1 / 33

618 326

618 326. Physics of electronic materials and devices I Lecture 9. Forward Bias. Forward Bias Under thermal equilibrium, numbers of electrons crossing in opposite direction are equal. Therefore, there is no net transfer of charge that leads to no current flow.

cheveyo-gimenez
Download Presentation

618 326

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. 618 326 Physics of electronic materials and devices I Lecture 9

  2. Forward Bias Forward Bias • Under thermal equilibrium, numbers of electrons crossing in opposite direction are equal. • Therefore, there is no net transfer of charge that leads to no current flow. • This is the same for the case of holes.

  3. Forward Bias • If a positive voltage VF is applied to the p-side with respect to the n-side, the p-n junction becomes forward-biased. • The flow of electrons from left to right is not affected, but the flow of them from right to left is certainly affected.

  4. Forward Bias For conclusion, forward bias causes • The depletion region decreases. • Barrier height decreases. qVbiq(Vbi - VF) • Net transfer of charges occurs as free electrons transported from right to left producing electron current flow and free holes transported from left to right producing hole current. VF I (Ie + Ih)  electrons: R  L Ie: L  R holes: L  R Ih: L  R

  5. Forward Bias

  6. Reverse Bias • If we apply positive VR to the n-side with respect to the p-side, the p-n junction is now reverse-biased. • This reverse bias causes • Depletion width increases • Barrier height increases qVbiq(Vbi + VR)

  7. Reverse Bias • Net transfer of charge occurs but just few free electrons are able to be transported from left to right causing electron current flow. • Also, few holes are transported from right to left causing hole current flow. I = Ie + Ih small current electrons: L  R Ie: R  L holes: R  L Ih: R  L

  8. Reverse Bias

  9. Depletion capacitance • When p-n junction is reverse biased, the presence of 2 layers of space-charge in the depletion region makes it look like a capacitor. • The solid line indicating charge and electric field distribution in the figure corresponds to an applied voltage V.

  10. Depletion capacitance • The dashed line represents the charge and electric field distribution when the applied voltage is increased by dV. • The incremental space charges on both n- and p-sides in the space-charge region are equal but with opposite charge polarity.

  11. Depletion capacitance • The incremental charge dQ causes an increase in the electric field by dE = dQ/s. • While the corresponding change in the applied voltage dV is approximately equal to WdE.

  12. Depletion capacitance The depletion capacitance per unit area is given by Or (1) Equation (1) is a good assumption for the reverse bias condition.

  13. Depletion capacitance • For forward bias, there is an additional term called diffusion capacitancedue to a large number of mobile carriers moving across the junction.

  14. Depletion capacitance Therefore, for a one-sided abrupt junction, we have (2) Consider a plot of 1/Cj2 versus V of (2). The slope gives the impurity concentration NB of the semiconductor and the intercept at 1/Cj2= 0 yields Vbi.

  15. Depletion capacitance • Whereas, the depletion layer capacitance in a case of linearly graded junction can be expressed by (3)

  16. Example 1 • For a silicon one-sided abrupt junction with NA = 2 x 1019 cm-3 and ND = 8 x 1015 cm-3, calculate the junction capacitance at zero bias and reversed bias of 4 V.

  17. Example 1 • Soln

  18. Current-Voltage Characteristics We now consider an ideal case of current-voltagecharacteristics based on these assumptions • the depletion region has abrupt boundaries • the low-injection condition • neither generation nor recombination current exists in the depletion region • the electron and hole currents are constant throughout the depletion region

  19. Current-Voltage Characteristics • At thermal equilibrium, the majority carrier density (nn0 or pp0) in the neutral regions is equal to the doping concentration. • The built-in potential can be written as (4)

  20. Current-Voltage Characteristics By using the mass action law pp0.np0 = ni2, equation (4) can be rewritten as (5) From equation (5), we have (6) (7)

  21. Current-Voltage Characteristics • We clearly see from (6) and (7) that the charge densities at the boundaries of the depletion region are related to the potential difference Vbi at thermal equilibrium. • If the voltage V is applied to the junction, the potential Vbi is changed and (6) becomes (8) where nn and npare the nonequilibrium densities at the boundaries of the depletion region in the n- and p-sides, respectively, with V positive for forward bias and negative for reverse bias.

  22. Current-Voltage Characteristics • In case of low-injection condition, the injected minority carrier is much smaller than the majority carrier density, so that nnnn0. Substituting this condition with (6) into (8) yields (9) (10) where np = the electron density at the boundary of the depletion region on the p-side at x = -xp.

  23. Current-Voltage Characteristics • Similarly, we have (11) (12) where pn = the hole density at the boundary of the depletion region on the n-side at x= xn.

  24. Current-Voltage Characteristics Depletion region, energyband diagram and carrier distribution. • (a) Forward bias. • (b) Reverse bias.

  25. Current-Voltage Characteristics • Since we assume that no current is generated within the depletion region, all currents come from the neutral regions. • In the neutral n-region, the steady-state continuity equation can be expressed as (13)

  26. Current-Voltage Characteristics • The solution of (13) with the boundary conditions of (12) and pn(x = )= pn0gives (14) where Lp is the diffusion length of holes in the n-region = .

  27. Current-Voltage Characteristics (15) Similarly, for the neutral p-region (16) (17) where Lp is the diffusion length of electrons in the p-region =.

  28. Current-Voltage Characteristics • The figure shows that the injected minority carriers recombine with the majority carriers as they move away from the boundaries. • (a) forward bias • (b) reverse bias

  29. Current-Voltage Characteristics • The hole and electron diffusion current will decay exponentially in the n-region with diffusion length Lp and p-region with diffusion length Ln, respectively. • The figure illustrates idealized currents. For practical devices, the currents are not constant across the space charge layer.

  30. Current-Voltage Characteristics • The total current is constant throughout the device and can be written as (18) This is called anideal diode equation. (19) where Js is the saturation current density.

  31. Current-Voltage Characteristics Ideal current-voltage characteristics. • (a) Cartesian plot. • (b) Semilog plot.

  32. Example 2 Calculate the ideal reverse saturation current in a Si p-n junction diode with a cross-section area of 2 x 10-4 cm2. The parameters of the diode are NA = 5 x 106 cm-3, ND = 1016 cm-3, ni = 9.65 x 109 cm-3, Dn = 21 cm2/s, Dp = 10 cm2/s, and τp = τn = 5 x 10-7 s.

  33. Example 2 • SolnFrom (19) and

More Related