PN Junction PDF

Lectures: Dr.Hesham Ibrahim Lecturer: Mariam Omer Hesham Ibrahim SCPD DEEE @ UofK 2/19/2024 PN junction can be fabricated by implanting or diffusing donors into a P-type substrate such that a layer of semiconductor is converted into N type. Converting a layer of an N-type semiconducto...

Lectures: Dr.Hesham Ibrahim Lecturer: Mariam Omer Hesham Ibrahim SCPD DEEE @ UofK 2/19/2024 PN junction can be fabricated by implanting or diffusing donors into a P-type substrate such that a layer of semiconductor is converted into N type. Converting a layer of an N-type semiconductor into P type with acceptors would also create a PN junction Hesham Ibrahim SCPD DEEE @ UofK 2/19/2024 A PN junction has rectifying current–voltage (I–V or IV) characteristics as shown in Fig. 4–2. As a device, it is called a rectifier or a diode. The PN junction is the basic structure of solar cell, light-emitting diode, and diode laser, and is present in all types of transistors Hesham Ibrahim SCPD DEEE @ UofK 2/19/2024 The interface separating the n and p region is referred to as the metallurgical junction Hesham Ibrahim SCPD DEEE @ UofK 2/19/2024 For simplicity, it is usually assumed that the P and N layers are uniformly doped at acceptor density Na, and donor density Nd, respectively. This idealized PN junction is known as a step junction or an abrupt junction in which the doping concentration in uniform in the p and n region and there is an abrupt change in doping at the junction. Hesham Ibrahim SCPD DEEE @ UofK 2/19/2024 Let us construct a rough energy band diagram for a PN junction at equilibrium or zero bias voltage First draw a horizontal line for because there is only one Fermi level at equilibrium Hesham Ibrahim SCPD DEEE @ UofK 2/19/2024 Far from the junction, we simply have an N-type semiconductor on one side (with Ec close to EF), and a P-type semiconductor on the other side (with Ev close to EF). Hesham Ibrahim SCPD DEEE @ UofK 2/19/2024 Finally, in we draw an arbitrary (for now) smooth curve to link the Ec from the N layer to the P layer. Ev of course follows Ec, being below Ec by a constant Eg. Hesham Ibrahim SCPD DEEE @ UofK 2/19/2024 As electron diffuse from n to p region, positively charged donor are left in the n region As holes diffuse from p to n region, negatively charged acceptor are left in the p region The two region are referred to as the space charge region The charges will induce electric field Hesham Ibrahim SCPD DEEE @ UofK 2/19/2024 Ec and Ev are not flat. This indicates the presence of a voltage differential. The conduction and valence band must bend through the space charge region. Vbi  Fn  Fp Hesham Ibrahim SCPD DEEE @ UofK 2/19/2024 Electron in the conduction band of the n region see a potential barrier when moving into the conduction band in the p region. This built-in potential barrier is denoted as eVbi Vbi  Fn  Fp Hesham Ibrahim SCPD DEEE @ UofK 2/19/2024 This built-in potential barrier maintain equilibrium between i.majority carrier electron in the n region and minority electron carrier in the p region ii.majority carrier holes in the p region and minority holes carrier in the n region Vbi  Fn  Fp Hesham Ibrahim SCPD DEEE @ UofK 2/19/2024  Built-in Potential Barrier The built-in potential barrier is the difference between the intrinsic Fermi levels in the p and n regions Vbi  Fn  Fp In the n region the electron concentration is given by  ( EC  EF )  no  NC exp    kT  which can also be written in the form  ( EC  EF )   EF  EFi  no  NC exp    ni exp    kT   kT  Hesham Ibrahim SCPD DEEE @ UofK 2/19/2024  Built-in Potential Barrier The built-in