The A M , f H , f t , f z and gain at high frequencies. Also, draw the bode plot for the given amplifier. A M = − 40 V/V f H = 72.34 MHz f t = 2.89 GHz f z = 31.83 GHz Fig: Bode plot Given Information: R L ' = 20 k Ω g m = 2 mA/V C g d = 10 fF C L = 100 fF Calculation: The expression for gain in the CS amplifier is given by, A M = − g m R L ' Where R L ' is the load resistance and g m is the trans-conductance of the amplifier. Plugging given values. A M = − 2 mA/V 20 k Ω = − 40 V/V Write the equation for 3-dB frequency in terms of capacitance C g d , C L and resistance R L ' as follows: f H = 1 2 π C L + C g d R L ' Plugging given values. f H = 1 2 π 100 fF + 10 fF 20 k Ω f H = 1 2 π 100 + 10 × 10 − 15 × 20 × 10 3 f H = 7.23 × 10 7 Hz f H = 72.3 MHz The equation for unity gain frequency in terms of trans-conductance g m , capacitance C g d , C L and resistance R L ' as follows: f t = A M f H f t = 40 V/V × 72.3 MHz f t = 2892 MHz f t = 2.892 GHz . The equation for the frequency of the transmission zero in terms of trans-conductance g m , and capacitance C g d as follows: f z = g m 2 π C g d Plugging value for parameters f z = 2 mA/V 2 π × 10 fF f z = 2 mA/V 2 π × 10 × 10 − 15 F f z = 31.83 GHz Gain at f H = 20 log 10 A M dB = 20 log 10 40 dB = 32.04 dB Gain at f t = 0 dB Gain Magnitude A ( s ) = − g m R ' L 1 − s C g d g m 1 + s C L + C g d R ' L as s → ∞ A = C g d C g d + C L = 10 f F 100 + 10 f F = 1 11 Gain at f z = 20 log 10 A dB = 20 log 10 1 11 dB = − 20.82 dB Fig: Bode plot

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Answer 1

Question

Chapter 10, Problem 10.23P

To determine

The AM, fH, ft, fz and gain at high frequencies. Also, draw the bode plot for the given amplifier.

AM=−40 V/VfH=72.34 MHzft=2.89 GHzfz=31.83 GHz

Fig: Bode plot

Given Information:

RL'=20 kΩgm=2 mA/VCgd=10 fFCL=100 fF

Calculation:

The expression for gain in the CS amplifier is given by,

AM=−gmRL'

Where RL' is the load resistance and gm is the trans-conductance of the amplifier.

Plugging given values.

AM=−2 mA/V20 kΩ=−40 V/V

Write the equation for 3-dB frequency in terms of capacitance Cgd, CL and resistance RL' as follows:

fH=12πCL+CgdRL'

Plugging given values.

fH=12π100 fF+10 fF20 kΩfH=12π100+10×10−15×20×103fH=7.23×107HzfH=72.3MHz

The equation for unity gain frequency in terms of trans-conductance gm , capacitance Cgd, CL and resistance RL' as follows:

ft=AMfHft=40V/V×72.3MHzft=2892MHzft=2.892GHz .

The equation for the frequency of the transmission zero in terms of trans-conductance gm , and capacitance Cgd as follows:

fz=gm2πCgd

Plugging value for parameters

fz=2 mA/V2π×10 fFfz=2 mA/V2π×10×10−15Ffz=31.83 GHz

Gain at fH=20log10AMdB=20log1040dB=32.04dB

Gain at ft=0 dB

Gain Magnitude A(s)= −gmR'L1−sCgdgm1+sCL+CgdR'Las s→∞A=CgdCgd+CL=10 fF100+10fF=111

Gain at fz=20log10AdB=20log10111 dB=−20.82 dB

Fig: Bode plot

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Answer 2

Question

Chapter 10, Problem 10.23P

To determine

The AM, fH, ft, fz and gain at high frequencies. Also, draw the bode plot for the given amplifier.

AM=−40 V/VfH=72.34 MHzft=2.89 GHzfz=31.83 GHz

Fig: Bode plot

Given Information:

RL'=20 kΩgm=2 mA/VCgd=10 fFCL=100 fF

Calculation:

The expression for gain in the CS amplifier is given by,

AM=−gmRL'

Where RL' is the load resistance and gm is the trans-conductance of the amplifier.

Plugging given values.

AM=−2 mA/V20 kΩ=−40 V/V

Write the equation for 3-dB frequency in terms of capacitance Cgd, CL and resistance RL' as follows:

fH=12πCL+CgdRL'

Plugging given values.

fH=12π100 fF+10 fF20 kΩfH=12π100+10×10−15×20×103fH=7.23×107HzfH=72.3MHz

The equation for unity gain frequency in terms of trans-conductance gm , capacitance Cgd, CL and resistance RL' as follows:

ft=AMfHft=40V/V×72.3MHzft=2892MHzft=2.892GHz .

