them to get a good understanding of how to use them. You don't necessarily need to know the inner workings of a transistor if you want to use them(although it can't hurt). This will help you in recognizing the basic building blocks of larger circuits that use transistors.

Learn the basic models and apply them to pratical, but small circuits. This will/should help you in analyzing transistor circuits and also to make your own.

For me, the fundamental problem comes from not understanding completely the inner workings of a diode. A transistor, after all, is basicaly a diode that can be controlled(like a vacuum tude) by a current. So if you can't understand a diode then you can't really understand a transistor(atleast on the fundamental atomic level).

I do understand the motion of electron flow and holes in a semiconductor and also the junction as these are basic physics... but theres something more that I can't seem to grasp.

My problem is that I insist on viewing it in terms of electron flow instead of electrons and holes... I feel the hole idea is just a "trick"(even though its equivilent it should work without it). But when I try to think about it with just electrons I get the fact that electrons can move either way so the diode has no problems conducting one way or the other(which is true but its not symmetric). (its also true that the electrons will flow easily in both directions at some point).

I can't figure out why for low voltages the current flow is more in one direction than the other except that it has something to do with the junction and electrons on one side not having enough energy to get over the junction barrier(but they are able to get over it in the reverse direction easily for some reason).

Anyways, to use a diode you don't have to understand its inner workings either.

The idea with a transistor is that its like a pipe:

|

--------| | |--------- | | |

--------| | |--------- |

where the line inbetween is some "value" that controlls the flow from one side to the other(could be water, light, etc...).

the left and right sides are the N junctions and the "value" is the P junction(for NPN)... by attaching a voltage to the P junction you can remove its effects(because after all it will resist flow but if you can remove it then electrons will flow very easily).

If you think about it somewhat you can see that the P junction acts like a barrier(just like a diode in a diode) but that since you have an N on the other side the electrons have to flow through the P to get to it... by controlling the effective "size" of the P region you can control how many electrons flow(its like a variable resister to some degree).

Obviously if you had no P region between the two N's then electrons would flow unimpeaded. The larger the region(corresponding to a larger distance in a vacuum tube) makes it harder and harder for the electrons to get across(meaning you have to have a larger voltage). By biasing the P-N region properly you can weaken its effects and make it much easier for the electrons to get across it. Hence you can use it as an "amplifier" by putting a small varying single on the "valve" and having a large current flowing through it.

There are other ways to use the transistors too but thats the basic idea for the transistor amplifier. I think it mainly rests on understanding a diode as its just two diodes(but different dopings) stuck back to back.

It would be nice if someone who really understood this stuff(and not just that they think they do but actually do) would make some animations(or even some simulations) of the inner workings. (I've seen some on the net but they just tend to be crap and expect you to assume a lot of stuff without telling you why).

Jon

Try this animated description:

More info here too:

Keep trying until you find a description that clicks!

Dave :)

The animations I've seen are 'crap'.

By the way, a place where I'm unsure is the idea of mobility of holes as compared to the mobility of electrons in an N-doped material, for example. (Not a diode, not a transistor.)

I believe I understand enough of valence for both the lattice atoms as well as the dopant atoms, the lattice conduction band, the proximity of the dopant's valance to the lattice's conduction band and the ease with which room temperature thermal agitation can move n-dopant electrons into the lattice conduction band, etc. All this seems sensible.

But I see the mobility of both holes and electrons described as different in the same material -- for example, in lower dopant situations where the mobility is limited by lattice collisions, the holes are about 3 times "slower" than electrons in a silicon lattice. This is where I am unsure about the exact explanation.

Almost all the holes are dopant holes, where the thermal agitation is enough to overcome the very slight eV required (30-50 milli-eV?) When an electron in the conduction band does collide after being drift accelerated by an E-field, it will most likely collide with a lattice atom. The collision energy may be absorbed via an energy band transition of one of that atom's electrons or else converted into phonon energy in the lattice and energy in the electron as it leaves again in a random direction. I don't suppose this necessarily often creates another conduction band electron and a lattice hole atom, though. And the only way a dopant hole "moves" is with recombination events, where a conduction band electron is recaptured for a moment?

So is it correct to see that the reason for the difference between the speed of electron motion and hole motion is due to the difference in these mechanisms -- that of the mean free path and the E-field acceleration for electrons and that of the frequency of recombination events for hole motion?

Or?

Jon

Andrew,

You are not alone!

