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EE233 lab report 1
The University of Danang Danang University of Science and Technology Contents Step Response of RC Circuits Objectives……………………………………………………… Reference……………………………………………………… Circuits………………………………………………………… Components and specifications……………………………… LAB REPORT Instructor : Nguyen Tri Bang Lab :1 Class : 15ECE2 Group members: Tran Viet Tu Nguyen Cong Thien Dinh Ngoc Tien Le Dinh Hoai Nam Danang 2017 Danang 2017 Step Response of RC Circuits Objectives Measure the internal resistance of a signal source (e.g an arbitrary waveform generator) Measure the output waveform of simple RC circuits excited by step functions Calculate and measure various timing parameters of switching waveforms (time constant, delay time, rise time, and fall time) common in computer systems Compare theoretical calculations and experimental data, and explain any discrepancies Reference The step response of RC circuits is covered in the textbook Review the appropriate sections, look at signal waveforms, and review the definition and formula for the time constant Review the usage of laboratory instruments Circuits Figure shows a simple circuit of a function generator driving a resistive load This circuit is used to illustrate and measure the internal resistance of a function generator Figure shows the first-order RC circuit whose step response will be studied in this lab Figure shows two sections of the first-order RC circuit connected in series to illustrate a simple technique to model computer bus systems (PCI bus, SCSI bus, etc.) Components and Specifications Quantity Descriptions Comments 50 Ω resistor Real value : 50.42 Ω 10 KΩ resistor Real value : 9.98 kΩ 27 KΩ resistor Real value : 27.30 kΩ 0.01 μF capacitor Real value : 0.01μF capacitor - - Experimental proceduces: 5.1 Instruments needed for this experiment An arbitrary waveform generator A multimeter A board and an oscilloscope 5.2 Effects of internal resistance of function generator Build the circuit in Figure using a 50 Ω resistor as load Set the function generator to provide a square wave with amplitude 400 mV, DC offset 0V, and frequency 100 Hz 2 Use the scope to display the signal Vout on channel 1, using DC coupling Set the horizontal Time base to display or complete cycles of the signal Use the scope to measure the amplitude of Vout Record this value in your report Is it the same as the amplitude displayed by the function generator? Explain any difference - The amplitude of is = 160 (mV) * It is smaller than the amplitude displayed The reason is that the voltage is dropped by the internal resistor Rs of the generator 4 Vary the square wave amplitudes from 400 mV to V, using 100 mV step size (e.g the amplitudes are 400 mV, 500 mV up to V) Repeat step to measure the amplitude of Vout on the scope for each setting Get a hardcopy for the case of 500 mV amplitude only Remove the 50 Ωresistor and replace it with a 27 KΩresistor Repeat the steps through above Observe and explain any difference insignal amplitudes when the loading on the function generator is changed from50 Ω to 27 KΩ Source Amplitu de - 400 401 500 539 600 610 700 715 800 820 900 925 1000 1020 When R1= 27 kOhm, we can see that the value of Vout equals to the value of Vin Source (mV) 400 500 600 700 800 900 1000 Amplitude (mV) 100 125 150 174 204 228 254 - When the loading on the function generator is changed to 27 , the inaccuracy of signal amplitudes is very small Thus, the output voltage is approximately the input voltage - 7.3 When using the formula in the prelab 1, since R1 >> R s, we can consider that Rs is approximately zero Therefore, the value of Vout is equal to the value of Vin In other words, when the value of R1 >> value of Rs, the value of Vout is not be affected by the value of the internal resistance of the generator Step response of first-order RC circuits Build the circuit in Figure using R = 10 KΩ and C = 0.01 μF Set the arbitrary waveform generator provide a square wave input as follows: a Frequency = 300 HZ (to ensure that T >> RC, T=1/f) This value of frequency guarantees that the output signal has sufficient time to reach a final value before the next input transition b Set the Amplitude from V to 5.0 V Note that you need to set the offset to achieve this waveform Use the oscilloscope to display this waveform on Channel to make sure the amplitude is correct We use this amplitude since it is common in computer systems c Set both channel and channel to DC coupling Use Channel of the oscilloscope to display the output signal waveform Adjust the timebase to display complete cycles of the signals Record the maximum and the minimum values of the output signal Vout Max=4.98V Vout = V 3.Use the capability of measurement the scope to measure the period T of the input signal, the time value of the 10%point of Vout, the time value of the 90%-point of Vout, and the time value of the 50%-point of Vout The time value of the 10%-point of Vout Period of input = 3.000 ms Rise time: the time interval between the 10%-point and the 90%-point of the waveform when thesignal makes the transition from low voltage (L) to high voltage (H) = 290 µs Fall time: the time interval between the 90%-point and the 10%-point of the waveform when thesignal makes the transition from high voltage (H) to low voltage (L) = 304 