potential barrier is the difference between the intrinsic Fermi levels in the p and n regions Vbi  Fn  Fp We can define potential  Fn in the n region as eFn  EFi  EF Thus, n0 may be written as  EF  EFi     eFn   no  ni exp    ni exp    kT   kT  Hesham Ibrahim SCPD DEEE @ UofK 2/19/2024  Built-in Potential Barrier Taking the natural log of both sides of where n0 = Nd    eFn   no  ni exp    kT  It becomes kT  N d  Fn  ln   e  i  n Hesham Ibrahim SCPD DEEE @ UofK 2/19/2024  Built-in Potential Barrier Similarly in the p region, the hole concentration is given as  ( EF  Ev )   ( EF  EFi )  po  N a  N v exp    ni exp    kT   kT  We can define potential  Fp in the n region as eFp  EFi  EF Thus, n0 may be written as  ( EF  EFi )   eFp  no  ni exp    ni exp    kT   kT  Hesham Ibrahim SCPD DEEE @ UofK 2/19/2024  Built-in Potential Barrier Taking the natural log of both sides of where n0 = Nd  eFp  no  ni exp    kT  It becomes kT  N a  Fp   ln   e  ni  Hesham Ibrahim SCPD DEEE @ UofK 2/19/2024  Built-in Potential Barrier Therefore, the built-in potential barrier becomes Vbi  Fn  Fp  kT  N d  kT  N a   ln   ln   e  ni  e  ni  kT  N a N d   Na Nd   ln  2   Vt ln  2  e  ni   ni  Hesham Ibrahim SCPD DEEE @ UofK 2/19/2024  Depletion Model Let’s divide the PN junction into three regions—the neutral regions at x > xP and x < –xN, and the depletion layer or depletion region in between, where p = n = 0.The charge density is zero everywhere except in the depletion layer where it takes the value of the dopant ion charge density Hesham Ibrahim SCPD DEEE @ UofK 2/19/2024  (a) P Depletion-Layer Width N Nd Na N eut ra l Re gion D eple tion La yer N e utral R egi on (b) N P –xnN 0 xpP V is continuous at x = 0  2 sbi  1 1  xP  x N  Wdep     q  Na Nd  If Na >> Nd , as in a P+NqN junction, d (c) xp 2 sbi Wdep  –xxn N xP  xN N d N ax qN d –qN a What about a N+P junction? 1 1 1 1 Wdep  2 s bi qN where    N N d N a lighter dopant density (d) –xn 0 Hesham Ibrahim SCPD DEEE @ UofK xp x 2/19/2024 EXAMPLE: A P+N junction has Na=1020 cm-3 and Nd =1017cm-3. What is a) its built in potential, b)Wdep , c)xN , and d) xP ? Solution: a) kT N d N a 1020 1017 cm6 bi  ln 2  0.026V ln 20 6 1V q ni 10 cm 1/ 2 b) W  2 sbi   2 12  8.85 10 1  14 dep  1.6 1019 1017   0.12 μm qN d   c) xN  Wdep  0.12 μm d) xP  xN N d N a  1.2  10 4 μm  1.2 Å  0 Hesham Ibrahim SCPD DEEE @ UofK 2/19/2024  Reverse-Biased PN Junction V + – N P Ec 2 s (bi  | Vr |) 2 s  potential barrier qbi Wdep   qN qN Ec Ef Ef Ev Ev (a) V = 0 1 1 1 1    Ec N N d N a lighter dopant density qbi + qV Efp Ev Does the depletion layer Ec qV Efn widen or shrink with increasing reverse bias? Ev (b) reverse-biased Hesham Ibrahim SCPD DEEE @ UofK 2/19/2024  Maximum Electric Field The maximum electric field at the metallurgical (The interface between the p- and n-doped regions ) junction is eN d xn eN a x p Emax   s s that yield  2e Vbi  VR   N a N d 12   Emax       s  Na  Nd   The maximum electric field in the pn junction can also be written as 2 Vbi  VR  Emax   W Hesham Ibrahim SCPD DEEE @ UofK 2/19/2024  summary At the p-n junction, the excess conduction-band electrons on the n-type side are attracted to the valence-band holes on the p-type side. The electrons in the n-type material migrate across the junction to the p-type material (electron flow) The electron migration results in a negative charge on the p-type side of the junction and a positive charge on the n-type side of the junction. The result is the formation of a depletion region around the junction. Hesham Ibrahim SCPD DEEE @ UofK 2/19/2024  Example 1 Calculate the built-in potential barrier, Vbi for