According to :     Note that the holes injected into Emitter are supplied from Base   electrode (Base current), whereas the electrons injected into the Base   are supplied from Emitter electrode (Emitter current). The ratio   between these currents is what makes BJT a current amplifying device -   small current at Base terminal can cause a much higher current at   Emitter terminal. The conventional current amplification is defined as   Collector-to-Base currents ratio, but it is the ratio between the   above currents which makes any current amplification possible.   First off, Why collector current increases as base current increase? Is the former causes the later, or the later causes the former, or something else (voltage on electrodes, maybe) causes both?  And here is my question, Why collector current always increases more than the increment of base current? Say after something changes, a extra holes are "injected into" emitter region, and b extra electrons are injected into base region. Then why b is greater to a?
122529
In fact, this question has been asked on the , but it's not well-answered. I suppose it might be more on-topic here.  According to :     Note that the holes injected into Emitter are supplied from Base   electrode (Base current), whereas the electrons injected into the Base   are supplied from Emitter electrode (Emitter current). The ratio   between these currents is what makes BJT a current amplifying device -   small current at Base terminal can cause a much higher current at   Emitter terminal. The conventional current amplification is defined as   Collector-to-Base currents ratio, but it is the ratio between the   above currents which makes any current amplification possible.   First off, Why collector current increases as base current increase? Is the former causes the later, or the later causes the former, or something else (voltage on electrodes, maybe) causes both?  And here is my question, Why collector current always increases more than the increment of base current? Say after something changes, a extra holes are "injected into" emitter region, and b extra electrons are injected into base region. Then why b is greater to a?
98958
I'm curious, when connecting my 2N2222 transistor's base to a PWM output pin on my Arduino, I know the Arduino can deliver a maximum of 40mA current. The 2N2222 only requires around 5 to 10 mA on it's base to switch the collector and emitter. Now, when calculating the base-resistor under "normal" conditions where I'm merely switching it directly from Vcc, the following applies:   However, when putting in a 500 ohm or 1K resistor in (1K for 5mA) directly it works great. But putting it in between the Arduino and the transistor, the transistor doesn't switch.  I suspect thus, that the resistance is too high of the base-resistor. Therefore, we may very well not need a base-resistor.  But, according to spec, the Arduino can deliver 40mA. So, am I correct in assuming the folling: 1) I should calculate the perceived resistance that the Arduino is giving, if it goes high to 5Volt and only gives 40mA, and then 2) Subtract that from the 500ohm resistor needed to make the transistor operate optimally.  Example:    So, I'm theorising that I have to then put in a 375 ohm resistor in to make the total resistance between the Arduino and the base resistor 500 ohm.  Somehow this doesn't feel right. Keep in mind, I'm not an electronic engineer, so I might have the cat by the tail here :P
123742
Here is what I have understood. Please correct me if I'm wrong.  The depth of induced channel depends on the voltage V. Since the voltage across the gate and  the path along the channel decreases from Vgs at the source to drain end to drain end, the channel is not uniform an is tapered. If the applied voltage Vds is increased to some value which causes the voltage between gate and drain end to be lesser than the threshold voltage, the channel is said to be pinched off.  If the channel is pinched off, how does the drain current flow? It's given in the book that the MOSFET enters saturation region and will have a constant value of drain current. But I don't understand how does the current flow when the channel no longer exists at the drain end.
