CC415 Chapter 2 Signal Conditioning

CC 415 Data Acquisition System

CHAPTER 2 Signal Conditioning Amplification and Filtering Prof. Fawzy Ibrahim Electronics and Communication Department Arab Academy for Science and Technology (AAST) 1 of 39

CC415 Ch.2 S. Conditioning

Prof Fawzy Ibrahim

Chapter Contents 2.1 Ideal Operational Amplifiers 2.2 The Inverting Configuration 2.3 The Noninverting Configuration 2.4 Difference Amplifiers 2.5 The Instrumentation Amplifier 2.6 Active Filters

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CC415 Ch.2 S. Conditioning

Prof Fawzy Ibrahim

2.1 Ideal Operational Amplifier (op–Amp) Introduction: • •

OP–Amps are utilized in hundreds of application. One can do almost anything with op–Amps. Therefore, op – amps are produced on IC’s. An IC op–amp is made of a large number of transistors (tens), resistors and usually one capacitor.

2.1.1 The op–Amp Terminals •

•

From signal point of view op–Amp has three terminals as shown in Fig. 2.1: – Two input terminals (Terminal 1 and Terminal 2 ) – One output terminal (Terminal 3) From operational point of view op–Amp has five terminals as shown in Fig. 2.2: – Two terminals for the inputs and one for the output. – Two terminals for dc power supplies: (Terminal 4 for the positive voltage VCC and Terminal 5 for the negative voltage – VEE)

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CC415 Ch.2 S. Conditioning

Prof Fawzy Ibrahim

2.1.1 The op–Amp Terminals (Continued) •

Other terminals are induced for frequency compensation and terminals for offset nulling.

Fig. 2.1 Circuit symbol for the op amp.

Fig. 2.2 The op amp shown connected to dc power supplies. 4 of 39

CC415 Ch.2 S. Conditioning

Prof Fawzy Ibrahim

2.1.2 Function and characteristics of the ideal op–Amp Refer to Fig. 2.3: 1. V3 = Vo = A (V2 – V1) where A is the amplifier gain (open-loop gain) 2. The ideal op – amp is not supposed to draw any input current: i1 = 0 and i2 = 0, therefore, the input impedance Zin = ∞ 3. V3 = A (V2 – V1) independent of the current that may be drawn from terminal 3 into the load impedance, therefore the out impedance Zout = 0. 4. V3 is in phase with V2 and out of phase with V1. For this reason Terminal 1 is called the inverting input terminal and is distinguished by “–” sign, while terminal 2 is called the noninverting input terminal and is distinguished by a “+” sign.

Fig. 2.3 Equivalent circuit of the ideal op amp.

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CC415 Ch.2 S. Conditioning

Prof Fawzy Ibrahim

2.1.2 characteristics of the ideal op–Amp (Continued) 5.

V3 = A (V2 – V1), the op–amp respond only to the difference signal (V2 – V1) and hence ignore any signal common to both inputs. This is called common – mode rejection (zero common mode gain or infinite common mode rejection) Op–amps are direct–coupled or dc amplifiers that amplify signal whose frequency is low or zero. The ideal op–amp has a very large gain and ideally A ∞. These characteristics are summarized in Table 2.1.

6. 7.

Table 2.1 characteristics of ideal Op – Amp. #

Characteristic

Description

1

Zin = ∞

Infinite input impedance

2

Zout = 0

Zero output impedance

3

Zero common mode gain

Infinite common mode rejection

4

A≈ ∞

Infinite open–loop gain

5

BW = ∞

Infinite bandwidth

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CC415 Ch.2 S. Conditioning

Prof Fawzy Ibrahim

2.2 The Inverting Configuration Refer to Fig. 2.4 • R2 is connected from the output terminal of the op–amp (Terminal 3) back to the inverting or negative input terminal (Terminal 1) R2 apply a negative feedback. • Input signal is connected terminal 1 through a Resistance R1 • Terminal 2 is grounded

Fig. 2.4 The inverting closed-loop configuration. 7 of 39

CC415 Ch.2 S. Conditioning

Prof Fawzy Ibrahim

2.2 The inverting configuration (Continued) 2.2.1 The Closed – Loop Gain Refer to Fig. 2.5 • The closed – loop gain G is defined as G = Vo/V1 Since Vo = (V2 – V1) A yields V2 – V1 = Vo/A = 0 (A  ∞) V2 = V1 or V1 = V2 • This is called a virtual short circuit means that whatever voltage at 2 will automatically appears at 1 because A  ∞

