Part V. (40 pts.) A diode is composed of an abrupt PN junction with N D = 10 16 /cm 3 and N A =10 17 /cm 3. The diode is very long so you can assume the ends are at x =positive and negative infinity. 1. Without doing any calculations, tell me if current at the junction is due mostly to holes or electron, give their approximate ratio and explain why? Holes (10x larger). Diode currents proportional to minority carrier densities on each side of the depletion region: J n n p0 = n i 2, J N p p n0 = n 2 i so for A N D N A =10xN D : J p =10xJ n. 2. Starting with the current and continuity equations, derive an expression for the minority carrier current in the N quasi-neutral region of the PN junction. Minority carrier in N quasi-neutral region is holes (p) so use the hole current and hole continuity equations: J p = qμ p pe qd p p x p J p t = 1 q x + G R Depletion approximation says E=0 in QN regions Steady state means: = 0 (true always in this class) t Gen/Recombination term proportional to excess concentration, where τ is the minority carrier lifetime: G R = p n τ Plug current equation into continuity equation, apply the above conditions, and rewrite p n (x) as p n (x) + p n0 : d 2 p n dx 2 = p n D p τ Define: L p 2 = D p τ Solution to differential equation: p n (x) = Ae x/l p + Be x/l p
Apply B.C. @ x=- : p n ( ) = 0 B = 0 Apply B.C. @x=-x n : From the definition of BI potential: p n0 = p p0 e φ 0 V T With applied potential: p n0 ( x n ) = p p0 e ( φ 0+V A ) V T Solve for p n ( x n ) using our def. of the substitution: p n0 ( x n ) = p p0 e ( φ 0+V A ) V T p n0 Use def. of BI potential to write p p0 in terms of p n0 : Solve for A: p n0 ( x n ) = p n0 (e V A V T 1) p n0 ( x n ) = p n0 (e V A V T 1) = Ae x n L p A = p n0 (e V A V T 1)e x n L p
Replace A to find our excess minority carrier concentration everywhere in the QN-n region: p n0 (x) = p n0 (e V A V T 1)e (x+x n) L p Finally, solve for the minority carrier current everywhere in the QN-n region (remember E=0 in QN regions): J p (x) = qd p p n x = qd p p n x = qd pp n0 (e V A V T 1)e (x+x n) L p L p Replacing p n0 with known value in terms of doping: J p (x) = qd 2 pn i (e V A V T 1)e (x+x n) L p L p N D x < x n 3. Without doing any more derivations, but by symmetry arguments only, provide an expression for the minority carrier current in the P quasi neutral region. Due to symmetry, replace all p with n and N D with N A. Then, because we are in the region x>x p, we need to replace the (x+x n ) term with (-x+x p ) because we have the + B.C. instead of the - one. 4. If there is no recombination or generation in the depletion region, mathematically show that the electron and hole currents must be constant in this region. No G-R term makes the continuity equations read as follows: J p x = 0 J n x = 0 The solutions to these equations are constants. 5. Describe in words and provide a sketch of where the capacitors in the PN junction come from. Junction capacitance C j is associated with the charge in the depletion region and how the depletion region width (and thus charge) changes with applied bias. Diffusion capacitances C diff-n and C diff-p are associated with the change in excess minority carrier concentrations in the QN regions due to applied bias (particularly forward bias).
