1. Many measuring devices depend on some basic electrical principal for their operation. Also, nearly all data gathering and analysis systems depend on electronic devices. Hence, it is crucial to discuss some electrical devices currently used to emphasize their use in the measurement process. Basic Electrical Measurements and Sensing Devices

2. If a charge moves through a conductor which exists in a magnetic field, a force (F) exerted on the conductor as a result of the interaction between the charge and the magnetic field. Forces of Electromagnetic Origin i F B

3. This force can be written as: F = BiL where: B: magnetic field density, in tesla (Weber/meter square) i : the current in the conductor L: the length of the conductor This equation is very important… It provides a bridge between a basic electrical quantity (i) and a mechanical property (F). It reduces the problem of measuring i into simply measuring F. Forces of Electromagnetic Origin

4. Consider the accompanying figure. As i flows through the conductor, the spring will stretch and develop the force required to balance the electromagnetic force. By equating the two forces: Kx = Bil i can be found as: i = (K/BL) x where K is the spring constant Forces of Electromagnetic Origin Fs = Kx F = Bil B i

5. In order to extend the previous concept into a more realistic model of actual current-measuring meters, we use a coil instead of a single-wire conductor. If the coil has N turns, then the exerted force can be given by: F = NBiL Basic Analog Meters

6. The D’ Arsonval movement may be used only for the measurement of direct current (dc). Two different types of movements are used for alternating current (ac) measurement; the iron-vane or moving-iron, and electrodynamometer. In the iron-vane instrument, the current is applied to a fixed coil. The iron vane is movable and connected to a restraining spring. The displacement of the vane is then proportional to the inductive force exerted by the coil. Basic Analog Meters

7. In order to measure a dc voltage, a voltmeter may be easily constructed by placed a resistor in series with a current meter. Thus, when the instrument is connected to a voltage source, the current in the instrument is an indication of the voltage. A series-resistor arrangement may also be used with the iron-vane. Study the electrostatic-voltmeter…(page 156). Basic Analog Meters

8. Basic Input Circuits The general method of gathering, processing, and displaying physical information includes the following steps: 1) Transducer: converts the physical property into an electrical signal. E.g.: microphone, thermocouple, solar cell… 2) Input circuit. 3) Signal conditioning: for noise reduction. 4) Transmission. 5) Processing: E.g.: AM radio receiver 6) Display: E.g.: cathode-ray screen, flash memory disks,…

9. Basic Input Circuits Transducer Display Processing Transmission Signal Conditioning Input circuit Signal Physical Property

10. Basic Input Circuits Consider a gas sensor resistance of which changes as a function of the gas concentration surrounding the sensor. The following figure represents a simple type of input circuit which uses the current flow through the sensor resistance as an indication of the value of the resistance. The current is given as: i = E/(R +R i )

11. Basic Input Circuits Ri i Ei Rm R

12. Basic Input Circuits-Balanced Bridges Bridge circuits are used to measure resistances, inductances, and capacitances. These bridges can be broadly classified into: balanced and unbalanced. The Wheatstone bridge is normally used for measuring resistances in the range of 1  to 1M , (see figure 4.24.)

13. Basic Input Circuits-Balanced Bridges The Wheatstone bridge is widely used for measuring the output resistance of various transducers, such as resistance thermometers, strain gages, and other devices that register the change in the physical variable as a change in output resistance. R 2 and R 3 are normally known. R 1 is a variable resistance, and R x is the unknown resistance associated with the transducer output.

14. Basic Input Circuits-Balanced Bridges The voltage E is applied to the bridge. The bridge may be balanced by adjusting R 1 until the sensing device (G) indicates no current flow. When this occurs, the voltage drop across R 2 must be equal the voltage drop across R 1 since this implies the difference between B and D must be zero. In this case:

15. Basic Input Circuits-Balanced Bridges E B = E D, or: i 2 R 2 = i 1 R 1 Also at balance: i 2 = i 3 = E/(R 2 +R 3 ), and: i 1 = i x = E/(R 1 +R x ), By eliminating the currents from these relations, the result is: R x = R 1 R 3 /R 2 Hence, R x can be found in terms of the other three resistances.

16. Basic Input Circuits-Unbalanced Bridges If the galvanometer does not have a null reading, we say that the bridge is unbalanced. Unbalanced bridges are particularly important for the measurements of dynamic signals where insufficient time is available for achieving balance conditions. Consider the bridge circuit shown in figure 4.25. R g is the galvanometer resistance; i b and i g are the battery and galvanometer currents, respectively. R b represents the resistance of the power supply.R 4 is the unknown resistance

17. Basic Input Circuits-Unbalanced Bridges If the bridge is slightly unbalanced, R b effect can be eliminated and i g can be written as: i g = E g /(R+R g ), where R is the effective resistance of the bridge circuit presented to the galvanometer and is given by: R = R 1 R 4 /(R 1 +R 4 ) + R 2 R 3 /(R 2 +R 3 ) The voltage presented at the terminals of the galvanometer E g is: E g = E[(R 1 /(R 1 +R 4 )-R 2 /(R 2 +R 3 )] Finally, substitute in the ig equation and solve for R 4.