Try this, it is much closer to the truth although, as with most explanations, an over-simplification.

++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++

The base and emitter form a diode.

Put a forward voltage across that diode and electrons will flow out of the emitter into the base (i.e. in a NPN transistor).

Now the base region is narrow and the base/collector junction is biased so as to attract electrons to it.

So, on the way to the base terminal, more than 90% of them are kidnapped by the collector and never get there. It acts like a narrow pipe with a big hole in it.

More voltage on the base, more emitter current and so more of it available to be diverted.

++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++

There are two hang-ups that can cause confusion -

1). Current flow in the collector is caused by a smaller current flow in the base.

But what causes that base current in the first place? It is of course the base to emitter voltage.

Essentially then, a transistor is a voltage (not a current) operated device although when working out biasing, it is convenient to stay with the 'current' model.

2). The collector-base region is reverse-biased.

Yes it is, but only for electrons trying to flow in from the collector. For the electrons in the base trying to get into the collector it is forward-biased.

Cheers - Joe

Joe McElvenney wrote:

Dear friends,

Once again, the "VICE" issue has reared its head ("voltage-current/chicken-egg" or "voltage-current/cause-effect" to some folks). When motors, generators, transformers, LED's incandescent light bulbs, transistors, etc. are under examination, the VICE monster often shows up. In a bjt, the base current Ib, and the base-emitter voltage Vbe,

**mutually coexist**. They are interdependent, interactive, concurrent, mutual, simultaneous, inclusive, joined at the hip, etc. In other words one cannot exist without the other, but at the same time, one is not "caused" by the other. Vbe does NOT "cause" Ib, and vice-versa. Examine the I-V curve of any p-n junction, whether it be a juction diode, gate-cathode terminals of an SCR, base-emitter junction of a bjt, LED, etc., and it is immediately obvious that the curve passes through the origin and does not touch either axis elsewhere. In other words, diode current Id is zero only when diode voltage Vd is zero, and vice-versa. If one is non-zero, so is the other. In spite of this, some have insisted, since day one, that current is "caused" by voltage. If that were the case (it isn't), then every electrical device in the universe would be "voltage operated" and nothing would be "current operated". We wouldn't even have to bother with these terms. A bjt is a *charge* operated device. Charge must be injected into the forward biased b-e jcn, and the reverse biased c-b jcn. It takes energy, or work in order to move the charge. Voltage doesn't move the charge, energy does. Likewise with FETs. It takes energy, or work, to charge the gate to source terminals. With FETs, this charge and energy can be provided by either a low-impedance constant-voltage source, or a high impedance constant-current source. To keep it brief, a low-Z constant voltage source is better suited for driving a FET gate due to the high-Z of the gate. The v-source provides both Vgs and Ig, both necessary for FET operation. With a bjt, Ib and Vbe are also both needed for operation. A low-Z constant voltage source, or a high-Z constant-current source could provide both Vbe and Ib. Again, for brevity, a high-Z constant-current source is better suited for driving the b-e junction due to its low-Z characteristic and temperature dependency of current. The constant i-source provides both Ib and Vbe, both of which are absolutely indispensable. All electrical devices require BOTH I and V working together in tandem in order to function. Some devices, depending on their terminal characteristics regarding impedance, temperature, etc., are more amenable to being driven by a constant I source vs. a constant V source, or vice versa. The constant I source provides BOTH I and V, as does a constant V source. Have I explained myself? Best regards.

Claude

yeah. To me too... I think ;)

Do you mean holes in a p doped material as compared to electrons in a n doped material?

I'm not entirely sure but what I think is the free(conduction/valences) electrons have enough energy to "ride" the lattice in the sense that they will not collide with any atom(in the sense as far as electrons and atoms "collide").

If, say, one free electron "colides" with a bound electron then atleast one must stay and the other must leave. I suppose the electron either doesn't have enough energy to dislodge a bound electron, has enough only to exchange places(which, as far as were concerned doesn't do much), or has enough to dislodege a bound electron and not take its place(i.e. the avalanche effect).

The hole concept though is really just a mathematical analogy of how bound electrons move through the lattice. I would expect that conduction electrons and holes are different but I have nothing to back it up. I suppose these are the things you learn in an advanced class in semiconductor theory.

