( )! N D ( x) ) and equilibrium


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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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