18. Amplifiers In many experimental measurements the signals are comparatively weak and must be amplified before they can be used to “drive” an output device. Typical situations where amplifiers are used is the sound systems. Consider the given amplifier schematic below. Without the feedback-attenuator loop, the amplifier gain (A) is given as: A = Eo/Ei

19. Amplifiers Attenuator, K Amplifier Gain = A Eo Ei E f =KEo E=Ei - E f

20. Amplifiers Now, the feedback system attenuates Eo by a factor k and feeds it back into the input. The feedback voltage is Ef = KEo, is subtracted from Ei resulting in an input to the amplifier of E = Ei-Ef. Now the gain of the amplifier with the feedback system is: Af = Eo/Ei = Eo/[(Eo/A) + kEo] = A/(1+Ak) If Ak is very large, Af  1/k

21. Amplifiers A differential amplifier is a device which provides for two inputs and an output proportional to the difference in the two input voltages. Generally, it has a gain of: A = Eo/abs(E i,1 – E i,2 ) A Eo E i,1 E i,2

22. Transformers Transformers are used to match impedance in many experimental situations. An ideal n-turn transformer is shown below ++ -- v1v1 v2v2 n:1 i1i2

23. Transformers For this transformer: v2 = nv1 i2 = (1/n)i1, therefore: Also, We may state that the output impedance Z 2 is related to the input impedance Z 1 by:

24. Noise can be defined as the unwanted part of a signal. It is present in all physical situations in which measurements are attempted. A priori knowledge of frequencies ranges in which the desired signal exits can significantly help the experimenter. For example, the human ear responds only to signals in the range of frequencies between 20 Hz and 18 kHz. Therefore, the processing of a sensor used for such sounds can consider all frequencies out of this range as noise. Signal Conditioning

25. Filters operate such that they allow only for a certain range of frequencies to pass through it. Various arrangements may be used for filter circuits, but they fall into three categories: 1)lowpass: permits the transmission of signals with frequencies below a certain cut off value, 2)highpass: permits the transmission of signals with frequencies above a certain cut off value, 3)banpass circuits: permits the transmission of signals with frequencies in a certain range. Signal Conditioning

26. Gain is a dimensionless quantity, but engineers speak of decibels of gain or loss as: Decibels = 10logP 2 /P 1 Decibels = 20logE 2 /E 1 Decibels = 20logI 2 /I 1 where P 2/ E 2/ I 2 and P 1/ E 1/ I 1 are the input and output powers/voltage/current, respectively. When the ratio of P2/P1 less than 1, the gain in decibels is negative. For such circuits, insertion loss is defined as: 10logP 1 /P 2 Signal Conditioning

27. E.g.: calculate the overall voltage amplification for: Sol. 20logE2/E0 = 87.92dB Signal Conditioning Gain 1000 Gain 25 Eo = 1mV E1 = 1VE2 = 25V

28. A simple RC circuit is to be used as a lowpass filter. It is desired that the output voltage be attenuated 3 dB at 100Hz. Calculate the required time constant (T=RC) value? Eo/Ei = -3 or: -3= 20logE 2 /E 1 Thus: Eo/Ei = 0.708 = where: So: T = 0.00159s. That is, the circuit can be built using a 1.0  F capacitor and 1.59k  resistor Signal Conditioning 1/(1+  2 T 2 ) 1/2  2 = (2  f) 2 =[2  (100)] 2

29. See figure 4.48. This is a distance-measuring transducer Working method: - An ac voltage is impressed in the primary coil - the output voltage from the two secondary coils depends on the magnetic coupling between the core and the coils. - this coupling is dependent on the position of the core. - thus, the output voltage is a measurement of the displacement of the core. That is, an Eo-displacement calibration curve can be drawn for this device. The Differential Transducer

30. See figure 4.52 The capacitance (in picofarads) for this arrangement is given by: C = 0.0885  A/d where: d : the distance between the plates, cm A : overlapping area, cm2  : dielectric constant (equals1 for air and 3 for plastics) Capacitate Transducer

31. Working method: - changing A, d, or  would result in a change in C. - C may be measured with bridge circuit. - this transducer can be used for displacement measurement through changing in A or d. - It is also used to measure liquid-levels by changing , (see figure 4.53). Capacitate Transducer