# Tag Info

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You might find the Yahoo "home_transistor" group a useful resource. There's also a series of videos on YouTube by Jeri Ellsworth including some where she makes transistors. In one, in particular, she takes the crystal out of a germanium point-contact diode and turns the crystal into a point-contact transistor (much like the Bell Labs transistor.) There ...

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In experimental physics it is required to use electronics as instruments. You must know how they work(amplifiers, ADC's, MCA's etc) in order to fully understand and design an experiment. Usually, you don't need too much electronics(filters, amplifiers, transistors, digital electronics-boolean algebra) is more often than not, more than enough. You need ...

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The $A/W$ units refer to the current (in Ampère) produced per Watt of light incident on the photodiode. This current-production happens when the diode operates in the so-called photoconductive mode. Since your question wasn't on the inner workings of a photodiode, I won't expand on this, but Wikipedia contains some more information if desired.

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The key difference between a Zener diode and a normal diode is that the Zener diode has a low breakdown voltage - typically in the few volts range. The breakdown voltage is low because the heavy doping means the depletion layer is very thin, and even at a low voltage the field strength over this thin depletion layer is very high. With a conventional diode ...

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It's surprisingly difficult to find a nice simple description of how a transistor works. This description is from my old physics book - I suspect this may be oversimplified and I'm sure a complete description would run to lots of equations! Anyhow, this is what an NPN transistor looks like: so as you say, the collector-base junction is reverse biased and ...

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The answer is that the whole circuit is full of electrons. I think you may be thinking along the lines of "if I switch a tap on, the water takes time $v/L$ to reach the end of a hose of length $L$. So, if I switch a light on, the electrons must take analogous time the reach the light". Because the circuit is full of electrons, the energy source shoves the ...

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Calling it a built-in voltage is something of a misnomer. People usually think of "voltage" as "what you measure with a voltmeter". So "voltage" is normally synonymous with "electrochemical potential of electrons" (in stat mech terminology) and with "difference in fermi level" (in semiconductor terminology). Under this definition, the built-in "voltage" is ...

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For a given circuit in a given technology, power increases at a rate proportional to $f^3$ or worse. You can see by looking at the graph in @Martin Thompson's answer that power is superlinear in frequency. $P=c V^2 f + P_S$ is correct, but only superficially so because $f$ and $P_S$ are functions of $V$ and $V_{th}$ (the threshold voltage.) In practice ...

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You've correctly deduced how the circuit works. This particular configuration is better known as a bridge rectifier and is often packaged as a single component containing 4 diodes. There are two uses for this - rectifying alternating current as depicted in your question, and creating circuits that can handle direct current with reversed polarity (for ...

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The complete explanation takes a few lectures - it is simply impossible to provide this amount of information as an answer. Very general explanation: Let's take a look at NMOS transistor (the one shown in the schematic attached to the question). It has 4 pins which you can force potentials on: Gate Bulk Source Drain In order to understand how the ...

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Your first question- Why aren't electrons being attracted by the positive charge region? Any free charge will move in response to an electric field created by some charge distribution. So it's important to see the electric field in the region. Well, the first thing you should do is find out the where the electric fields exist and where they don't. Electric ...

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The flash of blue light was probably due to ionisation of the air by the ionising radiation emitted as the experiment went briefly critical. It's no different from the blue light emitted from air in a lightening strike or the Aurora Borealis, so it can be photographed in the same way. Note that it is different from Cerenkov radiation, which is normally ...

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Any delicate equipment, including cameras, should be safe behind a thick wall of lead or other heavy shield. What, it can't see through the shield? Then point the camera sideways, and use a mirror to see around the edge of the shield. Light reflects, like you'd expect. Most of the obnoxious high energy radiation will pass through the mirror, or perhaps ...

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Very roughly there are two different processes at work: diffusion and drift. Drift is the motion of the charged particles induced by a field (either the applied voltage or the field in the depletion region.) Diffusion is a function of the width of the depletion region. Drift is not. Drift is a function of the number of minority charge carriers near the ...