These biases act in parallel so they ADD as C Tot = C j + C diff-n + C diff-p Part VI (35pts.) The following questions refer to the material with band structure given by the two equations below. (G gives the gap. CB and VB represent conduction and valence bands, respectively.) E(k) = Ak 2 + G E(k) = Bk 2 (CB) (VB) 1) Provide a simple criterion for the minimum frequency of the light that is capable of generating electron-hole pairs in this material in terms of the above band structure. (Hint, use the bandgap G). Light is only absorbed (making an e-h pair) if E ph E G where E G is the bandgap of the material in this case E G = G. Minimum energy corresponds to minimum frequency: E ph = ħω ph = 2πħf ph = G f ph = G 2πħ
2) If the scattering rate for both electrons and holes is 1/τ, find the electron and hole mobilities in terms of the band structure parameters above. (Hint from the band structure obtain the effective masses for electrons and holes, and then use that in the expression for mobility) Expressions for mobility: μ n = qτ m n μ p = qτ m p Solve for effective masses using band structure (use the positive version of the hole mass to get a positive mobility): m = ħ2 2 E m n = ħ2 k 2 2A m p = ħ2 2B μ n = 2qτA μ ħ 2 p = 2qτB ħ 2 3) What is the intrinsic carrier concentration for this material in terms of the above band parameters. (Hint, use the expression that relates bandgap to intrinsic carrier concentration) Intrinsic carrier concentration formula: n i = N C N V e E G/(2k B T) Effective density of states formulas: N C = 2 ( 2πm n 3 k B T 2 ) h 2 N V = 2 ( 2πm p 3 k B T 2 ) h 2 Plug in the effective masses we solved for in question 2) as well as G for the bandgap: n i = 2 ( 2πk 3 3 BT 2 h 2 ) ( ħ2 2A + ħ2 4 2B ) e G/(2k BT) 4) If a voltage V A is applied across a bar of length D made of this intrinsic material, and the bar has crosssection area A, give an expression for the current into and out of the bar. Write I in terms of applied voltage: I = V A /R Solve for resistance in terms of known quantities: (n=p=n i in intrinsic material) R = ρd A = D σa σ = q(μ n n + μ p p) = q(μ n + μ p )n i
Substitute mobilities and n i for their values we calculated in parts 2) and 3): I = q2 (A + B)2τ ħ 2 2 ( 2πk 3 3 BT 2 h 2 ) ( ħ2 2A + ħ2 4 2B ) e G/(2kBT) A V A D (I substituted the band parameter A for A so as to not confuse it with the area A) 5) If the material is now doped with N D donors and N A acceptors, and N A >> N D >> n i, Is this material N- type or P-type, why? Is there a net charge on this material or is it neutral, why? P-type because acceptor concentration N A dominates N D and n i so p N A. It is charge neutral because dopants do not add net charge. 6) Give an expression for the current I A through the doped bar if the bar has cross-sectional area A, length D, and voltage applied V A. Following the same steps as part 4), solve for resistance: R = ρd A = D σa σ = q(μ n n + μ p p) qμ p p Neglect electrons because (p N A ) (n n i 2 N A ) Substitute equation for mobility found in part 2) I A = V A R = 2q2 τbn A ħ 2 D AV A Part VII (33pts). A silicon PN junction diode is doped with 1017/cm3 donors and 1018/cm3 acceptors. The diode has a cross-section area of 100um x 100um. (um=micron). 1. Sketch a generic current vs. voltage curve for a PN junction diode for forward and reverse bias. Indicate the value of the voltage where the PN junction starts to turn on strongly on the curve.
2. What is the built-in voltage for this PN junction? φ 0 = V T ln ( N AN D 2 ) = 0.893 V n i 3. What are the two types of capacitors associated with a PN junction diode? Describe in words physically what gives rise to these capacitors? Same as Part V. 5) 4. For the circuit in Figure 1 below, VDC= 5.0V, and VAC = 1.0sin(ωt), write an expression for the output voltage as a function of the angular frequency ω. The output voltage is between the diode and ground. (Hint, calculate and use the appropriate capacitance from the previous question. Also, neglect diode reverse bias current. Replace the diode with a capacitor of C=C j because the 5V DC bias reverse biases the diode and thus the diffusion capacitances are small. Circuit becomes a low-pass filter with cutoff frequency ω c = 1 RC j. Small signal voltage is subject to the impedance divider of the R & C. So we write the output as: Vout(ω) = 5V + 1V sin(ωt + θ) 1 ωc j ( 1 2 ωc ) + (10kΩ) 2 j
Where sin(θ) = 1 ωc j ( 1 2 ) +(10kΩ) ωc 2 j To calculate the average value of C j we use the value of the DC bias around which the output voltage is changing. Using the formula for junction capacitance (where V R is the reverse bias, and A is the diode junction area): qε C j = A 2(φ 0 + V R ) ( N AN D ) = 3.575pF N A + N D ω c = 1 RC j = 27.97 MHz