Fig. 2.5 The inverting closed-loop configuration. 8 of 39

CC415 Ch.2 S. Conditioning

Prof Fawzy Ibrahim

2.2.1 The Closed – Loop Gain (Continued) Apply Ohm’s law we have:

V i1 

I

V 1

R1

For ideal op–amp Zin = ∞ Vo is given by:

V

o

V V V

• •

I o I

V

I

0

R1

V I R1

No current go into the op–amp therefore

 i1 R 2  0  V I R 2   V R1   R R

2 1

 G   R R

I

R2 R1

2 1

The closed–loop gain is the simply the ratio of the two resistor R2 and R1. The minus sign (–) means that the closed–loop gain amplifies provides a signal inversion (or the output signal is 180° phase– shifted with respect to the input. Therefore, this is called inverting configuration).

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CC415 Ch.2 S. Conditioning

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2.2.2 Input and Output Resistances • Assume ideal op–amp with infinite open–loop gain. • The input resistance of the closed – loop inverting amplifier of Fig.2.6 is given by:

Ri •

• • •

VI VI  R1 i1 V I / R 1

Fig.2.6 The inverting closed-loop configuration

It is required that the input resistance must be high, we have to select a high value of R1. However if the required gain R2/R1 is also high, then R2 could became impractically large. We may conclude that the inverting configuration suffers from a low input resistance. Refer to fig.2.6 (a) Vo = A (V2 – V1) ideal voltage source, it follows that the output resistance of the closed–loop amplifier Ro = 0 A solution to compromise between the requirement high Ri and high gain consider the following example: 10 of 39

CC415 Ch.2 S. Conditioning

Prof Fawzy Ibrahim

2.2.2 Input and Output Resistances Example 2.1 Assuming ideal op–amp, consider the amplifier circuit shown in Fig.2.7. Assume also that for practical reasons it is required not to use resistors greater than 1 MΩ. For this circuit do the following: a) Derive an expression for the closed–loop gain Vo/VI. b) Design this circuit to have a gain of G = 100 and Ri = 1 MΩ. c) Compare your design with that based on the inverting configuration of fig.2.5 Solution: V2 – V1 = Vo/A = -Vo/∞ = 0 V1 = V2 = 0 virtual ground Determination of the current i1

V  i1

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I

V

R1

1

V I R1

Figure 2.7 Circuit for Example 2.1 the circled numbers indicate the sequence of the steps in the analysis CC415 Ch.2 S. Conditioning

Prof Fawzy Ibrahim

2.2.2 Input and Output Resistances Example 2.1 (Continued) Determination of i2 i1 = i2 Since zero current flows into the inverting terminal i2 = i1 = VI/R1 The voltage at node x is given by:

V I R2    0    V x V 1 i2 R 2 V R2 R1 R1

I

Determination of i3:

i

3



0  V R 3

x

R

 R

1

2

R

V

I

3

Determination of i4:

V i4  i2  i3  R 12 of 39

I 1



R

2

R 1R

V

I

3

CC415 Ch.2 S. Conditioning

Prof Fawzy Ibrahim

2.2.2 Input and Output Resistances Example 2.1 (Continued) Determination of Vo:

V I  R2 R2 V o V x  i4 R 4   V I( V I) R 4 R1 R1 R1R 3 a) The closed – loop voltage gain is given by:

G  V o   R 2 (1  R 4  R 4 ) VI R1 R2 R3 b) Since the input resistance of 1 MΩ is required. Ri = R1 = 1 MΩ. From the gain equation G, the maximum possible value for the first factor R2/R1 = 1, thus select R2 = 1 MΩ To obtain G = -100, the second factor of the gain expression is 100, we select maximum allowed value of resistance to R4 = 1 MΩ The value of R3 is calculated as

(1  R3 = 10.2 KΩ 13 of 39

1M  1M   )  100 1M  R3

CC415 Ch.2 S. Conditioning

Prof Fawzy Ibrahim

2.2.2 Input and Output Resistances Example 2.1 (Continued) c) In comparison with inverting configuration shown in Fig. 2.8, use input resistance: Ri = R1 = 1 MΩ For G = -R2/R1 R2 = 100 MΩ Which is impractically large value. Figure 2.8 The inverting configuration