The equation for the frequency of the transmission zero in terms of trans-conductance gm , and capacitance Cgd as follows:

fz=gm2πCgd

Plugging value for parameters

fz=2 mA/V2π×10 fFfz=2 mA/V2π×10×10−15Ffz=31.83 GHz

Gain at fH=20log10AMdB=20log1040dB=32.04dB

Gain at ft=0 dB

Gain Magnitude A(s)= −gmR'L1−sCgdgm1+sCL+CgdR'Las s→∞A=CgdCgd+CL=10 fF100+10fF=111

Gain at fz=20log10AdB=20log10111 dB=−20.82 dB

Fig: Bode plot

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Answer 3

Question

Chapter 10, Problem 10.23P

To determine

The AM, fH, ft, fz and gain at high frequencies. Also, draw the bode plot for the given amplifier.

AM=−40 V/VfH=72.34 MHzft=2.89 GHzfz=31.83 GHz

Fig: Bode plot

Given Information:

RL'=20 kΩgm=2 mA/VCgd=10 fFCL=100 fF

Calculation:

The expression for gain in the CS amplifier is given by,

AM=−gmRL'

Where RL' is the load resistance and gm is the trans-conductance of the amplifier.

Plugging given values.

AM=−2 mA/V20 kΩ=−40 V/V

Write the equation for 3-dB frequency in terms of capacitance Cgd, CL and resistance RL' as follows:

fH=12πCL+CgdRL'

Plugging given values.

fH=12π100 fF+10 fF20 kΩfH=12π100+10×10−15×20×103fH=7.23×107HzfH=72.3MHz

The equation for unity gain frequency in terms of trans-conductance gm , capacitance Cgd, CL and resistance RL' as follows:

ft=AMfHft=40V/V×72.3MHzft=2892MHzft=2.892GHz .

The equation for the frequency of the transmission zero in terms of trans-conductance gm , and capacitance Cgd as follows:

fz=gm2πCgd

Plugging value for parameters

fz=2 mA/V2π×10 fFfz=2 mA/V2π×10×10−15Ffz=31.83 GHz

Gain at fH=20log10AMdB=20log1040dB=32.04dB

Gain at ft=0 dB

Gain Magnitude A(s)= −gmR'L1−sCgdgm1+sCL+CgdR'Las s→∞A=CgdCgd+CL=10 fF100+10fF=111

Gain at fz=20log10AdB=20log10111 dB=−20.82 dB

Fig: Bode plot

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Answer 4

Question

Chapter 10, Problem 10.23P

To determine

The AM, fH, ft, fz and gain at high frequencies. Also, draw the bode plot for the given amplifier.

AM=−40 V/VfH=72.34 MHzft=2.89 GHzfz=31.83 GHz

Fig: Bode plot

Given Information:

RL'=20 kΩgm=2 mA/VCgd=10 fFCL=100 fF

Calculation:

The expression for gain in the CS amplifier is given by,

AM=−gmRL'

Where RL' is the load resistance and gm is the trans-conductance of the amplifier.

Plugging given values.

AM=−2 mA/V20 kΩ=−40 V/V

Write the equation for 3-dB frequency in terms of capacitance Cgd, CL and resistance RL' as follows:

fH=12πCL+CgdRL'

Plugging given values.

fH=12π100 fF+10 fF20 kΩfH=12π100+10×10−15×20×103fH=7.23×107HzfH=72.3MHz

The equation for unity gain frequency in terms of trans-conductance gm , capacitance Cgd, CL and resistance RL' as follows:

ft=AMfHft=40V/V×72.3MHzft=2892MHzft=2.892GHz .

The equation for the frequency of the transmission zero in terms of trans-conductance gm , and capacitance Cgd as follows:

fz=gm2πCgd

Plugging value for parameters

fz=2 mA/V2π×10 fFfz=2 mA/V2π×10×10−15Ffz=31.83 GHz

Gain at fH=20log10AMdB=20log1040dB=32.04dB

Gain at ft=0 dB

Gain Magnitude A(s)= −gmR'L1−sCgdgm1+sCL+CgdR'Las s→∞A=CgdCgd+CL=10 fF100+10fF=111

Gain at fz=20log10AdB=20log10111 dB=−20.82 dB

Fig: Bode plot

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Answer 5

Chapter 10, Problem 10.23P

To determine

The AM, fH, ft, fz and gain at high frequencies. Also, draw the bode plot for the given amplifier.

AM=−40 V/VfH=72.34 MHzft=2.89 GHzfz=31.83 GHz

Fig: Bode plot

Given Information:

RL'=20 kΩgm=2 mA/VCgd=10 fFCL=100 fF

Calculation:

The expression for gain in the CS amplifier is given by,

AM=−gmRL'

Where RL' is the load resistance and gm is the trans-conductance of the amplifier.

Plugging given values.

AM=−2 mA/V20 kΩ=−40 V/V

Write the equation for 3-dB frequency in terms of capacitance Cgd, CL and resistance RL' as follows:

fH=12πCL+CgdRL'

Plugging given values.

fH=12π100 fF+10 fF20 kΩfH=12π100+10×10−15×20×103fH=7.23×107HzfH=72.3MHz

The equation for unity gain frequency in terms of trans-conductance gm , capacitance Cgd, CL and resistance RL' as follows:

ft=AMfHft=40V/V×72.3MHzft=2892MHzft=2.892GHz .

