# q05 - S r e 1 = − 9 167 × 10 − 3 v s = ⇒ i c 1 v s =...

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ECE 3050 Analog Electronics Quiz 5 September 23, 2009 Professor Leach Name Instructions. Print your name in the space above. Place a box around your answer. Express each numerical answer as a decimal number. Honor Code Statement: Ihave neither given nor received help on this quiz. Initials The f gure shows a CB/CE ampli f er. For each transistor, r x =100 , β =99 , α =0 . 99 , I E =1mA , r ic =100k ,and V T =25mV . The circuit element values are R S =82 , R C 1 =12k , R E 2 =50 ,and R C 2 =16k . Reference equations: g m = I C /V T , r π = V T /I B , r e = V T /I E , i 0 c = g m v π = βi b = αi 0 e , r 0 π = r x + r π +(1+ β ) R te , r 0 e =( R tb + r x ) / (1 + β )+ r e . First express your answers in symbolic form. Then evaluate them numerically. Draw a box around your answers. r e = V T I E =25 r 0 e 1 = r x 1+ β + r e =26 R tb 2 = R C 1 k r ic 1 =10 . 71k r 0 e 2 = R tb 2 + r x 1+ β + r e =133 . 143 The following solutions are based on the simpli f ed T model. (a) Solve for i 0 c 1 /v s . i 0 c 1 = αi 0 e 1 = α v s R

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Unformatted text preview: S + r e 1 = − 9 . 167 × 10 − 3 v s = ⇒ i c 1 v s = − 9 . 167 × 10 − 3 (b) Solve for v tb 2 /i c 1 . v tb 2 = − i c 1 R C 1 k r ic 1 = − 10 . 71 × 10 3 i c 1 = ⇒ v tb 2 i c 1 = − 10 . 71 × 10 3 (c) Solve for i c 2 /v tb 2 . i c 2 = αi e 2 = α v tb 2 R E 2 + r e 2 = 5 . 406 × 10 − 3 v tb 2 = ⇒ i c 2 v tb 2 = 5 . 406 × 10 − 3 1 (d) Solve for v o /i c 2 . v o = − i c 2 R C 2 k r ic 2 = − 13 . 79 × 10 3 i c 2 = ⇒ v o i c 2 = − 13 . 79 × 10 3 (e) Combine the above answers to solve for v o /v s . v o v s = i c 1 v s × v tb 2 i c 1 × i c 2 v tb 2 × v o i c 2 = − 7323 2...
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q05 - S r e 1 = − 9 167 × 10 − 3 v s = ⇒ i c 1 v s =...

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