My guess is though that conduction electrons don't get captured in any real size... even if they do it is a very smooth transition and for all pratical purposes we can consider them as "free"(i.e., they don't take part in any interactions on the lattice). I'm not sure how true this is or how well it models the actual problem. I suppose it is determined by how strongly bound the valence electrons are to the atom and the probability of an electron getting "close" enough to an atom to be associated with it.

Now hole movement may actually be the exact same thing in the sense that in a p doped material an "absense" of an electron(which basicaly means a potential to bind the electron into its atom). The reason is we have to know how strong the "hole" is. When an electron moves through a p doped material depending on how stronge the holes affinity for the electron is it may be completely captured by the atom(ofcourse this doesn't happen completely but). If the holes affinity to capture the electron is very small then obviously the electron acts like a free electron similar to in n doped material.

heh, I have no idea. I doubt anyone really does and these are just models we use that seems to work. Its surely a much more complex situation than think about electrons floating around in a well defined lattice, etc...

I suppose its up to you to dig deep enough until you are satisfied with it.

I think what I said above basicaly boils down to saying that the difference in speed due to hole flow and that due to free electron flow is due to the hole affinity for the electrons. applying a potential across both types of materials should result in the same mean free velocity if the atoms were exactly the same(obviously). Hence the difference must come about from the doping. N doping, AFAIK, is in effect lowering the free electrons attraction from the lattice atoms... p doping will have the opposite effect though.

You might think about it like and see if it makes any sense. Think of an electron in an n-doped material moving along through space. As it approaches an impurity there is a small repulsion due to the electron cloud in the atom. We could graph the repulsion with time/distance as sorta a convex parabola with maximum corresponding to the atom's nucleus(basicaly).

The "holes" are opposite as when the electron approaches an impurity in a p-doped material there is an attraction and it gets stronger as the electron gets closer corresponding to a similar, but concave, parabola(ofcourse not a real parabola but just an approximation).

As the electrons bounce around in the p doped material there will be attractions toward the valence shells of the impurity atoms.... but ofcourse since the valence shell is full(or empty) this electron either cannot be captured or will have to boot out another electron.

I have no idea how to tell though as it is impossible to fine what an atom is and what it means to be "captured"(one would probably try and define in terms of distance but thats only a mathematical concept).

I don't know and don't remember enough about nuclear physics to know whats really going on though. You could bet that its you'll never arrive at the answer though ;/ It sounds like you got the basic idea as far as I'm concerned ;)

For me, the real issue, and the more problematic is at the junction of two oppositely doped materials. There seems to be some "strange" things happening there. There has to be some non-symmetric aspect of this junction IMO to give rise to its non-symmetric behavior. Most will use the hole analogy to get the result but if you replace the holes with its "corresponding" electron flow then it gets much more difficult as electrons can "easily" flow in both directions.

What I have come up with in trying to understand it is that one gets two charge density functions for the two biasing types.

One looks like this

/\\ /\\ | / \\---/ \\| --- / |\\ / \\ / | \\/ \\/

/\\ | /\\ ---/ \\| ---/ \\ \\ / \\ / \\/ \\/

where the | represents the junction and the "bumps" are the potential/charage distribution.

The first one corresponds to a P-N that is reversed biased and the second to one that is forward biased. As "pressure"(i.e., potential) is applied to the second the far edge potentials get pushed closer to the center and eventually will cancel out leaving a constant potential(approximately) which means electrons flow easily through.

The first will just tend to increase the overall potential as more and more voltage is put across it.

I'm not sure if this is the right interpretation but I basicaly got it as thinking about how electrons would distribute themselfs in the two different materials and by considering the junction potential.

The problem is that you can view it in two ways in that if you apply enough "reverse" bias then electron flow will become easy too(you end up pulling the two potentials apart so much that they become negligible...) this correseponds to breakdown of the diode.

So the problem is not so much about symmetric as it is symmetric to some degree but about magnituide. If you look at a diodes IV curn it is actually pretty damn symmetric except the reverse current is scaled differently.

The reason is that it might just be harder for "hole" flow than conduction flow and hence this makes it easier for electrons to flow in one way than the other... but this isn't right either as it breaks physical symmetry laws(flipping the material shouldn't change its behavior).

I'm still searching for what is really going on too ;/ Maybe one day it will make sense.

Jon

Why are transistors so extremely important? Because there are millions upon millions upon millions upon millions etc in use. The number is so huge its beyond counting. They are used in almost every electronic device most people encounter on a daily basis, throughout the day.