The equation for the frequency of the transmission zero in terms of trans-conductance gm , and capacitance Cgd as follows:

fz=gm2πCgd

Plugging value for parameters

fz=2 mA/V2π×10 fFfz=2 mA/V2π×10×10−15Ffz=31.83 GHz

Gain at fH=20log10AMdB=20log1040dB=32.04dB

Gain at ft=0 dB

Gain Magnitude A(s)= −gmR'L1−sCgdgm1+sCL+CgdR'Las s→∞A=CgdCgd+CL=10 fF100+10fF=111

Gain at fz=20log10AdB=20log10111 dB=−20.82 dB

Fig: Bode plot

Answer 6

Chapter 10, Problem 10.23P

To determine

The AM, fH, ft, fz and gain at high frequencies. Also, draw the bode plot for the given amplifier.

AM=−40 V/VfH=72.34 MHzft=2.89 GHzfz=31.83 GHz

Fig: Bode plot

Given Information:

RL'=20 kΩgm=2 mA/VCgd=10 fFCL=100 fF

Calculation:

The expression for gain in the CS amplifier is given by,

AM=−gmRL'

Where RL' is the load resistance and gm is the trans-conductance of the amplifier.

Plugging given values.

AM=−2 mA/V20 kΩ=−40 V/V

Write the equation for 3-dB frequency in terms of capacitance Cgd, CL and resistance RL' as follows:

fH=12πCL+CgdRL'

Plugging given values.

fH=12π100 fF+10 fF20 kΩfH=12π100+10×10−15×20×103fH=7.23×107HzfH=72.3MHz

The equation for unity gain frequency in terms of trans-conductance gm , capacitance Cgd, CL and resistance RL' as follows:

ft=AMfHft=40V/V×72.3MHzft=2892MHzft=2.892GHz .

The equation for the frequency of the transmission zero in terms of trans-conductance gm , and capacitance Cgd as follows:

fz=gm2πCgd

Plugging value for parameters

fz=2 mA/V2π×10 fFfz=2 mA/V2π×10×10−15Ffz=31.83 GHz

Gain at fH=20log10AMdB=20log1040dB=32.04dB

Gain at ft=0 dB

Gain Magnitude A(s)= −gmR'L1−sCgdgm1+sCL+CgdR'Las s→∞A=CgdCgd+CL=10 fF100+10fF=111

Gain at fz=20log10AdB=20log10111 dB=−20.82 dB

Fig: Bode plot

Answer 7

Chapter 10, Problem 10.23P

To determine

The AM, fH, ft, fz and gain at high frequencies. Also, draw the bode plot for the given amplifier.

AM=−40 V/VfH=72.34 MHzft=2.89 GHzfz=31.83 GHz

Fig: Bode plot

Given Information:

RL'=20 kΩgm=2 mA/VCgd=10 fFCL=100 fF

Calculation:

The expression for gain in the CS amplifier is given by,

AM=−gmRL'

Where RL' is the load resistance and gm is the trans-conductance of the amplifier.

Plugging given values.

AM=−2 mA/V20 kΩ=−40 V/V

Write the equation for 3-dB frequency in terms of capacitance Cgd, CL and resistance RL' as follows:

fH=12πCL+CgdRL'

Plugging given values.

fH=12π100 fF+10 fF20 kΩfH=12π100+10×10−15×20×103fH=7.23×107HzfH=72.3MHz

The equation for unity gain frequency in terms of trans-conductance gm , capacitance Cgd, CL and resistance RL' as follows:

ft=AMfHft=40V/V×72.3MHzft=2892MHzft=2.892GHz .

The equation for the frequency of the transmission zero in terms of trans-conductance gm , and capacitance Cgd as follows:

fz=gm2πCgd

Plugging value for parameters

fz=2 mA/V2π×10 fFfz=2 mA/V2π×10×10−15Ffz=31.83 GHz

Gain at fH=20log10AMdB=20log1040dB=32.04dB

Gain at ft=0 dB

Gain Magnitude A(s)= −gmR'L1−sCgdgm1+sCL+CgdR'Las s→∞A=CgdCgd+CL=10 fF100+10fF=111

Gain at fz=20log10AdB=20log10111 dB=−20.82 dB

Fig: Bode plot

Answer 8

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Answer 9

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Answer 10

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Answer 11

Ch. 10.1 - Prob. 10.1E Ch. 10.1 - Prob. 10.2E Ch. 10.2 - Prob. 10.6E Ch. 10.2 - Prob. 10.7E Ch. 10.2 - Prob. 10.8E Ch. 10.2 - Prob. 10.9E Ch. 10.3 - Prob. 10.11E Ch. 10.3 - Prob. 10.12E Ch. 10.3 - Prob. 10.13E Ch. 10.3 - Prob. 10.14E

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