Current Amplifier: Refer to Fig. 2.9, marking R3 < R2, if (R3 = R2/K, where K > 1) i2 = i1 and i3 = K i1 i4 = (K+1) i1 • The circuit can be used as a current amplifier • The current i4 is independent of R4. The current amplifier of Fig. 2.9 delivers its output current to R4. It has a current gain of (1+R2/R3), a zero input resistance, and an infinite output resistance. The load(R4), however, must be floating (i.e., neither of its two terminal can be connected to ground). Figure 2.9 a current amplifier 14 of 39

CC415 Ch.2 S. Conditioning

Prof Fawzy Ibrahim

2.2.3 The Weighted Summer Amplifier •

•

•

A very important application of the inverting configuration is the weighted summer amplifier as shown in Fig. 2.10. V2 – V1 = Vo/A = Vo/∞ = 0 V2 = V1 = 0 Virtual Ground

Apply ohm’s law to have: Figure 2.10 weighted summer amplifier i1 = V1/R1, i2 = V2/R2, …… in = Vn/Rn i = i1 + i2 + i3 +……+in since op–amp input current = 0

V o  0  i R f  i R f   ( • •

Rf Rf Rf   ....  V1 V2 V n) R1 R2 Rn

The output voltage is a weighted sum of the input signals V1, V2, … Each summing coefficient can be independently adjusted by the corresponding feed in resistors (R1, R2, … or Rn). 15 of 39

CC415 Ch.2 S. Conditioning

Prof Fawzy Ibrahim

2.2.3 The Weighted Summer Amplifier (Continued) •

For summing signals with opposite signs, two ideal op–amps can be used as shown is Fig.2.11 and is given by:

Ra Rc Ra Rc Rc Rc V o  V 1 ( )( )  V 2 ( )( )  V 3 ( )  V 4 ( ) R1 R b R2 Rb R3 R4

Figure 2.11 A weighted summer capable of implementing summing coefficients of both signs. 16 of 39

CC415 Ch.2 S. Conditioning

Prof Fawzy Ibrahim

2.3 The Noninverting Op Amp Configuration As shown in Fig.2.12, the input is connected to input (+ve) terminal of the op-amp and one terminal of R1 is connected to the ground.

2.3.1 Closed – loop Gain

Fig.2.12 The Noninverting configuration

Refer to Fig. 2.13 • The closed loop – gain • G = Vo/V1

Fig.3.13 Analysis of the noninverting circuit. The sequence of the steps in the analysis is indicated by the circled numbers. 17 of 39

CC415 Ch.2 S. Conditioning

Prof Fawzy Ibrahim

2.3.1 Closed – loop Gain (Continued) • •

∞ V1 = V2 = VI

Assuming that the op-amp is ideal A VId = V2 – V1 = Vo/A = 0

i

V

1

o

 V

I

R

V

I

 0 1

 V R

 i2 R 2  V

I

I

 i

2

1

 V I . R 2  V I (1  R 2 ) R1 R1

which yields:

G  V o  1  R 2 V I R 1 2.3.2 Characteristics of the Noninverting Configuration The gain G is positive • The input impedance Zin = ∞ • The output impedance Zout = 0 18 of 39

VI/I = VI/0 = ∞ ideal voltage source

CC415 Ch.2 S. Conditioning

Prof Fawzy Ibrahim

2.3.4 The Voltage Follower •

In many applications, a buffer amplifier is needed as an impedance transformer or power amplifier that has Rin = ∞, Rout = 0 and G = 1 or the output just follow the input which is called voltage follower.

•

Fig.2.13 can be utilized as voltage follower by making R2 = 0 and R1 = ∞ as shown in Fig. 2.14.

Fig. 2.14 (a) The unity-gain buffer or follower amplifier. (b) Its equivalent circuit model. 19 of 39

CC415 Ch.2 S. Conditioning

Prof Fawzy Ibrahim

2.4 Difference Amplifiers •

•

•

Difference amplifier is one that responds to the difference between the two signals applied at its input and ideally rejects signals that are common to the two inputs. As discussed before, the representation of signals in terms of their differential and common – mode components as shown in Fig. 2.15 Fig. 2.15 Representing the input signals to a differential amplifier in terms of their differential and common-mode components.