Regrading the above, the confusion may come from pondering "how it works" versus thinking about "what it does". How it works gets into physics - how material behaves at the atomic level. It's fascinating and interesting etc - but not needed to understand *what* a transistor does.

So, look at what a transistor does.

It can be used as a switch that turns things on or off. It can be used as an amplifier.

In both cases a small electrical source controls the transistor at its input. And in both cases, the transistor output controls a separate and much stronger electrical source.

Damn clever engineers create a wide variety of circuits that use these two functions of a transistor to do miraculous things - display your heartbeat on an EKG machine, guide rockets to the moon, control your microwave oven, create an ABS system to prevent brake lockup on your car, allow you to play games or surf the internet on your PC and on and on and on.

All based on a device that is controlled by a very weak source, which in turn controls a separate and much stronger source. Yes, it is important!

Ed

So your saying without a voltage(difference) the charge would still move?

Um.. Voltage is defined as the change in potential energy between two points needed to move a "test" charge in an electric field between two points.

Hence it has everything to do with motion. One can argue that there is no potential difference between two points unless an electron is moving.

If an electron is moving in an electric field then it is doing so because there is a potential difference.

there are three explicit points involved in the definition: Distance, charge, and the electric field. Since current measures the rate of change of w.r.t to time past some point and since the speed of charge in a conductor is approximately constant then current approximately measures the amount of charge passing through a point at any given time.

If we consider a uniform electric field, say between two points a distance d apart and electrons are flowing between the two points with a constant velocity then

V = -q*E*d and

I = a*q

hence

V = I*E*d/a

or if since E,d,a are all constants in this problem we see that V is proportional to I.

i.e., ohms law in an ideal resistor.

in a transistor its the same thing except E and q are changing due(and they are related).

But defintely there is a relationship and you can't have one without the other. Voltage and energy are synonymous in the sense that you can't really have a potential difference without moving a charge. When you measure a voltage with your voltmeter you are actually measuring current flow on an extremly small scale and for all pratical purposes you assume there is no current flow.

Its true though that you can't have voltage without current flow but the opposite is true too.

Yeah. I think you just need to understand that ultimately voltage and energy related. They are exactly related by

V = U/q where U is the change in potential energy and q is charge.

Otherwise everything else seems to make sense(which you happen to point). Just seems that you are implying in some cases that you can have current without a voltage.

I think ultimately current and voltage are one in the *same* but we view them from different perspectives. You can't have one without the other and there is a relationship between them. This relationship is sorta a transformation that depends on the physical constraints we impose on the electrons... they are related by an electric field and in some sense they the same manifestations of some singular phenomena. (sorta like the wave particle duality idea or special theory of relativity or even how the electric and magnetic fields are manifestations of the same thing(and are equivilent in some sense depending on your "perspective"))

Anyways,

Jon

(snip)

John,

I've seen quite a few descriptions of transistors, but this one has given me a better feel for what's going on than any of the others. (And I'm no slouch when it comes to math & physics.) Thank you.

Mark

(snip)

That's the kind of thinking that finally allowed me to understand photodiodes. Forget about electrons, holes, and bandgaps, and just think in terms of current and voltage characteristics. (Still working on my understanding of transistors.)

Mark

I'm glad it helped you. That makes the effort worthwhile. I like this story, because it can be extended to cover so many different cases.

It covers the case of collector to base low reverse and even forward bias. The street is only slightly tilted, so some of the bums stagger back up on the curb, or the street is a little higher than the curb, so bums from the other side show up on the curb.

It also covers base thinning (curb narrowing) from increased collector to base reverse voltage (Early effect). A narrower curb decreases the chance that any bum will make it to the end of the curb, so the current gain goes up slightly.

But mainly, the first part of the description illustrates that junction transistors are thermal devices. Without the random drift produced by heat, there is no collector current or current gain.

While I don't completely agree with Claude's post I can shed some light on one concept you took issue with; the idea of a charge moving with no applied voltage. It is indeed possible for a charge to move with no voltage. Superconductivity is one situation where charges move with no applied voltage. Granted, a voltage is initially required to get the charges moving but the voltage is not required to keep charges moving at a constant rate in a superconducting material.

One special case of this is an electron moving at some constant velocity in a vacuum. Voltage is initially required to accelerate the electron but momentum will carry it on at a constant velocity if no extraneous fields are allowed to act on the charge.

This may seem to be minor point to some, but physical concepts such as momentum, inertia and energy are just as important in electronics as they are in mechanical systems.

