LAB2final.pdfprelab2LAB2: Electronic ScaleStrain GagesIn this lab we design an electronic scale. The device could equally well be used as an orientation sensor for anelectronic camera or display, or as an acceleration sensor e.g. to detect carcrashes. In fact, similar circuits to theone we build are used in all these applications, albeit using technologiesthat allow much smaller size.For our scale we use the fact that metal bends if subjected to a force. In the lab we use an aluminum band with one endattached to the lab bench. If we load the other side, the band bends down. As a result, one side of the band gets slightlylonger and the other one correspondingly shorter. All we need to do to build a scale is measure this length change.How can we do this with an electronic device?It turns out we need to look no farther than to simple resistors. Aresistor is similar to a road constriction, such as a bridge or tunnel. The longer the constriction, the higher the“resistance”. Cars (or electrons) will back up. Increasing the width on the other hand reduces the resistance.If we glue a resistor to our metal band its value will increase and decrease proportional to the length change. Thepercent change is called the “gage factor” GF and is approximately two (since an increase in length isaccompanied by a corresponding decrease in width due to conservation of volume): a 1 % change in lengthresults in a 2 % change in resistance. Mathematically we can express this relationship as∆RRo=GF∆LLo(1)where Loand Roare the nominal length and resistance, respectively, and ∆L and ∆R are the changes due to appliedforce. The nominal length and value of the resistor, Loand Ro, can be measured. If we further determine ∆R wecan calculate ∆L, and, with a bit of physics, determine the applied force.Assuming you can measure resistance with a resolution of 0.1 Ω, what is the minimum length change that you candetect for Ro=Ω and Lo=mm? Use GF=2 for this and all subsequent calculations.1 pt.0In the laboratory, attach the metal band with attached strain gage to the bench. Measure the the nominal resistanceRowithout any extra weight applied to the band. Then determine ∆R for one, three, and six weights. Report yourresults in the table below:Ro1 pt.1∆R, 1 weight1 pt.1∆R, 3 weights1 pt.1∆R, 6 weights1 pt.1The small changes may be difficult to resolve if the display of the meter flickers. Use the bench top meter (nothandheld device), and make sure the connections are reliable. Poor connections can contribute several Ohmsofresistance, and small changes in the setup (e.g. a wire moved) can result in big resistance changes. Also, as for allmeasurements, keep wires short.Half Bridge CircuitUsing an Ohm-meter to evaluate the output of our scale is not very practical. Typically we prefer a voltage outputfor sensors. Voltages can easily be interfaced for example to microcontrollers (small computers), which in turn canbe connected to a display or other appropriate device. In this lab we focus on getting a voltage out of our sensorand leave the microcontroller interface for later.1February 12, 2009 LAB2 v64421882UC Berkeley, EECS 100 LabB. BoserNAME 2:SID:NAME 1:SID:Figure 1 Strain gage in a half bridge circuit configuration.Resistors, in combination with a voltage source, come to the rescue here also. Figure 1 shows a so-called half bridgeconfiguration where the strain gage resistor Rstrainis connected to a reference resistor Rrefand a balanced supply.An important objective is to achieve nominally zero output voltage vowhen no strain (weights) are applied to thescale. This is achieved in the circuit when Rrefis set equal to the nominal value Roof the strain gage resistor andthe supply voltages Vddand Vssare equal.Under these conditions, and for Vdd=Vss=V, what is the value of Voif Rstrainincreases by 1 % from itsnominal value?1 pt.1Build the circuit in the laboratory using a solderless breadboard (download the guide from the manual section ofthe website). Choose Rrefas close as possible to Ro. Set up the laboratory supply for Vdd=Vss=5 V. Then takethe following measurements of vo:no weights1 pt.2-- Now adjust Vsssuch that vo=0 V --no weights1 pt.21 weight1 pt.26 weights1 pt.2Ask the teaching assistant to verify the circuit operation.Full Bridge CircuitThe half bridge circuits has several drawbacks. While doing the measurements you may have noticed how difficultit is to accurately set the null point and keep it stable. Any change of the supply voltage directly affects theoutput of the circuit. In practice such changes occur frequently, e.g. as the result of a sudden increased currentconsumption of a different part of the circuit such as an amplifier or microcontroller. The need for balancedsupply Vdd=Vssis a further drawback of the half bridge circuit.The full bridge configuration, shown in Figure 2 on the next page, results in significant improvements. Fourresistors are used, all with nominally equal value Ro. The output voltage vois the difference of vaand vband onlya single supply Vsis needed.Let’s investigate the full bridge’s ability to reject supply voltage variations. For our analysis, let’s assume that thebridge is balanced, i.e. vo=0 V. Now let’s say the supply voltage is initially Vs=V, but then drops suddenlyby 10 %. Calculate the resulting change of vo.1 pt.2This property of the full bridge significantly reduces its offset voltage in practical situations where e.g. supplyvariations are common.In practice of course the reference and nominal strain resistance will not be exactly equal. As a consequence, vo2February 12, 2009 LAB2 v64444.1Figure 2 Full bridge configuration.Figure 3 Potentiometer: Equivalent circuit diagram (left) andsymbol (right).has an offset, i.e. its value is not zero when no load is applied to the scale. Rather than tweaking the referenceresistor values, we use a potentiometer for offset adjustment. Figure 3 shows the circuit diagram and symbol fora potentiometer. The sum RT=Ra+Rbof the values of resistors Raand Rbis constant, e.g. 1 kΩ(potentiometersare available with many different values). A knob or screw terminal is used to adjust values of Raand Rbsuch thatRa=β RT(2)Rb= (1−β)RT(3)with 0<β≤1, depending on the setting of the adjustment knob. Calculate the values of Raand Rbfor apotentiometer with RT=kΩ and its adjustment knob set such that β=/10:Ra=1 pt.3Rb=1 pt.4Ra+Rb=1 pt.5If we add a potentiometer in parallel with Rref2and Rref3as shown in
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