Si, Ge and GaAs pn junctions if they each have the following dopant concentrations at T=300K a)Nd= 1014 cm-3 Na= 1017 cm-3 b)Nd= 51016 cm-3 Na= 5 1016 cm-3 c)Nd=1017 cm-3 Na=1017 cm-3 Hesham Ibrahim SCPD DEEE @ UofK 2/19/2024  Example 1 Calculate the built-in potential barrier, Vbi for Si, Ge and GaAs pn junctions if they each have the following dopant concentrations at T=300K a)Nd= 1014 cm-3 Na= 1017 cm-3 b)Nd= 51016 cm-3 Na= 5 1016 cm-3 c)Nd=1017 cm-3 Na=1017 cm-3 Si : ni  1.5 1010 cm Ge : ni  2.4 1013 cm GaAs : ni  1.8 106 cm  Na Nd  Vbi  Vt ln  2   ni  Hesham Ibrahim SCPD DEEE @ UofK 2/19/2024  Example 1 Calculate the built-in potential barrier, Vbi for Si, Ge and GaAs pn junctions if they each have the following dopant concentrations at T=300K a)Nd= 1014 cm-3 Na= 1017 cm-3 Then Si: 0.635V Ge: 0.253V GaAs: 1.10V Hesham Ibrahim SCPD DEEE @ UofK 2/19/2024  Example 1 Calculate the built-in potential barrier, Vbi for Si, Ge and GaAs pn junctions if they each have the following dopant concentrations at T=300K b)Nd= 51016 cm-3 Na= 5 1016 cm-3 Then Si: 0.778V Ge: 0.396V GaAs: 1.25V Hesham Ibrahim SCPD DEEE @ UofK 2/19/2024  Example 1 Calculate the built-in potential barrier, Vbi for Si, Ge and GaAs pn junctions if they each have the following dopant concentrations at T=300K c)Nd=1017 cm-3 Na=1017 cm-3 Then Si: 0.814V Ge: 0.432V GaAs: 1.28V Hesham Ibrahim SCPD DEEE @ UofK 2/19/2024  Example 2 An abrupt silicon pn junction at zero bias has dopant concentration of Na=1017 cm-3 and Nd=51017 cm-3. T=300K a)Calculate the Fermi level on each side of the junction with respect to the intrinsic Femi level b)Sketch the equilibrium energy-band diagram for the junction and determine Vbi N from the diagram and the results of the part (a) c)Determine xn and xp and the peak electric for this junction Hesham Ibrahim SCPD DEEE @ UofK 2/19/2024  Example 2 An abrupt silicon pn junction at zero bias has dopant concentration of Na=1017 cm-3 and Nd=51017 cm-3. T=300K a)Calculate the Fermi level on each side of the junction with respect to the intrinsic Femi level Vbi  Fn  Fp eFn  EFi  EF eFp  EFi  EF EFi  EF EFi  EF Vbi   e n e p Hesham Ibrahim SCPD DEEE @ UofK 2/19/2024  Example 2 An abrupt silicon pn junction at zero bias has dopant concentration of Na=1017 cm-3 and Nd=51017 cm-3. T=300K a)Calculate the Fermi level on each side of the junction with respect to the intrinsic Femi level N-side  Nd   5 1015  EF  EFi  kT ln     0.0259  ln  10   0.3294 eV  ni   1.5 10  P-side  Na   1017  EFi  EF  kT ln     0.0259  ln  10   0.4070 eV  ni   1.5 10  Hesham Ibrahim SCPD DEEE @ UofK 2/19/2024  Example 2 An abrupt silicon pn junction at zero bias has dopant concentration of Na=1017 cm-3 and Nd=51017 cm-3. T=300K b)Sketch the equilibrium energy-band diagram for the junction and determine Vbi from the diagram and the results of the part (a) Vbi  Fn  Fp  0.7364 V Hesham Ibrahim SCPD DEEE @ UofK 2/19/2024  Example 2 An abrupt silicon pn junction at zero bias has dopant concentration of Na=1017 cm-3 and Nd=51017 cm-3. T=300K c)Determine xn and xp and the peak electric for this junction 1/2  2 s Vbi  N a  1  xn        0.426  m  e  d  N a  N d N  1/2  2 s Vbi  N d  1  xp        0.0213  m  e  a  N a  N d N   max  eN d xn  1.6 10  5 10  0.426 10  19 15 4 s 11.7  8.85 10  14  3.29 104 V/cm Hesham Ibrahim SCPD DEEE @ UofK 2/19/2024  Example 3 An abrupt silicon pn junction at T=300K zero bias has impurity doping concentration of Na=51016 cm-3 and Nd=1015 cm-3. Calculate a)Vbi b)W at (i) VR=0 and (ii) VR=5V c)Emax at (i) VR=0 and (ii) VR=5V Hesham Ibrahim SCPD DEEE @ UofK 2/19/2024  Example 3 An abrupt silicon