what is meant by amplification factor of n-p-n common base configuration?
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.
804195
Product of Powers Property: This property states that to multiply powers having the same base, add the exponents. \n\nFor example, take x² times x³.\n\nx² is x times x.\n\nx³ is x times x times x.\n\nx² times x³ is therefore (x times x) times (x times x times x), or x to the fifth power.\n\nSo when you multiply two numbers with the same base, you add the exponents.
527802
N(Nitrogen), P(Phosphorus), K(Potassium)
219925
Look at it this way. When you configure an amplifier as a common emitter, the emitter goes to ground. When a signal enters the base, it increases the emitter current. This causes a corresponding increase in the collector current. This makes the collector resistor drop _more_ voltage, producing less voltage in the load.\n\nSo consider the extreme cases. If no signal is present on the base, the transistor is effectively "off" and all the voltage at the collector resistor is avaialble to the load.\n\nThe other extreme case is if so much signal is at the base, the transistor saturates "on" -- then the collector will be at, say, 0.2V or so and the collector resistor will drop almost all the voltage, leaving nothing for the load.\n\nSo you can see that a lot of signal produces a little voltage, and no (or a little) signal produces a lot of voltage. Phase inversion!
840888
PCI5 +4H20---->H3PO4 +5HCI..thats it ...ve any more doubts do mail me
728472
samples or sample size; usually designated "n"
464163
FORMATION CONSTANT IS :\n\nkf = 1.5*10^7
586314
ñ = n - iκ\n\nHere, n is the refractive index indicating the phase velocity as above, while κ is called the extinction coefficient, which indicates the amount of absorption loss when the electromagnetic wave propagates through the material. Both n and κ are dependent on the frequency (wavelength).
884757
I assume that you mean for p, the only parameter.\n\nIt's k/n where k = number of successes in n trials.
593819
It appears that the constants are calculated at one condition (dry,wet,etc) whereas the effective is closer to reality.  See\n\nhttp://www.pnas.org/cgi/content/full/102/45/16293
867397
The next number is 6 (and every subsequent number is 6 as well).\n\nIt has to do with converting bases.  If you have a_n=6, then n has to be greater than or equal to 2.  Therefore, \n\n6 in base 2 = 110\n6 in base 3 = 20\n6 in base 4 = 12\n6 in base 5 = 11\n6 in base 6 = 10\n6 in base 7 = 6\n\nand so on....\n\nHope that helps!
788892
OF2 (~104), NF3(~107), CF4 (~109.5),  BF3 (120), BeF2(~180)
845828
Here you go...\n\nhttp://home.planet.nl/~skok/techniques/laboratory/pka_pkb.html
336152
I think you mean by\nn[a_n, a_(n-1), ..., a_2, a_1]\na number p_n^a_n * ... * p_2^a_2 * p_1^a_1, where a_i is a power of i-th prime p_i.\n\nSo, we have:\n4[1,1,1,2] = 7 * 5 * 3 * 2^2 = 420\n4[1,3,1,2] = 7 * 5^3 * 3 * 2^2 = 10500\nand the rate is 10500/420 = 25.\n2[5,0] = 3^5 = 243\nSo, the last number is\n3[2,5,0] = 5^2 * 3^5 = 6075.\n\nAnd the answer is 3[2,5,0].
605008
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
201802
a molecule of dna contains many nucleotides which contain a nitrogen containing base (adenice,guanine, cytosine or thymine), a deoxribose sugar, and a and a phosphate group
795345
Same number of protons\n\ndifferent number of neutrons