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31

An alternative way to generate random numbers, that truly is quantum, and also quite easy: put a small radioactive source near a Geiger counter. Radioactive decay is a truly random event in the quantum sense, and is basically not subject to thermal noise at all. For maximum visual impact, replace the Geiger counter with a cloud chamber. That way you can ...


16

Corresponding wavelength is 22.11 meters long, but we want also to emit our EM waves into the environment. This means if we get a nice half-wave dipole antenna we would need it about 11 meters in length, $\lambda/2$. Which is quite large for mobile device. Ok, lets reduce size by using quarter-wave antenna as in WiFi, based on ideas of quarter-wave ...


11

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 ...


11

If I understood correctly, what you are trying to build is a hardware based random number generator, where you want to use some quantum mechanics-based mechanism to supply the randomness. I'm no experimentalist, thus, take my comments with a grain of salt. Your suggestion is to use Schottky noise from a illuminated photodiode. I believe that it's a pretty ...


10

To add to Nathaniel's Answer because (1) it is a good answer and (2) I get nervous recommending radioactive materials handling to anybody I don't know: I would really think about the cloud chamber idea, especially since you're a software guy with a math background. It would need to be inside a darkened container, but you could run a webcam to show what is ...


9

I'll give the answer to this question using an unusual method that showed up in the American Mathematical Monthly's problem section perhaps in the late 1970s. This is not necessarily the easy way to solve the problem, but it works out nicely from an algebraic point of view. The way most people solve most resistance problems is to use series and parallel ...


9

I found a general, qualitative answer in David Blackstock's book Physical Acoustics, on page 46: Impedance is often described as the ratio of a "push" variable $q_p$ (such as voltage or pressure) to a corresponding "flow" variable $q_f$ (such as current or particle velocity). I also received a nice answer to this question on another Q&A site ...


9

What makes it a good idea to use RMS rather than peak values The rms value, not the peak value, is the equivalent DC value that gives the same average power. Recall that power is the product of voltage and current: $p(t) = v(t) \cdot i(t)$ For a resistor, we have: $p(t) = R[i(t)]^2$ To find the average power, we must take the time average of both ...


7

I'll take it step by step here. First I'll write the answer for the first few cases with circuit analysis. Then I'll apply a reduction to show the pattern that the problem arrives at. N=1 $$Z = R+R=2R$$ N=2 $$Z = R+\frac{1}{\frac{1}{R}+\frac{1}{R + R}} = R \left( 1+\frac{1}{1+\frac{1}{1 + 1}} \right)=\frac{5}{3} R$$ N=3 $$Z = ...


7

Dims is almost correct, in that you would only see resonant tunnelling effects at very low temperatures. In other words, at very low temperatures the electrons will sit at very well defined energy levels within the transistors. Under certain biases (voltages), the energy levels on either side of the thin barriers between devices will line up, and electrons ...


6

Capacitors, as used in electric circuits, do not store electric charge. When we say a capacitor is charged, we mean energy is stored in the capacitor and, in fact, energy storage is one application of capacitors. Now, for an ideal capacitor in a circuit context, the current through is proportional to the rate of change of the voltage across: $$i_C = C ...


6

For any given $n$, you can work it out via the rules for series and parallel resistors, but to get a general formula, valid for all $n$, doesn't look easy to me. The best way I know of is to get a recursive relationship giving the resistance of an $n$-step ladder in terms of an $(n-1)$-step ladder. If I'm not mistaken, the $n$-step ladder can be thought of ...


6

As suggested by Manishearth, one can perform a $Y$-$\Delta$ transform from $Y$-resistances $R_1$, $R_2$ and $R_3$, to $\Delta$-conductances $G_1$, $G_2$ and $G_3$ (using a $123$ symmetric labeling convention), cf. Fig.1 below. A x----x------x-----[3]-----x------x----x B | | | | [4] [2] [1] [5] | ...


6

Use a star-delta transform to simplify part of the circuit. You may also use the principle of superposition.


5

While I agree with Alfred Centauri's answer, I am not sure it gives a direct answer to your specific question for the following reason. If there is a receiving antenna somewhere, there is always some reflection, however minute. If the antenna is connected to a circuit, the conditions of reflection will change, and the transmitter can "notice" that. In ...


5

Well, believe it or not, eels have been used to power Christmas trees (youtube link), so powering an electric motor isn't quite out of the question. However, Eels emit that 400 V at 1 A = 400 Watts (though the youtube video says that the eel was emitting 800 Watts). An electric car requires over ten thousand Watts of power to operate. So in theory, you ...


5

The way every electric car works is by converting electrical energy into kinetic energy. The externally released energy of a discharge of Electrophorus electricus has been studied and has been found to be around $17 \mu\mathrm J$ per discharge. Since kinetic energy is $E_k = \frac{1}{2}mv^2$, if we assume a combined driver+vehicle weight of 1000 kg, then ...


4

As an intermediate step, consider a sinusoidal source driving an infinite transmission line with some characteristic impedance $Z_0 = 50\Omega$. The source "sees" a real impedance of $50\Omega$ and so, power is delivered to the line and, since the TL is infinitely long, the power is transported down the line, via an electromagnetic wave, without reflection. ...


4

Every electric motor runs using the Lorentz force, so there is no difference in principle between the motor in the hard drive and an electric car motor. There are commercial 60 hp electric motors that run electric cars, so the answer is yes. It's power consumption is, well, 60 hp. That's 45000 watts, plus a little more for heating the engine and so on, so ...


4

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.


4

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 ...


3

I'm not sure why the resistor to ground from B is there, but you are incorrect at point D, the capacitor doesn't pass the DC level as you've indicated. It's a high-pass filter with C and R, so basically you need to move the DC-level on the Vd plot to ground - but keep the two transients like you've plotted them. That is, the curve should start at ground and ...


3

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 ...


3

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 ...


3

As pointed out by Akash in a comment: when a battery is charged electric energy (potential difference) is converted to potential chemical energy. In an ideal battery only reversible chemical reactions occur, so that it can deliver many discharge/recharge cycles. However, it's not possible to design a battery in such a way that no unwanted reactions occur at ...


3

Before I start, I should say, I am yet to meet a physicist who has a clue about the practical aspects of loudspeaker design. Likewise about audio engineers ;-) so the fact you are asking this, means you are well set up for making some good speakers. John Bird has written an electronics textbook which could be very useful for you which covers all the ...


3

Here is how I would do it, following the method outlined by kleingordon in a comment. This method is less cool but more general than Carl Brannen's answer, because it will work even in the case where there are crossing wires and you can't rearrange it into a single sheet of resistive material. Let the electric potential at $A$ be $V_A$ and that at $B$ be ...


3

Uh... I liked this question. And a quick look at the Google or Google scholar did not give me much. So is written here more what I grasp people understand about impedance. Like Noldorin was saying I take impedance as how much is impeding power to be transferred from one place to another. In a homogeneous medium power is transferred as wave unimpeded but when ...


3

A good example of impedance matching outside the world of transmission lines is acoustic horns, which match acoustic impedance between a sound source (like a vibrating string or reed) and the air.


3

In many applications we are interested in the power. For example your electricity bill is based on the power you consume. For a DC source the power is: $$ W = VI = \frac{V^2}{R} $$ and for an AC source (assuming a resistive load so the voltage and current stay in phase): $$ W = V_{rms}I_{rms} = \frac{V_{rms}^2}{R} $$ So using the RMS values makes the ...



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