The output voltage V0 is given by: V0 = Ad VId + ACm VICm Where Ad denotes the amplifier differential amplifier gain. ACm denotes the common – mode gain. 20 of 39

CC415 Ch.2 S. Conditioning

Prof Fawzy Ibrahim

2.4 Difference Amplifiers (Continued) •

The efficiency of a differential amplifier is measured by the degree of its rejection of the common–mode signals in preference to differential signal which is referred to as Common – Mode Rejection Ratio (CMRR) and is defined as: CMRR  20 log

• •

•

Ad A Cm

The gain of the noninverting Configuration (1+ R2/R1) and the gain of inverting configuration (-R2/R1). To get the difference between the two inputs, we have to make the two gain magnitudes equal in order to reject the common mode signals. Fig.2.16 is used to alternate (1+ R2/R1) to (R2/R1). The proper ratio of the voltage divider can be determined by:

R 4 (1  R 2 )  R 2 R 4 R 3 R1 R1 21 of 39

or

R4  R2 R 4 R 3 R 2 R1

CC415 Ch.2 S. Conditioning

Prof Fawzy Ibrahim

2.4 Difference Amplifiers (Continued) R4  R2 R 3 R1

•

This condition is satisfied

•

The output can be determined by superposition as shown in Fig.2.17

•

Refer to Fig.2.17 (a) V

•

Refer to Fig.2.17 (b) V

•

I2

 0 V

I1  0  V

o1

o2

  R 2 .V R1

V

I2

I1

R 4 (1  R 2 )  R 2 V  R4 R3 R1 R1

The output voltage, by applying the superposition principle

R V o  V o1  V o 2  R 2 (V I 2  V I 1 )  2 V Id R1 R1 R2  • The difference amplifier gain A d R1 •

We can select R3 = R1 and R4 = R2

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CC415 Ch.2 S. Conditioning

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I2

2.4 Difference Amplifiers (Continued)

Fig. 2.16 A difference amplifier.

Fig. 2.17 Application of superposition to the analysis of the circuit of Fig. 2.16. 23 of 39

CC415 Ch.2 S. Conditioning

Prof Fawzy Ibrahim

2.4 Difference Amplifiers (Continued) •

Refer to Fig.2.18 the common – mode gain ACm can be calculated as follows:

i1 

1 R1

[V

ICm



R4 V  R4 R3

ICm

]V

ICm

R3  1  i2  R 4 R 3 R1

R4 R4 R2 . R3 R4 (1 R2 . R3)      Vo V ICm i2 R2 V ICm V ICm V ICm R4  R3 R4  R3 R1 R4  R3 R4  R3 R1 R4 Thus

A Cm When

 V o ( R 4 )( 1  R 2 . R 3 )  0 V ICm R 4  R 3 R1 R 4 R2  R3 R1 R 4

A ICm  0  CMRR 24 of 39

 

CC415 Ch.2 S. Conditioning

Prof Fawzy Ibrahim

2.4 Difference Amplifiers (Continued) •

Refer to Fig. 2.19 the input resistance

V Id  R 1i1 0  R1i1 R id  2 R1 • •

V Id  R id i1

R1 is required to be large and R2/R1 is desired to be large This configuration has two disadvantages: 1. Low input resistance 2. It is difficult to vary Ad

Fig. 2.18 Analysis of the difference amplifier 25 of 39

Fig. 2.19 Finding the input resistance of the difference amplifier. CC415 Ch.2 S. Conditioning

Prof Fawzy Ibrahim

2.5 The Instrumentation Amplifier •

•

The low input resistance problem of the difference amplifier can be solved by using two voltage followers with gain as shown in Fig. 2.20(a) A1 and A2 which connected as noninverting configuration with gain (1  R 2 ) R1 The difference amplifier A3 operates on the difference signal (V

R 2 )  (1  R 2 ) )( 1  V Id and provides the output I1 R1 R1  R 4 ( 1  R 2 ) V Id R 3 R 1 Therefore, the differential gain  R 4 (1  R 2 ) R 3 R1

I2 V

V

o

A

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d

CC415 Ch.2 S. Conditioning

Prof Fawzy Ibrahim

2.5 The instrumentation Amplifier (Continued) • •

ACm = 0 and high differential gain The disadvantages of 2.20 (a) are: 1. VICm is amplified in the first stage and may saturate the 2nd state 2. A1 and A2 must be matched to have the same gain. 3. To vary Ad, we have to change R1 (up with A1) and R1 (with A2) and have to be perfectly matched (very difficult). • Disconnect the node X between the two resistors R1 and R1 as shown in Fig.2.20.(b) will solve the three problems. - Refer to Fig.2.20 (b) assuming ideal op–amps as shown in Fig.2.20 (c)

