Hot answers tagged

126

1. Capacitors There are a lot of misconceptions about capacitors, so I wanted to briefly clarify what capacitance is and what capacitors do. Capacitance measures how much energy will be stored in the electric field generated between two different points for a given difference of potential. This is why capacitance is often called the 'dual' of inductance. ...


55

For example, why do I see it suggested to use ceramic caps for power decoupling per microprocessor & a larger electrolytic capacitor per board? why not use electrolytic all around? The three main types have differing characteristics - I suggest you do some research on them but the main things to look for are self-resonant frequency (brought about ...


46

There are several reasons for this. First of all, memory takes up a lot of silicon area. This means that increasing the amount of RAM directly increases the silicon area of the chip and hence the cost. Larger silicon area has a 'double whammy' effect on price: larger chips mean less chips per wafer, especially around the edge, and larger chips means each ...


32

It's difficult to come up with an analogy because the usual analogies for electrical systems are fluid systems. A great thing about fluid systems is that the working fluid is also good at cooling things, and most people's practical experience with fluid systems involves rates of flow where heating is not very significant. So let's try a different analogy: a ...


25

Think of it as "drawing" extra breath whilst jogging as opposed to walking. A circuit under normal conditions will appear as a certain impedance. For instance, a DC motor operating without a mechanical load will spin at a rate determined by the number of its windings, contacts, permanent magnets etc. As a load is applied to the shaft, the rotor decelerates, ...


23

If it needs 500 mA then it will take 500 mA, even when you provide a 3000 mA capacity. If you're standing at the bottom of the Niagara falls with a 10 liter bucket you can fill it till it contains 10 liter, even though the waterfall has the capacity to provide much more.


19

No, they are completely orthogonal concepts. The probability distribution says nothing about the frequency content, and the power distribution across frequency says nothing about the sample probability distribution. You have to specify both.


17

This is generally true for incandescent lamps, motors, other things made of coils, and most electronics that predates semiconductors. It is also generally true for lots of integrated circuits, which draw from their power rails as needed. It is specifically false for LEDs and bipolar transistors, both of which can easily draw enough current to self-destruct ...


17

You've done the calculations correctly, but the entire basis for the calculations is just a rough approximation of transistor behavior. As you see, \$V_{BE}\$ isn't really exactly 0.7V (most of the time) and the \$\beta\$ isn't exactly 225 (most of the time). This kind of biasing is very sensitive to variations in transistor parameters so it isn't used much ...


17

Your measurements are impossible if the circuit shown is correct. 12V / 20kΩ = 0.6mA. Therefore there cannot be 8.03mA going into the Base. However the slightly high Base-Emitter voltage suggests there is, so the '20k' resistor isn't 20kΩ. According to your measurements, the voltage at the Emitter should be 200Ω*(8.03mA+0.5114mA) = 1.708V, ...


16

The common mode voltage is a voltage offset that is "common" to both the inverting and noninverting (i.e. "+" and "-") inputs of the instrumentation amp. An instrumentation amplifier is set up as a difference amplifier, so it measures the difference between these two inputs and so rejects any voltage that is common to the two. In other words, if you have two ...


16

The best reason I've heard is to avoid this: - V = 2 V (which of course is meant to say "voltage = 2 volts") U = 2 V sounds more sensible after all we use a different symbol for current (I) and also amps. Voltage is a bit on its own - we wouldn't say "amps = 2 amps" or "current = 2 currents". It seems to me this is the sensible reason for choosing U over ...


16

There's a forward voltage drop of a couple volts across the LED. This Voltage drop times the Current is the Power dissipated in the device. It creates light, but also heat. It's the heat that kills the LED.


16

As transformers are usually used with AC rather than with DC, there is what is known as inductance \$L\$, which is a property of a conductor to "resist" the changes in the current flowing in it due to the magnetic fields induced by that current (self-inductance). The magnetic field is "resisting" due to the fact that alternating magnetic field is in turn ...


16

The obvious difference is that electrolytics are much bigger than ceramics. 1mm by 0.5mm ceramics are common garden variety, your electrolytic cans are far larger. Then, as others have pointed out already, electrolytics do not do that well in high frequencies so they're not suitable for bypassing "high" frequencies, it can't keep up with 1MHz chip, let ...


