EE233 lab report 3
The University of Danang Danang University of Science and Technology D I Objectives Read IC component specifications and get data from them for circuit analysis and design Analyze and measure characteristics of circuits built with opamps Use the op-amp as a component in the design of simple circuits Analyze the effect of open fault in manufacturing II Equipment used LAB REPORT Oscilloscope: Function: display the waveform of the Vin and Vout in the electrical circuit Multi-meter : Function: measuring the value of resistors Instructor : Nguyen Tri Bang Lab :3 Class : 15ECE2 Group members: Tran VietinTu Function: electric source, we use this device order to generate the Nguyen Cong Thien waveform Dinh Ngoc Tien Le Dinh Hoai Nam Arbitrary waveform generator : Section 6: Pre-lab Section 6.1 Designs of simple amplifiers Danang 2017 Danang 2016 For the circuit in figure with power supplies VCC = 12.0 V, VEE = -12.0 V, and assuming the op-amp is ideal, answer the following questions: Design an inverting amplifier using one op-amp and two or more resistors Design it such that it has a gain of -10 (this gain is negative) Pick resistor values that you have in the lab kit Include a schematic of this circuit with the component values labeled with your completed prelab assignment Set V=V+=V-=0 (V+ and V- : voltage at non-inverting and inverting node respectively.) Using KCL: Gain is -10 so we choice R2=10KΩ and R1=1KΩ Simulate this inverting amplifier circuit with SPICE to make sure the circuit works as designed Design a non-inverting amplifier such that it has a gain of +11 (this gain is positive) Pick resistor values that you have in the lab kit Include a schematic of this circuit with the component values labeled with your completed pre-lab assignment The op-amp is ideal: V+=V-=V3=Vi Apply KCL: Simulate this non-inverting amplifier circuit with SPICE to make sure the circuit works as designed Vi =1V and Vo=11V, the gain is 11 therefore the circuit works as designed Section 6.2 Analysis of integrators and differentiators For the circuit in Figure with power supplies VCC = 12 V, VEE = -12 V, and assuming the op-amp is ideal, answer the following question Derive the time-domain equation for Vo (t) in terms of Vi (t) Show that the circuit performs the function of an integrator Vc = Q/C, Vc = Vx – Vout = - Vout ∆ - (d Vout )/(dt) = (dQ)/(Cdt) We have dQ/dt is electric current and the node voltage of the integrating op-amp at its inverting input terminal is 0, X = 0, the input current Iin flowing through the input resistor, Rin is given as: Iin = (Vin-0)/Rin= Vin/ Rin The current flowing through the feedback capacitor C: If = (CdVout)/ (dt) = (CdQ)/(Cdt) = (dQ)/(dt) = (dVoutC)/(dt) Assuming we have ideal op-amp, no current flows into its terminal Hence, the nodal equation at the inverting input terminal: Iin = If = Vin/ Rin = (dVoutC)/(dt) ∆ Vin/ Vout x (dt)/ RinC = The ideal voltage output for the OP-amp Integrator as: Vout = -1/( RinC) (Vin dt) = - (Vin dt)/ ( RinC) Or this can also be re-written as: Vout = -(Vin)/(jωRC) If , then For the circuit in Figure with power supplies VCC = 12 V, VEE = -12 V, and assuming the op-amp is ideal, answer the following question Derive the time-domain equation for Vo(t) in terms of Vi(t).Show that the circuit performs the function of a differentiator Iin = IF and IF = - Vout / RF The charge on the capacitor equals Capacitance x Voltage across the capacitor Q = C x Iin The rate of change of this charge is (dQ)/(dt) = C (dVin)/(dt) dQ/dt is the capacitor current i Iin = (CdVin)/(dt) = IF ∆ - Vout/RF = (CdVin)/(dt) We have an ideal voltage output for the op-amp differentiator is given as: Vout = - RF(CdVin)/(dt) Hence, the output voltage Vout is (-Rf.C x Vin) with respect to time The minus sign indicates a 180o phase shift because the input signal is connected to the inverting input terminal of the operational amplifier For the circuit in Figure with power supplies VCC = 12 V, VEE = -12 V, and assuming the op-amp is ideal, answer the following questions: What is the low-frequency gain of this circuit? Using KCL: At low frequency: Av = Low frequency gain: Av = -R2/R1= -20 Cut off frequency: For frequencies ω>> 1/(RbC), show that the circuit performs the function of an integrator The circuit performs nearly like the circuit in figure which is an integrator circuit Use SPICE transient analysis to simulate