1. Signal Conditioning Elements (SCE)

2. 6/13/2016Measurement & Transducers2 1. Voltage dividers Example :Potentiometer circuit

3. Deflection bridges Deflection bridges are used to convert the output of resistive, capacitive and inductive sensors into a voltage signal Amplifiers Amplifiers are necessary in order to amplify low-level signals, e.g. thermocouple or strain gauge bridge output voltages, to a level which enables them to be further processed Oscillators and resonators

4. Thévenin equivalent circuit for a deflection bridge

5. 6/13/2016Measurement & Transducers5 a. Range of output Bridge Parameters

6. 6/13/2016Measurement & Transducers6 b. Sensitivity c.Maximum power dissipation

7. 6/13/2016Measurement & Transducers7 d. Non linearity

8. Design of resistive deflection bridges Output voltage for resistive deflection bridge R 1 = R I, and R 2, R 3 and R 4 are fixed resistors

9. Relationship between resistances in a balanced Wheatstone bridge Often we require V MIN = 0, i.e. the bridge to be balanced when I = I MIN ;

10. Output voltage for single-element strain gauge bridge Four-element strain gauge bridge

11. strain-gage arrangements in a Wheatstone bridge

12. 6/13/2016Measurement & Transducers12 Case 1. -utilizing a single active gage in position R1 - it is often employed for both static and dynamic strain-gage measurement if temperature compensation is not required. -The resistance R1 = Rg and the other three resistances are selected to maximize the circuit sensitivity while maintaining the balance condition R1R3 = R2R4.

13. 6/13/2016Measurement & Transducers13 The sensitivity S s of the strain-gage—Wheatstone bridge system is defined as the product of the sensitivity of the gage S g and the sensitivity of the bridge circuit S. Thus,

14. 6/13/2016Measurement & Transducers14 -the dummy gage is inserted in arm R4 of the bridge instead of arm R2. -The active gage remains in arm R1 - fixed-value resistors are used in arms R2 and R3. - With this positioning of the dummy gage -the system sensitivity is the same as that given by case 1. - Temperature compensation is achieved in the same manner that was illustrated in Case 2, but without loss of circuit efficiency. - When a dummy gage is to be used to effect temperature compensation, arm R4 of the bridge is the preferred location for the dummy gage.

15. 6/13/2016Measurement & Transducers15 Case 4. - Four active gages are used in this Wheatstone bridge arrangement: - - it is used to measure transverse and axial strain

16. 6/13/2016Measurement & Transducers16 Load Cell : Force measurement Link-type Load Cell

17. 6/13/2016Measurement & Transducers17 Beam-type load cells

18. 6/13/2016Measurement & Transducers18 Ring-type load cell.

19. Output voltage for single-element Thermoresistor bridge Output voltage for double -elements Thermoresistor bridge

20. Output voltage for cantilever and torque elements

21. Output voltage for Pillar load cell

22. Design of reactive deflection bridges Bridge for capacitive level sensor

23. Thus in order to get :E Th = 0 at minimum level h MIN, we require C 0 = C hMIN (R 3 /R 2 ), giving: Output voltage for capacitance level bridge if R3 /R2 is made large compared with 1, this approximates to the linear form:

24. This gives: Output voltage for capacitance push-pull bridge

25. Output voltage for inductive push-pull bridge This gives:

26. Amplifiers

27. Why do we need Amplifiers?  Signal Amplification (I,V,,P)  Signal processing  Inverting  Buffering  Filtering  Compression  Integration  Differentiation  Converters * (How)

28. Ideal operational amplifier characteristics

29. typical operational amplifier characteristics (Ideal vs. OPA27)

30. Transfer characteristics of Op Amp

31. Inverting amplifier

32. Since V+ = V− = 0 Also giving The output voltage of inverting Amplifier

33. Non-inverting amplifier

34. Since i + = 0, V + = V IN The output voltage of Non-inverting amplifier, R F and R 1 form a potential divider, we have Also since V+ = V− we have

35. Voltage follower.

36. Differential amplifier

37. Strain gauge bridge connected to differential amplifier

38. Instrumentation amplifier High input impedance High common mode rejection ratio Low input offset voltage Low temperature coefficient of offset voltage.

39. Voltage adder.

40. Parameters influence the d.c. performance of the amplifier  Input offset voltage V OS The existence of input offset voltage V OS means that V OUT is unequal to zero when both V− and V+ = 0 volts, i.e. Some operational amplifiers have facilities for adjusting V OS to zero, i.e. for obtaining V OUT = 0 when V+ = V− = 0. Where A OL -D.C. open-loop gain

41. The effect of V OS on inverting amplifier

42.  Appropriate temperature coefficient V OS is dependent on the temperature T E °C of the amplifier environment Example If V OS is set to zero at T E = 15 °C; then if TE subsequently increases to 25 °C, the resulting input offset voltage is γ (25 − 15), i.e. ≈ 6 μV, which causes a change of approximately 0.6 × 10 6 μV, i.e. 0.6 V in the output of the open-loop operational amplifier

43.  Common mode voltage Common mode voltage affects on Vout where A CM is the common mode gain

44. Common Mode Rejection Ratio (CMRR)

45. The equivalent circuit for an open-loop amplifier

46. a.c. performance of a practical operational amplifier Gain–frequency relation for open-loop amplifier where f B = 1/2πτ is the −3 dB cut-off frequency

47. Typical gain–frequency characteristics for operational amplifier

48. Instrumentation amplifiers High input impedance High common mode rejection ratio Low input offset voltage Low temperature coefficient of offset voltage.

49. Oscillators and resonators  Oscillators  Inductive Oscillators

50.  Capacitive Oscillators

51.  resonator

52. Mathematical model of resonator

53. Examples of resonators Vibrating plate element

54. Vibrating tube element.

55. Other types of Op. Amps

56. 6/13/2016Measurement & Transducers56 Example Find vo for the following circuit. It follows that and

57. 6/13/2016Measurement & Transducers57 Therefore From KVL, we have and

58. 6/13/2016Measurement & Transducers58 Example. Find vo for the following circuit. With the noninverting input connected to ground, we have vp = 0 = vn. From KVL and it follows that

59. 6/13/2016Measurement & Transducers59 Since no current flows into the op amp, i C = i R.With and we have

60. 6/13/2016Measurement & Transducers60 Logarithmic amplifiers - When a sensor’s output dynamic is of a high amplitude (10 mV to 10 V, for example), it can be useful to compress the signal by using a logarithmic amplifier. - After amplification and digitization, the signal can be easily transmitted across a transmission line. At reception, it is enough to carry out the reverse operation to restore the measurement signal. This principle allows us to lower noise sensitivity. - Logarithmic amplifiers also help us “linearize” sensors, carry out multiplications, divisions, elevations in the square, and extractions of the root squared. To construct this type of amplifier, we use the feature of a P-N junction with an equation (Ebres-Moll equation) in the following form: where q is the electron charge k the Boltzmann’s constant T is the absolute temperature U is the direct voltage and i 0 is the flow of reverse current

61. 6/13/2016Measurement & Transducers61 Schemata of logarithmic amplifier principle