1 / 30

ENTC 4350 BIOMEDICAL INSTRUMENTATION I

ENTC 4350 BIOMEDICAL INSTRUMENTATION I. BASIC DIFFERENTIAL AMPLIFIER. Introduction. The differential amplifier can measure as well as amplify small signals that are buried in much larger signals. There are two input terminals, labeled (  ) input, and (+) input. Superposition.

lucio
Download Presentation

ENTC 4350 BIOMEDICAL INSTRUMENTATION I

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. ENTC 4350BIOMEDICAL INSTRUMENTATION I BASIC DIFFERENTIAL AMPLIFIER

  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 E1is replaced by a short circuit, E2sees an inverting amplifier with a gain of m. • Therefore, the output voltage due to E2is  mE2.  mE2

  5. Now let E2be short-circuited: • E1divides between R and mR to apply a voltage of E1m/ (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 E1is the divided voltage: • E1m/(1 + m)times the noninverting amplifier gain, (1 + m),which yields mE1. mE1

  7. Therefore, E1is amplified at the output by the multiplier m to mE1. • When both E1and E2are present at the (+) and () inputs, respectively. • Vois mE1mE2.

  8. The output voltage of the differential amplifier, Vo, 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 E1 = E2the output voltage is 0. • To put it another way, when a common (same) voltage is applied to the input terminals, Vo = 0.

  10. 22 kW Rf 2.2 kW ' Ri RS ' VOUT 4.7 kW ─ RD + Lab 6_Differential Amplifier • The gain of the amplifier below can be determined using the Superposition Principle.

  11. Inverting Amplifier • Forcing V2 to 0 develops an inverting amplifier with an output, VOUT of: 22 kW Rf V1 2.2 kW ' ─ Ri RS ' + VOUT 4.7 kW RD

  12. 22 kW Rf 2.2 kW ' ─ Ri RS ' + VOUT 4.7 kW RD Non-inverting Amplifier • Forcing V1 to 0 develops a non-inverting amplifier. V2

  13. 22 kW Rf 2.2 kW ' ─ Ri RS ' + VOUT 4.7 kW V2 RD • Applying Thevenin’s Theorem:

  14. The output of the non-inverting amplifier is: 22 kW Rf 2.2 kW ' ─ Ri RTh ' + VOUT

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

  16. So the balanced condition yields • and the differential gain Ad is

  17. 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.

  18. 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.

  19. The common-mode rejection ratio CMRR is defined as the magnitude of the ratio of the differential voltage gain Adto the common-mode voltage gain Ac.

  20. Adequals VOUTdivided by V1when node 2 is grounded, and V1is applied to node 1. • Also, ACequals VOUTdivided by V1when node 1 is connected to node 2, and V1is applied again.

  21. In practice the CMRR is measured in the following steps: • Ground V2, and apply a voltage V1 to the upper terminal. • Measure the resulting VOUT. • Lift V2from ground and short the two input leads, then apply the same value of V1. • Measure the resulting VOUT. • To compute CMRR, divide the results of step 2 by the result of step 4, and take the magnitude.

  22. The CMRR is a voltage ratio, and therefore in decibel units we may define CMRRdb as

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

  24. For such a connection, the input voltage is called the common-mode input voltage, ECM.

  25. Now Vowill be 0 ifthe resistor ratios are equal (mR to R for the inverting amplifier gain equals mR to R of the voltage-divider network.)

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

  27. The potentiometer is trimmed until Vois reduced to a negligible value. • This causes the common-mode voltage gain, Vo/ECMto 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.

  28. 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.

More Related