What are the conditions for saturation in a p-channel MOSFET?
461326
After the Mosfet is past it's linear region, 2-3 volts on the gate, it is saturated and no different from an N-Channel.
971113
i have my difficulties to determine what you mean with PLC  but in case you talk about chip technology or the standard transistor in semiconductors the answer could be:\nthere is semiconducting material which is mixed with specific conducting material. \nout of this process you can have material that prefers negative charging or positive charging, N or P material.\nnow the function of a transistor is that you put 3 layers of such material together in a NPN or PNP configuration, connect the base of the transistor to the material in the middle, and two other connections to that material to both 'ends'.\nnow if you put some voltage to the outer connections nothing much will happen except you start filling up the middle section with a negative or in other configuration a positive charge. In this case the whole setup becomes 'conducting' and you can observe current running from one side to the other.\nthis can be used as a switch, or a amplifier in electronics\nnow the difference is which kind of charge you prefer to make the transistor switch.\n\ndoes this solve your question ?
843449
maybe this would help\nhttp://encarta.msn.com/encnet/refpages/search.aspx?q=semiconductors
567883
Quite well, although today it's been replaced with integrated circuits.  (It's basically a switch, and either off or on, which comes in handy when you are working with base 2 binary numbers, as in a computer.)\n\nThere are two types, NPN and PNP, depending of which way current flows through them.  They are devices which allow, or limit current flow.\n\nThe difference between PNP and NPN transistors is that NPN use electrons as carriers of current and PNP use a lack of electrons (known as "holes"). Basically, nothing moves very far at a time. One atom simply robs an electron from an adjacent atom so you get the impression of "flow". It's a bit like "light pipes". In the case of "N" material, there are lots of spare electrons. In the case of "P" there aren't. In fact "P" is gasping for electrons. Clear as mud isn't it?\n\nOK, bear in mind that the Base is only a few atoms in thickness - almost a membrane - so any electrons allowed into the base "membrane" act as a catalyst to allow other electrons to break through from emitter to collector.\n\nImagine a pool of water near the edge of a table. It rests there with surface tension holding it in place. Now put one tiny drop of water on the table edge and let it touch the pool of water. Suddenly, the pool drains onto the floor as gravity takes over! Your tiny drop provided the catalyst to get it moving. So the base electrons do a similar job for the "pool" of electrons in the emitter - helped by the "gravity suction" of the power supply voltage on the collector.\n\nA transistor doesn't "increase" current. It simply allows power supply current to pass from collector to emitter* - the actual amount depends on the (small) current allowed to flow into its base. The more electrons you allow into the base, the more (x 100) that flow from collector to emitter . I put "x 100" because that is the typical gain (amplification factor) of a transistor. For example, one electron put into the base could allow 100 to escape from collector to emitter.\n\n* The purist might argue that current flows from emitter to collector - dependent on whether we are discussing electron flow or "hole" flow. I don't want to get involved in the physics of current flow. You don't need to know this to design a circuit.\n\n* This discussion relates to Bipolar transistors. Other types of transistor such as "FETs" (Field Effect Transistors) are in common use and work in a slightly different way in that the voltage applied to the "gate" terminal controls the current flowing from "cathode" terminal to "anode" terminal. In effect, a FET is simply a semiconducting (one-way) resistor whose value is controlled by the voltage applied to its "gate".\n\n* OK, having told you all that, I now have to point out that the above description is basically WRONG! What I've described is based on what is called "the beta model" of a transistor. A transistor actually relies on base voltage input - the current input is incidental. If you are at college, your teacher will explain this fully with lots of mathematical equations that will let you design anything at all. However, my description will let you design simple circuits with a minimum of effort and they will almost certainly work. Unless you are going to take up design as a profession, this is all you need to know.