this circuit in the time domain using a sine wave input with amplitude 300 mV and frequency 300 Hz From the SPICE output plot of the input and output waveforms, confirm that this circuit is an integrator At frequency of 300Hz, voltage gain is reduced and can be seen significantly Figure: Output (blue) and Input (green) signals Try to explain the function of the resistor Rb in this circuit If Rb increases, the voltage gain also increase Hence, Rb has a purpose to increase the voltage gain Rb is used to prevent the capacitor from storing charge because of small offset current and voltage at input When Rb increases, the current through capacitor is gradually decreased, then capacitor spends more amount of time to reach the steady state Moreover, a trade-off between voltage gain and the time taking for circuit operates in steady state However that amount of time is really not clear We can observe the longer amount of time for circuit to operate in steady state in the case of Rb = 100kΩ Rb = 20kΩ Rb = 100kΩ Section 6.3 Analysis and simulation of an active low-pass filter Derive the equation for the amplitude of Vo(t) in terms of the input amplitude A, input frequency ω, and circuit components R and C Do not use any numerical values Note: there is no need to derive the output phase equation then the expression of Vout is the same as the case in figure The ratio of the output amplitude over the input amplitude is called the gain of the circuit From item above, what is the equation for the circuit gain in terms of input frequency ω and circuit components R and C? Do not use any numerical values We have that CH1 is Vin CH2 is Vout 7.3 Integrators Build the circuit in Figure with power supplies ±12 V Apply a sine wave input signal with amplitude 300 mV and frequency 300 Hz Display the input signal on channel of the oscilloscope Display Vo on Channel and adjust the time base to display to complete cycles of the signals Get a hardcopy output from the scope display with both waveforms to confirm that the circuit is an integrator Turn this hardcopy in as part of your lab report Real values: R = 1.034 kΩ Rb = 20.247 kΩ C = 0.2 µF Change the input signal to a square wave with the same 300 mV amplitude (300 mV to +300 mV) and frequency 300 Hz Get a hardcopy output from the scope display with both waveforms to confirm that the circuit is still an integrator Turn this hardcopy in as part of your lab report Repeat item above using a triangular input signal with same amplitude and frequency Turn this hardcopy in as part of your lab report Figure: Output (blue) and input ( yellow) waveforms with triangular input signal Now change the input back to a sine wave as in item Remove the resistor Rb What happens to the output signal? Explain the phenomenon you observe on the oscilloscope Reinsert the resistor Rb and verify that the circuit functions as designed Figure: Output (blue) and input (yellow) signals when Rb is removed - Explain: when Rb is removed out of the circuit, there is no DC current go through negative feedback that makes the DC offset become unstable The output signal will oscillate corresponding to the instability of the DC offset 7.4 Low-pass filters Build the circuit in Figure with power supplies ±12 V Use a sine wave of amplitude 100 mV as an input signal (see item below for frequency) and display both the input and output signals on the oscilloscope (2 to complete cycles) Vary the input signal frequency in 1-2-5 sequence from Hz to 100 KHz At each frequency, measure the gain of the circuit, using the data from the oscilloscope display Keep this data in a table for later plotting Real values: R1 = 1.173 kΩ R2 = 10.514 kΩ Gain (ratio) Frequency (Hz) V in (Vpp - V) V out (Vpp - V ) 0.182 1.84 10.11 0.184 1.86 10.108 0.182 1.80 9.89 10 0.184 1.82 9.891 20 0.184 1.86 10.108 50 0.182 1.84 10.11 100 0.180 1.83 10.167 200 0.182 1.80 9.89 500 0.180 1.76 9.778 1K 0.182 1.54 8.462 2K 0.180 1.14 6.333 5K 0.182 0.63 3.461 10K 0.184 0.25 1.359 20K 0.182 0.17 0.934 50K 0.182 0.15 0.824 100K 0.180 0.13 0.722 Data analysis: 8.1 Inverting Amplifiers Inverting amplifiers 1.1 Compare the experimental gain measured in section 7.2 item with the calculated gain in the pre-lab and with the gain as simulated by SPICE Explain any difference between these values The calculation gain is 10 and it is nearly 10 with small error Measured gain Calculated gain %Error