User:Friscosembel/draft

Drift current is charged particle motion in response to an applied electric field. Drift current in silicon has two components-current due to moving free electrons, and current due to moving holes. When an electrical field E is established in a semiconductor crystal, holes are accelerated in the direction of E, and free electrons are accelerated in the direction opposite to that of E. In fact holes behave as if they are positively charged particles, with a charge equal to that of an electron which have negative charge. When the charge start moving, they produce current. The motion of each carrier can be described as a constant drift velocity, $$v_{\it drift}$$.

Derivation
When an electrical field E is established in a semiconductor crystal, holes are accelerated in the direction of E, and free electrons are accelerated n the direction opposite to that of E. This situation is illustrated in the right figure. The hole velocity $$v_{\it p-drift}$$ given by


 * $$v_{\it p-drift} = \mu_{\it p} \cdot E$$

where $$\mu_{\it p}$$ us a constant called the hole mobility: it represents the degree of ease by which holes move through the silicon crystal in response to the electrical field E. Since velocity has the units of centimeters per second and E has the units of volts per centimeter, we see from the equation above that the mobility $$\mu_{\it p}$$ must have the unit of centimeter square per volt-second (cm^2/V.s) For intrinsic silicon $$\mu_{\it p}$$=480 cm^2/V.s

The free electrons acquire a drift velocity $$v_{\it n-drift}$$. given by


 * $$v_{\it n-drift} = \mu_{\it n} \cdot E$$

where the results is negative because the electrons move the the direction opposite to E. here $$\mu_{\it n}$$ is the electron mobility, which for intrinsic silicon is about 1350 cm^2/V.s. Note that $$\mu_{\it n}$$ is about 2.5 times $$\mu_{\it p}$$, signifying that electrons move with much greater ease through the silicon crystal that do holes.

Let us now return to the single-crystal silicon bar shown in figure above. Let the concentration of holes be p and that of free electron n. We wish to calculated the current component due to the flow of holes. Consider a plane perpendicular to the x direction. In one second, the hole charge that crosses that plane will be ($$Aqp v_{\it p-drift}$$) coulombs, where A is the cross-sectional area of the silicon bar and q is the magnitude of electron charge. This then must be the hole component of the drift current flowing through the bar ,


 * $$I_{\it p} = Aqpv_{\it p-drift}$$

Substituting for $$v_{\it p-drift}$$ from the above equation, we obtain


 * $$I_{\it p} = Aqp mu_{\it p} E$$

We are usually interested in the current density $$J_{\it p}$$, which is the current per unit cross-sectional area,


 * $$J_{\it p}=\frac {I_{it p}}{A}=qp mu_{\it p} E$$

The current component due to the drift of free electrons can be found in the similar manner. Note, however, that electrons drifting from right to left result in a current component from left to right. This is because of the convention of taking the direction of flow of negative charge. Thus,


 * $$I_{\it n} = -Aqpv_{\it n-drift}$$

Substituting for $$v_{\it n-drift}$$, we obtain the current density $$J_{\it n} = \frac {I_{it n}}{A}$$ as


 * $$J_{\it n}=\frac {I_{it p}}{A}=qn mu_{\it n} E$$

The total drift current density can now be found by summing $$J_{\it n}$$ and $$J_{\it p}$$,


 * $$J = J_{\it n} + J_{\it p} = q \left (p \mu_{\it p} + n \mu_{\it n} \right) E$$

Drift current and equilibrium
In addition to the current component Id due to majority-carrier diffusion, a component due to minority-carrier drift exist across the junction. Specifically, some of the thermally generated holes in the n material move toward the junction and reach the edge of the depletion region. There, they experience the electric field in the depletion region, which sweeps them across that region into the p side. Similarly, some of the minority thermally generated electrons in the p material move to the edge of the depletion region and get swept by the electric field in the depletion region across that region into the n side. These two current components-electrons moved by drift from p to n and holes moved by drift form n to p-add together to form drift current Is, whose direction is from the n side to the p side of the junction. Since the current Is is carried by thermally generated minority carriers, its value is strongly dependent on temperature; however, it is independent of the value of the depletion-layer voltage Vo. This is due to the fact that the drift current is determined by the number of minority carriers that make it to the edge of depletion region; any minority carriers that manage to get to the edge of the depletion region will be swept across by E irrespective of the value of E, or correspondingly, of Vo. See figure on the left. Under open circuit conditions no external current exist; thus the two opposite current across the junction must be equal in magnitude:

This equilibrium condition is maintained by the barrier voltage Vo. Thus, if for some reason Id exceeds Is, then more bound charge will be uncovered on both sides of the junction, the depletion layer will be widen, and the voltage across it (Vo) will increase. This in turn causes Id to decrease until equilibrium condition is achieved with Id=Is. On the other hand, if Is exceeds Id, then the amount of uncovered charge will decrease, the depletion layer will narrow, and the voltage across it (Vo) will decrease. This causes Id to increase until equilibrium is achieved with Id=Is..

Drift current vs diffusion current
Unlike diffusion current which composed of majority carrier electrons on the n side surmounting the potential energy barrier to diffuse to he p side and holes surmounting their barrier form p to n, drift current is relatively insensitive to the height of the potential barrier. The reason for this is the fact that the drift current is limited not by how fast carriers are swept down the barrier, but rather how often. For example, minority carrier electrons on the p side which wander into the transition region will be swept down the barrier by the electric field, giving rise to he electron component of drift current. However, this current is small not because of the size of the barrier, but because there are very few minority electrons in the p side to participate. Every electron on the p side which diffuses to the transition region will be swept down the potential energy hill, whether the hill is large or small. The electron drift current does not depend on how fast an individual electron is swept from p to n, but rather on how many electrons are swept down the barrier per second. Therefore, the electron and hole drift currents at the junction are independent of the applied voltage.