= ;7:05 10 ;13 The collector current is ead b n bo I C = sinh WB L B A = (1:6 10;19 C)(10 ; cm )(0 cm s ;1 )(: 10 3 cm ;3 ) (1:41 10 ;3 cm)(7:1 10 ; )

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1 Prof. Jasprit Singh Fall 001 EECS 30 Solutions to Homework 8 Problem 1: In a pnp silicon transistor at 300 K, the base doping is cm ;3. The base width is 1.0 m and = 10.0 m. What is the total minority carrier charge in the base (a) at V EB = 0:5 V V BC = 1.0 V, (b) at V EB =0:7 V V BC =.0 V. The area of the device is 10 ; cm. The minority charge in the base under equilibrium conditions in p bo = n i N = :5 100 cm ; cm ;3 =4: cm ;3 The excess charge of holes injected into the base is given by the linear approxi mation since W b. The excess charge for case (a) is Q = ep boa eveb + ; ev BC = (1:6 10;19 c)(4: cm ;3 )(10 ; cm ) : = 8:1 10 ;10 C For case (b) the injected charge is essentially zero. Problem An Si npn transistor at 300 K has an area of 1 mm, base width of 1.0 m, and dopings of N de =10 18 cm ;3 N ab =10 17 cm ;3 N dc =10 16 cm ;3. The minority carrier lifetimes are E =10 ;7 = B C =10 ;6 s. Calculate the collector current in the active modefor(a)v BE = 0.5 V, (b) I E =:5 ma, and (c) I B =5 A. The base diusion coecient isd b =0cm s ;1. The diusion length in the base is given by =(D b B ) 1= =(0 10 ;7 ) 1= =1:41 10 ;3 cm The emitter current isgiven by (n bo =: 10 3 cm ;3 ), I E = ead bn bo coth Wb = ; (1:6 10;19 C)(10 ; cm )(0 cm s ;1 )(: 10 3 cm ;3 )(14:1) (1:41 10 ;3 cm) 1

2 = ;7:05 10 ;13 The collector current is ead b n bo I C = sinh WB L B A = (1:6 10;19 C)(10 ; cm )(0 cm s ;1 )(: 10 3 cm ;3 ) (1:41 10 ;3 cm)(7:1 10 ; ) = 7:03 10 ;13 The base current is I B = ead bn bo tanh Wb A = (1:6 10;19 C)(10 ; cm )(: 10 3 cm ;3 )(3: ; ) 1:41 10 ;3 cm) = 8:85 10 ;17 Note that we did not nd I B by simply taking the dierence of I E and I C because of round-o errors. For part (a), we have V BE = 0:5V, so that the collector current is I C = 7:03 10 ;13 : = 1:58 10 ;4 A For part (b), the emitter current is.5 ma. The base emitter bias is then, The collector current becomes V BE = kb T :5 10 ;3 `n e 7:05 10 ;13 = 0:5717V I C =: ;3 A For part (c), the base current is5a. The base emitter bias is V BE = kb T 5 10 ;6 `n e 8:85 10 ;17 I C = 3:97 10 ; A =0:6437

3 Problem 3 Plot the dependence of the base transport factor in a bipolar transistor as a function of W b = over the range 10 ; W b = 10. Assume that the emitter eciency is unity. How does the common-emitter current gain vary over the same range of W b =? The base transport factor is given by B = 1 cosh Wbn The common emitter current gain is given by (under the condition of high emitter eciency) = I C I B = We have the following values: 1 sinh Wbn tanh =10 ; B =0:99995 Wbn = L b W bn =10 ;1 B =0:995 = 10 4 =1 =10 =00 B =0:648 =1:85 B =910 ;5 =910 ;5 We see that as W b increases the properties of the transistor start to degrade. Problem 4 The mobility of holes in silicon is 100 cm /Vs. It is required that a BJT be made with a base width of 0.5 m and base resistivity of no more than 1.0 -cm. It is also desired that the emitter injection eciency be at least Calculate the emitter doping required. The various device parameters 3

4 are = 10 m L e = 10 m D e = 10 cm =s D b = 0 cm =s What is the current gain of the device? Assume = W b. The resistivity ofamaterialisgiven by (for a p-type material) = m pe where p is the hole density (=N a ). For our device we get p = N a = m e Also, from the mobility ression we have so that = m e Using this value, = :84 10 ;14 s = e m = (0:5 0:91 10;30 kg)( ;4 m =V ; s) (1:6 10 ;19 c) p = N a = (0:5 0:91 10 ;30 kg) (1:0 10 ;! ; m)(1:6 10 ;19 c) (:84 10 ;14 s) = 6:6 10 m ;3 =6: cm ;3 You could get the resistivity directly from the mobility as well and avoid the unnecessary calculations for scattering time. Since W b, we can use the simple approximation developed in the text for the emitter eciency This gives e =1; p eod e n bo D b L e =0:999 4

5 Using = Wb p eo = (0:001)n bod b L e D e n bo = n i N ab = (:5 100 cm ;6 ) (6: cm ;3 ) =3: cm ;3 p eo = 10;3 (3: cm ;3 )(0cm =s)(10 10 ;4 cm) (10cm =s)(0:5 10 ;4 cm) = 1:44 10 cm ;3 The emitter doping density is n eo = N d = N de = n i p eo = (:5 100 cm ;6 ) (1:44 10 cm ;3 ) The current gain is given by, = 1: cm ;3 = 1 ; = (0:999) 1 ; (0:5 10;4 cm) (10 10 ;4 cm) = 0:99775 = 0: ; 0:99775 =443 Problem 5 The punchthrough voltage of a Ge pnp bipolar transistor is 0 V. The base doping is cm ;3, and the emitter and collector dopings are cm ;3. Calculate the zero bias base width. If B =10 ;6 s, what is the of the transistor at a 10 V reverse bias across the collector-base junction at 300 K? The hole diusion coecient in the base is 40 cm s ;1. Since the base is very lightly doped with respect to the collector and the emitter, essentially all the depletion region is in the base. We have, (Vbi + V pt ) W b = en db 1= The built-in voltage is given by (n i =5:3 106 cm ;3 for Ge at 300 K) V bi = `n pco = `n p bo 5 5: =0:436V

6 This gives W b = (16: 8:84 10 ;14 F=cm)(0:436 V ) = 1:91 m (1:6 10 ;19 C)(10 16 cm ;3 ) 1= The value of for the case where doping in the emitter is much larger than the base doping is The diusion length in the base is =1; W bn L b L d =(D b b ) = (40 10 ;6 ) 1= =6:3 10 ;3 cm The depletion length at a bias of 10 volts across the base collector junction is The neutral base width is thus Thus we have W b =1:36m (V BC =10V)=1:91 :36 = 0:55m =1; (0:55 10;4 cm) =1:0 ; 3:9 (6:3 10 ;3 10;5 cm) The value of is veryclosetounity. 6