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19

Yes. That is the operating principle of this device, among many others:


15

In the early days of radio, the resonance of the antenna in combination with its associated inductive and capacitive properties was indeed the item which "dialed in" the frequency you wanted to listen to. You didn't actually change the length of the antenna, but by changing the inductor (a coil) or capacitor connected to the antenna you tuned the resonance. ...


15

The physical 'meaning' of the imaginary part of the impedance is that it represents the energy storage part of the circuit element. To see this, let the sinusoidal current $i = I\cos(\omega t)$ be the current through a series RL circuit. The voltage across the combination is $$v = Ri + L\frac{di}{dt} = RI\cos(\omega t) - \omega LI\sin(\omega t)$$ The ...


7

A diode consists of two materials known as p-type and n-type semiconductors, connected in series which allows current to flow through them differently. In the n-type semiconductor, electrons travel with enough energy such that they're not attached to an atom and are said to be in the conduction energy band. For the p-type semiconductor, electrons "hop" from ...


7

"Ground" refers to a particular voltage, generally taken to be "zero", or the voltage of the earth. A "virtual ground" is a wire in a circuit whose voltage is held to be zero not because it is directly connected to the true ground, but instead because it is actively driven to that voltage typically by feedback mechanisms. Here's an example of a virtual ...


7

Virtual ground refers to a circuit element not directly connected to ground, held at a reference voltage. This reference voltage need not be the same voltage as ground either. For example many op-amp circuits were originally designed for dual power supplies (say +12V and -12V) and could handle filtering or modification of a signal that was oscillating ...


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

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 ...


6

Power consumption is about linear with frequency. The processor contains millions of complementary FETs as shown. When the input goes low the small capacitance gets charged and it will hold a small amount of energy. A same amount is lost during the charging. When the input goes high again the charge will be drained to ground and be lost. So with each ...


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

There is a physical meaning behind the imaginary component of the impedance. You can re-cast the complex impedance $Z = R + jX$ (using engineering's notation $j$ for the imaginary unit) in polar form to get $Z = |Z|\exp(j\phi)$. $|Z|$ is the magnitude of the impedance, and scales the amplitude of the current to get the amptlitude of the voltage. $\phi = ...


5

Nowadays, the answer is negligibly so. Video cameras now digitise the image as pixels in parallel using charge coupled device technology. Former technologies, however, would emit appreciable bremstrahlung from decelerating electron beams, as I now describe. Before the coming of CCD arrays, the main video technology was the scanned photocathode, also called ...


5

This is a part answer - my (published) research has been in detecting UVA and UVB using smartphone cameras, but there has been research and a successful app made to detect gamma radiation using a smartphone camera by the Australian Nuclear Science and Technology Organisation, as reported on their webpage Smartphone radiation detector app tests positive. ...


5

To add to the "linear with frequency" point, there is also an additional factor. As that "dynamic power" increases, the temperature of the die will increase and this will also increase the leakage current through the millions of transistors, which will cause more dissipation (termed "static power") There's a long Anandtech thread taking lots of values and ...


5

First of all, an electric spark is moving of the electric charge through the air. This is somehow curious, as air itself is an electric insulator and does not conduct charges. However, when the electric field in air exceeds certain value, air gets ionized and highly conductive, enabling the movement of the charge. The next step to understand is why we get ...


5

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 ...


5

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 ...


4

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 ...


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

I spent many years working in the video graphics design industry. One of our problems is the opposite problem, that is, we have limits on how much electromagnetic radiation our products can produce. Before we can ship any new design, we have to test it to make sure it meets the limits. (The tests are done on samples, not on every item shipped.) The limits ...


4

If we believe this measurement shown by Omen, smartphone cameras are basically useless below Dose rates of 10uSv/h. The max. exposure limit for a human who is not a radiation worker is 1mSv/year, which translates into roughly 0.11uSv/h. In other words, the camera chip in a phone would have to be 100 times more sensitive to pick up relevant amounts of ...


4

Your eye has a lens in it. Without a lens, the light is all spread out and overlapping, just like you say. The light from any given pixel goes out in all directions, but a lens can make it re-converge back to a point. Hold up a sheet of white paper. Is there an image on it? No, of course not. It has light on it---light coming from each object in the ...


4

From "The Transistor, A Semi-Conductor Triode", by J. Bardeen and W. H. Brattain, Phys Rev. 74(2), 230-231 (1948): "The device consists of three electrodes placed on a block of germanium as shown schematically in Fig. 1. Two, called the emitter and collector, are of the point-contact rectifier type and are placed in close proximity (separation ~0.005 to ...


4

… an ideal power source capable of providing infinite current with no drop in the voltage it supplies. … Let's ignore the effects of current density on superconductors for now. … In these phrases is the explanation for the contradictory possibilities you have computed: you have supposed an impossible circuit. As a mathematical model, the behavior of ...


3

As Kevin Reid aptly explains, the circuit you have drawn is not realizable. But, let's take the closest physical thing you could build, assuming: your voltage source can supply enough energy that we don't hit its limits like all physical things, this apparatus has non-zero size Then, the circuit you actually built is this: simulate this circuit ...


3

The electrons need to get from the top to the bottom without any interference from any gas molecules that might be in the channels. If nothing else, collisions with gas molecules will degrade performance. At atmospheric pressure, I don't think the device would work at all. You can blow a hole through an MCP with over-voltage, but I'm not sure how this ...


3

A capacitor is often used for "decoupling". The wires into any electrical appliance have inductance (because they are long and thin). This means that if there is a sudden increased demand in current, there will be a significant voltage drop. A capacitor can act as a "tiny battery" that briefly supplies this current while the main supply catches up. A fan ...


3

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 ...


3

From the specifications of your battery, that is 1.5V and 2700mAh, you can compute that there is $14580$ Joules of energy stored in your batteries. The formula $P=U\cdot I$ relates power to voltage and current. You battery specs give voltage and capacity (that is total charge stored). The former is in Volt, the latter in milli-Ampère-hour. The product is ...


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 ...



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