Добавил:
Опубликованный материал нарушает ваши авторские права? Сообщите нам.
Вуз: Предмет: Файл:
Analysis and Application of Analog Electronic Circuits to Biomedical Instrumentation - Northrop.pdf
Скачиваний:
205
Добавлен:
10.08.2013
Размер:
4.41 Mб
Скачать

292

Analysis and Application of Analog Electronic Circuits

To obtain the symmetric APF, let R1 = R; R5 = 2R4; and R6 = R5. The desired transfer function is then:

Vap

 

s2R 2C2

s R CR R

+ 1

 

3

3

2

 

V

(s) = −

s2R 2C2

+ s R CR R

+ 1

 

S

3

3

2

 

The natural frequency of poles and zeros is:

ωn = 1/R3C r/s ξ = R/2R2

Kvo = −1, −• ≤ ω ≤ •

(7.36)

(7.37A)

(7.37B)

(7.37C)

The purpose of all-pass filters is to generate a phase shift with frequency with no attenuation; the phase of the APF described earlier can be shown to be:

ϕ = −π − 2 tan

−1

 

ω R3CR R2

˘

 

 

 

 

˙ radians

(7.38)

 

1− ω2R 2C3

 

 

 

3

˚

 

7.2.4 Generalized Impedance Converter AFs

Often one wishes to design AFs that work at very low frequencies, e.g., below 20 Hz. Such filters can find application in conditioning low-frequency physiological signals such as the ECG, EEG, and heart sounds. If a conventional biquad AF is used to make a high-pass filter whose fn = 1 Hz, for example, it is necessary to obtain very large capacitors that are expensive and take up excessive volume on a PC board. For example, the capacitor required for fn = 1 Hz and R3 = 100 kΩ is:

C =

1

= 1.59 μF

(7.39)

2 π f R

 

n 3

 

 

The GIC circuit, shown in general format in Figure 7.8, allows the size of a small capacitor, e.g., 0.001 μF, to be magnified electronically, transformed

into a large, equivalent, very high-Q inductor, or made into a D-element that can be shown to have the impedance, ZD(ω) = 1/(ω2 D), where D is deter-

mined by certain Rs and Cs as shown next.

Find an expression for the driving point impedance of the general GIC circuit, Z11 = V1/I1. Clearly, from Ohm’s law:

V1

 

Z11 = V1 I1 = (V1 V2 ) Z1

(7.40)

© 2004 by CRC Press LLC

Analog Active Filters

293

I1

Z1

V1

V2 IOA

Z2 I2

V1

Z3 I3

IOA V3

Z4 I4

V1

Z5 I5

FIGURE 7.8

General architecture for a generalized impedance converter (GIC) circuit. The Zk can be resistances or capacitances (1/jωCk), depending on the filter requirement.

Also, from the ideal op amp assumption and Ohm’s law, it is possible to write the currents:

I2 = (V2 V1 ) Z2 = I3 = (V1 V3 ) Z3

(7.41)

I4 = (V3 V1 ) Z4 = I5 = V1 Z5

(7.42)

From the Equation 7.40, Equation 7.41, and Equation 7.42, it is easy to show:

V3 = V1(1+ Z4 Z5 )

 

(7.43)

 

Z

Z

5

Z

Z

4

ˆ

 

V2

= V1

3

 

2

 

˜

(7.44)

 

Z3Z5

 

 

 

 

 

 

 

and the GIC driving point impedance is given compactly as:

Z11 = V1/I1 = Z1 Z3 Z5/(Z2 Z4) complex ohms

(7.45)

© 2004 by CRC Press LLC

294

 

 

 

 

 

 

 

 

 

 

 

Analysis and Application of Analog Electronic Circuits

GAIN

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

GICINDUC

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

PHASE

DB

Temperature = 27

 

 

 

 

 

 

 

 

 

 

 

 

 

Case = 1

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

DEG

130.00

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

90.0

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

104.00

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

0.0

78.00

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

−90.0

52.00

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

−180.0

26.00

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

−270.0

0.00

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

−360.0

 

 

 

 

 

 

1K

 

 

 

 

10K

 

 

 

 

 

100K

 

 

 

 

1M

 

 

 

 

 

 

 

100

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

10M

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Frequency in Hz

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Frequency

= 100.00000E + 05

Hz

Gain

= 89.003 Db

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Phase angle = −158.263

Degrees

Group delay = 0.00000E + 00

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Gain slope

= −114.36240E − 01 Db/Oct

Peak gain

= 128.556 Db/F = 107.01700E + 04

1: Another run 2: Analysis limits

 

 

3: Quit

4: Dump

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

FIGURE 7.9

Magnitude and phase of Z11(f) looking into a GIC emulation of a 0.1-HY inductor. In this MicroCap simulation, TL072 op amps were used. Smooth line is magnitude.

