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Actually the holes are in a lower energy band than the electrons. And if you could just join a p-type and n-type material then probably what would happen is the electrons would diffuse across to the P-side and fall into the holes. Similarly the holes would diffuse to the N-side and recombine with electrons. Then the process stops, or rather reaches a dynamic equilibrium, because the uncompensated dopants have a charge which creates an electric field, which creates a drift current to balance out the diffusion. In the process it forms the built-in barrier.
That is pretty much the way I understand it. I'm wondering, why you put this into subjunctive. One thing I want to add is, that the depletion region formed is not completely empty of charge carriers due to thermal statistics, which keeps some electrons above the lowest energy states. AFAIK the Shockley law describes the upper tail of the electron and hole (thermal) energy distribution, i.e. the quantity of charge carriers in the depletion region. These carry the current.
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Joined: Fri Oct 31 2008, 08:12PM
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Heh, I had to look up "subjunctive" since I haven't studied that since Spanish class in high school. The reason I said "if you could" is because when you fabricate a PN junction you aren't literally slapping a piece of P and N type material together, so it's more of a thought problem. But it does turn out that some people are working on wafer bonding methods where they may literally slap two semiconductors together (carefully), so it's not necessarily so far from reality.
But the usual ways of making PN junctions aren't so different. The junctions are formed by diffusing or implanting dopants, or growing epitaxially, so you can imagine as dopants diffuse inward the depletion region moves deeper and deeper as the charge profile changes. Or as the n-type regions are grown on top of p-type the electrons begin diffusing over to the p-side.
True, the depletion region is not empty. In fact one of my books introduces it as the "transition region" which is empty of charge in the "depletion region approximation". But the concentration tails off exponentially with the band bending so for the purposes of electrostatics it's usually valid to treat it as empty, but it's those few charges in the region which carry all of the current so that charge is very important.
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Joined: Sun Nov 14 2010, 05:05PM
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Mattski wrote ...
Good points Ash
wrote ...
I also agree that as much energy is expended (and heat created) by trying to pull holes out of the crystal lattice as energy is expended (and heat it created) trying to energise the electrons to move.
The phrasing is a little weird here so I'm going to clarify although this may already be your understanding. When holes move, they are not pulled out of the lattice, nor do you need to pull an electron out to create a hole. Holes are formed by other methods (dopants, thermal energy, light) and then they move under an electric field. Also I feel like "energizing" an electron to move is implying that it needs some minimum energy to move, like a forward voltage, but it does not.
Well yes, I'll clarify. The electrons are certainly mobile, that's why it's the electrons that move, they just lack 'direction' before the potential (voltage) is applied. They are just moving randomly (brownian motion?), but the equilibrium is maintained. Once the potential is applied, they start migrating.
But the potential is initially across the power supply. This potential across the power supply, I would suggest, exerts forces on the electrons either side of the power supply, which start moving. This movement of electrons, or 'current flow' then causes a PD to develop across the emitter-base junction within the transistor, and once enough current has flowed for the PD to increase above the 'foreward voltage' of the device, the 'conduction band' electrons on the emitter start to flow through the base to the collector, except those that undergo 're-combination', and replace the 'valence band' electrons removed previously by the base current. As long as sufficient base current flows to maintain the potential across the base-emitter junction above the foreward voltage, collector current will continue to flow (ignoring 'switch off' times. The reason you have a 'switch on' time is that current has to start flowing, and 'valence band' electrons have to be removed from the base in order to 'create' the potential at the junction.
This supports the 'widely held' opinion that it's the base current that 'drives' the transistor, and as long as base current is flowing, and valence band electrons are being removed from the base at a sufficient rate, conduction band electrons will continue to flow as collector current.
I've been puzzling over this for the best part of four decades, about WHY base current is DIFFERENT to collector current. I knew it had to be something simple like base current is valence band electrons and collector current is conduction band electrons, but I'd never been able to put it into words..... I have my answer now, pretty much.
Mattski wrote ...
wrote ... The only 'effective' mechanism you have to 'create' the 'potential gradient', or whatever you want to call it is by 'moving electrons' (same as in a capacitor, for example). I'm not trying to challenge traditional models that have proved 'easy to understand' and 'produce credible results', I'm just trying to understand it in terms that I find easy to relate to.
True, only the motion of electrons can create an electric field in a semiconductor (unless you do something like shoot positive ions at a semiconductor). But the positive charge of the nuclei usually does play a role in the electrostatics even though they are immobile. Holes are a strange concept which take some getting used to, but it's a good way to do analysis. You could get the same results without holes but you'd get weird things like conductivity of p-type regions being inversely proportional to the electron concentration.
