{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

chapter10sm

# chapter10sm - SOLUTIONS CHAPTER 10 10.1 For an npn...

This preview shows pages 1–5. Sign up to view the full content.

SOLUTIONS CHAPTER 10 Anderson & Anderson 1 January 19, 2005 Chapter 10: Solutions 10 DE Nc 10.1 For an npn transistor with '1 9 3 m = 7 3 21 0 AB m 51 0 DC 0.13 E , , , Wm 6 3 m µ = m , and 0.15 B W = under the bias conditions of 20 B I A = and , find 2.5 BC VV =− a) β b) I C c) r π d) g m e) () BE CC f) C 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, . Then * 50 g Em e ∆= V * ' g E Cn B E kT AB pE B ND W e NDW −∆ = 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 BE kT AB pE B e e = × = × =

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
() ( ) ( ) ( ) 1 2 1 19 14 17 3 16 3 2 72 17 16 -14 2 1.6 10 11.8 8.85 10 / 2 10 5 10 81 0 2 2 10 5 10 0.83 2.5 2.5x10 0.025 AD jBC b ia qNN CC A NNVV Fcm cm cm Fp F µ ε −− ⎡⎤ == ⎢⎥ +− ⎣⎦ ×× × × ×+ × where we found the value of V biBC from '' 17 16 22 10 2x10 5 10 ln 0.026ln 0.83 1.08 10 bi i NN kT VV q n ⎛⎞ ⋅× ⎜⎟ = ⎝⎠ × g) f co The cutoff frequency for this transistor is, from Equation (10.37), 13 14 11 462 2 2 1300 2.4 10 2.5 10 co be f MHz rC C F πµ π = + Ω×+ × h) f T From Equation (10.38), 69 83 462 10 38 10 Hz 38GHz TD C c o ff 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 6 83 1 1 462 10 c b co i i f f f + + × 2 and plot against frequency. Anderson & Anderson 3 January 19, 2005 Chapter 10: Solutions

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
Note that from the plot, 38 T f GHz = in agreement with the result from Problem 10.1h. 10.3. An npn BJT with uniformly doped emitter, base, and collector has β =95, base width W B of 0.15 µ m, an electron diffusion coefficient of 10 cm 2 /s, an r π of 1000 , and a CB junction capacitance of 0.05 pF. What is its cut-off frequency?
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 10

chapter10sm - SOLUTIONS CHAPTER 10 10.1 For an npn...

This preview shows document pages 1 - 5. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online