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Chapter 7: Electronic Measuring Instruments




RICHLAND COLLEGE School of Engineering Business & Technology Rev. 0 – W. Slonecker Rev. 1 – (8/26/2012)– J. Bradbury Rev. 2 – (3/13/2015)– J. Bradbury INTC 1307 Instrumentation Test Equipment Teaching Unit 7 Electronic Measuring Instruments Unit 7: Electronic Measuring Instruments OBJECTIVES: 1. To define the operation of the EVM (Electronic Volt Meter) 2. To draw the schematic of a Difference Amplifier. 3. To state the theory of operation of a Difference Amplifier. 4. To design a range switch for an EVM. 5. To compare analog and digital voltmeters. The volt-ohm-milliammeter (VOM) is a rugged, accurate meter. Its principal problems are that it lacks both sensitivity and high input impedance (Zin). The electronic voltmeter (EVM) can be designed to have a high input impedance that remains constant for all ranges. The EVM loads the circuit under test less than a VOM. An EVM can be analog or digital, depending on the readout. An analog meter uses a meter movement with a dial and needle to indicate values. The digital uses a numerical readout. Both meters perform the same functions, such as measuring voltage, current, and resistance. The digital meters will be discussed in later chapters. The Difference Amplifier The difference amplifier or differential RD1 RD amplifier has two input signals and one output signal. If the two input voltages are different, an output VD1 (error) signal is produced between X and Y. This error can drive a meter circuit or an external control D circuit for automation purposes. The diff amp can G Q1 use bipolar junction transistors or field effect transistors (FETs) for Q1 and Q2. FETs provide S much higher impedance than BJTs and are the V1 preferred devices for instruments. If the input voltages to both FETs are equal, and RD1 = RD2 and Q1 is identical in all respects to Q2, then the voltage at X equals the voltage at Y, and no output occurs. Figure 1. The configuration is called a difference amplifier because it amplifies differences between V1 and V2. If input signals V1 and V2 are both ac signals of the same amplitude and phase as in Figure 1, then the difference is 0 and VX equals VY, making the output 0 volts. If V1 and V2 are equal but 180 degrees out of phase, as in Figure 2, then as V1 increases, V2 decreases. At the maximum positive value of V1, V2 is at its maximum negative value. The voltage across RS is constant but the current through RS is no longer evenly shared by the transistors. Q1 current increases as V1 increases, and Q2 current VDD + RD2 DR X Y VD2 D G Q2 S RS V2 VEE VDD - Page 1 of 3 decreases. At the V1 maximum, VD1 is at minimum value due to more current through RD1, and VD2 is at maximum value due to less current through RD2. The error signal can be quite large. If Q2 has its gate grounded, and the input to Q1 is 0 volts, both Q1 and Q2 have equal current flow. When a positive rising signal is applied to the gate of Q1, Q1 current increases. Since more of the RS current goes through Q1, the Q2 current decrease. Diff amp transistors steer the almost constant RS current to either RD1 or RD2 or both resistors when the two gate voltages are equal. VDDpos Balance Pot RZ 50% RD1 RD2 Calibration Pot Disp RC 50% Input Q1 Q2 Rin 9MOhm RS Since the difference amplifier requires a very good match between VEEneg transistors Q1 and Q2, as well as resistors Figure 3. RD1 and RD2 to function correctly, the diff amp in Figure 3 has two potentiometers added for balance and calibration. The balance pot equalizes the drain resistance for each transistor to obtain a zero output for a grounded input. The calibration pot is then adjusted for gain calibration of the display device. VDDpos An EVM makes use of the extremely high input impedance of FET devices. Balance Pot The VOM had the problem of changing the input Z whenever the RZ range switch was changed. 50% Input 1.0v Remember that the power to drive the RD1 RD2 VOM came from the circuit being Calibration measured. The lower the input Z, the Pot more power was required and the Disp 6MOhm RC more the loading occurred. 3.3v 50% To lessen this problem, the EVM uses 10v Q1 Q2 an FET as the input amplifier. By 33v 2.1MOhm switching the amplifier input to taps on a voltage divider, the amplifier 100v input voltage can be kept within range 600kOhm of the amplifier. This range switch RS provides a constant input impedance of 9 Meg ohms no matter what 210kOhm 90kOhm VEEneg position the range switch is in. The maximum voltage that can be applied to the voltage divider is 100v full scale. The maximum voltage that can Figure 4 be applied to the difference amplifier is 1.0 volt. Any input above 1.0 volt will cause distortion and an erroneous (false) reading. The input signal is always applied at the top of the divider. The input voltage to Q1 is taken from the tap on the voltage divider corresponding to the voltage being measured. Page 2 of 3 For example, if a 10 volt signal was to be measured, the 10 volt tap on the input divider would be connected to the input of the amplifier as shown in Figure 4. Remember, the input to the FET should not exceed 1.0 volt, no matter what the voltage level applied to the input of the divider. If vin is 1v or less, it is applied directly to the Q1 gate. If vin = 10v and the maximum signal to the input amp is 1 volt, the divider must be used. Using Series Circuit Voltage divider Law: R tap V Q 1 in =V Div in Rtap is the resistance from the tap to ground. Rtotal is 9M. R total ( ) VDiv in is voltage at the top of the divider. VQ1 in is the tap voltage. Rtap= V Q 1 in⋅Rtotal V Div in Rtap= Calculate tap resistance from Rtotal, input voltage and tap voltage. 1 v⋅9 M Ω =900 k Ω 10 v Using an EVM To Measure Current and Resistance An EVM can be expanded to provide accurate current and ohmic measurements. Ohm’s Law converts voltage to current if a precise resistance is known. All that is required for an EVM to measure current is a precisely known shunt resistance and an amplifier instead of a divider. The amplifier is needed so that the shunt can have a low value of resistance, perhaps 10 ohms. For a 1mA current through a 10 ohm shunt, 0.01 volts would result. An amplifier with a gain of 100 would supply 1v to the EVM transistor input for a full-scale 1.0mA reading. By adding a precision current source to the instrument, a known current can be injected V meas into a resistance and the voltage determined by the EVM. RUnk = A current of 1.0mA I known through a 10k resistor would give 10v. The meter readout would be calibrated to read 10k instead of 10v when set for ohmic measurements. Remember that conventional ohmmeters had very non-linear scales since current was being measured and current is in the denominator of the equation above. Voltage has a linear relationship to resistance. Precision current sources are not difficult to make from bipolar junction transistors. Page 3 of 3