13740
For those of you that uses Advanced Design System (ADS), from this circuit:   Fundamental Frequency for transient simulation tool is set to 1GHz.  Yet, I'm getting this output:   Main question:  Since this is Common Drain, VOUT should at most have the same amplitude as VIN. Probably even negative shift, since I didn't change the voltage threshold (I haven't got to tweaking parameters yet). So what's up with the positive DC bias?? Why is the amplification significantly less than unity?  Other questions:  1.) Also, what's that Transient Response at the very first cycle of VOUT? Why would a semiconductor have a Transient Response? Is it because of the "inductance" on the drain and source of a MOSFET?  2.) Lastly, why is it that this simulation does not work on very low frequencies? I first started with 1Hz scale for everything (just change all units from 1GHz below to 1Hz). The M1 just acted like a 1kOhm constant resistor.  UPDATE:  Because of recent response, I bothered to simulate this in the 1kHz scale (replaced all units from GHz in the pics to kHz, did the same with Fundamental Frequency) -- really should've done that in the first place (stupid me). This is the output:    Even here, the DC bias is nearly the same and the amplification is much, much less. It also lags by 90°. Why?  I'm starting to think it's either just using the wrong component or there was a set of parameters to fill-in for this component to get this to work.  UPDATE:  the continuation of this discussion:
17804
I had questions about holes in semiconductor, when i checked the net it is said holes are said to be equivalent positive charge and they say because the hole moves from one place to another when it is occupied by electron the electron leaves holes and etc... But i couldn't find the clear answer in " P type " semi conductor it is written the holes in the valence band increase, my problem is i see that lags and reduces the current not increases it because the holes somehow will attract electrons and get them from conduction band to valence band and since the conduction band is the band that does current , isn't that will reduce current not increase it ?
71646
So can anyone explain me how the collector voltage(V (output)) will vary with variation in base current with respect to the following figure?         &ndash; Schematic created using
101022
Sinking and sourcing current refers to current going into and out of the MCU (respectively), right? If you configure a GPIO pin as an output, why would current ever go into the pin (since it's configured as an OUTPUT)? I suppose it depends what the output circuit looks like, if it's tied to the voltage supply rail or ground... maybe I just answered my own question...?
93662
I have been using and making LED flashlights for years now and I am VERY familiar with all of the circuitry and technology inside of them, but I just got my first LED house light bulb that runs off of 110v AC electricity.  It is very similar to a flashlight - it has a large heatsink, and 1 or more LEDs inside of it (brand and specs of emitters, I do not know).  The part that I am most curious about is how the electricity from the socket gets to the emitter.  In a flashlight the power comes from a DC battery through a current regulator (LED driver) and into the LED itself.  This is not difficult because the voltage of common LED emitters is about the same voltage as Lithium batteries (3.7v, 7.4v, etc) which makes the current limiting relatively simple coming from a one or two cell light.  How does this work for a high voltage AC bulb?  Is there a transformer and rectifier inside of EVERY bulb (I sure hope not), or are the emitters manufactured to run off of 110v AC power and just use half of the wave cycle?
114085
The electron drift speed is estimated to be very low.How could there is current almost the instant a circuit is closed??   By the discussions it is known that The information about beginning of the flow of current is transmitted through the propagation of electromagnetic waves(electric impulse)and not with the drift velocity of the electrons. But I want any one to explain how this process takes place.CURIE:)
102360
We know that the electric field inside resistor produced by battery is the actual thing that pushes the electrons in it to move and thus producing some current. We also know that strength of electric field can be described by the electric flux which is the total amount of fields passing through an area. That flux when combined(multiplied) with charge of electron gives the amount of force on it. Now comes our two resistors R1 and R2. Assuming a number of field lines coming out of a terminal going towards another terminal through these resistors, the field lines should get divided into the resistances R1 and R2. Therefore giving a reduced flux through each resistor than in a series. Which will then give a lower voltage for both resistors. Wasn't the electric fields strength divided across resistors in parallel? What is the main cause of voltage? How electrons get a force to produce current even when the wire is very long in? I think it is better to understand with reality than analogy. Please don't give misleading pictures
95284
Now this might be a very stupid question for many readers, but I am getting really confused.   I was reading about electricity and current, and I read this:  Suppose we connect a wire to the ends of a battery, and a bulb in the circuit, current flows and lights up the bulb. The battery is responsible for doing work on the positive charges to move them from negative terminal to positive terminal, that is moving against the field. Now, these energised positive charges move through the external circuit, and within the circuit they give some of their energy to light the bulb. I understood this very well.  Now, reading further, it was written that "Electrons are actually responsible for current flow". It was written that there is concentration of energised negative charges on the negative terminal, and these electrons then move through the external circuit from negative to positive terminal. This is really confusing.     So my question: What actually flows? The negative charge or the   positive charge?   Further reading, I also read that in semi-conductors, fluorescent light bulbs etc., both charges move through the circuit.     Now, one more doubt: If both charges move through the circuit in this   case, shouldn't the charges stop in the middle of the wire, because   opposite charges will attract each other, hence stopping their   movement?
