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1 ECE 66: SOLUTIONS: ECE 66 Homework Week 8 Mark Lundstrom March 7, 13 1) The doping profile for an n- type silicon wafer ( N D = 1 15 cm - 3 ) with a heavily doped thin layer at the surface (surface concentration, N S = 1 cm - 3 ) is sketched below. Answer the following questions. 1a) Assume approximate space charge neutrality ( n x! N D ( x) ) and equilibrium conditions and compute the position of the Fermi level with respect to the bottom of the conduction band at x = and as x!. n ( ( x) = N C e E F!E C ) k T B N D x E F! E C ( x) = k B T ln N x D ' N C ' E F! E C = k B T ln N D N C ' ' E F E C N C = 3.3!1 19 cm - 3 = k B T ln = +1.13k B T ECE- 66 1

2 ECE 66: = k B T ln N x D E F! E C x N C 1 E F E C ( x ) 15 = k B T ln = 1.4k B T ' ) () 1b) Using the above information, sketch E C ( x) vs. x. Be sure to include the Fermi level. 1c) Sketch the electrostatic potential vs. position. 1d) Sketch the electric field vs. portion. ECE- 66

3 ECE 66:, in terms of ( x). HINT: Use the electron current 1e) Derive an expression for the position dependent electric field, E x the position- dependent doping density, N D equation and assume equilibrium conditions. J n = nqµ n E + k B T µ n dn dx = E = k B T q! E = k T B q 1 dn n dx = k T B q N D 1 x N D dn D dx 1 x ( x) dn D x dx ( Another way is to begin with n N D = N C e E F E C ) k T B and differentiate. ) A silicon diode is symmetrically doped at N D = = 1 15 cm - 3. Answer the following questions assuming room temperature, equilibrium conditions, and the depletion approximation. a) Compute. = k T B q ln! N D! 13 =.6ln 1 =.6 V =.6 b) Compute x n,x p and W. x n =! S q N D + N D ( '( 1/ =.65 µm x n = x p =.65 µm (because N and P regions are symmetrical) W = x n + x p = 1.5 µm ECE- 66 3

4 ECE 66: c) Compute V ( x = ) and E ( x = ). By symmetry: V ( ) = =.3 V or use V ( x = ) = q x! S p E ( x = ) = q x! S p = E ( ) =! V/cm d) Sketch!( x) vs. x. ρ N = +qn D = C/cm 3! P = q = C/cm 3 3) Your textbook (Pierret, SDF) presents the classic expressions for PN junction electrostatics. Simplify these expressions for a one- sided P + N junction for which >> N D. Present simplified expressions (when possible) for: 3a) The built- in potential,, from Pierret, Eqn. (5.1). = k BT q ln! N D no simplification possible ECE- 66 4

5 ECE 66: 3b) The total depletion layer depth, W, from Pierret, Eqn. (5.31). ) W =! S + q * + N D N D ' ( V, bi. - 1/ >> N D W =! S qn D ( ' 1/ 3c) The peak electric field, E ( ), from Pierret, Eqn. (5.19) or (5.1). E ( ) = W = q N D! s + N D ' ( E = qn D! s 3d) The electrostatic potential, V ( x) from Pierret, Eqn. (5.8) V ( x) =! qn D ( x n! x) V ( x) = V S bi qn D ( W x) κ S ε Now use the expression for W above to find: = 1 ( 1 x W ) V x 4) A silicon diode is asymmetrically doped at = 1 19 cm - 3 and N D = 1 15 cm - 3 Answer the following questions assuming room temperature, equilibrium conditions, and the depletion approximation. 4a) Compute. = k T B q ln! N D =.6ln! 15 ' =.84 V =.84 ECE- 66 5

6 ECE 66: 4b) Compute x n,x p and W. x p! x n! W = S qn D ' ) ( 1/ = 1.5 µm W = 1.5 µm (depletion region mostly on the N- side, the lightly doped side) 4c) Compute V ( x = ) and E ( x = ). V ( )! V E ( ) = qn D W = V/cm! S E ( ) = 1.6!1 4 V/cm (plus sign assumes N regios on the left) 4d) Sketch!( x) vs. x. The charge on the P- side is essentially a delta function with the total charge in C/cm equal in magnitude and opposite in sign to the charge on the N- side. ECE- 66 6

7 ECE 66: 5) Repeat problem 4) using the exact solution to PN junction electrostatics. V N = + k BT q ln V P =! k T B q ln N D 115 =.6 ln 1 1 = ' =.6ln 1 1 ' =!.54 = V N!V P =.84 V =.84 V ( ) = C! C N P a N! a P a N = N D = 1 15 a P =! =!1 19 C N = a N V N! ( k B T q)cosh( qv N k B T ) C N = 1 15!.3!1 1 (.6)cosh 11.5 =.74!1 14 C P = a P V P! ( k B T q)cosh( qv P k B T ) C P =!1 19! 1 1 (.6)cosh!.7!.54 = V ( ) = C! C N P =!.518 a N! a P V ( )! V P =!.54!.51 =.8 k B T q The potential drop across the heavily doped side is about kbt/q. E = q! S k B T q Putting in numbers, we find: E ( )! V/cm V/cm ( e qv () kbt + ( k B T q)e qv () kbt a N V() + C N ) 1/ which is about 1X the electric field we found in prob. 4. ECE- 66 7

