If you need assistance, please send an email to forum at 4hv dot org. To ensure your email is not marked as spam, please include the phrase "4hv help" in the subject line. You can also find assistance via IRC, at irc.shadowworld.net, room #hvcomm.
Support 4hv.org!
Donate:
4hv.org is hosted on a dedicated server. Unfortunately, this server costs and we rely on the help of site members to keep 4hv.org running. Please consider donating. We will place your name on the thanks list and you'll be helping to keep 4hv.org alive and free for everyone. Members whose names appear in red bold have donated recently. Green bold denotes those who have recently donated to keep the server carbon neutral.
Special Thanks To:
Aaron Holmes
Aaron Wheeler
Adam Horden
Alan Scrimgeour
Andre
Andrew Haynes
Anonymous000
asabase
Austin Weil
barney
Barry
Bert Hickman
Bill Kukowski
Blitzorn
Brandon Paradelas
Bruce Bowling
BubeeMike
Byong Park
Cesiumsponge
Chris F.
Chris Hooper
Corey Worthington
Derek Woodroffe
Dalus
Dan Strother
Daniel Davis
Daniel Uhrenholt
datasheetarchive
Dave Billington
Dave Marshall
David F.
Dennis Rogers
drelectrix
Dr. John Gudenas
Dr. Spark
E.TexasTesla
eastvoltresearch
Eirik Taylor
Erik Dyakov
Erlend^SE
Finn Hammer
Firebug24k
GalliumMan
Gary Peterson
George Slade
GhostNull
Gordon Mcknight
Graham Armitage
Grant
GreySoul
Henry H
IamSmooth
In memory of Leo Powning
Jacob Cash
James Howells
James Pawson
Jeff Greenfield
Jeff Thomas
Jesse Frost
Jim Mitchell
jlr134
Joe Mastroianni
John Forcina
John Oberg
John Willcutt
Jon Newcomb
klugesmith
Leslie Wright
Lutz Hoffman
Mads Barnkob
Martin King
Mats Karlsson
Matt Gibson
Matthew Guidry
mbd
Michael D'Angelo
Mikkel
mileswaldron
mister_rf
Neil Foster
Nick de Smith
Nick Soroka
nicklenorp
Nik
Norman Stanley
Patrick Coleman
Paul Brodie
Paul Jordan
Paul Montgomery
Ped
Peter Krogen
Peter Terren
PhilGood
Richard Feldman
Robert Bush
Royce Bailey
Scott Fusare
Scott Newman
smiffy
Stella
Steven Busic
Steve Conner
Steve Jones
Steve Ward
Sulaiman
Thomas Coyle
Thomas A. Wallace
Thomas W
Timo
Torch
Ulf Jonsson
vasil
Vaxian
vladi mazzilli
wastehl
Weston
William Kim
William N.
William Stehl
Wesley Venis
The aforementioned have contributed financially to the continuing triumph of 4hv.org. They are deserving of my most heartfelt thanks.
Registered Member #3414
Joined: Sun Nov 14 2010, 05:05PM
Location: UK
Posts: 4245
Steve Conner wrote ...
Can anyone explain along these lines why the mobility of holes is different to the mobility of electrons? Surely if a hole is just an "absence of electron", then moving one hole is the same as moving one electron, so it ought to have the same mobility.
But in practice the mobility of holes is less, and P-channel MOSFETs, where the current is carried by holes, never perform quite as well as their N-channel counterparts.
I've been thinking about this for most of the day, and doing a bit of relevant googling, and I'm rapidly coming to a conclusion something along these lines:
In an NPN device, where the P region is connected to the base, once you establish a suitable PD, there are plenty of electrons available to move, whereas in a PNP device, where there are less available sites for electrons in the base region, once you establish the PD, there are much fewer electrons available to move, assuming both packages are the same size, therefore PNP devices don't work as well as NPN devices. They have fewer available electrons in the base region at any one time, and therefore can't conduct as much current as an NPN device can.
BigBad wrote ...
Personally I think there's only one electron, and it fills the universe with a quantum mechanical wave.
I agree, but it's sometimes more convenient, depending on what you are observing, to quantify it as 'lots of little electrons', each with their own centre of mass and centre of charge, although they are all nodes of the same 'universal negative charge'.
Registered Member #1792
Joined: Fri Oct 31 2008, 08:12PM
Location: University of California
Posts: 527
Steve Conner wrote ...
Can anyone explain along these lines why the mobility of holes is different to the mobility of electrons? Surely if a hole is just an "absence of electron", then moving one hole is the same as moving one electron, so it ought to have the same mobility.
