Microelectronics: Circuit Analysis and Design
Microelectronics: Circuit Analysis and Design
4th Edition
ISBN: 9780073380643
Author: Donald A. Neamen
Publisher: McGraw-Hill Companies, The
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Chapter 15, Problem D15.38P

a.

To determine

The expression for the frequency of the oscillation.

a.

Expert Solution
Check Mark

Answer to Problem D15.38P

The oscillation frequency is f0=12π1C( L 1 + L 2 ) .

Explanation of Solution

Given:

The circuit for the Hartley oscillator is given as:

  Microelectronics: Circuit Analysis and Design, Chapter 15, Problem D15.38P , additional homework tip  1

  Transconductance, gm=30mA/VForward biased junction diffusion resistance, rπ

Forward biased junction diffusion resistance is rπ

Drawing the small signal equivalent of the Harley oscillato:

  Microelectronics: Circuit Analysis and Design, Chapter 15, Problem D15.38P , additional homework tip  2

The output resistance r0 of the transistor is included in R .

Applying the nodal analysis at the output node ( V0 ):

  V0sL1+V0R+gmVπ+V01sC+sL2=0..........(1)

And the result by the voltage divider:

  Vπ=(sL2sL2+1 sC)V0..............(2)

Solving equation (2) and equation (1):

  V0sL1+V0R+gm[( s L 2 s L 2 + 1 sC )V0]+V01 sC+sL2=0V0[1s L 1+1R+gm( s L 2 s L 2 + 1 sC )+1 1 sC+s L 2]=0V0[1s L 1+1R+1s L 2+ 1 sC( g ms L 2+1)]=01sL1+1R+1sL2+1 sC(gmsL2+1)=01sL1+1R+sCs2L2C+1(gmsL2+1)=0..........(3)

Substituting s= jω in the equation (3):

  1jωL1+1R+jωC ( jω )2L2C+1(gm( jω)L2+1)=0j1ωL1+1R+jωCω2L2C+1(gm( jω)L2+1)=0j1ωL1+1Rgmω2L2C1ω2L2C+jωC1ω2L2C=0(1R g m ω 2 L 2 C 1 ω 2 L 2 C)+j( ωC 1 ω 2 L 2 C1 ω L 1 )=0.........(4)

The condition for the oscillation indicates that the real and the imaginary components of the equation 4.

should be zero.

From the imaginary component, the oscillation frequency:

  ω0C1ω2L2C1ω0L1=0ω0C1ω2L2C=1ω0L1ω20L1C=1ω20L2Cω02(L1C+L2C)=1ω02=1L1C+L2Cω02=1C( L 1 + L 2 )ω0=1 C( L 1 + L 2 )2πf0=1 C( L 1 + L 2 )f0=12π1 C( L 1 + L 2 )

Hence, the oscillation frequency is f0=12π1C( L 1 + L 2 ) .

b.

To determine

To show: The condition for the oscillation is gmR=L1/L2 .

b.

Expert Solution
Check Mark

Explanation of Solution

Given:

The circuit for the Hartley oscillator is given as:

  Microelectronics: Circuit Analysis and Design, Chapter 15, Problem D15.38P , additional homework tip  3

  Transconductance, gm=30mA/VForward biased junction diffusion resistance, rπ(1R g m ω 2 L 2 C 1 ω 2 L 2 C)+j( ωC 1 ω 2 L 2 C1 ω L 1 )=0.........(4)

The real component of the equation 4 should be zero:

  1Rgmω2L2C1ω20L2C=0gmω2L2C1ω20L2C=1Rgmω20L2C=1ω20L2CgmR=1ω2L2Cω20L2CgmR=1ω20L2C1gmR=1( 1 C( L 1 + L 2 ) )L2C1(Sinceω0= 1 C( L 1 + L 2 ) )gmR=C( L 1 + L 2 )L2C1gmR=( L 1 C+ L 2 C)L2CL2CgmR=L1CL2CgmR=L1L2

Therefore, gmR=L1L2

c.

To determine

To design: The circuit for oscillation at the given frequency.

c.

Expert Solution
Check Mark

Explanation of Solution

Given:

The circuit for the Hartley oscillator is given as:

  Microelectronics: Circuit Analysis and Design, Chapter 15, Problem D15.38P , additional homework tip  4

  Transconductance, gm=30mA/VForward biased junction diffusion resistance, rπ(1R g m ω 2 L 2 C 1 ω 2 L 2 C)+j( ωC 1 ω 2 L 2 C1 ω L 1 )=0.........(4)

Frequency of oscillation is f0=750kHz

Inductors is L1=L2=50μH

The condition for oscillation:

  gmR=L1L2R=L1gmL2

Substituting L1,L2 and gm in the above equation:

  R=50× 10 6( 50× 10 6 )( 30× 10 3 )R=33.33Ω

Hence, R=33.33Ω

The frequency of oscillation:

  ω0=1 C( L 1 + L 2 )ω20=1C( L 1 + L 2 )(2πf0)2=1C( L 1 + L 2 )C=1 (2π f 0 )2( L 1 + L 2 )C=1 ( 2π( 750× 10 3 ) )2( 50× 10 6 )+( 50× 10 6 )C=1( 2.221× 10 13 )( 100× 10 6 )C=450.25×1012

Hence, C=450.3pF .

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Chapter 15 Solutions

Microelectronics: Circuit Analysis and Design

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