1. Consider the formation of a p-n junction by placing two n- doped and p-doped crystals side-by-side:  (doping density) NdNd NaNa e-e- Positive charges occur when donors on the n-side are ionized during the electron transfer to the p-side. n-type p-type 0 x Negative charges form when holes are ionized by the capture of electrons. Electrons (e - ) flow to the p-side holes (h + ) flow to the n-side. Consider a p-n junction in equilibrium: Assume that non-degenerate conditions hold for all x. “Abrupt junction model” (1) There is a transfer of charge between the n- and p-type regions in order to equalize the Fermi-level (or chemical potential) on both sides. h+h+ Before: p-siden-side ECEVECEV   After: p-side n-side  ( E F ) ECEVECEV

2. In order to describe the spatial variation in the band edges, E C and E V, we introduce a potential function,  (x), such that each level is shifted by -e  (x). That is, E C  E C - e  (x) and E V  E V - e  (x). Therefore, the carrier concentrations can be expressed by Consider limits at x =  (2)

3. We can solve for the difference of the two potentials at  The electric field is related to the potential by Gauss’s Law: for our 1D problem. We assume that donor and acceptor impurities are fully ionized at all x. (3) We can substitute Eqs. (1) and (2) into Eq.(3) to get a difficult nonlinear differential equation.

4. We can make some important approximations to simplify and solve: Assume that the potential change  occurs in a finite region near the junction that is defined by -d p  x  d n. This is referred to as the depletion region. Also n << N d and p << N a in this region since  is closer to the middle of the gap (i.e., midgap). We can solve this problem assuming an abrupt junction model which will give a simple linear second order ordinary differential equation: Note that we have first used the boundary conditions at the edges of the depletion region (see graph on next slide) to solve for  (x): The potential must be its limiting value at the boundaries of the depletion region. The electric field must be zero outside the depletion region for equilibrium to hold.

5. NdNd NaNa Carrier density n c (x) p v (x) n-typep-type -dp-dp dndn 0 Depletion layer Charge density  (x) eN d -eN a dndn -dp-dp x x dndn -dp-dp (x)(x)  ()()  (-  ) x Graphical illustration of the Abrupt-junction model.

6. dndn -dp-dp (x)(x)  ()()  (-  ) x x Comparison of the potential and electric field in the abrupt-junction model: There are important relations that we get by imposing continuity of  (x) and  ’(x). Continuity of the derivative gives:

7. Secondly, continuity of  (x) gives Together with the derivative condition: N d d n =N a d p this gives For N a,d ranging from 10 14 to 10 18 cm -3  d n,p ranges from 10 4 to 10 2 Å The total depletion width is L D = d n +d p We can also write this in a numerically convenient form: The maximum electric f ield within the depletion layer is of the order of  / (d n+a d p ) and ranges from 10 5 to 10 7 volts/m.

8. Charge density,  (x) -eN a dndn -dp-dp x dndn -dp-dp x dndn -dp-dp x (x)(x)  o  o - |V 1 |  o + |V 2 | 1(x)1(x) 2(x)2(x) Consider the effect of an applied voltage on the depletion width of a p-n junction: p n V = 0 “zero-bias” V 1 > 0 “Forward-Bias” p n V 2 < 0 “Reverse-Bias” The potential simply changes according to  =  o  |V| - sign: “Forward bias” + sign: “Reverse bias” We are changing the size of the depletion region by applying an external voltage.

9. The formula for the depletion width can easily be modified to include the effects of a voltage bias: Zero-bias depletion length Consider rectification by a p-n junction. The symbol J is used for # of carriers /(area·time). The lower case symbol j is more commonly used for current density and has units j = coulombs/ (area·time). Note j e = -e J e and j h = e J h. p-side n-side  ( E F ) ECEVECEV Consider electron and hole generation currents. These carriers are generated by thermal excitation as we saw before according to Note: The generation currents involve minority carriers. e-e- h+h+

10. In a band diagram, electrons fall down hill and holes “float” uphill (both towards lowest energy). In the process of the generation currents, holes (from n-side)  p-side, electrons (from p-side)  n-side. Secondly, consider another kind of current, Recombination current. With this kind of current holes (from p-side)  n-side. Electrons from the n-side  p-side. This current involves majority carriers, and it is made possible by thermal excitation over the barrier, as shown below: p-side n-side  ( E F ) ECEVECEV  o - |V 1 | e-e- h+h+ Consider the hole recombination current. The current is approximated well by assuming thermionic emission over the barrier: The middle equation is a statement that no net current can flow during equilibrium when V=0.

11. The total hole current from the p to the n-side is given by the recombination current minus the generation current: The last equation includes the currents of both holes and electrons, since the same analysis will also apply to electrons. It is obvious to understand the rectifying properties of a p-n junction (diode) from this equation. p n V V j Rectifying behavior of a p-n junction (diode). Saturation Reverse bias (V<0)Forward bias (V>0) We still need to calculate this term in terms of fundamental parameters describing transport. Note: In our notation J h and J p will mean the same thing.

12. In a more general treatment, it is not necessary to separate into generation and recombination currents. Note that to solve for the following five fundamental quantities (J e (x), J h (x), n(x), p(x), and  (x)), we need five equations. In the equilibrium case with V=0, J e (x)= J h (x)=0, we need three equations to solve the three unknowns, including Poisson’s equation for the potential. In the presence of a field E and  n and  p, we can write the electron and hole current densities as: Where  n and  p are the electron and hole mobilities in units of cm 2 /V·s, and D n and D p are the electron and hole diffusion constants in units of cm 2 /s When dn/dx=0, From the Dude theory (m  v=F  t) recall Therefore, Which expresses  in terms of the collision time and masses, based on the Drude model.