8 ECE 66: = e qv = e!qv n p! k BT = 7 cm - 3 = q p q !! + k BT = cm - 3 n ( ) + N D = q.37 ' ( q.37 '1 19 ' ( (depletion approximation would give! = q p ( ) n = q ! + = q.37 ' ' (depletion approximation would give! + q 1 15 ' ) q 1 19 ' ) 6) Semiconductor devices often contain high- low junctions for which the doping density changes magnitude, but not sign. The example below shows a high- low step junction. Answer the questions below. ECE- 66 8

9 ECE 66: 6a) Sketch an energy band diagram for this junction. 6b) Sketch V ( x) 6c) Sketch E ( x) ECE- 66 9

10 ECE 66: 6d) Sketch!( x) vs. x. 6e) Name the charged entities responsible for!( x) in 6d). For x <, the charge is a depletion charge. Mobile electrons leave the heavily doped side of the junction leaving behind a concentration, ND1, of ionized donors. For x >, the charge is due to the additional mobile electrons that have spilled over from the heavily doped side. This is NOT a depletion region. 6f) Explain why the depletion approximation cannot be used for this problem. Because, as explained above, there is a depletion region on only ONE side of the junction. We could use the depletion approximation there, but not on the lightly doped side. 6g) Calculate for this high- low junction assuming silicon at room temperature. First, consider the two sides of the junction separately: ( n 1 = N C e E F 1!E C ) k B T ( n = N C e E F!E C ) k B T n 1 ( = e E F 1!E F ) k B T n The built- in potential develops to align these two Fermi levels: ( E F1! E F ) = q = k B T ln n 1 ' = k T B q ln N D1 N D n ECE- 66 1

11 ECE 66: 7) Consider an N + P diode with the length of the quasi- neutral P- region being, WP. Answer the following questions assuming that recombination the space- charge region can be neglected. 7a) Derive a general expression for I D ( V A ) valid for a P region of any length, WP. In HW7, problem 1c, we solved the minority carrier diffusion equation for a region of any length and found:!n( x)=!n sinh ( W x) / L P n sinh W P / L n Let x = be the edge of the neutral P- region. The electron current is: J n = +qd n d!n dx x= = q D n L n!n cosh ( W L P n ) (minus sign means that the electron sinh W P L n current is flowing in the minus x direction. Since this is a one- sided junction, and we are ignoring recombination the space- charge region, this is the total diode current, ID. Let s define the forward biased current to be positive. I D =! AJ n = qa D n L n n cosh ( W L P n ) sinh W P Finally, use the Law of the Junction for the boundary condition:!n = n i to find: e qv A k BT 1 L n! I D = qa D n L n cosh W P sinh W P ( L n ) kbt '1 eqva L n 7b) Simplify the expression derived in 7a) for a long diode. Explain what long means (i.e. WP is long compared to what?) A long diode is one with the quasi- neutral regios much longer than the diffusion length, W P >> L n. ECE

12 ECE 66: cosh( x)! ex sinh ( x )! ex! I D = qa D n L n e qv A k B T '1 and we find 7c) Simplify the expression derived in 7a) for a short diode. Explain what short means. A short diode is one with the quasi- neutral regios much shorter than the diffusion length, W P << L n. cosh x! 1 sinh( x)! x and we find! I D = qa D n W P e qv A k B T '1 8) Consider a P + N diode that is illuminated with light, which produces a uniform generation, GL, of electron- holes pairs per cm 3 per second. The N- regios long compared to a diffusion length. 8a) Consider first a uniform, infinitely long N- type semiconductor with a uniform generation rate and solve for the steady- state excess minority carrier density,!p. We have solved this problem before, in HW7. The answer is:!p = G L p 8b) Now consider the illuminated P+N diode. What are the boundary conditions at!p n ( x n ) and!p n ( x )? Assume that the Law of the Junction still applies.!p n x n = n i N D e qv A k BT 1!p n ( x ) = G L n ECE- 66 1

13 ECE 66: 8c) Use the boundary conditions developed in 8b), neglect recombination- generation in the SCR and in the P+ layer, and solve for I D ( V A ) for this illuminated diode. Having solved the MDE so many times, we can see that the solutios:!p x = Ae x/ L p + G L p This satisfies the b.c. for x! = A + G L p A = G L! p p( ) so the solutios: =!p( )e x/ L p + G L p 1 e x/ L p!p!p x The current is: dp J p =!qd p = q D p p( )! q D p G dx x= L p L L p p Use the Law of the Junction: J p = J D = q D n p i e qv A k BT L p! q D p G L L p p Note that the first term is just the diode current in the dark, J DARK and the second term is the photo- generated current, which is bias- independent and what we measure under short circuit conditions. J D = J DARK ( V A )! J SC J DARK ( V A ) = q D n p i J SC = q D p L p G L! p L p e qv A k BT 1 This result is the classical way of describing a solar cell the approach is called superposition we add the dark current and the current due to collection of photo- generated carriers. Note that superposition assumes that the collected photocurrent is independent of bias and that the Law of the Junctios valid under illumination. ECE

14 ECE 66: 8d) Sketch I D ( V A ) for G L =, G L = G and G L = G. ECE

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