15

Memory probably takes up the most silicon space, and RAM being very fast to use is volatile - and uses power constantly to keep its state. Unless you need lots of RAM, it's not useful for many other applications. If an embedded system designer needs more RAM, they merely get an external RAM chip and use peripheral memory interfaces that microcontrollers ...


15

Remember decoupling has several purposes. On a load which draws transient, spiky currents like a CPU, decoupling caps store energy locally and close to the load, so it is available quickly (ie, with low inductance). The idea is that on each clock cycle, the cpu will gobble up a certain amount of charge (coulombs), which means the caps must both have large ...


14

It's possible to "turn a magnet off" without requiring continuous power in either position- simply shunt the field lines away from the open end. Here is an example of a magnet with an on-off switch: And here is how it works:


14

Besides the excellent points brought up in the other answers, another reason for limited RAM is the architecture of the microcontroller. For example, take the the Microchip PIC10LF320, which has only 448 bytes of program (flash) memory and 64 bytes of RAM. But it probably costs only 25ȼ (or less) in large quantities. The limited size of the PIC10 ...


14

A 100BASE-TX device transmits on 1 pair (2 wires) and receives on another pair (2 more wires). Because of the 4B5B encoding -- for each 4 data bits, it transmits a series of 5 symbols -- 100BASE-TX has a 125 MHz symbol rate (sometimes called the baud rate) on each pair. Each symbol carries slightly less than 1 bit of information. 1000BASE-T -- Gigabit ...


13

In general, yes, higher frequencies attenuate more the further distance they travel. There are two effects that are responsible for this. First, higher frequency radio waves tend to be absorbed more readily by objects (ie: the penetration depth in the material is shorter). You may have noticed this effect with your home WiFi network. If you have a dual-band ...


12

If you put a minus sign in front of the number "5" it becomes "-5". Try and look at this differently. Try thinking that it rotates the number "5" (tied to the origin by a piece of string of length 5) through 180 degrees to become "-5" OK so far? Negative signs are the same as rotating thru 180 degrees... Why not extend this further to produce something ...


12

TL;DR: Flowing current creates heat, and for LEDs, heat kills the part. Whenever electrons flow through a conductor, Joule heating happens. This is partly because of what heat really is, particles that makes up the object moving around, and having electrons pulled through it guarantees that some electrons will collide into something and have its energy ...


12

Well one it slows down the rise time of your signal which reduces the high frequency content. So that could help you if you didn't need your edges to be that fast. It also lowers the current flowing through the trace and back through its return path which would lower the strength of the field created around it (and radiated out). I would add that I've ...


12

You have done the calculations correctly. However, your biassing scheme will only work well with one set of transistor parameters, it's very sensitive to variations in beta. A better biassing scheme will maintain a good bias point even with transistor variations. A more stable one is shown below. simulate this circuit – Schematic created using ...


12

Capacitance is given by: This is the formula applicable for free space or vacuum. If you are using a dielectric material between the plates, you multiply the resultant capacitance with the relative permittivity of the dielectric material. For poly-ethylene, relative permitivitty is 2.25. For simplicity, lets start with flat sheets of aluminium separated by ...


12

If the signal is represented as a voltage \$v(t)\$ or a current \$i(t)\$ and it is connected to a (1 ohm) resistor, the power dissipated in the resistor is proportional to \$v^2(t)\$ or \$i^2(t)\$. Apart from that, defining power as a positive, increasing function of the signal amplitude has useful mathematical properties.


10

I'd argue the challenge isn't technical, it's economical. A normal lightning strike transfers about 500 megajoules of energy (says Wikipedia). Wikipedia also says a single wind turbine comes as big as 7.5 megawatts these days. Thus, a single bolt of lightning is equivalent to running a big wind turbine for: $$ \frac{500MJ}{7.5MW} \approx 67 \text{ seconds} $...


10

In pure maths we use \$ i \$ to represent the prime square root of \$-1\$. The other square root of \$-1\$ being \$-i\$. If you imagine a number line with real numbers placed horizontally. We can now add a second number line going vertically containing the imaginary numbers. We have now created a system of complex where every point on the plane is ...


10

Zero volts doesn't mean zero current. Assume your circuit looks like this: simulate this circuit – Schematic created using CircuitLab When the switch turns on (connects to 10V), current flows to the right and charges the capacitor up to 10V. Once that happens, the current stops*. When the switch turns off (connects to ground/0V), current flows to the ...


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