Now if Z2 or Z4 = 1/jωC (a capacitor) and the other elements are resistors, Z11 has the form:

Z11 = jω [C4 R1R3R5/R2] reactive (inductive) ohms

(7.46)

that is, the GIC emulates a low-loss inductor over a wide range of frequency. The inductance of the emulated inductor is:

Leq = [C4 R1R3R5/R2] Hy

(7.47)

Figure 7.9 illustrates the magnitude and phase response of an inductive Z11 vs. frequency. Using MicroCap‘, 20 log Z11(f ) and – Z11(f ) vs. f are

plotted. The circuit used in this simulation is the same as in Figure 7.8, with Z1 = Z3 = Z4 = Z5 = 1 kΩ and Z2 = 1/jωC2; C2 = 0.1 μF.

Equation 7.47 shows that the simulated inductance is 0.1 Hy. (Op amp output current saturation effects are not included in this simple simulation.) Note that Z11(f ) increases linearly with frequency until about 180 kHz, at which its phase abruptly goes from the ideal +90 to –270, and 20 log Z11(f ) exhibits a tall peak at approximately 1 MHz. Clearly, the circuit model emulates a 0.1-Hy inductor below 180 kHz. A simple nonsaturating, two-time- constant model of the TL082 op amp was used in the MicroCap simulation.

Because the GIC inductive Z11 is referenced to ground, it must be used in active filters that require inductors with one end tied to ground. Figure 7.10(A) illustrates an inductive GIC used in a simple high-Q, quadratic BPF.

© 2004 by CRC Press LLC

Analog Active Filters

295

 

 

IOA

R

 

V1

 

 

+

 

 

I1

+

 

Z11

 

 

VS

C

Vo

GIC

 

 

 

 

INDUCTOR

A

R

V1 = Vo

 

+

I1

+

 

 

 

 

C

 

VS

Leq

Vo

 

 

B

FIGURE 7.10

(A) Circuit of an RLC band-pass filter using the GIC inductor. The GIC circuit allows emulation of a very large, high-Q inductor over a wide range of frequencies and is particularly well suited for making filters in the subaudio range of frequencies. (B) The actual BPF.

Figure 7.10(B) illustrates a simple R–L–C BPF suitable for circuit analysis. A node equation gives the BPF’s transfer function:

V

(s) =

sLeq R

o

 

 

 

V

s2C L

+ sL R + 1

S

 

eq

eq

In this filter:

ωn = 1 CLeq rs

Q = 1 (2ξ) = R CLeq

VVo (n ) = 10

S

Leq is given by Equation 7.47.

(7.48)

(7.49A)

(7.49B)

(7.49C)

© 2004 by CRC Press LLC

296

Analysis and Application of Analog Electronic Circuits

R

V1

IOA

+

 

 

 

I1

 

+

 

 

C

 

Vo

VS

GIC

 

 

 

D-ELEMENT

 

 

 

 

FIGURE 7.11

Circuit showing how a GIC “D” element can make a low-frequency, low-pass filter.

The GIC “D element” Z11 is a frequency-dependent negative resistance (FDNR) that can be made by putting capacitors in the Z1 and Z3 positions in the GIC circuit. Thus, the driving point impedance is real and negative:

Z

=

(1 jωC1)(1 jωC3 )R5

=

R5

 

 

=

−1

ohms

(7.50)

 

 

 

 

 

11

 

R R

ω2C C R R

ω2D

 

 

2

4

 

1

3

2

5

 

 

 

 

 

 

 

D = C1C3R2R4

R5

 

 

 

 

 

(7.51)

Figure 7.11 shows that the D element can make a quadratic LPF. Write the node equation for Vo in terms of the Laplace complex variable, s:

V

 

[G + sC + s2

D] = V

S

G

(7.52)

o

 

 

 

 

 

 

 

Solving for Vo yields:

 

 

 

 

 

 

 

 

 

 

V

(s)

 

 

1

 

 

(7.53)

 

o

 

 

 

 

V

s2DR

+ sCR + 1

 

 

S

 

 

 

 

 

 

 

Clearly, for the GID FDNR LPF,

 

 

 

 

 

 

 

 

ωn = 1

DR r s

 

 

(7.54A)

 

 

 

(

)

R D

 

 

(7.54B)

 

 

 

ξ = C 2

 

 

 

By using relation Equation 7.44 for V2/VS, it is possible to realize notch and all-pass filters from the basic GIC circuit. Figure 7.12 and Figure 7.13 illustrate examples of these filters. Readers can develop their transfer functions as exercises.

© 2004 by CRC Press LLC