0195136047.pdf

8.5 FREQUENCY RESPONSE OF AMPLIFIERS 413

From Equations (8.5.7) and (8.5.8),

CD=

1

RL 2 ωL 2

=

1

RL 2 ωL/ 10

=

10

1900 ( 200 π)

= 8. 37 μF

whereωL 2 is chosen to beωL/10 so that the break frequency of the capacitor is at least 10 times

smaller thanωL.

CG=

1

RL 3 ωL 3

=

1

RL 3 ωL/ 10

=

10

79,780( 200 π)

= 0. 2 μF

whereωL 3 is chosen to beωL/10 so that the break frequency of the capacitor is at least 10 times

smaller thanωL.

In order to determineωH, we first find

RA=RS‖RG=

2000 (77,780)

79,780

= 1969. 9 

RD‖RL=

900 ( 1000 )

1900

= 473. 68 

RDL=ro‖RD‖RL=

15,000( 473. 68 )

15,473.68

= 459. 2 

From Equation (8.5.17),

ωH=

1012

1950[3+ 1 +( 6 × 10 −^3 )( 459. 2 )+ 459. 2 /1950]

= 73. 26 × 106 rad/s

or

ωH

2 π

= 11 .655 MHz

Midband gain from Equation (8.5.3) is

Av 0 =

−( 6 × 10 −^3 )(77,780)( 900 )( 1000 )

( 2000 +77,780)( 900 + 1000 )

=− 2. 77

Amplifiers with Feedback

Almost all practical amplifier circuits include some form of negative feedback. The advantages
gained with feedback may include the following:

• Less sensitivity to transistor parameter variations.


• Improved linearity of the output signal by reducing the effect of nonlinear distortion.


• Better low- and high-frequency response.


• Reduction of input or output loading effects.
  The gain is willingly sacrificed in exchange for the benefits of negative feedback. Compen-
  sating networks are often employed to ensure stability. A voltage amplifier with negative feedback
  is shown by the generalized block diagram in Figure 8.5.3. The negative feedback creates a self-
  correcting amplification system that compensates for the shortcomings of the amplifying unit.
  The gain with feedback can be seen to be

Af=

vout

vin

=

A

1 +AB

(8.5.19)

Also noting thatvd=vout/A, it follows