what is miller capacitance?
406751
The Miller Effect means that if the circuit configuration takes an input to a transistor/triode at the base/grid and develops an output voltage signal inverted and amplified at the collector/anode, the large output voltage signal will suck very much more current from the input through this small capacitance than would be otherwise expected.  That is to say, it lowers the input impedance at high frequency.\nI assume the Miller Capacitance is this capacitance, between base and collector of a transistor, or grid and anode of a triode.
58713
Measure its capacitance.  Many multimeters have this capability.  If the body is not a two-terminal device (e.g., a capacitor), measure the capacitance from the body to a ground plane.  Then measure its voltage with a very high-impedance voltmeter.  The act of measuring will drain off some charge, so you have to take a reading quickly.  The charge in Coulombs is V*C, where V is voltage in volts, C is capacitance in Farads.
1029882
The Farad is the unit of capacitance, but it does not have the unit of distance.\n\nq = CV where q is charge, C is capacitance, and V is voltage.\n\nC = q/V   q is expressed in Coulombs and V is Joules/Couloumb\n\nso the units of capacitance are:\n\n1F = Col/(J/Col) = Col/1 x Col/J = Col^2/J\n\nCouloumb is a fundamental unit like the meter or second.\n\n1F = Col^2/N-m   probably about as far as you want to take it, but you see it does not have units of distance.
365027
The capacity of a capacitor is measured in farads (F). Ususally they range from several picofarads (pF) to several microfarads (µF). For the measurement, look on the capacitor itself or if it is a disk capacitor then there is a system for figuring that out (look it up).
736497
The capacitance causes leakage as the other answer says.\n\nThis capacitane creates problem in your calculations. Say, if there were only resistance in junction, then you can determine the performance of the junction at different operating voltages easily [hold temperature at fixed value].\n\nBut, capacitance causes the frequency dependance. So, the junction impedance becomes non-linear with the operating frequency. So, calculations are complicated.
836917
WL = h/(mv)\n\nh is the planck constant\nm is the mass of the elecron\nLook for them in you book.  Plug the numbers.
1178857
Did it ever have a high temperature of heat? (The normal Temp is around 50 up to 53). If yes, Aopen as i experienced it, always get capacitors defection when it get heated,so try to replace capacitors. take care that Mp may went failure So check it. and dont expose it to more heat.
924272
the amount of voltage initially on a capacitor will be whatever voltage you charge it to. It will never be any higher unless there's other circuitry involved.  The current (Amps) will flow at a rate decided by the resistance it is emptied into.  To use the common water pipe analogy, the voltage is the amount of water pressure, and the amperage is the size of the pipe.\n\nA capacitor is like a water tower.  There's lots of voltage, because the water's trying to get to the ground.  If you put a small hole in the water tower, you have a high resistance path to ground.  Your voltage will slowly fall as more of the water drains out.  the current will start fairly slow, and gradually fall along with the voltage(there's less pressure pushing it).\nIf you were to blast a big hole in it with a stick of dynamite, you would have a huge amount of flow(current/Amps), but the voltage would fall quickly, and the current will follow.\n\nit all relates to ohms law.  Voltage = Current * Resistance\n\nor you could re-arrange it to calculate your current, based on the voltage you've charged the capacitor to, and the resistance you're draining it through.  Current = Voltage/Resistance\n\nI knew a physics teacher who would demonstrate this to his classes by shorting a very large capacitor, which was charged to 12V with a screwdriver.  It would make a huge shower of sparks, and often partially weld the screwdriver to the top because of the huge currents.  He'd then charge it up again and would ask if anyone wants to put their finger on top of it.  Nobody would of course, but he would then do it himself and get no shock at all.  The resistance of skin is high, so with only 12V there's almost no current flow, even though there's a huge current capability if you give it a low resistance.\n\nIf you want the capacitors you're playing with to last a long time, though, you're best charging and discharging through a resistor.  Very quick charges and discharges create hot spots inside the capacitor, and will eventually damage them.
867378
A way to make sure the capacitance doesn't change with temperature.  This can be done by selecting a suitable dialectric material to make the capacitor, as with "NP0" capacitors (NP0 means "negative-positive zero," meaning no - or + temp. coefficient).  One could also design a circuit to compensate for capacitance changes due to temp, such as, using a thermistor and transistor to provide a voltage that varies with temperature, and applying it to a vari-cap placed in parallel with the capacitor.
