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Analysis and Application of Analog Electronic Circuits to Biomedical Instrumentation - Northrop.pdf
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Operational Amplifiers

253

Equation 6.40 is substituted into Equation 6.33 to find the frequency response of the closed-loop amplifier:

Vo = −(VsG1

+ VoGF )

Ωo

jωτ

a

+ 1

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

¬

 

 

 

 

 

V

=

−G1Ωo (1+ GF Ωo )

 

o

jωτa

(1+ GF Ωo )+ 1

 

 

Vs

 

Again, the closed-loop break frequency depends on the size of RF:

fbcl

=

(1+ GF Ωo )

 

Ωo

Hz

2πτa

2πτaRF

 

 

 

 

The GBWP is:

(6.41)

(6.42)

(6.43)

GBWP =

(1+ GF Ωo )

[G1Ωo (1+ GF Ωo )]=

Ωo

 

Hz

2πτ

a

2πτ R

 

 

 

a

1

 

Curiously, the closed-loop break frequency depends only on closed-loop dc gain is

Vo/Vs −RF/R1.

(6.44)

R1 and the

(6.45)

6.4.3Limitations of CFOAs

CFOAs cannot be directly substituted into most conventional voltage feedback OA circuits. For example, consider the CFOA “integrator” circuit shown in Figure 6.7(A); the node equation on the vinode is written:

Ic = (0 Vs)G1 + (0 Vo) jωC

(6.46)

Equation 6.46 for Ic is substituted into Equation 6.33 for Vo:

Vo = −(VsG1

+ Vo

jωC)

Ωo

 

(6.47)

jωτ

a

+ 1

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

¬

 

 

 

 

 

 

 

Vo

=

 

−G1Ωo

 

 

 

 

(6.48)

 

 

(τa + ΩoC)+ 1

 

 

 

 

Vs

 

 

 

© 2004 by CRC Press LLC

254

Analysis and Application of Analog Electronic Circuits

 

 

 

 

C

 

R

 

Vi

 

 

 

 

In

Vo

 

 

 

CFOA

 

 

 

 

Vs

 

 

Vi

 

 

 

 

 

 

 

 

 

A

 

 

 

 

RF

R

vi1

C

 

vi2

 

In2

 

 

 

 

 

 

 

CFOA2

+

 

 

 

R1

 

 

 

 

Vs

Vo

CFOA1

In1

B

FIGURE 6.7

(A) A CFOA connected as a conventional integrator. The circuit does not work. (B) Two CFOAs connected to make a near ideal inverting integrator.

Clearly, Equation 6.48 is the frequency response of a low-pass filter, not that of an operational integrator. The closed-loop time constant will be long, however. If Ωo = 107 ohms and C = 10−6 F, the closed-loop time constant will be approximately 10 sec, which means that signals with frequencies above approximately 40 mHz will be “integrated.” This is a poor integrator.

By way of comparison, find the frequency response of a conventional voltage-feedback OA integrator. The VCVS op amp’s open-loop gain is given by:

Vo = −Vi

Kvo

OLG

jωτ

a

+ 1

 

 

 

 

 

 

The node equation for the summing junction is:

(Vi′ − Vs)G1 + (Vi′ − Vo) jωC = 0

from which,

Vi′ =

VsG1 + Vo jωC

G1

+ jωC

 

(6.49)

(6.50)

(6.51)

© 2004 by CRC Press LLC

Operational Amplifiers

255

Equation 6.51 for Viis substituted into Equation 6.49 to find the frequency response function for the OA integrator:

Vo

() =

−Kvo

(6.52)

Vs

()2 RLC τa + jω(τa + KvoR1C)+ 1

The practical integrator is seen to be a two-pole low-pass filter with dc gain,

−Kvo.

To appreciate what happens to the integrator’s poles, substitute numerical values for the parameters and factor the quadratic denominator. Let R1 = 106 Ω; C = 1 μF; Kvo = 105; and τa = 15 MS. The numerical frequency response function for the integrator is found to be:

V

(

 

)

 

 

 

 

2

1 E5

 

 

Vo

 

 

=

 

 

 

 

 

 

s

 

 

(

)

 

[

]

+ 1

 

 

 

 

 

 

 

 

0.015 + jω

0.015 + 1E5

 

 

 

 

 

 

 

 

 

6.667

E6

(6.53)

 

 

 

 

=

 

 

 

 

 

 

 

 

 

 

()2

+ jω 6.667

E6 + 66.667

 

 

 

The right-hand fraction of the frequency response function is in Laplace form and is most easily factored to find its roots (poles). Thus, the integrator’s frequency response function can be written in factored form:

Vo

6.667 E6

 

() = (jω + 3 E12)(jω + 6.667 E6)

Vs

(6.54)

= 3 E11

(3.333 E11+ 1)(1.5 E7 + 1)

Note that the integrator time constant is enormous (3.333 E11 sec) in this case and the high-frequency pole is at 6.667 E6 r/s, which is the GBWP of this voltage feedback OA.

As a closing note, it is possible to design an effective integrator using two CFOAs, as shown in Figure 6.7(B); however, this integrator is noninverting. If R is replaced with impedance Z1 and C with impedance Z2, then it can be shown that the circuit’s transfer function is:

Vo

(s)

Z2

(s)RA

(6.55)

V

Z

(s)R

s

 

1

F

 

providing that Ωo1 = Ωo2; τa1 = τa2 0; Ωo1Go1 1; and RF > RA.

© 2004 by CRC Press LLC