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You can find the position of Fermi level in two cases: $$n_0=n_i\exp\left[\frac{E_F-E_{Fi}}{k_BT}\right]$$ $$p_0=n_i\exp\left[-\frac{E_F-E_{Fi}}{k_BT}\right]$$ Rearranging the two relations, We get: $$\rightarrow E_F-E_{Fi}=k_BT\ln\left(\frac{n_0}{n_i}\right)$$ $$\rightarrow E_{Fi}-E_{F}=k_BT\ln\left(\frac{p_0}{n_i}\right)$$ For an $n$-type semiconductor, $... 1 Then how does one say they constitute a current if they just diffuse by any means to p side Movement of charge is electric current. This is moving charge, therefore it is current. ANY movement of charge is current. If you put a bunch of electrons on a baseball and threw it across the field, that's an electric current. Its not a very constant current, but it ... 0 A current source is a complex (and probably non-existent) power supply with internal resistance which can adjust its terminal voltage (or resistance) so that the current flowing through it is always at a predetermined level. 2 Yes. The name is “current source.” Since you say the video already mentions this, I am not entirely certain what you’re looking for here. 0 Potentiometer wire has high resistance relative to what? If it was meant that relative to conducting wires, it must be high then thats because, in a potentiometer you just connect a wire to a voltage source$V$. A low resitance wire provides no load and would drive huge currents without a limiting resistance in the circuit. wont the current skip ... 1 Potentiometer wire has high resistance so that by turning the control knob to "add" more wire in series, the measured resistance increases by a convenient amount. If the "pot" were wound instead with copper wire, which has very low resistance, it would take a huge increase in the number of wraps in the pot to get a measurable increase in ... 0 Dynamic resistance is defined as the differentiation of the voltage with respect to the current in a VI curve at the operating point. It is an interesting phenomenon that the dynamic resistance is the resistance that the AC voltage sees, since dynamic resistance,$r=\frac{\Delta V}{\Delta I}\$ and the AC voltage is the voltage that is changing. To speak more ...

3

To enlarge slightly upon DKNguyen's answer: Indeed, the thing that causes air to break down is the strength of the electric field in the gap between the switch terminals, as measured in volts per meter. For small values of the gap, even a small number of volts will produce a strong field. Furthermore, if there are any sharp points or asperities on the ...

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I assume you're asking about a mechanical switch and not a solid-state switch. Yes, there is a spark when you flip a mechanical switch closed because doesn't just close cleanly. It closes, then bounces open, then closed, then bounces open, etc. until it finally stays closed. The voltage producing the arc is not the voltage from the supply. It is the voltage ...

2

A block of semiconductor has some concentration of mobile charge carriers (electrons and holes) and their charge is exactly cancelled out by fixed charges in the form of ionized dopants (donor and acceptor ions) The depletion region is, unsurprisingly, depleted of mobile charge carriers (electrons and holes). When depleted, what remains are the fixed ionized ...

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There are different ways to control power input to an electric device. One is, indeed, to input stabilized voltage using a source with very small output resistance. Regardless of the load resistance (within engineering limits), the voltage on the output terminals of such a source will be constant. But sometimes it's useful to know how a device performs when ...

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The current in a diode depends roughly exponentially on the voltage. So the only real difference from one to the other is the distribution of the measurement points. If you do a voltage sweep at equidistant points, you will be sampling the current in exponential intervals. If you sample in equal distances in current, you will be taking samples at logarithmic ...

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Actually you are right, the most easy way is to (carefully) increase voltage. However, the problem is that above the threshold voltage the characteristic curve rises up very steeply. Meaning if you are not super carefully you can end up with very large currents that destroy your LED or laser diode. This is why usually LEDs and lasers schould be operated &...

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First, note that your circuit diagram shows the flow of conventional current which can be very confusing to a student trying to understand the operation of a diode, never mind a transistor. Reversing the arrows of your diagram to show electron flow shows more clearly the physics of what's going on: Electrons flow from the connected negative battery terminal ...

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Issue#1. There are various aspects to consider here, let me skip for a moment. Issue#2. Ok, I think I see what your instructor means but it might be indeed a bit confusing. First of all: outside the depleted region at the EB and BC junctions, in the non-depleted portion of the base (I don't see it in you sketch but it is there, at least in normal conditions)...

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You won't run out of electrons and holes. There are two main processes through which you will get more. Thermal energy will naturally generate electron/hole pairs in your semiconductor. That's where they come from in a semiconductor at thermal equilibrium. However, this process is slower than we would like for an LED, or any diode. Fortunately, we can help ...

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The emitter is very heavily doped, and the base is very, very thin and very lightly doped. The collector is moderately doped. Typically the emitter is the most heavily doped, the collector is the most lightly doped, and the base is somewhere in the middle. current doesn't move by means of recombination of electrons and holes! Current moves by means of flow ...

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No, the current won't stop flowing as the battery maintains constant electromotive force. Some free electrons will lose their kinetic energy and become gridlocked. You can view this phenomenon, that is, free electrons losing their kinetic energy, as free electrons losing kinetic energy to a resistance, say a light bulb.

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Yes they do. And as the energy of electron and holes are different the recombination leads to emission of light of various colors depending on the band gap of the material of the diode. After recombination the site where the pair was created becomes neutral but fresh electron -holes pair get created and the process continues.

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Please note, I'm no electrical engineer, but a similar question was posted here: https://electronics.stackexchange.com/questions/349677/what-happened-to-depletion-region-when-biasing-is-off-after-applying-reverse-bia. Basically, the diode behaves like a capacitor. If you isolate it from a circuit after extending the depletion zone by applying a reversed ...

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