Lecture 4: Signal Conditioning

1. Lecture 4:Signal Conditioning

2. These slides are prepared by Dr. ChokKeawboonchuay, Assumption University.

3. Signal conditioning is the operation performed on the signal to convert it to a form suitable for interfacing with other elements in the process.

4. Before discussing signal conditioning, it is important to understand the loading effect.

5. Concept of Loading Connecting a sensor or circuit to a load introduces uncertainty in the measurement (amplitude of output voltage) without load: Vy = Vx with load: Vy < Vx

6. The output voltage is calculated using voltage division as • The output voltage is reduced by the voltage drop over the internal resistance of the sensor RX. • To reduce the uncertainty (i.e. to keep VyVx), RL RX

7. Example An amplifier outputs a voltage that is 10 times the voltage on its input terminals. It has an input resistance of 10 k. A sensor outputs a voltage proportional to temperature with a transfer function of 20 mV/°C. The sensor has an output resistance of 5.0 k. If the temperate is 50 °C, find the amplifier output. Sensor Amplification 50 °C ? V

8. Answer The ideal situation (with zero sensor o/p impedance and infinite amplifier i/p impedance) is that: • The sensor should give 50*20 mV = 1V. • This is then amplified 10 times to give 10V.

9. Answer However, due to the non-zero sensor output resistance (5kΩ) and finite amplifier input resistance (10kΩ), the actual situation is that: • The sensor deliver only V = 0.667V. • This is amplified 10 times to give 6.67V.

10. The buffer circuit • To minimize the loading effect, we must look for an amplifier that has infinite (very large) input impedance to take the whole sensor output voltage. • Furthermore, as this amplifier is probably going to drive other circuits, it should have zero (very small) output impedance. • A device having these two properties is called a buffer. One example is the voltage follower circuit.

11. Voltage Follower • The input impedance of the voltage follower is very high.

12. Difference Amplifier In many situations, it is required to find the difference between two signals. This can be achieved using the following difference amplifier circuit.

13. The instrumentation amplifier • The input impedances of the difference amplifier can be relatively low and, hence, tend to load the sensor output. • To have high input impedance, the difference amplifier is preceded by two voltage follower circuits to form the so-called instrumentation amplifier.

14. One disadvantage of this differential circuit is that changing gain requires changing 2 pairs of resistors • A more common differential amplifier in which the gain can be adjusted using one resistor (RG) is shown below.

15. Signal Conditioning Signal conditioning can be categorized into the following types • Signal-range and offset changes • Linearization • Conversions • Filtering

16. Signal-range and offset (bias) Example Design a circuit to achieve the following voltage conversion. Signal conditioning circuit 0.2 V – 0.7 V 0 V – 5 V

17. Answer • It is clear that we need to subtract 0.2V, then multiply the signal by 10. Zero shift Amplification 0.2 V – 0.7 V 0 V – 0.5 V 0 V – 5 V • This looks like a differential amplifier with a gain of 10 and a fixed input of 0.2 volts to the inverting side. The following circuit shows how this could be done using an instrumentation amplifier.

18. Note the voltage divider, which is used to provide the 0.2V offset. The zener diode is used to keep the bias voltage constant against changes of the supply.

19. Example 2-20 (Johnson, page 89) A sensor output a range of 20 to 250 mV as a variable varies over its range. Develop signal conditioning so that this become 0 to 5 V. The circuit must have very high input impedance. Answer Let us develop an linear equation for the output in terms of the input Vout= aVin+b where a and b are to be found.

20. For the two conditions we have in this problem we can write 0 = a(0.020)+b 5 = a(0.250)+b • Solving these two equations gives a = 21.7and b = -0.434. • Hence, the required equation is Vout = 21.7Vin-0.434 • This can also be written as Vout= 21.7(Vin-0.02) • This looks like a differential amplifier with a gain of 21.7 and a fixed input of 0.02 volts to the inverting side. The following circuit shows how this could be done using an instrumentation amplifier.

22. Example 2.21 A bridge circuit for which R4 varies from 100  to 102 . Show how an instrumentation amplifier (the circuit with RG) could be used to provide an output of 0 to 2.5 V. Assume that R2 = R3 = 1 k and that R1 = 100 k.

23. Answer • Clearly, the bridge is at null when R4=100 Ω. • When R4 = 102 Ω the bridge offset voltage is found as mV. • To get an output of 2.5V at 102 Ω means that we need a differential gain of (2.5 V/24.75 mV) = 101. • We have () () This gives RG = 2 kΩ.

25. Linearization • Often, the characteristic of a sensor is nonlinear • Special circuits were devised to linearize signals • Modern approach is to use computer software for linearization.

26. Conversion • Is to convert one form of signal or physical values into another form. • Example: resistance to voltage • Typical conversion is to convert resistance or voltage to 4 to 20 mA and convert back to voltage at the receiving end. • Thus, voltage-to-current and current-to-voltage circuits are essential.

27. Current to voltage converter Vout = - R1 IIN

28. Voltage to current converter

29. The op-amp forces its positive and negative inputs to be equal; hence, the voltage at the negative input of the op-amp is equal to Vin. • The current through the load resistor, RL, the transistor and R is consequently equal to Vin/R. • We put a transistor at the output of the op-amp since the transistor is a high current gain stage (often a typical op-amp has a fairly small output current limit).