But in practice the mobility of holes is less, and P-channel MOSFETs, where the current is carried by holes, never perform quite as well as their N-channel counterparts.
It's not that holes have lower mobility, it's that mobility in the valence band is lower than in the conduction band. Mathematically the electrons (and holes) in a crystal are quantum-mechanically restricted to specific states on an energy-momentum (E-k) surface, and the curvature of this surface versus momentum tends to be smaller (flatter) for the valence band than the conduction band. That means for a particular value of energy a hole in the valence band will need more momentum to exist at that energy (because that's the only available state). If it has more momentum but the same energy it is heavier (effective mass, not real mass because electrons/holes in a crystal are not free particles) and a larger effective mass means that it has a lower mobility.
I do recall that solving for a band diagram in a 1D crystal with the Kronig Penney model gives a solution which in the long run follows the parabolic shape which you'd expect for a free electron (E=E0+p^2/(2m), where momentum p is h_bar*k in quantum) interrupted by the various bandgaps, so it seems reasonable that in general the curvature of the band minima/maxima at the bandgaps increases with energy since the overall shape of the E vs k curve is steeper at that point and so higher bands will generally have lower effective masses. *
*I am not a solid state physicist but this sounds halfway reasonable to me which probably means it's way too simple of an explanation
Registered Member #1792
Joined: Fri Oct 31 2008, 08:12PM
Location: University of California
Posts: 527
Ash Small wrote ...
In an NPN device, where the P region is connected to the base, once you establish a suitable PD, there are plenty of electrons available to move, whereas in a PNP device, where there are less available sites for electrons in the base region, once you establish the PD, there are much fewer electrons available to move, assuming both packages are the same size, therefore PNP devices don't work as well as NPN devices. They have fewer available electrons in the base region at any one time, and therefore can't conduct as much current as an NPN device can.
PNP transistors actually have more electrons in the base than NPN transistors because the base is doped n-type. This increases the number of valence band electrons, and that also reduces the number of valence band holes (which increases the number of valence band electrons).
But conduction in a PNP transistor is determined by holes which are injected by the emitter into the base and diffuse across, so the less favorable performance comes from the lower hole mobility. You could come up with a PNP transistor which was the exact analogue of an NPN, where the doping levels were the same but the opposite type of dopant, and so the majority and minority carrier concentrations were the same in each section, but the PNP would still be worse because of the lower hole mobility.
Registered Member #3414
Joined: Sun Nov 14 2010, 05:05PM
Location: UK
Posts: 4245
Mattski wrote ...
PNP transistors actually have more electrons in the base than NPN transistors because the base is doped n-type. This increases the number of valence band electrons, and that also reduces the number of valence band holes (which increases the number of valence band electrons).
The general impression I got from the reading I did yesterday (and I may be mistaken) is that the P type doping results in holes being created, in which mobile electrons can 'sit' until the voltage is sufficient for them to become mobile, and that N type doping results in electrons being 'bonded' in place, allowing more ' mobile holes' to sit until the voltage is sufficient for them to move. The impression that I got was that, while the N type contains more electrons, fewer are free to move once the voltage is sufficient to allow them to do so. I'll have to go back and re-read.
I think the general consensus here is that the holes don't move, it's just convenient for us to quantify it that way sometimes. I could supply a couple of quotes from earlier in the thread to support this if required.
Registered Member #3414
Joined: Sun Nov 14 2010, 05:05PM
Location: UK
Posts: 4245
I've done some more reading this evening, and I'm still saying there are fewer 'mobile' electrons in the base region of a PNP type device. It's also true to say, I believe, that fewer electrons are entering the base region from the collector. I believe the electrons find it more difficult to cross from the positive doped collector to the negative doped base, compared to electrons in an NPN device, which find it relatively easier to cross from the negative doped emitter to the positive doped base.
The 'forward voltage drop' of the device is the voltage required to get the electrons in the base to become mobile, so I assume there is very little base current at voltages below the 'foreward voltage drop'.
Because of the negative doping in the base of a PNP device, there are not as many 'mobile electrons' present, less 'mobile electrons, gives the 'appearance' of 'more holes'. There appears to be more holes because there are fewer electrons.
This is a simplistic explanation of why PNP devices don't perform as well as their NPN counterparts.
I can see why a lot of people prefer the 'hole' theory, though, but this does explain why 'holes are less mobile'.
There is another fundamental difference between the two devices, and that is that electrons enter the base in a PNP device, and exit the base in an NPN device. I've not fully worked through the implications of this yet, but it's likely to affect the respective efficiencies in some way. I'll work on that later.
