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q(VAB/w) qtVAB in .NET Paint barcode code 128 in .NET q(VAB/w) qtVAB

q(VAB/w) qtVAB using .net vs 2010 toencode ansi/aim code 128 with asp.net web,windows application Modified Plessey (3-50). Since all of the quantities i n the right-hand side of Eq. (3-50) can be measured, the Hall effect can be used to give quite accurate values for carrier concentration..

3 . If a measurement of resistanc e R is made, the sample resistivity p can be calculated: p(n-cm) = Rwt VCD/IX L L/wt. 3 -> Since the conductivity a = 1/ p is given by q\ypPo, the mobility is simply the ratio of the Hall coefficient and the resistivity:. I/P _RH i-5.7 qpo q(UqR ) P M .net vs 2010 barcode 128 easurements of the Hall coefficient and the resistivity over a range of temperatures yield plots of majority carrier concentration and mobility vs.

temperature. Such measurements are extremely useful in the analysis of semiconductor materials. Although the discussion here has been related to p-type material, similar results are obtained for n-type material.

A negative value of q is used for electrons, and the Hall voltage VAB and Hall coefficient RH are negative. In fact, measurement of the sign of the Hall voltage is a common technique for determining if an unknown sample is p-type or n-type..

M< = EXAMPLE 3-8. Referring t o Fig. 3-25, cons ider a semiconductor bar with w ~ 0.1 mm, ! t = 10 urn, and L = 5 mm.

For 2 = 10 kg in the direction shown (1 kG = 10~5 Wb/cm 2 ) and a current of 1 mA, we have VAB = - 2 rnV, VCD - 100 mV. R n d the type, concentration, and mobility of the majority carrier. % = 10" 4 Wb/cm 2 From the sign of VAB, we can see that the majority carriers are electrons:.

SOLUTION 0 =. (io-3)(io-4). 19 3 3. QK-VAB) I.6 x io- (io- )(2 x code 128b for .NET ixr ) R = VcpUx fe 0.

1/10"*1 j = 0.002 0 - c m L/wt ~ Llw\ " 0.5/0.

01 x 10" 3 1 ll7> (0.002)(1.6 x 10~19)(3.

125 x 101"). 3.125 X 10 17 c m - 3. 1 P-n = P4 o 10,000 c m ^ V " s r 1 3.5 INVARIANCE OF THE FERMI LEVEL AT EQUILIBRIUM In this chapter we have discu ssed homogeneous semiconductors, without variations in doping and without junctions between dissimilar materials. In the following chapters we will be considering cases in which nonuniform doping occurs in a given semiconductor, or junctions occur between different semiconductors or a semiconductor and a metal. These cases are crucial.

Energy Bands and Charge Carriers in Semiconductors to the various types of elect ronic and optoelectronic devices made in semiconductors. In anticipation of those discussions, an important concept should be established here regarding the demands of equilibrium. That concept can be summarized by noting that no discontinuity or gradient can arise in the equilibrium Fermi level EF.

To demonstrate this assertion, let us consider two materials in intimate contact such that electrons can move between the two (Fig. 3-26). These may be, for example, dissimilar semiconductors, n- and p-type regions, a metal and a semiconductor, or simply two adjacent regions of a nonuniformly doped semiconductor.

Each material is described by a Fermi-Dirac distribution function and some distribution of available energy states that electrons can occupy. There is no current, and therefore no net charge transport, at thermal equilibrium. There is also no net transfer of energy.

Therefore, for each energy E in Fig. 3-26 any transfer of electrons from material 1 to material 2 must be exactly balanced by the opposite transfer of electrons from 2 to 1. We will let the density of states at energy E in material 1 be called NX(E) and in material 2 we will call it N2(E).

At energy E the rate of transfer of electrons from 1 to 2 is proportional to the number of filled states at E in material 1 times the number of empty states at E in material 2: rate from 1 to 2 N^E^E) N2(E)[l - f2(E)] (3-53). where f(E) is the probability barcode 128 for .NET of a state being filled at E in each material, i.e.

, the Fermi-Dirac distribution function given by Eq. (3-10). Similarly, rate from 2 to 1 a N2(E)f2(E) N,( )[l - ( )] At equilibrium these must be equal: N^EME) N2(E)[1 - f2(E)) = N2(E)f2(E) #,( )[! - fx(E)] (3-55) (3-54).

Figure 3 - 2 6.
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