Dorian

I suspect this is not so good phrasing, as it might imply a dU divided by some finite q. In which case, the result would be a dV, not a V.

I prefer to think of V, especially when using the word "change" in phrasing, as the infinitesimal form: V = dU/dq (joules/coulomb) or else just dU = V*dq = dV*q. Otherwise, if U and q are both finite, they represent some meaningful average finite value or the differences between two meaningful finite values, which is too often less "exact" in my mind.

Jon

I trust you feel better now :-)

Cheers - Joe

"Dorian McIntire" wrote in message news:T_KGf.10427$ snipped-for-privacy@bignews3.bellsouth.net...

Technically if there is an electric field then there is charge moving then there is work being done. Obviously there are electrons "moving" all the time without a "field"(well, theres always a field though) but we don't consider a voltage moving them. The reason, Ig uess, is that there is no mean free movement and/or the voltages are extremely small as to be inconsequential.

About the SC material. The problem with that is that one cannot do any work with it(hence no voltage) else one has a perpetual motion device. If you try to measure the current flowing the electrons will stop flowing(or decrease depending on how much work was used to get them to move). Its similar to the uncertainty principle. So we can only talk about it in a theoretical way.

Yes, but you are forgetting that voltage is basicaly the work/energy supplied or given up by the moving charge. The electron has energy(you said momentum) so technically it has a voltage. Voltage is an abstract definition as is an electric field. As far as nature is concerned theres no such thing as a E field or voltage. So you say that when we apply a voltage to accelerate an electron and then remove that voltage it is gone... but it is not. We have just transfered it to the electron(as far as you can transfer voltage(really transfering energy)).

I think the problem is that you are thinking of voltage in macroscopic terms then applying it to the microscopic. (sorta like wave-particle duality). I.e., if you could just look at one electron at the microscopic level then you wouldn't be able to see a "voltage" directly. If you were watching the electron and you saw it move then you could hypothesize that there was a voltage applied to it by some electric field. But you wouldn't be able to really "see" the voltage.

Basicaly these are mathematical concepts applied to natural so we have to be careful what we really been by these ideas. Although I can see what you mean when we "apply" a voltage then stop "applying" it but you have to ask yourself "What does it really mean to "apply" a voltage?". Then you might realize that a voltage is really looking at the boundry conditions of some even... i.e., measuring the energy at one point in time then at another point in time and if they are different then something must have happened(i.e., a "voltage").

For an analogy I could create a "field" that discribes how people move. We could think of people as point charges and then plot there velecity vectors. We would see areas in the field that look like there are "paths"(like a highway with all the cars going in one direction or in an elevator, etc...)... We could then hypothesize there is some force that makes the "people" go in that common direction. In some locations it would be disorganized or random and we could say there is no field there. We could come up with some way to measure this "field" and create a whole theory around it. In actuality there is probably not a field though and it is just a "tool"/"concept" used to help us understand why the people seem to be moving as they do. We could say "These people over here are experiencing a force that makes them move along this path" and "These people here have some forces that are holding them together"(such as at a football game), etc...

Hell, who knows ;) Basicaly the definition of voltage is independent of what happens between two measurements so we can't really talk about what is going between them(just like a particle as voltage is a measurement on particles)... If we do talk about them we end up arriving at the problems we do.

True, but they are not well defined physical terms. They are mathematical concepts applied to reality. Any time you start to dive into this stuff all kinds of little problems start creeping up and then you have to start using the quantum mechanical definition of these terms to get around these problems... but then even stranger things start happening. I do agree that it helps and is usually a good idea but you gotta be careful in how you use them too. Our knowledge about reality is very limited and is only decent approximation. We could actually be way off and just by happenstance happen to have a model that approximates it well. That is, our model's axioms could be completely wrong in that the "real" model has totally different axioms but it turns out that the end results(large theorems) happen to agree with each other. I have a feeling this is what is going on and our concepts of reality is way off(but in some sense parallel to reality). This is why we have so many little problems in the sciences today that don't make much sense and it seems that we keep on having to modify them to get it to work and introduce things that make less and less sense. I doubt thats how reality works but it could be. My evidence is it seems everyone we find a problem in nuclear physics it seems we have to create/find a new particle to make everything right again... It could be this way or we are just "forcing" it to work.

Anyways,

Jon

But thats what it is depending on how you look at it ;)

It doesn't matter to much if you use a dV or V as long as you are consistant.