Anyway, back to winding one of the inductors for the circuit above. The core is 9cm^2 N27 material, because it's what I had to hand, wire is ~1.5mm^2, I think, just normal 7 strand house-wire from CEF, with 75 turns on the first layer, second layer is half done. I think I'll get around five layers on, so maybe 300+ turns altogether. I still can't measure any significant resistance with my DMM, but not tried my AVO yet. It's a real killer winding a core of this size by hand, though. It gets the wrists and thumbs, mostly.
Registered Member #1792
Joined: Fri Oct 31 2008, 08:12PM
Location: University of California
Posts: 527
Let's take this bit by bit. I'm going to recommend also my preferred online reference materials for semiconductor physics since you may benefit from these if you're not already reading them: (Good reference, skips over some areas but has a ton of useful info) (The famous Britney Spears Guide to Semiconductor Physics, actually it has good info)
wrote ... The general impression I got from the reading I did yesterday (and I may be mistaken) is that the P type doping results in holes being created, in which mobile electrons can 'sit' until the voltage is sufficient for them to become mobile,
I don't think your understanding is correct yet. Mobile electrons are electrons in the conduction band and mobile holes are holes in the valence band. P type dopants take up electrons from the valence band, which creates holes in the valence band. The electrons on the P-type dopant sites are now fixed, immobile electrons. If you want to think of valence electrons instead of holes then what you can say is that the p-type dopant traps some of the electrons from the valence band which relieves the valence band electron gridlock and allows more of the electrons to move. For every hole there is one electron which can now move in the valence band. There is no minimum voltage for the holes/electrons to move, as soon as a field is applied they will move.
wrote ...
and that N type doping results in electrons being 'bonded' in place, allowing more ' mobile holes' to sit until the voltage is sufficient for them to move. The impression that I got was that, while the N type contains more electrons, fewer are free to move once the voltage is sufficient to allow them to do so. I'll have to go back and re-read.
N-type doping is dopants that have extra electrons which are easily excited to the conduction band. So the number of mobile conduction band electrons electrons are increased. Fixed holes are left behind on the dopant site leading to a fixed positive charge to balance the extra mobile electron charge.
Under any applied field the electrons and holes will move. You can think about valence electrons instead of holes if you really want to, but either way a homogeneous doped semiconductor never has a minimum voltage for the carriers to move.
wrote ... I think the general consensus here is that the holes don't move, it's just convenient for us to quantify it that way sometimes. I could supply a couple of quotes from earlier in the thread to support this if required.
All physics is just a convenient way of quantifying observed phenomena Electrons don't really even behave as discrete particles in semiconductors, it just so happens that we can pretend that they do by coming up with an "effective mass" which works in many but not all situations. Since treating the electron as a moving particle is a convenient fiction it's no major stretch to extend the same courtesy to holes.
For your understanding of PNP and NPN transistors I think you need to make sure you understand band diagrams and PN junctions first. The forward voltage of the base-emitter junction doesn't come from a minimum voltage required to make electrons mobile, it comes from the potential barrier which naturally forms when a P-type region is next to an n-type region.
Registered Member #3414
Joined: Sun Nov 14 2010, 05:05PM
Location: UK
Posts: 4245
Thanks for the links, Mattski, I'll look at them later. And thanks for having the patience and perseverence to continue replying.
What I understand so far goes something like this:
The 'base current' consists of those electrons that undergo 're-combination' with the 'valence holes'. The electrons that get across the base region without re-combining form the collector current.
I expect you are aware of the 'model' of a transistor that consists of two opposing diodes in series, depending on whether the diodes point towards each other or away from each other it's a model of an NPN or a PNP device. The base can be represented by the junction between the two diodes, although in a 'real' transistor you wouldn't have a wire between the two halves of the 'base region'. This model represents a 'symmetrical planar' transistor, I think, which has unity gain (Hfe=1), ie only the base current flows, there is no collector current (these are used, among other purposes, as electrostatic protection in uchips).
In order for a transistor to amplify, or have a 'gain function', it needs to be asymetrical. In an NPN device, the surface area of the junction between the collector and base is much larger than the surface area of the junction between the base and emitter, the ratio of these two surface areas being proportional to the 'gain function'. It is this asymmetry that enables the 'overshoot' into the collector region of electrons before they re-combine. The electrons that 'overshoot' into the collector region form the collector current, the electrons that re-combine form the base current. PNP devices are the reverse, in that the emmitter is larger than the collector.
