In a meter bridge, the null point is found at a distance of l 1 cm from A. If now a resistance of X is connected in parallel with S, the null point occurs at l 2 cm. Obtain a formula for X in terms of l 1, l 2 and S.
If one side of crystal pure semiconductor Si(silicon) or Ge(Germanium) is doped with acceptor impurity atoms and the other side is doped with donor impurity atoms , a PN junction is formed as shown in figure.P region has high concentration of holes and N region contains large number of electrons.
As soon as the junction is formed, free electrons and holes cross through the junction by the process of diffusion.During this process , the electrons crossing the junction from N- region into P-region , recombine with holes in the P-region very close to the junction.Similarly holes crossing the junction from the P-region into the N-region, recombine with electrons in the N-region very close to the junction. Thus a region is formed, which does not have any mobile charge very close to the junction. This region is called the depletion layer of pn junction.
In this region, on the left side of the junction, the acceptor atoms become negative ions and on the right side of the junction, the donor atoms become positive ions as shown in figure.
An electric field is set up, between the donor and acceptor ions in the depletion layer of the pn junction .The potential at the N-side is higher than the potential at P-side.Therefore electrons in the N- side are prevented to go to the lower potential of P-side. Similarly, holes in the P-side find themselves at a lower potential and are prevented to cross to the N-side. Thus, there is a barrier at the junction which opposes the movement of the majority charge carriers. The difference of potential from one side of the barrier to the other side of the barrier is called potential barrier.The potential barrier is approximately 0.7V for a silicon PN junction and 0.3V for germanium PN junction. The distance from one side of the barrier to the other side is called the width of the barrier, which depends on the nature of the material.
1/ RP=1/X +1/S
FOR METRE BRIDGE
R/l1 = RP/l2
X/XS + S/XS =l1 /Rl2
1/S+1/X = l1 /Rl2
1/X= l1 /Rl2 - 1/S
1/X=Sl1-Rl2 / RSl2
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this is the answer of your previous question
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