V I2 V

I1  V

V Id  i  Id 2R1

2R 2)  ( 1  V V o 2  V o1 2R1 27 of 39

Id

CC415 Ch.2 S. Conditioning

Prof Fawzy Ibrahim

2.5 The Instrumentation Amplifier (Continued)

Fig. 2.20 A popular circuit for an instrumentation amplifier: (a) Initial approach to the circuit; (b) The circuit in (a) with the connection between node X and ground removed and the two resistors R1 and R1 lumped together. This simple wiring change dramatically improves performance; (c) Analysis of the circuit in (b) assuming ideal op amps. 28 of 39

CC415 Ch.2 S. Conditioning

Prof Fawzy Ibrahim

2.5 The Instrumentation Amplifier (Continued) •

The overall difference voltage gain is given by:

V o  R 4 (1  R 2 ) Ad  V Id R 3 R1 Example 2.3: Design an instrumentation amplifier circuit (Fig.3.20(b)) to provide a gain that can be varied over a range of 2 to 1000 utilizing 100 KΩ variable resistances or potentiometer (pot). Solution: • The first stage is used as gain stage. • The second stage is used as a difference stage of gain 1 therefore select R3 = R4 = 10 KΩ. • Choose 2R1 as a series of two resistors R1f (fixed) and R1V = 100KΩ pot. as shown in Fig.2.21 • difference voltage gain is Fig. 2.21 Choice of 2R1

R 4 (1  2 R 2 ) Ad  R3 2R1

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CC415 Ch.2 S. Conditioning

Prof Fawzy Ibrahim

2.5 The Instrumentation Amplifier (Continued) Example 2.3: (Continued) • To realize an adjustable gain over a range of 2 to 100, we can write:

•

(1 

2R 2 ) = 2 to 1000 R 1 f  R 1V

(1 

2R 2 )  2 R 1 f  R 1V

and

(1 

2 R 2 )  1000 R 1 f  R 1V

Solve these two equations gives R1f = 100.2 Ω and R2 = 50.05 KΩ

We can choose the standard resistors R1f = 100 KΩ and R2 = 49.9 KΩ

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CC415 Ch.2 S. Conditioning

Prof Fawzy Ibrahim

2.6 Active Filters •

A Filter circuit can be constructed using passive components: resistors and capacitors. An active filter additionally uses an amplifier to provide voltage amplification and signal isolation or buffering.

2.6.1 Low-Pass Filter An ideal low-pass filter provides a content output from the dc up to cutoff frequency fOH and then passes no signal above that frequency. The ideal response of low-pass filter is shown in Fig.2.22. • The first order low-pass filter using a signal resistor and capacitor is shown in Fig.2.23 (a). The output voltage is given Fig.2.22 The idea response of LPF by:

Rf Rf 1 / Sc1 1 V o ( s )  Vi ( s )( ).(1  )  Vi ( s )(1  ).( ) 1 / Sc 1  R1 RG RG 1  Sc 1 R1

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CC415 Ch.2 S. Conditioning

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2.6.1 Low-Pass Filter (Continue) • The filter transfer function is defined as H(s) = Vo(s) / Vi (s) and is given by:

Rf Vo ( s ) 1 H ( s)   (1  )( ) Vi ( s) RG 1  s R1 C1 • The filter frequency response is determined by replacing s by j in the filter transfer function H(s) which leads to H(j) = Vo(j) / Vi (j) and is given by:

Rf Vo ( j  ) 1 H ( j )   (1  )( ) Vi ( j  ) RG 1  j /  oH

1 Av (1  j /  oH )

Where Av is the filter voltage gain and OH is the cutoff frequency which calculated as:

woH  2f oH 32 of 39

1 1   f oH  C1 R1 2R1C1

CC415 Ch.2 S. Conditioning

Prof Fawzy Ibrahim

2.6.1 Low-pass Filter (Continued) •

The Bode-plot of H(j) is shown in Fig.2.23 (b)

AV  1 

Rf RG

w oH  2 f oH 

1 1  f oH  C 1 R1 2 R1C 1

Example: • Calculate the cutoff frequency of a first order low-pass filter when

R 1  1.2 K

and

1 C  0.02µF foH   6.63kHz 3 6 2 (1.210 )(0.0210 )

Fig.2.23 First-order low-pass active filter (a) the circuit; (b) the Bode plot. 33 of 39