Due to the above, two diodes can be used to model the base current flowing in a transistor. In fact, the base current can be modelled using the diode that forms the representation of the base-emitter junction of an NPN device, and by the diode that represents the base-collector junction in a PNP device, so you can use a diode to model the base current of a transistor.
If you then connect this NPN 'model' to a variable voltage source, and gradually increase the voltage to the 'foreward voltage' of the device (model), and current starts to flow, you've modelled the mechanism by which the transistor 'switches on'. Now, are you 'injecting holes' into the 'base region' of the model, or just 'applying a voltage across the junction'?.....I'd suggest you are increasing the energy of the electons sufficiently for them to move away from the junction itself, thereby creating 'holes' (or is 'leaving' holes more accurate?) that are filled by electrons migrating from the 'emitter region of the 'model', and that, as long as you maintain the voltage above the 'foreward voltage' of the device, the electrons in the 'base region' will continue moving away from the junction, allowing more electrons to migrate from the emitter region into the base region.
EDIT: I've started reading the first link above, and the first statement I've noticed is this "It is important to understand that one could deal with only electrons if one is willing to keep track of all the electrons in the "almost-full" valence band. After all, electrons are the only real particles available in a semiconductor."
EDIT: This diagram looks like it's proved my point:
It shows that, under high injection conditions, there are very few electrons in the few microns adjacent to the junction, this rises to a peak at ~10 microns, and then drops as a result of diffusion through the N region.
Unfortunately, the article does not seem to show a corresponding diagram of electron density under low injection conditions.
Does anyone have a link to an article that shows diagrams of electron density near the junction in these two cases?
Registered Member #1792
Joined: Fri Oct 31 2008, 08:12PM
Location: University of California
Posts: 527
wrote ... The 'base current' consists of those electrons that undergo 're-combination' with the 'valence holes'. The electrons that get across the base region without re-combining form the collector current.
Correct
wrote ...
I expect you are aware of the 'model' of a transistor that consists of two opposing diodes in series, depending on whether the diodes point towards each other or away from each other it's a model of an NPN or a PNP device. The base can be represented by the junction between the two diodes, although in a 'real' transistor you wouldn't have a wire between the two halves of the 'base region'. This model represents a 'symmetrical planar' transistor, I think, which has unity gain (Hfe=1), ie only the base current flows, there is no collector current (these are used, among other purposes, as electrostatic protection in uchips).
This transistor can have current gain. The model of two diodes in anti-series can probably describe some of the operation of a BJT but definitely not all, especially not forward active with current gain.
wrote ...
In order for a transistor to amplify, or have a 'gain function', it needs to be asymetrical. In an NPN device, the surface area of the junction between the collector and base is much larger than the surface area of the junction between the base and emitter, the ratio of these two surface areas being proportional to the 'gain function'. It is this asymmetry that enables the 'overshoot' into the collector region of electrons before they re-combine. The electrons that 'overshoot' into the collector region form the collector current, the electrons that re-combine form the base current. PNP devices are the reverse, in that the emmitter is larger than the collector.
Assymetry is not required for current gain, and I don't know if transistors are even made with asymmetric base-collector and base-emitter junctions. Overshoot is not the right way to visualize it. What happens when the base emitter is forward biased it increases the concentration of conduction band (CB) electrons at the edge of the depletion region in the base. Naturally the p-type base has very few CB electrons. On the collector side this is reverse biased, so any electron reaching the collector is swept away by the high electric field in the depletion region. So on the emitter side of the base you have an excess of electrons which were provided by the emitter, and on the collector side of the base you have (essentially) zero electrons. So by random thermal motion the electrons diffuse across because of the concentration gradient, same as perfume molecules will diffuse a across a room to areas of less concentration. The base is made very short so that very few of the electrons will recombine with holes.
The "forward voltage" iss a function of the PN junction barrier height which forms between P and N regions because of the charged depletion region.
wrote ... If you then connect this NPN 'model' to a variable voltage source, and gradually increase the voltage to the 'foreward voltage' of the device (model), and current starts to flow, you've modelled the mechanism by which the transistor 'switches on'. Now, are you 'injecting holes' into the 'base region' of the model, or just 'applying a voltage across the junction'?.....I'd suggest you are increasing the energy of the electons sufficiently for them to move away from the junction itself, thereby creating 'holes' (or is 'leaving' holes more accurate?) that are filled by electrons migrating from the 'emitter region of the 'model', and that, as long as you maintain the voltage above the 'foreward voltage' of the device, the electrons in the 'base region' will continue moving away from the junction, allowing more electrons to migrate from the emitter region into the base region.