64205
I was studying pn junctions and came across the concepts of diffusion currents. From what I understand is, when the junctions come together, since there is an excess of donors on the n side and an excess of a accdeptor impurities on the p side the concentration gradients force a cross junction diffusion phenomena. Now what I would like to know is that what are the particles that move to create the diffusion currents. The notes from where I'm learning says holes and electrons. Now the actual movement of electrons make sense, because atomic ionization is a pretty ordinary event, but what does it mean when they say that a "hole" crosses the junction from the p side to the n side? Does the entire acceptor atom separate itself from the crystal lattice of the semiconductor and cross the junction? It doesn't really make sense to me? Or is it some exotic effect which I'm unaware of?
37990
According to Ohm's law, $V=IR$ (voltage equals current times resistance).  So if the voltage increases, then the current increases provided that the resistance remains constant.  I know that Voltage or potential difference means work done per unit positive charge in bringing that charge from one point to another.  So according to Ohm's law, if the work done per unit charge increases then current will increase. How can this be true? Point out my mistakes.
105960
I see every day high voltage, high current; high voltage, low current. I rarely see low voltage, high current; why?  I know that I can take a high voltage, high current signal, send it through a step-down transformer and theoretically get a low voltage high current (with the same power output, of course), but I never see this done.  Can someone give me a reason as to why I wouldn't want a low-voltage high-current scenario?
77283
I know this question sounds silly, as if there was a potential difference a current would be created when the terminals are connected together and this would mean energy has come from somewhere.  The reason I ask this though is that from my understanding of the depletion region and built in potential of a diode it seems like if you connected a voltmeter across the whole diode it would show the value of the built in potential.  This is explained in the image below:    At first, electrons flow from the n type to the p type because there are a higher concentration in the n type, and holes do vise versa. This is called the diffusion current. The first electrons and holes to cross the pn boundary are the ones which are closest to it; these carriers recombine when they meet each other and are then no longer a carrier. This means there is a depletion region of no carriers near the pn boundary. because electrons have left the n type material, and holes have left the p type material, there is a surplus of positive and negative charge on the n and p side of the pn boundary respectively. This causes an electric field that opposes the diffusion current, and so no more electrons or holes cross the boundary and combine. In short, only the electrons and holes near the boundary combine, because after they have done that an electric field is formed that prevents any more carriers from crossing. The current due to this electric field is called drift current, and when in equilibrium this will equal the diffusion current. Because there is an electric field at the boundary (pointing from the positive charge to the negative charge) there is an associated voltage. This is called the built in potential.  If you sample the electric field at each point along the diode from left to right, you would start with 0 in the p region because there are an equal number of protons and electrons. As you approach the depletion region you would see a small electric field pointing back towards the p region, caused by acceptor impurities which now have an extra electron (due to recombination) and therefore now have a net negative charge. This electric field would increase in strength as you get closer to the boundary, and then die away as you get further away.  This electric field means there is a voltage, as shown in graph (d). The p side is at an arbitrary potential, and the n side is at a potential higher than this because there is an electric field between them. This means there is a potential difference across the depletion region; this is known as the built-in potential.  But why when I connect a volt-meter across the whole diode will I not see this built in potential?
30478
I learned about the coronal discharge, and the common explanation is because the electric field is strong where radius of curvature is small. But I haven't found anything yet that explains why electrons like to crowd at the peaks, and escape from the holes. My intuition suggests electrons try to distribute on the surface as uniform as they can, but they don't. Why?
26903
Op amp circuits are designed to achieve a specific gain regardless of the differences between individual op amps.  One very common circuit has a gain of -R2/R1.  Here's a (corrected) schematic:    Another common configuration has a gain of R2/R1+1 and is non-inverting:    What I can't see is why on earth anyone would use the inverting one, except for the odd case where you actually want inversion. The non-inverting one has high input impedance without an extra input stage, and almost the same gain. Is there any advantage to the first example?  Also, since the first example does not have high input impedance, it can take significant current to drive.  So, often a source follower is placed before the amplifier.  For the second configuration, is there any reason why a source follower would ever be necessary?