Why? in either case you arrive at the same result and also V is really a difference in itself(but on a global scale). They are really relative terms and we tend to use concepts like dV -> ^v(delta V) -> V -> limit V->oo to move through the different "resolutions".

I do see what you mean. Normally it should be written xV = xU/q where x represents the "resolution". But since we normally take V to have 0 reference at infinity it doesn't matter in the end result.

Cause with your logic one could even say that q should be dq and we must write dV = dU/dq but it depends on how we are treaing q. i.e., if q = int(dq). if dq is not changing with position or time then we would normally write q but if we are setting up an equation to handle varying q we have to write dq and integrate).

I guess what I mean is that dV = dU/dq implies that we are dealing with infintesimal quantities that will be used as part of a larger picture(i.e. to integrate over). But if we are only dealing with "point charges" then we can think of dq as = q and write a "simplified" notion of dV = dU/q. Now since U and V does not depend on the path taken we can write it as Va - Vb = (Ua - Ub)/q which in the standard notation delta(V) = delta(U)/q. But if our reference point for V is taken to be Vb = 0 then we have either Va = delta(U)/q or Va = U/q depending on how we look at it. delta(U) means change in energy and U means change in energy w.r.t to an infinit point.

So you end up getting the same result since Va = Va - Voo. i.e., the potential difference ALWAYS represents a "change" as does the potential energy. So while notationally it might be a bit confusioning in general we take a "short" cut to have to write one or two less characters as long as we know that we mean that they represent differences else they make no sense.

so you can write it as dV = dU/q, V = dU/q delta(V) = delta(U)/q or V = delta(U)/q. If you want to be precise depending on the problem then ofcourse you have to choose the appropriate notation. (i.e., if you are dealing writing a differential equation for the problem to find out is macroscopic characteristics then you need to use the dV = dU/dq version).

heh. I guess what I'm saying is that its true if we use dU then we need to use dV. In actuallity the real difference is simply that its always dV (or delta(U) and delta(V)) because there is no such thing as V... but we interpret V to be V - V_oo and make V_oo = 0 so we are left with just V.... so even though if we use delta(V) it is always equal to V with the above criteria.

So mathematically you are correct but since we have that little criterion it makes it *ok* for basic use.

i.e., when you measure voltage you are actually measuring between two points... but we almost always use gnd as an implicit reference so we can talk about the potential at a point(w.r.t to ground implied). I don't see it as being technically wrong if everyone is in agreement about the implicit reference used.

yeah, but if V = dU/dq then mathematically it makes no sense if V is implied to be a real number. dU/dq are not real numbers(or even complex) so we can't get a real number from there ratio. Hell, differentials themselfs are not really mathematically sound anyways and really are just short cut(as most mathematics is anyways) to writing difference equations and then taking the limit. As long as you are consistant in what you mean then it should be ok... after all, they are just symbols anyways and they could mean anything you want them to. (Ofcourse mathematicians have defined standard meanings for them so everyone can communicate easily)

Jon

No, I couldn't logically say that for a physical system of finite functions of U and q.

Jon

What do you mean by finite functions? Discrete functions? bounded functions?

If you mean that q can only take on discrete values(and even U) then you can say it pretty easy. Differentials exist for all for generalized functions which can be used to sorta make discrete functions act like continuous functions(I mean that mathematically you can use the same notation and theorems).

i.e, say you have 10 point charges you can make this function into a "normal" function(I mean one that behaves in mathematically similar way to normal functions that you see every day) by using dirac delta functions. You can then integrate and differentiate this function just like "normal" functions and arrive at valid results just like normal functions.

You have to remember that this is mainly all notation and depending on what concept you attach to the notation could make it "mean" something completely different.

If you want to know what I mean just look up stieltjes integrals and it has a pretty clear idea of what I mean. Basicaly lets you integrate over a discontinuous "differential"(notationally anyways).

Ultimately in the standard definition derivatives you are right but physicists always try and bend the rules to make things easier to work with so one has all kinds of extentions of notations and theorems that ordinarly would seem odd to most people.

for example, check out fractional differentiation/calculus. You will seed a notation that looks like an ordinary derivative but instead of the "nth" derivative its something like the "rth" derivative where r is not an integer. You could call it an abuse of notation but really its just an extension made by someone that fits logically into the frame work in most cases and sometimes can even be helpful.

Jon