ch9

# ch9 - SOLUTIONS CHAPTER 9 9.1 For each of the transistors...

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SOLUTIONS CHAPTER 9 9.1. For each of the transistors in Figure P9.1, indicate the mode of operation (forward active, cutoff, saturation, etc.) 0.7 V -5 V 9 V -9 V 0.7 V 0.5V 4.9 V 4.1V 4.8V 9 V 5.0V 9 V 0.2 V (a) (b) (c) (d) a) This is an npn transistor. The collector voltage is higher than the base voltage so the CB junction is reverse biased. The base voltage is higher than the emitter voltage so that junction is forward biased. Thus this transistor operates in forward active mode. b) Also npn. The EB junction is forward biased by 0.2 V, which is not enough to turn it on. The CB junction is reverse biased, so this transistor is in cutoff. c) This one is pnp. The EB junction is forward biased and the CB junction is reverse biased, so this one is in forward active mode. d) Also pnp; both junctions forward biased- this one is in saturation. Anderson & Anderson 1 November 19, 2007 Solutions Chapter 9

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9.2. For each area in Figure P9.2, identify how the region should be doped to make a good pnp transistor (n + , n, intrinsic, p, p + ). and indicate the reason(s). Indicate where the actual pnp transistor occurs. E B C a b c d e f transistor a) p + . The emitter should be heavily doped to maximize β and also to make a good contact with the metal. b) n-type base should be more lightly doped than emitter c) p-type collector should be lightly doped. d) this layer should be p + to make it highly conductive. It transfers the collected holes to the collector contact. e) n + to ensure a good contact with the base contact. f) p + to conduct collected holes easily and to provide good contact with the collector contact. 9.3. Draw the energy band diagram for a pnp transistor at equilibrium and under forward active bias. The drawing should reflect the relative dopings as well: Anderson & Anderson 2 November 19, 2007 Solutions Chapter 9
p+ em itter n base p collector p+ collector equilibrium E ,E f C E V active m ode bias F n F p I n I p -W W x E BM E ,E f C E V E ,E f V C E 9.4. From Equation (9.13), α = γα T M, where M is the carrier multiplication factor in the base collector junction. For small base-collector voltage, M=1 and α = γα T and 1 α β α = - . Show that for avalanche breakdown, 1 1 M β = + . Solution: 0 0 1 1 M M α α β α α = = - - Anderson & Anderson 3 November 19, 2007 Solutions Chapter 9

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For breakdown, or 1 M β α = = But and 1 1 1 M M M β β α α β β = = = + + Solving for M , 1 1 M β = + 9.5. Using the prototype model, (i.e. ignore the apparent band-gap narrowing and its effect on injection efficiency) find α and β for an npn BJT with 19 3 17 3 17 3 10 , 2 10 , 10 . DE AB DC N cm N cm and N cm - - - = = × = Indicate clearly all your steps. Let V BE =0.8V V CB =2.0 V. The metallurgical widths are W EM = 0.2 μm and W BM = 0.2 μm. We can find the minority carrier diffusion constants from Figure 3.11: D pE =3.8 cm 2 /s and D nB =15 cm 2 /s The built-in voltages for the two junctions are from Equation (5.13): ( 29 ( 29 ( 29 19 17 ( ) 2 2 10 3.1 10 2 10 ln 0.026 ln 1.00 1.08 10 V AB bi EB i N N kT V V q n × × = = = × (one-sided junction) and ( 29 ( 29 ( 29 ( 29 17 17 2 2 10 10 2 10 ln 0.026ln 0.85 1.08 10 DC AB bi CB i N N kT V V q n × = = =
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