200572
Do you think those would help?\n\nhttp://scholar.google.com/scholar?q=design%20of%20a%20Lithium%20Niobate%20modulator%2040%20gbps&num=20&hl=en&lr=&safe=off&rls=GGLR,GGLR:2005-38,GGLR:en&oi=scholart
501507
Here is the equation :  \n  \nI0 = Is*(exp(V0/nUT) - 1)\n\nwhere :  - I0 is the current (function of V0, the applied\n                              voltage)\n         - The others parameters should be given by the   \n           constructor:\n            -    1 < n < 2\n            -    UT = 26mV at 27°C\n            -    Is is the reverse saturation current\n\nPS : I'm not sure I'm using the right english vocabulary as it is not my mothertongue. Sorry for that...
6099
I may be living in the past but I see transistors every day.  The 2-diode analogy is pretty close to the mark.  They meet at the base, pointing in for PNP and out for NPN.  Check them that way with your multimeter set on diode setting.  If the reading is inconclusive you may have to de-solder them from the circuit.
450183
this might prove to be useful:\nhttp://islab.oregonstate.edu/koc/ece575/02Project/Mor/
524948
Today's electronics systems require lower voltages, higher currents, higher power density, and higher efficiency than their predecessors needed. These demands necessitate the use of paralleled power MOSFETs or large-die MOSFETs. Having to use these high capacitance devices has given some designers an uneasy feeling about which approach to use.\n\nThe successful design of robust gate-drive circuitry for high voltage, very high current MOSFETs is no different than the driver circuitry designed for smaller devices. It depends on an understanding of the drive requirements, a good model to predict the results, careful attention to the physical layout of the driver board, and the selection of the proper circuit topology.\n\nFigure 1 represents a generic gate driver circuit where the switching speed is controlled by the total gate impedance (RG), the total source inductance (LS), the load resistance, and the device's characteristics. Once the load is defined and the power-switching device is selected, the fast switching speed is achieved by minimizing the two remaining parameters, RG and LS.\n\nhttp://www.techonline.com/community/ed_resource/feature_article/5652
714718
Short for complementary metal oxide semiconductor. Pronounced see-moss, CMOS is a widely used type of semiconductor. CMOS semiconductors use both NMOS (negative polarity) and PMOS (positive polarity) circuits. Since only one of the circuit types is on at any given time, CMOS chips require less power than chips using just one type of transistor. This makes them particularly attractive for use in battery-powered devices, such as portable computers. Personal computers also contain a small amount of battery-powered CMOS memory to hold the date, time, and system setup parameters.\n\nYes
858123
voltage gain is just that.  volts out divided by volts in should give a voltage amplification factor or voltage gain.\n\nsaturation typically means the device is conducting all it can.  any more increase in input results in no more increase in output.  the stage is saturated.
930646
You probably want alpha, the common base current gain. In general, current gain is designated "beta," "alpha," or "gamma," for transistors connected in the common emitter, common base, or common collector configurations, respectively.  Beta = di(c)/di(b), the rate of change of collector current with respect to change in base current.  Alpha =di(c)/di(e), rate of change of collector current with respect to change in emitter current.  Gamma = di(e)/di(b), rate of change of emitter current with respect to change in base current.  Also, alpha = (beta)/(beta - 1), beta = (alpha)/(alpha - 1), and gamma = beta + 1.  For a detailed explanation, check out : http://www.tpub.com/neets/book7/25f.htm\n\nAnother possible answer: "amplification factor" for any amplifier circuit, regardless of configuration, or whether it uses tubes, BJTs, FETs, etc, refers to its voltage gain, dv(out)/dv(in), the change in output voltage with respect to a change in input voltage.
568078
The driving current of a transistor is proportional to the width/length ratio (thus the area, as minimum length is defined by the technology) and also the mobility of the carrier. Electrons have higher mobility than holes (~ 2 to 3 times). The carriers in p and n channels are holes and electrons respectively. A narrower (2-3 times smaller in area) n-channel transistor can deliver the same amount of current as the p-channel one. Therefore, n-channel is more suitable to drive the load as it has larger driving power per unit area. (Area is expensive in silicon!)
800315
visit\nhttp://www.allaboutcircuits.com/vol_3/chpt_7/5.html
338305
Simulation Program for Integrated Circuits Emphasis.
663500
Both, check your CMOS settings.
700558
Sounds like your CMOS battery may have failed.\n\ntry replacing it.\n\n///