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Modified Block Diagram

Part 5: op-amp circuits

Before building a circuit to condition the signal, you will build a couple of simple op-amp circuits to get afeel for what to do.

5.1 inverting circuit

The inverter is one of the simplest op-amp circuits. This circuit simply changes the sign of a signal. Theschematic diagram of an inverter is shown in Figure 6.

  • Build the inverter using two 10 k ohm resistors and an LM741 op-amp.
  • Input the signal from the function generator to the circuit.
  • Branch the input cable with a BNC T-connector. Connect this input to Channel 0 on Module 1 of the SCXI.
  • Connect the output signal from the circuit to the Channel 1 on module 1 of the SCXI.
  • When you run the VI, you may need to adjust the scales on the X and Y axes.

You should see that the output is an inverted version of the input.

Inverter and Inverting Gain Circuits

5.2 inverting gain circuit

The inverting gain circuit changes the sign of the signal and multiplies the amplitude. Increasing the amplitudeof a signal can be very important in distinguishing subtle signal characteristics. Quantization errors result when a signalcharacteristic is too small to be detected by a measurement system. (You will learn more about quantization errors in lecture, and onpages 249-250 in your text.)

  • Replace the feedback resistor (R2 as shown in Figure 5) in your inverter circuit with a 100 k ohm resistor. This will causethe input signal to be inverted and amplified by a factor of ten.
  • Run the VI again.

With the amplification, the input offset of 800 mV has increased to an 8 V output.

Note: Using the data acquisition card to view the signal, you may saturate the ADC with the amplified signal. Allyou will be able to see is a flat line near +5 V or–5 V.

Part 6: bias correction

For the amplified signal to be read correctly, the 800 mV bias must be removed. A two step process will be used toremove the bias:

  • A Trim Potentiometer will be used to step down a 12V DC signal to 800 mV.
  • A Differential Amplifier circuit will be built to subtract this stepped down voltage from the sine wave.

The result will be an 80 mV sine wave with zero mean.

6.1 trim potentiometer adjustment

Use the 10 k ohm trim potentiometer provided to build a voltage divider as shown below.

  • Connect one end lead of the potentiometer to the +12 V bus.
  • Connect the other end lead to the–12 V bus.
  • Connect a voltmeter across the wiper lead and ground to measure the output.
  • Adjust the potentiometer resistance until the voltage on the wiper lead is 800 mV.
Voltage divider circuit

6.2 differential amplifier

You will build a differential amplification circuit to subtract 800 mV from the sensor signal. A differentialamplifier is shown in Figure 8 below.

  • Use four 10 k ohm resistors to subtract the voltage divider output from the sensor signal. It is unlikely that the resultingoutput signal will be a zero-mean sine wave as desired.
  • Adjust the screw on the potentiometer to "trim" the offset of the sine wave to zero.
  • Amplify the difference between the signals by replacing the R2 resistors so that R2>R1. Replace the R2 resistors with 100 k ohm resistors.

Questions & Answers

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The answer is neither. The function, 2 = 0 cannot exist. Hence, the function is undefined.
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At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light.
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Source:  OpenStax, Introduction to mechanical measurements. OpenStax CNX. Oct 18, 2006 Download for free at http://cnx.org/content/col10385/1.1
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