The current density at a forward bias of 0.9 V is J( V) = 8:91 10 ;13 exp 0:06 = 9: :39=961:4 Acm ; 1: 10 ;8 exp 0:05 The current is dominated b

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1 Prof. Jasprit Singh Fall 000 EECS 30 Solutions to Homework 6 Problem 1 Two dierent processes are used to fabricate a Si p-n diode. The rst process results in a electron-hole recombination time via impurities in the depletion region of 10 ;7 s while the second one gives a time of 10 ;9 s. Calculate the diode ideality factors for the two cases near a forward bias of 0.9 V. Use the following parameters: N a = N d =10 18 cm ;3 n = p =10 ;7 s D n = 5 cm =s D p = 8 cm =s The built-in voltage for the diode is The prefactor for the ideal current is V bi =4 V J o = ; 1:6 10 ; :9 10 ;4 1:58 10 ;3 = 8:91 10 ;13 Acm ; The depletion width for the diode at 0.9 V forward bias is W =1 10 ;6 cm For the case where impurity assisted recombination is 10 ;7 s, the G-R current prefactor is J o GR = ; 1:6 10 ;19 ; 10 ;6; 1: ;7 = 1: 10 ;8 Acm ; The current density at a forward bias of 0.9 V is J( V) = 8:91 10 ;13 exp 0:06 = 1: 10 ;8 Acm ; 1: 10 ;8 exp 0:05 1

2 The current density at a forward bias of 0.9 V is J( V) = 8:91 10 ;13 exp 0:06 = 9: :39=961:4 Acm ; 1: 10 ;8 exp 0:05 The current is dominated by the ideal part and the ideality factor is unity. For the second case ( =10 ;8 s) current is J( V)=9613:9 =964:9 At a bias of 0.8 V the current density is1.13acm ;. Using these two values to get the ideality factor we nd that the ideality factor is We see that for high forward bias the diode is essentially ideal for both cases. Problem The critical eld for breakdown of silicon is V/cm. Calculate the n-side doping of an abrupt p n diode that allows one to have a breakdown voltage of 30 V. The critical eld for the silicon is F crit = V=cm The doping concentration needed to allow a breakdown voltage of 30 volts is N d = ef crit ev BD = (11:9 8:85 10;14 F=cm)( V=cm) (1:6 10 ;19 C)(30V ) = 9: cm ;3 Problem 3 Consider a Si p n diode with a long base. The diode is forwardbiased (at 300 K) at a current of ma. The hole lifetime in the n-region is 10 ;7 s. Assume that the depletion capacitance is negligible and calculate the diode impedance at the frequency of 100 KHz, 100 MHz, and 500 Mhz. We have the following small signal relation, i s =(G s j!c dif f )v s The diode impedance is where v s i s G s = = G s ; j!c dif f G s! C dif f ei k B T C dif f = ei p k B T

3 For this problem The impedance is now G s = 10;3 0:06 C dif f = 10;3 10 ;7 0:06 =7:69 10 ; A=V =3:85 10 ;9 F Z(100 KHz) = (13:0j6:5 10 ; ) Z(100 MHz) = (0:49 ; j:498) Z(500 MHz) = (:07 10 ; ; j0:5) Problem 4 parameters: Consider a Si p-n diode at room temperature with following N d = N a =10 17 cm ;3 D n = 0 cm =s D p = 1 cm =s n = p =10 ;7 s Calculate the reverse saturation current for a long ideal diode. Also estimate the storage delay time for the long diode. Now consider a narrow diode made from the structure given above. The thickness of the n-side region is 1.0 m. The thickness of the p-side region is also 1.0 m. Calculate the reverse saturation current in the narrow diode at a reverse bias of.0 volt. Also estimate the storage delay time for this diode. The built-in voltage for the diode is V bi =0:817 V The depletion width for the diode is (V bi V are in volts) W n = W p =8:13 10 ;6p V bi ; V For the long diode the reverse saturation current density is J o = ; 1:6 10 ; :09 10 ;3 1:41 10 ;3 = 9:05 10 ;1 Acm ; The storage delay time depends upon the details of the switching condition, but is approximately equal to n or p, i.e., 10 ;7 s. For the short diode the reverse current is bias dependent even for the ideal diode case. The depletion width 3

4 changes the neutral region width of the n and p diodes by about 10%. The reverse saturation current atareverse bias of.0 V is J o = ; 1:6 10 ; :6 10 ;5 8:6 10 ;5 = 1:34 10 ;10 Acm ; The storage delay time is approximately the transit time across the neutral n or p regions. For the p-side the transit time is 4:10 ;10 s. This is a very short time. Problem 5 Consider a Si long p-n diode with the following parameters: n-side doping = cm ;3 p-side doping = cm ;3 minority carrier lifetime n = p =10 ;7 s electron diusion constant = 30 cm =s hole diusion constant = 10 cm =s diode area = 10 ;4 cm =s Calculate the diode current at a forward bias of 1.0 V at 300 K. An electron comes from the p-side into the depletion region and is swept away by the eld to the n-side. Estimate thetimeittakes the electron to cross the depletion region at zero applied bias and a reverse bias of 1.0 volt. To calculate the current ow we need to calculate several quantities. We have for the structure given p n = :5 10 cm ;3 n p = :5 10 cm ;3 V bi = k BT e `n :5 10 =37 For a forward bias of say 0.5 V, the total voltage across the depletion region is V. This gives for the n-side depletion region W n = (11:9 8:84 10 ;14 F=cm)(0:437) cm ;3 = 1: ;6 cm 1:6 10 ;19 C 1= We can use this number and scale it to nd the depletion width at any bias. 4

5 The electron diusion length on the p-side is L n =(30 10 ;7 ) 1= =1:73 10 ;3 cm The hole diusion length on the n-side is The prefactor for the current is now I o = (1:6 10 ;19 C)(10 ;4 cm) = 9: ;17 A L p =(10 10 ;7 ) 1= =1:0 10 ;3 cm 10 :5 10 1:0 10 ;3 Using this prefactor the forward current at 1.0 V is I =4:97 A 30 :5 10 1:73 10 ;3 To nd the time for electrons to drift across the depletion width we need to nd the depletion widths for the zero bias and V r =1:0 volt. We can nd this from our earlier calculation by simply scaling the voltage levels. This gives W n (V =0)=:48 10 ;6 cm W n (V r =1:0 V)=3:57 10 ;6 cm The total depletion width is twice the values found above. Sice the electric eld in the depletion region is large ( 10 kv/cm), the electrons can be assumes to move with saturation velocity whichis10 7 cm/s. The transit times are then and t tr (V =0)= 4:964 10; =4: ;13 s t tr (V r =1:0 V)=7:14 10 ;13 s We see that the electrons drift very fast across the depletion region. 5

Problem 4 Consider a GaAs p-n + junction LED with the following parameters at 300 K: Electron diusion coecient, D n = 25 cm 2 =s Hole diusion coecient

Problem 4 Consider a GaAs p-n + junction LED with the following parameters at 300 K: Electron diusion coecient, D n = 25 cm 2 =s Hole diusion coecient Prof. Jasprit Singh Fall 2001 EECS 320 Homework 7 This homework is due on November 8. Problem 1 An optical power density of 1W/cm 2 is incident on a GaAs sample. The photon energy is 2.0 ev and there is