Probably quite a lot of energy is needed to pull a nuclei out, so much that things are already going very wrong in your semiconductor device :)
I understand how the nuclei plays a part, and about the electro-static forces being 'equal and opposite', but if you start trying to pull a nuclei out of a lattice, the lattice is just going to 'expel' valence band electrons in the opposite direction, because it takes a lot less energy to do so.
I agree that the 'hole model' is nice and simple and easy to balance (and works), but it's never satasfactorally explained the fundamental difference between base current and collector current, and that one is conducted in the valence band and one in the conduction band. It's genarally accepted as defining 'transistor theory', but it doesn't explain why there are 'two different currents', or what they are, at least, not in a way that made much sense to me.
Now, back to winding those inductors.
EDIT: I've deliberately not included any maths whatsoever in my argument. This isn't supposed to be 'an alternative mathematical model', "hole theory" is the 'mathematical model'.
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Ok, here's the larest simulation. I delayed posting while I checked everything as best I could.
It requires using 'supercaps' across the LED, but it effectively eliminates 'significant' spikes from surges.
The final circuit doesn't need to be as extreme as this, I can reduce values a lot, but I now undestand how it works. The current is a bit low too, 10mA, but I can increase that easily without affecting the basic circuit operation. I'll post again when I'm sober
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OK, now you're starting to get the hang of this transient suppression thing. Add I(C3) to the things you are tracing, and simulate with different phases of mains switch-on.
When you've done that, delete C1, make C2 a more sensible value, put a just large enough resistor in series with C3, so that its time constant with C2 gives you the LED protection, and double C3 to get the 20mA you want. Simulate again.
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Dr. Slack wrote ...
OK, now you're starting to get the hang of this transient suppression thing. Add I(C3) to the things you are tracing, and simulate with different phases of mains switch-on.
When you've done that, delete C1, make C2 a more sensible value, put a just large enough resistor in series with C3, so that its time constant with C2 gives you the LED protection, and double C3 to get the 20mA you want. Simulate again.
Ok, Here are the traces for the first part of what you suggest, ie I(C3).
The first case is the 'transient' case, where it switches on mid-cycle, with 250V across C3, the second is the 'steady state' case. Even in the 'steady state' case it takes nearly half a second to settle.
You can see that, after it settles, the current through C3 = current through the LED, Any 'spikes' will not have any 'significant' effect on current through the LED. Current through C3 = current through LED, which suggests, to me, that it's pretty efficient.
I still say that adding a resistor in series with C3, although it deals with transients, just wastes power in the 'steady state', and should therefore be avoided.
(I still plan to 'play around' with the values, but I want to wait until the inductors are finished, and I can input their actual valeas, then work from there.)
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Steve Conner wrote ...
You have two inductors with a value of just "1". Is that uH or H? What do you think? What does LTSpice think?
It's 1H (one Henry).
I've been doing some reading on various 'radio ham' forums, and winding inductors of several Henries is considered to be quite a common practice, especially in the 'Audiophile' world.
I've not done the maths to determine what 300 odd turns on a 9cm^2 N27 core works out at, I thought I'd just 'wind it and see', using what I had to hand as a starting point.
I'm sure lower values would work, but one idea I have is to adjust the inductance and capacitance to give a resonant frequency of 50Hz. (just one idea I'm looking into.)
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Your peak transient current only reaches 100mA. I don't see any components in the circuit to limit it to that small value. Unless as Steve hints, those might be 1H rather than 1uH inductors. <calc, calc> resonance for 2H with 330nF = 160Hz, which is 8ish cycles in 50mS <\calc, calc>. I see 8 cycles of ringing in the first 50mS, so I think Steve's worst fears are grounded.
So, for transient protection, what do you think is best? Two 1H inductors, at 1kg+ each, free if you rip them out of old flourescent fittings, or a 100 ohm resistor, which dissipates 40mW at 20mA LED current, so costs 1/3rd of a kWhour per year in excess dissipation?
<edit> oops, caught in the door there with Ash
Oh well, so it doesn't look like a double post of mine -
wrote ... one idea I have is to adjust the inductance and capacitance to give a resonant frequency of 50Hz.
what do you get if you have a series connection of an L and a C, which are resonant at 50Hz, in series with a low impedance load connected to 50Hz mains? No need to simulate this one, just wait until November 5th* and try it for real.
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This is fast turning into a "Worst LED driver ever" contest!
I have an entry of my own, a 2 watt Joule Thief with current regulation by saturable reactor. It worked, but it ended up less efficient than connecting the LED straight to the battery with a ballast resistor.
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Driving a led with mains.
