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ch10

ch10 - SOLUTIONS CHAPTER 10 10.1 For an npn transistor with...

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SOLUTIONS CHAPTER 10 10.1 For an npn transistor with ' 19 3 10 DE N cm - = , ' 17 3 2 10 AB N cm - = × , ' 16 3 5 10 DC N cm - = × , 0.13 E W m μ = , and 0.15 B W m μ = under the bias conditions of 20 B I A μ = and 2.5 BC V V = - , find a) β b) I C c) r π d) g m e) ( ) BE C C π f) C μ g) f co h) f T Note that band gap narrowing should be accounted for, and that both sides of the E-B junction are short. Let the area of the emitter junction be A E =2.5 × 10 -7 cm - 2 and the area of the collector junction be A C =8 × 10 -7 cm -2 . a) To find β , we use Equation (9.43) and from Figure 2.25, * 50 g E meV = . Then * ' g E C nB E kT AB pE B N D W e N D W β - ∆ = From Figure 3.11, we find D pE 3.8, and D nB 13 cm -2 /s. The widths are W E =0.13 μ m and W B =0.15 μ m. * ' 0.05 19 0.026 17 2.86 10 13 0.15 2 10 3.8 0.13 83 g E C nB E kT AB pE B N D W e N D W e β - ∆ - = × = × = Anderson & Anderson 1 November 19, 2007 Chapter 10: Solutions
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b) Ι C Since , C B I I β = 83x20 1.7 . C I A mA μ = = c) r π 6 0.026 1300 20 10 B kT r qI π - = = = × d) g m From Equation 10.15 83 64 1300 m g mS r π β = = = e) C π For a BJT operating in the active mode, scEB jEB C C ? and we can approximate scBE C C π 2245 . From Equation 10.30 for constant base doping, the stored charge capacitance is ( 29 ( 29 ( 29 ( 29 2 4 2 13 2 0.13 10 83 2.4 10 0.24 3 3 15 / 1300 B sc nB cm W C F pF D r cm s π β - - × = = = × = f) C μ The capacitance C μ is just the BC junction capacitance. This is not a one- sided junction, so the result is: Anderson & Anderson 2 November 19, 2007 Chapter 10: Solutions
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( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 1 2 1 19 14 17 3 16 3 2 7 2 17 16 -14 2 1.6 10 11.8 8.85 10 / 2 10 5 10 8 10 2 2 10 5 10 0.83 2.5 2.5x10 0.025 A D jBC A D bi a q N N C C A N N V V F cm cm cm cm F pF μ ε - - - - - = = + - × × × × = × × + × - - = = where we found the value of V biBC from ( 29 ' ' 17 16 2 2 10 2x10 5 10 ln 0.026ln 0.83 1.08 10 A D bi i N N kT V V q n × = = = × g) f co The cutoff frequency for this transistor is, from Equation (10.37), ( 29 ( 29 ( 29 13 14 1 1 462 2 2 1300 2.4 10 2.5 10 co be f MHz r C C F π μ π π - - = = = + × + × h) f T From Equation (10.38), ( 29 ( 29 6 9 83 462 10 38 10 Hz 38GHz T DC co f f β = = × = × = 10.2. For the transistor of Problem 10.1, plot i c i b as a function of frequency. What is the unity-gain frequency? We use 2 2 6 83 1 1 462 10 c b co i i f f f β = = + + × and plot against frequency. Anderson & Anderson 3 November 19, 2007 Chapter 10: Solutions
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Note that from the plot, 38 T f GHz = in agreement with the result from Problem 10.1h.
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