895782
Xc=1/2π f C\nAs you can plainly see, capacitive reactance is a function of both capacitance and frequency.\nIf you work the arithmetic, you will see that the reactance decreases as capacitance increases while freqency remains stable, thus is inversely proportional.\nYou can work through the various permutations and chart the various relationships.
744231
Well, mv = h/λ\nm = h/(λv)\n= 6.626E-34 / (3.31E-15 x 0.01 x 3E8)\n= 6.673E-26\nConvert this to Da\n= 6.673E-26 / 1.66E-27\n= 40.2\nYou know, this is Ca. (look for it in the periodic table)
921926
mg + ion\no -ion\nk +ion\ns -ion\nwhen they are 'added' together (MgO and K^2S) the charges are balanced, that means that the net charge is 0.
898015
Nominal= 36ohms\nTolerance=5% \nMultiply nominal by .05 , then add that figure to nominal for Max,.subtract it from nominal for Min.,,and study your ohm's LAW!! ;-)=
158584
Try,\n\nhttp://www.personal.psu.edu/faculty/d/x/dxh40/MBLemma.pdf
1095597
Reactance, like resistance, tells you how a capacitor responds (reacts) to a current or voltage. From the reactance, you can find the V/I relationship. It is defined as X=1/wC where w is the radian frequency and C is the capacitance of the circuit.\n\nIt will be better if you can let us know your education level and we can describe according to your need.\n\nhttp://www.kodkodkod.com/database.htm\nhttp://www.kodkodkod.com/index_files/submitquestion.htm\n\nRegards,\n\nKodKodKod Learning Consulting
631888
ahdunno is right in that mfd is microfarad.  (a 800 millifarad cap would be huge - still, 800 microfarad is a very large capacitance, so be extremely careful with that thing.  the energy stored in a capacitor is 1/2 C * V^2, which for 450V and .0008 Fd = 81 joules.  that amount of electrical energy can easily kill you.)\n\nanyway, if you connect a conductor across the capacitor terminals, the amount of current flow will be an exponential decay.  the time constant of the decay will be determined by the resistance of the conductor (as well as the internal resistance of the cap) so it's difficult to say exactly how much "current" will flow because it depends on several parameters that we don't know, and because one would need to specify *when* during the discharge you are measuring the current.  \n\nthere is a little applet here that you can use to calculate what the discharge will look like:\n\nhttp://hyperphysics.phy-astr.gsu.edu/HBASE/electric/capdis.html#c2\n\nyou'll need to find out the internal resistance (aka the "equivalent series resistance") of the capacitor and add it to the resistance of the conductor you use.  for the case of a "very rapid discharge" you'll have to use a very low resistance conductor, and there exists a limit to how fast you can discharge the cap, which is determined by its internal resistance.  the PEAK current which is at the very beginning of the discharge is simply I = 450V / R where R is just the total resistance of the discharge path.\n\nfinally, let me stress the safety point again.  if you are trying to get a "very rapid discharge" then you will need a very low resistance.  for example, 18 gauge copper wire has a resistance of 0.007 ohms / foot.  a typical ESR for an electrolytic is in the range of something like 0.05 - 0.01 ohms.  this means that the wire resistance itself is negligible.  if you go to that applet and plug your numbers in, for a resistance of 0.01 the peak current at discharge is 45,000 amps.\n\nread that again, and think about it.  45,000 amperes.  the wires in your house will melt and catch fire if you put 50 amperes through them.  an 18 gauge copper wire will easily vaporize at that current, and so will the capacitor.  so unless you want some serious burns and probably missing fingers, *don't* mess around with trying to get a "very fast discharge" with a big cap like this, unless you are very sure of what you're doing.  \n\nbelieve me - i've seen the insides of cabinets sprayed with molten copper and had their doors blown off when something shorted across a capacitor.  also, once we found a squirrel which got across the leads of a big cap and it had been cooked down to a hairy charcoal lump.  so please, please don't become a statistic.\n\ncheers !
241115
Electrolytic capacitors consist of two elements, the isolation material Aluminium oxide as one plate and an electrolyte-soaked paper (some modern types are even virtually solid) as the other plate.\n\nA second sort are Tantalum capacitors\nTantalum electrolytic capacitors exploit the tendency of tantalum to form a protective oxide surface layer, using tantalum foil as one plate of the capacitor, the oxide as the dielectric, and an electrolytic solution as the other plate. Because the dielectric layer can be very thin (thinner than the similar layer in, for instance, an aluminium electrolytic capacitor), a high capacitance can be achieved in a small volume.