There are some problems with this theory. The base already has plenty of holes at zero bias because its P-doped. The electrons in the emitter (which there are plenty of) stay in the emitter because there is a potential energy barrier, until this barrier is lowered by the applied Vbe. These electrons are injected into the conduction band of the base (which is initially empty).
wrote ... I've started reading the first link above, and the first statement I've noticed is this "It is important to understand that one could deal with only electrons I've started reading the first link above, and the first statement I've noticed is this "It is important to understand that one could deal with only electrons if one is willing to keep track of all the electrons in the "almost-full" valence band. After all, electrons are the only real particles available in a semiconductor.". After all, electrons are the only real particles available in a semiconductor."
Key part is the "if one is willing to keep track of all the electrons in the "almost-full" valence band".... you're free to try this but people use holes because it makes things easier and produces physically correct results ;)
I'd recommend you start going back to the basics and reviewing PN junctions (diffusion of carriers leading to a carrier depleted region and potential barrier, bias lowering this barrier and allowing current to flow, how the carrier concentrations look at zero bias and under bias). Once you understand PN junctions well then BJT's are a lot more straightforward.
Registered Member #3414
Joined: Sun Nov 14 2010, 05:05PM
Location: UK
Posts: 4245
The two quotes below are from here:
"(2) Forward bias (P positive with respect to N) narrows the depletion region and lowers the barrier to carrier injection. The diffusion component of the current greatly increases and the drift component decreases. In this case the net current is rightward in the figure of the p–n junction. The carrier density is large (it varies exponentially with the applied bias voltage), making the junction conductive and allowing a large forward current.[3] The mathematical description of the current is provided by the Shockley diode equation. The low current conducted under reverse bias and the large current under forward bias is an example of rectification."
"Electric field in depletion layer and band bending[edit]
Associated with the depletion layer is an effect known as band bending. This occurs because the electric field in the depletion layer varies linearly in space from its (maximum) value at the gate to zero at the edge of the depletion width:[7]
where A is the gate area, = 8.854×10−12 F/m, F is the farad and m is the meter. This linearly-varying electric field leads to an electrical potential that varies quadratically in space. The energy levels, or energy bands, bend in response to this potential."
Foreward biasing 'reduces' the size of the depletion region. This means it increases the 'voltage gradient'. I'm still of the opinion the mechanism here is that electrons are removed from the depletion region in the P region (base), leading to the depletion region shrinking and having a higher voltage gradient. When voltage is increased to the 'foreward voltage', the depletion region is so small, and the voltage gradient so large that electrons diffuse to the base from the emitter. (injecting holes is another way to 'look at' removing electrons.)
Registered Member #1792
Joined: Fri Oct 31 2008, 08:12PM
Location: University of California
Posts: 527
Ash Small wrote ...
Foreward biasing 'reduces' the size of the depletion region. This means it increases the 'voltage gradient'. I'm still of the opinion the mechanism here is that electrons are removed from the depletion region in the P region (base), leading to the depletion region shrinking and having a higher voltage gradient. When voltage is increased to the 'foreward voltage', the depletion region is so small, and the voltage gradient so large that electrons diffuse to the base from the emitter. (injecting holes is another way to 'look at' removing electrons.)
What is it that I'm missing here?
The voltage gradient is simply the electric field (E=-grad(V)). Forward bias _reduces_ the electric field in the depletion region. But it so happens that very little current in a forward-biased base-emitter diode is due to carriers moved by an electric field (the drift current), because the field is such that it's trying to move electrons from the base to the emitter and holes from the emitter to the base, and each of these carriers is in very short supply in those regions. Under forward bias most of the current is diffusion which increases as the electrostatic barrier is lowered. The drift current is the -Is in the diode equation, which is also the reverse bias current.
Electrons diffuse from the N-type emitter to the P-type base because there are tons of electrons in the emitter and very few in the base. Holes diffuse from the P-type base to the emitter again because the base has tons of holes.
(I'm also wondering if sometimes when you talk about electrons you're referring to electrons in the valence band instead of holes. I don't recommend this since it's nonstandard and you'll confuse people, much like you'd confuse people if you used electron potential and electron current instead of voltage and standard current in circuit analysis.)
This site is powered by e107, which is released under the GNU GPL License. All work on this site, except where otherwise noted, is licensed under a Creative Commons Attribution-ShareAlike 2.5 License. By submitting any information to this site, you agree that anything submitted will be so licensed. Please read our Disclaimer and Policies page for information on your rights and responsibilities regarding this site.
Driving a led with mains.
