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Presentation on theme: "ENTC 4350 BIOMEDICAL INSTRUMENTATION I BASIC DIFFERENTIAL AMPLIFIER."— Presentation transcript:


2 Introduction The differential amplifier can measure as well as amplify small signals that are buried in much larger signals.

3 There are two input terminals, labeled (  ) input, and (+) input.

4 Superposition If E 1 is replaced by a short circuit, E 2 sees an inverting amplifier with a gain of  m. Therefore, the output voltage due to E 2 is  mE 2.  mE 2

5 Now let E 2 be short-circuited: E 1 divides between R and mR to apply a voltage of E 1 m/ (1+ m) at the op amp’s (+) input.

6 This divided voltage sees a noninverting amplifier with a gain of (m + 1). The output voltage due to E 1 is the divided voltage: E 1 m/(1 + m) times the noninverting amplifier gain, (1 + m), which yields mE 1. mE 1

7 Therefore, E 1 is amplified at the output by the multiplier m to mE1. When both E 1 and E 2 are present at the (+) and (  ) inputs, respectively. V o is mE 1  mE 2.

8 The output voltage of the differential amplifier, V o, is proportional to the difference in voltage applied to the (+) and (  ) inputs. Multiplier m is called the differential gain and is set by the resistor ratios.

9 When E 1 = E 2 the output voltage is 0. To put it another way, when a common (same) voltage is applied to the input terminals, Vo = 0.

10 Lab 6_Differential Amplifier The gain of the amplifier below can be determined using the Superposition Principle. '' ' ' ─ + 2.2 k  4.7 k  22 k  RSRS RiRi RfRf V OUT RDRD

11 Inverting Amplifier Forcing V 2 to 0 develops an inverting amplifier with an output, V OUT of: '' ' ' ─ + 2.2 k  4.7 k  22 k  RSRS RiRi RfRf V OUT RDRD V1V1

12 Non-inverting Amplifier Forcing V 1 to 0 develops a non-inverting amplifier. '' ' ' ─ + 2.2 k  4.7 k  22 k  RSRS RiRi RfRf V OUT RDRD V2V2

13 Applying Thevenin’s Theorem: '' ' ' ─ + 2.2 k  4.7 k  22 k  RSRS RiRi RfRf V OUT RDRD V2V2

14 The output of the non-inverting amplifier is: '' ' ' ─ + 2.2 k  22 k  R Th RiRi RfRf V OUT

15 The total output is the sum: To balance the circuit, we set the coefficients to add to zero.



18 So the balanced condition yields and the differential gain A d is

19 Common-mode rejection of 60 cycle power line interference in medical instrumentation which measures difference potentials on the body is a fundamental problem. Power-line interference may exceed the level of the signal being measured. This bad news is often cancelled by the fact that the interfacing signal appears equally intense at both input terminals of the diff amp, and is therefore called a common-mode signal.

20 If the diff amp is not perfectly balanced, as is always the case in the real world, then the common-mode signal input will cause an output signal that then constitutes interference with the desired amplified signal. Since one of the functions of the diff amp is to reject the common-mode signal, we define a figure of merit, the common-mode rejection ratio (CMRR), which measures how well the rejection occurs.

21 The common-mode rejection ratio CMRR is defined as the magnitude of the ratio of the differential voltage gain A d to the common-mode voltage gain A c.

22 A d equals V OUT divided by V 1 when node 2 is grounded, and V 1 is applied to node 1. Also, A C equals V OUT divided by V 1 when node 1 is connected to node 2, and V 1 is applied again.

23 In practice the CMRR is measured in the following steps: 1.Ground V 2, and apply a voltage V 1 to the upper terminal. 2.Measure the resulting V OUT. 3.Lift V 2 from ground and short the two input leads, then apply the same value of V1. 4.Measure the resulting V OUT. 5.To compute CMRR, divide the results of step 2 by the result of step 4, and take the magnitude.

24 The CMRR is a voltage ratio, and therefore in decibel units we may define CMRR db as

25 Common-Mode Voltage The simplest way to apply equal voltages is to wire inputs together and connect them to the voltage source.

26 For such a connection, the input voltage is called the common-mode input voltage, E CM.

27 Now V o will be 0 if the resistor ratios are equal (mR to R for the inverting amplifier gain equals mR to R of the voltage- divider network.)

28 Practically, the resistor ratios are equalized by installing a potentiometer in series with one resistor.

29 The potentiometer is trimmed until V o is reduced to a negligible value. This causes the common-mode voltage gain, V o /E CM to approach 0. It is this characteristic of a differential amplifier that allows a small signal voltage to be picked out of a larger noise voltage.

30 It may be possible to arrange the circuit so that the larger undesired signal is the common-mode input voltage and the small signal is the differential input voltage. Then the differential amplifier’s output voltage will contain only an amplified version of the differential input voltage.


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