CC415 Ch.2 S. Conditioning

Prof Fawzy Ibrahim

2.6.2 High–Pass Filter •

•

A filter that provides or passes signals above cutoff frequency FOL is high-pass filter The ideal response of this filter is shown in Fig.2.24. The first-order high-pass filter circuit is shown in Fig.2.25 (a). The output voltage is given by:

R R1 ).( 1  f ) Fig.2.24 The idea response of HPF R1  1 / Sc 1 RG Rf Vo ( s ) SC 1 R1 SC 1 R1 H (s)   (1  )( )  Av ( ) The filter transfer function Vi ( s ) RG 1  SC 1 R1 1  SC 1 R1 V o ( s )  V i ( s )(

Vo ( j ) H ( j )   Vi ( j )

j / OL Av (1  j / OL )

The filter frequency response

• The Bode-plot of H(j) is given in Fig.3.25 (b)

AV  1  34 of 39

Rf RG

and

w oL  2 f oH

1 1   f oL  C 1 R1 2 R1C 1

CC415 Ch.2 S. Conditioning

Prof Fawzy Ibrahim

2.6.2 High–Pass Filter (Continued) Example: •

Calculate the voltage gain and the cutoff frequency of a high-pass filter shown in Fig.2.25 for R1 = 2.1 KΩ, C1 = 0.05 µF and RG=10KΩ and RF = 50 KΩ

Solution:

AV  1 

Rf RG

1

50 6 10

(The voltage gain)

The cutoff Frequency:

1 1 foL    1.5KHz 3 6 2R1C1 2 (2.110 )(0.0510 )

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CC415 Ch.2 S. Conditioning

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2.6.2 High–Pass Filter (Continued)

(b) (a) Fig. 2.25 High-pass filter: (a) the circuit; (c) response Bode plot. 36 of 39

CC415 Ch.2 S. Conditioning

Prof Fawzy Ibrahim

2.6.3 Band-Pass Filter •

•

The Band-pass filter circuit passes signals above one ideal cutoff frequency fOL and below a second cutoff frequency fOH as shown in Fig.2.26. Fig.2.27 shows a band-pass filter using two stages, the first a high-pass filter and the second a low-pass filter. The combined response represent the filter.

Fig.2.26 The idea response of BPF

Rf Rf R1 SR 1C 1 V o 1 ( s )  V i ( s )( ).( 1  )  V i ( s )( )(1  ) R1  1 / Sc 1 RG 1  SR 1C 1 RG Rf Rf 1 / SC 2 1 V o ( s )  V o1 ( s )( ).(1  )  V o1 ( s )( ).( 1  ) R 2  1 / Sc 2 RG 1  SR 2 c 2 RG • Combine the above equations we get the filter transfer function as:

Rf

SR 1C 1 1 Vo ( s ) V o ( s )  V1 ( s ).(1  ) ( )( )  H (s) RG 1  SR 1C 1 1  SR 2 c 2 Vi ( s ) 37 of 39

2

CC415 Ch.2 S. Conditioning

Prof Fawzy Ibrahim

2.6.3 Band-Pass Filter (Continued) • The filter frequency response is determined by replacing s by j in the filter transfer function H(s) which leads to H(j) = Vo(j) / Vi (j) and is given by: j /

H ( j ) 

AV  (1  • •

Rf RG

)2

 A 1  j / 

OL

v

OL

1 FoL  2R1C1

x

1 1  j /  OH and

1 FoH  2R2C2

The Bode-plot is shown in Fig.2.27 (b). Example: Calculate the cutoff frequencies of the band-pass filter if R1 = R2 =10kΩ, C1 = 0.1 µF and C2 = 0.002 µF Solution:

1 1 FoL    159.15Hz 3 7 2R1C1 2 (1010 )(10 ) 1 1 F oH    7 . 96 KHz 4 5 2 R 2 C 2 2   10  2  10 38 of 39

CC415 Ch.2 S. Conditioning

Prof Fawzy Ibrahim

2.6.3 Band-Pass Filter (Continued)

Fig. 2.27 Bandpass active filter (a) the circuit; (b) response Bode plot.

39 of 39

CC415 Ch.2 S. Conditioning

Prof Fawzy Ibrahim