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The device is either extremely hot ioniZing the air right in front of the lens or the air inside the car is ionized.


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But I wanted to know whether we can charge a capacitor while it is in use If, by "while it is in use", you mean while the capacitor is discharging, i.e., energy is flowing out of the capacitor to some load, then the answer is no since, by definition, if a capacitor is charging, energy is flowing into the capacitor. Put another way, a capacitor cannot ...


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There are no known transistors or oscillators yet produced that can handle 10^19 frequancies, we are still experimenting at terahertz range 10^12 or thereabouts so the simple answer is no.


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The answer suggesting "Art of Electronics" is spot on -- no argument. However, it is also spot on expensive. An alternative is Practical Electronics For Inventors which is now in its 4th edition and an excellent low priced book that allows you to move through the material more quickly. The scope of coverage for "Practical Electronics For Inventors" is ...


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Art of Electronics, now in its 3rd edition by Horowitz and Hill has always been a classic. Comprehensive and easy to read with an emphasis on practice rather than deep theory. I am a professional electronics engineer and I have used it (I transitioned from physics) for decades


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One way to think about it is similar to a chemical equilibrium. Electron-hole pairs are spontaneously generated every now and then from random thermal fluctuations, and when an electron collides with a hole they annihilate with each other (some fraction of the time). The frequency with which electrons and holes collide is $np$. In steady-state, this ...


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"Why is $np$ always equal to $n_i^2$ ?" Well, first of all, the easy way to answer your question "Why is $np$ always equal to $n_i^2$ ?" would be simply to notice that $np$ is independent of the Fermi level $E_F$, and thus independent on the fact that the semiconductor is doped or not. In the case of a non-degenerate semiconductor (i.e. when $E_F$ is ...


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Electrons and holes occupy their states according to the Fermi-Dirac distribution, which has a single parameter $E_f$, the Fermi level (assume a fixed temperature). Provided $E_f$ is in within the band gap and far from the band edges, the (energy integral of) Fermi-Dirac takes an exponential form $\propto e^{E_f}$ for electrons and $\propto e^{-E_f}$ for ...



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