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I have a couple courses in the atomic nature of analog and digital electronics next year. I'm in Engineering Physics, so I'm not exactly sure which side that confirms, though the courses are in the physics department. Eng Phys has its own courses as well, so if the physics department has these courses then I assume physics majors also have some use for ...

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The fundamental reason for this is that light is quantized. If it wasn't for this, it would in theory be possible to scale down any camera to an arbitraily small size. However, as a camera is scaled down, less light will enter it and therefore fewer photons. As the number of photons hitting a pixel goes down, the shot noise goes up (the signal-to-noise-ratio ...

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A pn junction is one piece of a semiconductor that receives n-type doping in one section and p-type doping in an adjacent section. If you simply stick two p-type and n-type semiconductors to each other by hand, it will not behave as a diode. The main reason that a pn junction can behave as a one-directional device is it's built-in potential. Upon formation ...

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I believe that the authors, of the reference you provided, explain the reason behind introducing these so-called quasi-Fermi levels at the end of section 4.3.3. For simplicity, let me just repeat it here; perhaps explaining it in different words, with a little more elaboration, would help. I’m sure you're aware of the fact that in $n$- ($p$-) doped ...

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After reading the wikipedia page and this article, I understand your question a little better. A CMOS sensor detects colors by letting light pass through three layers of detector, each with a certain spectral response and absorption. I don't think that each layer is specifically red, blue, and green, rather that their responses overlap and that the processor ...

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You can work through the derivation, but I think you are after a more intuitive answer to the question. Here is the way I think about it. Why exp()? You will know that the I-V curve a resistor is $V=IR$. That is to say that the when you put a voltage across a resistor the current is linearly related to the voltage simply through a constant of ...

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I do believe that yes. The most hard part will be to obtain the materials. If you manage to get a good piece of n-type (or p-type) Silicon, big enough to allow you to work with home tools, you'll "just" have to do local oxidation (with temperature for example) and make some soldering. Of course, the quality of that transistor would be very doubtful as it ...

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Since you mention batteries in a comment: The voltage source + resistance model works well in many circumstances. At the extremes: high frequency / fast rise time waveforms: There will always be some inductance, from wiring if nothing else, which appears as an inductance, an easy add to the model. low frequency / long durations: As the battery ...

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There are actually two slightly different versions of Thevenin's theorem. I think what you are describing is the weaker of the two: you can replace any circuit with a single voltage/current source and a single resistor. That version holds for any two-terminal network made up only of voltage/current sources and ohmic resistors. It fails as soon as you add ...

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if there was no central tapping, what would be the change? Consider this circuit diagram, from the answer you're not satisfied with: There is no "central tapping" so the two diodes are connected in series. In a series connected circuit, the current through each circuit element is identical. Now, note that D1 allows only a clockwise current while D2 ...

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Reverse Bias of P-N junction When the voltage is applied this way round it tends to pull the free electrons and holes apart, and increases the height of the energy barrier between the two sides of the diode. As a result it is almost impossible for any electrons or holes to cross the depletion zone and the diode current produced is virtually zero. A few lucky ...

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FORWARD BIAS OF A P-N JUNCTION As the electrons move towards the positive terminal and the holes towards the negative, they will come to the depletion layer. This is a very narrow layer around the junction (i.e. around the interface of the two semiconductors.) In the depletion layer, electrons and holes can recombine, but the recombination rate is not high ...

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Conductivity of intrinsic semiconductor is due to their own internal charge carriers. The bonding between between two electrons of two neighboring atoms is covalent, therefor at NTP, there is no free charge carrier for conduction. When it is heated, some covalent bonds break due to heat and thus some electron get free for conduction. As soon as one electron ...

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This isn't a great answer since all the papers I could find are behind paywalls, but ferric oxide is a semiconductor and depending on it's exact stochiometry can be n type or p type. If you know someone in academia with access have a look at: http://jes.ecsdl.org/content/126/3/419.abstract http://jes.ecsdl.org/content/131/8/1777.abstract

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This is very easy to understand why centeraltap transformer is needed in a full wave rectifier. Let us assume that we have a simple transformer, and there are two diodes and the central wire coming out from the transformer is not present there which is obvious since we are not using centeraltap transformer. So now see the figure In first case let A be at ...

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