I already know the conditions for a n-channel MOSFET but I wanted to know if p-channel saturation conditions are any different (i.e. inequalities switched)
208461
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.
161412
Nor gate\n\nRemember this fact\n\nNOR gate is the exact opposite of OR gate.\n\nSince OR gate has Low (0) output only when both the inputs are low =>NOR gate has High (1) output only when both inputs are low.So with this we can say that for NOR gate the following holds true.\n\n0 nor 0 gives 1\n0 nor 1 gives 0\n1 nor 0 gives 0\n1 nor 1 gives 0\n\nHope this answres ur question.
43820
No. you may count yourself lucky to have these conditions because with these you are able to get an unique solution ψ.\nBut there are 4 main conditions:\n\nNorm: |ψ| := 1\n\nContinuity of ψ\n\nContinuity of ψ'    (only for finite potentials)\n\nLimit: ψ(x) -> 0 for x -> ∞
655099
http://www.allaboutcircuits.com/vol_6/chpt_7/5.html\nhttp://hyperphysics.phy-astr.gsu.edu/Hbase/electronic/nandlatch.html
390944
try\nhttp://www.dir.ca.gov/title8/2395_42.html\nsection b\n\nGrounds are also tied to the cold water pipes and gas pipes.\nmake sure all circuits are dry and checked with a insulation resistor tester (megger).Readings above 2 Mega Ohms should be OK.
375299
Assume it's a PNP transistor. (if it's NPN, all polarities are reversed and the result is the same). A + increase in the base voltage will increase the current flow through the  collector - which will cause the voltage drop to increase across the load resistor - which means the collector end of the load will fall in voltage- exactly 180 degree phase reversal from the base if the load is purely resisitve (if it is inductive or capacitive, it will NOT be 180 degree out of phase with the base, since the base voltage is controlling the collector CURRENT flow.
368301
u lucky....it's a scsi standard. better than normal scsi, sometimes referred as ultra-scsi, has less limitations of cable lenght and higher input/output speeds.\n\nlvd (low voltage differential)
341723
No, the voltages are different
197982
In certain low volume and high speed circuits, transistors will always have their places. ALthough their are now many more radio chips.\nMIKE
288266
In a p-n-p diode, the base is n-type and the emitter is p-type. If you forward bias the transistor, it means that V_base < V_emitter (e.g. V_base = 4.3V and V_emitter = 5V). Since the emitter to base is just a p-n diode, you expect such biasing will give you a forward current as in a p-n diode. That means current flows from the emitter to the base.\n\nHowever, if you really want to understand the source of the current, you have to use the concept of electron and hole. (Let me know if you don't know what is hole http://www.kodkodkod.com/index_files/submitquestion.htm). In a p-n junction, holes are injected from p to n and electrons are injected from n to p. They together contribute the the current. Similarly, in a p-n-p transistor, the holes from the emitter are injected from p-emitter to the n-base and electrons are injected from the n-base to the p-emitter. However, most of the holes from the emitter cannot be "collected" by the base and indeed they are "collected" by the Collector. That's why you find that I_E and I_C >> I_B. Therefore, the base current is basically due to the electrons from the base.\n\nOf course, in a non-ideal transistor, some holes from the emitter are collected by the base and there is also reverse current between the base and the collector. As a result, the base current is a summation of all the aforementioned components.\n\nI think my answer might not really tailor-made to your need. It's better to let us know how much you know about the transistors (or your education in EE). The answer to the same question you ask can be very different from high school to research level. Let us know if you need more help!\n\nKodKodKod Learning Consulting - http://www.kodkodkod.com
670767
yeah you can use a switch (voltage switch) or if you can use binary logic, need to know how much voltage/current you are dealing with
496183
Well the second guy is half right, you can of course install them in different sizes.  However to call the performance gain "amazing" is a bit of a stretch.  Plus, you would be forfeiting dual channel mode if you install in different sizes, which is a bad thing.  My suggestion is to save up until you can afford the full upgrade to 2GB.
226625
It's going to be one or the other, it doesn't change during function.\n\nThe junctions are determined when the transistor is manufactured, it's part of the physical structure of the device, so it's polarity does not change.
298860
There's both an explanation as well as a diagram\nof a 2 input AND gate built just from transistors at:\n\nhttp://tams-www.informatik.uni-hamburg.de/applets/cmos/cmosdemo.html
386079
------/:=====:x:=====:&#92;-----@-------\nL     2Wy                Int             2Wy           Load          N\n\n(Does this answer your question??)\n\nL=Live, 2Wy is a 2 way light switch, int is an intermediate switch, and @ is a light??\n\nConfused?? (never heard of tunnel switching!!?)
550146
no, unfortunately not. they changed the pin layout slightly. socket AM2 is keyed differently than socket 939.\nhttp://techreport.com/reviews/2006q2/socket-am2/index.x?pg=1
136911
Odds are you will need an adapter.\n\nIf I recall correctly, the 1.0 GHz chips were the last ones electrically compatible with the older boards.\n\nTo be honest, the odds are pretty good the the P3 actually benchmarks equally if not better.\n\nI'd test it before messing around with it.  Benchmark them with something like FreshDiagnose.