pn junction at T=300K zero bias has impurity doping concentration of Na=51016 cm-3 and Nd=1015 cm-3. Calculate a)Vbi   5  1016 1015   Vbi   0.0259  ln    1.5  1010  2    0.6767V Hesham Ibrahim SCPD DEEE @ UofK 2/19/2024  Example 3 An abrupt silicon pn junction at T=300K zero bias has impurity doping concentration of Na=51016 cm-3 and Nd=1015 cm-3. Calculate b)W at (i) VR=0 and (ii) VR=5V  2 s Vbi  VR   N a  N d 1/2   W     e  Na Nd   For VR =0  2 11.7   8.85 10   0.6767   5  1016  1015   1/2 14 W      5  10 10    19  1.6 10 16 15 W  0.9452  m Hesham Ibrahim SCPD DEEE @ UofK 2/19/2024  Example 3 An abrupt silicon pn junction at T=300K zero bias has impurity doping concentration of Na=51016 cm-3 and Nd=1015 cm-3. Calculate b)W at (i) VR=0 and (ii) VR=5V  2 s Vbi  VR   N a  N d 1/2   W     e  Na Nd   For VR =5  2 11.7   8.85 10   0.6767  5   5  1016  1015   1/2 14 W      5  10 10    19  1.6 10 16 15 W  2.738  m Hesham Ibrahim SCPD DEEE @ UofK 2/19/2024  Example 3 An abrupt silicon pn junction at T=300K zero bias has impurity doping concentration of Na=51016 cm-3 and Nd=1015 cm-3. Calculate c)Emax at (i) VR=0 and (ii) VR=5V 2 Vbi  VR   max  W For VR =0 2  0.6767   max  4  1.43  10 4 V/cm 0.9452  10 Hesham Ibrahim SCPD DEEE @ UofK 2/19/2024  Example 3 An abrupt silicon pn junction at T=300K zero bias has impurity doping concentration of Na=51016 cm-3 and Nd=1015 cm-3. Calculate c)Emax at (i) VR=0 and (ii) VR=5V 2 Vbi  VR   max  W For VR =5 2  0.6767  5   max  4  4.15  10 4 V/cm 2.738 10 Hesham Ibrahim SCPD DEEE @ UofK 2/19/2024  Junction Breakdown Junction Breakdown – Zener Breakdown – Avalanche Breakdown Hesham Ibrahim SCPD DEEE @ UofK 2/19/2024  Junction Breakdown  As the reverse voltage increases the diode can avalanche-breakdown (zener breakdown).  Zener breakdown occurs when the electric field near the junction becomes large enough to excite valence electrons directly into the conduction band and generate carriers Hesham Ibrahim SCPD DEEE @ UofK 2/19/2024 Junction Breakdown Hesham Ibrahim SCPD DEEE @ UofK 2/19/2024  Avalanche Breakdown  The avalanche process occurs when the carriers in the transition region are accelerated by the electric field to energies sufficient to free electron-hole pairs via collisions with bound electrons. Hesham Ibrahim SCPD DEEE @ UofK 2/19/2024  Junction Breakdown The breakdown voltage can be given as  s Ecrit 2 VB  2eN B Where NB is the semiconductor doping in the low-doped region of the one sided junction while Ecrit is actually Emax at breakdown Hesham Ibrahim SCPD DEEE @ UofK 2/19/2024  Example 5 Consider a silicon n+p junction diode. The critical electric field for breakdown in silicon is approximately Ecrit= 4 105 V/cm. Determine the maximum p-type doping concentration such that the breakdown voltage is a)40V b)20 V Hesham Ibrahim SCPD DEEE @ UofK 2/19/2024  Example 5 Consider a silicon n+p junction diode. The critical electric field for breakdown in silicon is approximately Ecrit= 4 105 V/cm. Determine the maximum p-type doping concentration such that the breakdown voltage is a)40V b)20 V s  crit 2 VB  or 2eN B s  2 11.7  8.85 10  4 10 14  5 2 NB   crit 2eVB 2 1.6 10 19   40  N B  N a  1.294 1016 cm -3 Hesham Ibrahim SCPD DEEE @ UofK 2/19/2024  Example 5 Consider a silicon n+p junction diode. The critical electric field for breakdown in silicon is approximately Ecrit= 4 105 V/cm. Determine the maximum p-type doping concentration such that the breakdown voltage is a)40V b)20 V s  crit 2 VB  or 2eN B s  2 11.7  8.85 10  4 10 14  5 2 NB   crit 2eVB 2 1.6 10 19   20  N B  N a  2.59 1016 cm -3 Hesham Ibrahim SCPD DEEE @ UofK 2/19/2024