13. Now, recall that At equilibrium, These last two relations are known as the Einstein relations. Consider continuity equations for transport of charge: If V  0 and carriers are conserved. We have to include two other processes which act as a source and drain for carriers: (i) Generation by thermal excitation (ii) Recombination (electron  hole) ECEVECEV (i)(i)(ii) As a result the continuity equations need to include these additional terms:

14. The g-r terms act to restore the system to equilibrium when the system deviates from equilibrium. These terms can be further described by electron and hole lifetimes (  n and  p ): where n o and p o are equilibrium concentrations as determined by the law of mass action. Note that  n >>  n col and  p >>  p col ; typically  n ~10 -9 s and  n col ~10 -13 s Very often we deal with a steady state condition in which This gives p n V For V  0, we are not in equilibrium but we have a steady state for V = const. Note that J e = J h = 0 for V = 0  n = n o and p = p o. Now suppose that E  0, i.e., the electric field is negligible. Then Fick’s Law for Diffusion

15. Therefore, further give the following 2 nd order ordinary differential equations which are referred to as diffusion equations: Note n o and p o are equilibrium values. The solutions are easily written as: The diffusion lengths L p and L n are given by Note that when E  0, the majority carrier density is constant and the carriers in the diffusion equation are for the minority carriers. For example if we have a p-type material pn = n i 2 and p >> n so that electrons (n p ) are minority carriers. The notation is slightly changed to be more precise: p-typen-type

16. Examine the meaning of the diffusion lengths, L n and L p. Since From the equipartion theorem and Drude approximation: ½ mv th 2 = 3/2 k B T Further, the mean free path l n is given by l n =v th  col where v th is the thermal velocity of a carrier. The interpretation here is as follows: When a carrier undergoes N collisions before recombination, the net displacement will be The factor of 3 comes from the number of degrees of freedom (3D). This is basically a “random walk” problem. 1 2 3 4 Note that  n >>  n col and  p >>  p col ; typically  n ~10 -9 s and  n col ~10 -13 s  N  3000 as the number of collisions before recombination in GaAs.

17. We need to improve our understanding of transport across the p-n junction. Consider an equivalent description of Boltzmann statistics at thermal equilibrium: Remember that where n i is the intrinsic carrier concentration E F =  is the Fermi-level (or chemical potential), and E i is the position of the chemical potential for the intrinsic case. Let us define potentials  = -E i /e and  = -E F /e. At thermal equilibrium pn=n i 2 (V=0 across the junction). For the case of V  0, we can write the carrier concentrations as Where  n and  p are called quasi-Fermi levels or imrefs (imaginary references). The advantage of this formalism is that the potential difference V across the junction is just the difference in the quasi-Fermi levels, i.e.

18. E (eV) ECEC EVEV E F (  ) -e  p -e  n p-siden-side -d p 0 d n Potential pp nn -d p 0 d n  V log(n,p) -d p 0 d n p po p no n po n no pnpn npnp nini Forward-Bias Conditions -e  p -e  n -d p 0 d n p-siden-side Reverse-Bias Conditions ECEC EVEV E F (  ) pp nn -d p 0 d n -V log(n,p) -d p 0 d n p po p no n po n no pnpn npnp nini xx  x x xx Potential E (eV) Note that (d p +d n ) for reverse bias is greater than (d p +d n ) for forward bias.

19. For a forward bias V =  p -  n > 0 and pn > n i 2 in the junction. The opposite is the case for a reverse bias V =  p -  n < 0 and pn < n i 2. Now consider the current density and write in terms of  and  : Note that Our previous notation was Similarly, In the depletion layer –d p < x <d p since n and p are sharply decreasing functions in the depletion layer. Also, we assume that passage of carriers across the junction is very fast so that generation and recombination currents in the depletion region are negligible. This leads to assumption that J e and J p are constant in the depletion region –d p < x <d p For  p and  n constant, J e and J p are 0.

20. Consider n at x = -d p (on the p-side): Note that n po is the electron density on the p-side at x = - . Similarly, at x = +d n And p no is the hole density on the n-side at x = +  These equations serve as boundary conditions for the I-V equation of the ideal p-n junction. Recall that just outside of the depletion region (i.e. |x| > d n, d p the region is neutral, E  0, and this is called the diffusion region. In these regions, the following diffusion equations apply for the respective minority carrier densities: (1) (2) Using boundary conditions (1) and (2), the solutions are: Note that the expected limits are observed: n p  n po at x = -  while p n  p no at x = + 

21. We can now calculate the electron and hole currents entering the diffusion region where E  0. At x = d n for the n-side. for the p-side. The total current is given by J = J e + J p where This result is the well known Shockley equation. Previously we wrote the current in terms of electron and hole generation currents: Note that (on n-side)(on p-side) and Contains fundamental parameters describing the rectifying behavior.

22. Examine the minority carrier densities and current densities for forward and reverse bias: n,p -d p 0 d n p no n po pnpn npnp x J -d p 0 d n x J=J n +J p JnJn JpJp p-siden-side n,p -d p 0 d n p no n po pnpn npnp x p-siden-side J -d p 0 d n x J=J n +J p JnJn JpJp Note that (d p +d n ) for reverse bias is greater than (d p +d n ) for forward bias. Forward Bias (V >0)Reverse Bias (V<0)

23. Possible to examine details of diffusion, using the Haynes-Shockley Experiment Consider changes in minority carrier density with time: For E = 0, the solution follows the solution of the diffusion equation: Where N= number of holes/Area generated (laser light pulse). If E  0, x  x -  p Et h 0 L Sample t V Oscilloscope V t 1 (E = 0) t 2 t 3 t 1 t 2 (E > 0)  Et 1  Et 2