Exercise Solns Chapter1
2
©R. C. Jaeger and T. N. Blalock 08/09/10 1 Microelectronic Circuit Design Fourth Edition Solutions to Exercises CHAPTER 1 Page 11 V LSB = 5.12V 2 10 bits = 5.12 V 1024 bits = 5.00 mV V MSB = 5.12V 2 = 2.560V 1100010001 2 = 2 9 + 2 8 + 2 4 + 2 0 = 785 10 V O = 785 5.00 mV ( ) = 3.925 V or V O = 2 −1 + 2 −2 + 2 −6 + 2 −10 ( ) 5.12V = 3.925 V Page 12 V LSB = 5.0V 2 8 bits = 5.00 V 256bits = 19.53 mV N = 1.2V 5.00V 256bits = 61.44bits 61 = 32 + 16 + 8 + 4 + 1 = 2 5 + 2 4 + 2 3 + 2 2 + 2 0 = 00111101 2 Page 12 The dc component is V A = 4V. The signal consists of the remaining portion of v A : v a = (5 sin 2000πt + 3 cos 1000 πt) Volts. Page 23 v o = −5cos 2000πt + 25 o ( ) = −−5sin 2000πt + 25 o − 90 o ( ) [ ] = 5sin 2000πt − 65 o ( ) V o = 5 ∠− 65 o V i = 0.001 ∠0 o A v = 5 ∠− 65 o 0.001 ∠0 o = 5000∠− 65 o Page 25 A v = − R 2 R 1 | − 5 = − R 2 10kΩ → R 2 = 50 kΩ
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Transcript of Exercise Solns Chapter1
©R. C. Jaeger and T. N. Blalock08/09/10
1
Microelectronic Circuit DesignFourth Edition
Solutions to ExercisesCHAPTER 1
Page 11
€
VLSB =5.12V
210 bits=
5.12V1024bits
= 5.00 mV VMSB =5.12V
2= 2.560V
11000100012 = 29 + 28 + 24 + 20 = 78510 VO = 785 5.00mV( ) = 3.925 V
or VO = 2−1 + 2−2 + 2−6 + 2−10( ) 5.12V = 3.925 V
Page 12
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VLSB =5.0V
28 bits=
5.00V256bits
=19.53 mV N =1.2V5.00V
256bits = 61.44bits
61= 32 +16 + 8 + 4 +1= 25 + 24 + 23 + 22 + 20 = 001111012
Page 12The dc component is VA = 4V. The signal consists of the remaining portion of vA: va = (5 sin 2000πt + 3 cos 1000 πt) Volts.
Page 23
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vo = −5cos 2000πt + 25o( ) = − −5sin 2000πt + 25o − 90o( )[ ] = 5sin 2000πt − 65o( )Vo = 5∠− 65o Vi = 0.001∠0o Av =
5∠− 65o
0.001∠0o= 5000∠− 65o
Page 25
€
Av = −R2
R1
| − 5 = −R2
10kΩ→ R2 = 50 kΩ
©R. C. Jaeger and T. N. Blalock08/08/10
2
Page 26
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vs = 0.5sin 2000πt( ) + sin 4000πt( ) +1.5sin 6000πt( )[ ]The three spectral component frequencies are f1 =1000 Hz f2 = 2000 Hz f3 = 3000 Hz
a( ) The gain of the band - pass filter is zero at both f1 and f3. At f2, Vo =10 1V( ) =10V ,
and vo =10.0sin 4000πt( ) volts.
b( ) The gain of the low - pass filter is zero at both f2 and f3. At f2, Vo = 6 0.5V( ) = 3V ,
and vo = 3.00sin 2000πt( ) volts.
Page 27
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39kΩ 1− 0.1( ) ≤ R ≤ 39kΩ 1+ 0.1( ) or 35.1 kΩ≤ R ≤ 42.9 kΩ
3.6kΩ 1− 0.01( ) ≤ R ≤ 3.6kΩ 1+ 0.01( ) or 3.56 kΩ≤ R ≤ 3.64 kΩ
Page 29
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P =VI
2
R1 + R2
Pnom =152
54kΩ= 4.17 mW
Pmax =1.1x15( )
2
0.95x54kΩ= 5.31 mW Pmin =
0.9x15( )2
1.05x54kΩ= 3.21 mW
Page 33
€
R =10kΩ 1+10−3
oC−55− 25( )oC
= 9.20 kΩ R =10kΩ 1+
10−3
oC85− 25( )oC
=10.6 kΩ
