# Tag Info

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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 ...

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"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 ...

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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 = ... 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 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 ... 5 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 ... 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 ... 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 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 ... 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 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 ... 3 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 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 ... 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 For the most part cosmic rays do nothing to consumer electronics. This is not to say that they can't flip bits or even damage elements, but the rate for such effects is very, very low. Radiation effects are routinely observed in electronics placed in accelerator experimental halls (where the radiation levels are at lethal-dose-in-minutes levels when the ... 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 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 ... 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 ... 2 In fact, Schottky Diodes have the lowest forward voltages. This means, that this is not a question of band gap "voltage" (this is a energy difference originally!) but of technology. Second are Germanium point contact diodes with gold wires. "Diodes" made from Galena maybe are very low too, but due to the wiggely properties I would not dare to write ... 2 A flip-flop (bistable multivibrator) is, in simple terms, two transistors wired together in such a way that there are two stable conditions: (1) one transistor is full "on", while the other if full "off" (2) vice versa If the circuit happens to be in a state "in-between" these two states, it will, due to positive feedback, very rapidly move towards one of ... 2 Cosmic rays do have noticeable affect on electronics. The most prevalent effect is from memory bit flips (known as "soft errors"). The degree of significance of the effect depends on the application. A typical soft error rate for static RAM is in the region of 400 FITs/Mbit [1]. (Failures in time=failures per billion device hours) So if you have 1 Gb of ... 2 I would start from power consumption. It's a little unclear to me if you're talking about electron use in all of the components or the CPU, but since 3 Volts was used, I'll take the discussion to be limited to CPU. Let's say 30 W of power consumption.$$I=\frac{P}{V}=\frac{30 W}{3 V} = 10 A$$Now, for the electron flux (denoted \phi), we can just ... 2 You have (1.5V)(2.7A)(3600s) = 14580 Watt-seconds = 14580 Joules of energy in the battery. BUT you can only get that much energy out if you extract it the same way the manufacturer did it during their optimized tests, which is typically at a much lower amperage than what you want to drive your motor. Also, remember that there are friction and other losses. ... 2 If the weight is F in pounds, the coefficient of friction \mu and the speed of v in mph then the power W required to maintain this motion in Watts is$$ W \approx 2.0 \mu \cdot F \cdot v $$The coefficient of 2.0 comes from the conversion into metric units. To move 200 lbs at 10 mph with a coefficient of friction of \mu=0.4 is$$ W \approx ...

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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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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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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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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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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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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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