nearly 9.5 10 About 5% to 7% The difference is due to - The real values of resistors are not equal the values in calculations The real value of resistor R1= 1.173 Ω and R2 = 10.514 Ω - The op-amp is supposed to be ideal but actually it is not 1.2 From the table of data in section 7.2 item 1, plot the gain of this circuit as dB versus frequency, using the technique described in the Discussion section Frequency 10 Gain(ratio) Gain(dB) 9.8936 19.907 20 9.8369 19.857 50 9.9444 19.952 100 9.6808 19.718 200 9.6739 19.712 500 9.5555 19.605 1000 10 2000 9.89 19.903 5000 8.222 18.299 10K 7.174 17.115 20K 7.167 17.107 50K 7.055 16.969 20 100K 6.429 16.162 200K 3.9706 11.977 500K 2.0192 6.103 1M 1.568 3.907 Gain vs Frequency 25 20 15 10 10 100 1000 10000 100000 1000000 8.2 Integrators: Explain any difference between the SPICE output in section 6.2 items and the experimental data in section 7.3 item Figure: Input (green) and Output (blue) signal in section 6.2.5 Difference: There is only small different between two pictures at the amplitudes Explanation: In the oscilloscope, the signals on the screen only display stable form of the output signal after time while the LTspice shows us the output waveform from the beginning when the output voltage needs a small time to reach the stable state With the experimental observation in section 7.3 items 6, explain the function of the resistor Rb With high frequency of Vi, Rb restricts the value of Vo Without Rb, the Value of Vo will increase greatly until it reaches the value of Vcc 8.3 Low-pass filters 1.From the data in section 7.4 item 2, plot the gain (in dB) of the circuit as function of frequency(using the technique described in the Discussion section) and compare it with the plots in section 6.3 item and in section 6.3 item (SPICE plot) Explain any differences between these plots Frequency (Hz) 10 20 50 100 200 500 Gain Gain (ratio) (dB) 10.11 20.09 10.10 20.09 9.89 19.90 9.891 19.90 10.10 20.09 10.11 20.09 10.16 20.14 9.89 19.90 9.778 19.80 1K 8.462 2K 6.333 5K 3.461 10K 1.359 18.54 16.03 10.78 1.649 20K 0.934 -0.593 50K 0.824 -1.681 100K 0.722 -2.829 Gain vs Frequency 25 20 15 10 -5 10 100 1000 10000 100000 The table of Gain vs Frequency in prelab 6.3.1 Frequency (Hz) Gain (Ratio) 10.0000 10.0000 10.0000 10 9.9998 20 9.9992 50 9.9951 100 9.9803 200 9.9219 500 9.5403 1000 8.4673 2000 6.2268 5000 3.0332 10000 1.5718 20000 0.7933 50000 0.3181 100000 0.1591 Gain (dB) 20.0000 20.0000 20.0000 19.9998 19.9993 19.9957 19.9829 19.9319 19.5912 18.5549 15.8843 9.6380 3.9280 -2.0113 -9.9487 -15.9666 Gain vs Frequency 25 20 Gain (dB) 15 10 -5 10 100 1000 -10 -15 -20 Frequency (Hz) 10000 100000 There are some small differences between these graphs since: - The real values of resistors and capacitor are not equal the theoretical ones - The op-amp in fact is not ideal - Some errors in the measuring instruments Conclusion: This report has discussed about characteristics of circuits built with op-amps It also pointed out how important of using ltspice simulation is to verify circuit design, to test whether the designed circuits work properly, then we can adjust our circuits to work more accurately Through this lab we will : Design an inverting amplifier for a given gain Using LTspice to simulate and analyze electrical circuits Understanding about the operation of integrator, differentiator, low-pass filter and the functions of some of their components such as the feedback resistor Explaining some differences between calculated and real data.The error in experiments is the result of many subjective reasons and objective reasons The differences between the theoretical and practical values is the major reason which leads to error and mistakes in measurement process and some mechanical causes are also lead to error ... 6 .3. 1 Frequency (Hz) Gain (Ratio) 10.0000 10.0000 10.0000 10 9.9998 20 9.9992 50 9.9951 100 9.98 03 200 9.9219 500 9.54 03 1000 8.46 73 2000 6.2268 5000 3. 033 2 10000 1.5718 20000 0.7 933 50000 0 .31 81... 1. 83 10.167 200 0.182 1.80 9.89 500 0.180 1.76 9.778 1K 0.182 1.54 8.462 2K 0.180 1.14 6 .33 3 5K 0.182 0. 63 3.461 10K 0.184 0.25 1 .35 9 20K 0.182 0.17 0. 934 50K 0.182 0.15 0.824 100K 0.180 0. 13. .. 20.09 10.11 20.09 10.16 20.14 9.89 19.90 9.778 19.80 1K 8.462 2K 6 .33 3 5K 3. 461 10K 1 .35 9 18.54 16. 03 10.78 1.649 20K 0. 934 -0.5 93 50K 0.824 -1.681 100K 0.722 -2.829 Gain vs Frequency 25 20 15 10