µs = 68.0 µs = 56.0 µs Save a screenshot from the display with both waveforms and the measured values Measure the rise time of Vout, the fall time of Vout, and the two delay times tPHL and tPLH between the input and output signals Save a screenshot from the display with both waveforms and the measured values Measure the voltage and time values at 10 points on the Vout waveform during one interval when Vout rises or falls with time (pick one interval only) Note that the time values should be referred to time t = at the point where the input signal rises from V to V or falls from V to V Record these 10 measurements Time 16 μs 20 μs 22 μs 36 μs 42 μs 48 μs 108 μs134 μs190 μs240 μs Voltage 638 mV 790 mV 1.1 V 1.4 V 1.6 V 1.8 V 3V 3.4 V V 4.32 V 7.4 Step response of cascaded RC sections Build the circuit in Figure 3, using identical resistors R = 10 KΩ and identical capacitors C =0.01 µF Use the same square input as in item 1, section 7.3 above and display it on Channel Display Vout on Channel and adjust the time base to display complete cycles of the signal Use the scope measurement capability to measure the two delay times tPHL and tPLH between the input and output signals tPHL = 220 μs = 260 µs 7.5 Manufacturing test time and test cost considerations The more points you measure on a waveform, the more accurate the measured results but this also takes more time and increases the test cost This is an important tradeoff in measurement accuracy and test cost Given the circuit in Figure 2, ten data points per waveform were collected in section 7.3 item “Good” estimate means the estimated value is within 10% of the correct value (from computation or simulation) We should collect more than 10 points to extract a “good” estimate of the rise or fall time of the circuit The minimum number of data points you need to collect to get a good estimate is 15 Other teams collect fewer points We think their results are not “better” than ours 2500 2000 1500 1000 500 0 500 1000 1500 2000 2500 3000 3500 Analysis, calculation and results (Tabulated data) and answers to questions: 8.1 Extracting internal resistance of an arbitrary waveform generator Vout = R1/(R1+Rs) * Vs R1 = 50 Ω Vout = 50/(50+50) * 500 = 250 (mV) R1= 27 kΩ Vout = 2700/(2700+50) * 500 = 499.07 (mV) This value doesn’t agree with the recorded data in the lab From the data recorded in section 7.2 : Rs=50Ω The values for Vs (as displayed by an arbitrary waveform generator panel) and the measured values on the scope are not the same.Because of the resistor of the wire and the generator is not ideal 8.2 R = 10.013 kΩ C=0.01 µF Vout = Vs (1 - e –t/RC) Vs = (V) when t = T Voutmin= (mV) when t = Voutmax(measured) = 4.96V < 5V due to the existence of sources internal resistance 2 Calculated value 10.53 μs 0->10%: Δt1 = -RC.ln 0.9 0->50%: Δt2 = -RC.ln 69.31 μs 0.5 0->90%: Δt3 = -RC.ln 230.25 μs 0.1 Error: 0->10% : 5.03% Measured value 10 μs 80 μs 280 μs 0->50% : 15.42% 0->90% : 21.6% Because the source has internal resistance so R eq = Rs + R > R so t in practical is larger than t in theory tfall= trise = Δt3 - Δt1 Calculated value 219.72μs tPHL=tPLH= Δt2 69.31μs Measured value trise=270 μs, tfall= 260 μs 80 μs The internal resistance is the sources of errors leading to the differences Error is committed by neglecting the internal resistance of the arbitrary waveform generator: t rise: 22.88% tfall :18.33% tPHL and tPLH :15.42% Time 16 μs 20 μs Voltag 720 e mV 22 36 42 48 108 134 190 240 μs μs μs μs μs μs μs μs 880 1.4 1.6 1.8 4.32 1V V 3.4 V V mV V V V V (Ƭ is time constant) Ƭ=RC=0.0001s Ƭ measured:0.000104658s The difference in percent: 4.658% (Ƭ is time constant) Ƭ=RC=0.0001 Ƭ measured : 0.000110494s The difference in percent with 4: 10.494 % greater than 4.658% Ƭ=RC therefore C= Ƭ/R= 0.011035 µF this value greater than the marked value 8.3 For the measurements in section 7.4 item3, the delay time for the cascaded circuit in Figure ( of identical RC sections) are not twice as large as the delay times for the simple RC circuit The delay time scale with the number of sections We have Vc1 = VR + Vc2 Taking the derivative both sides, we obtain: dVc1/dt = dVc2/dt We also have: Is = Ic1 + Ic2 Vs/R = C1 dVc1 / dt + C2dVc2 / dt C1 = C2 = C => Vs= 2RC * dV2/dt = 2RC dVout/dt => Ƭ =1/2 RC The expression of t: t = -2RC * ln(1 – Vout/Vs) In section 7.3, we have: t = -RC * ln(1 – Vout/Vs) So it is easy to show that the delay time for the cascaded circuit (in Figure 3) is double the delay time for the simple RC circuit Similarly, we can show this problem for n sections: t =n* ( -RC * ln(1 – Vout/Vs)) The number of cascaded RC sections so that the propagation delay time is about T/2 Tdelay = T/2 n* ( -RC * ln(1 – Vout/Vs)) = T/2 =1/(2f) n = -1/(2RCf * ln(1 - Vout/Vs)) With delay time, we have Vout/Vs = 0.5 in section 7.3, we have R = 10kΩ, C=0.01µF n = 24 sections According to the delay time measured at the Figure 5: tdelay.1 = 76 µs of one RC section (in Figure 1) With n sections, we have tdelay.n= n*tdelay.1 n = tdelay.n/tdelay.1 With tdelay.n = T/2=1/2f=1/600(s) Therefore, n=1/(600*76*10-6)=22 %errors = 8.3% < 10%.So this is a good estimate CONCLUSION: This report has discussed the output waveform of simple RC circuits excited by step functions, timing parameters of switching waveform conclude time constant, delay time, rise time, and fall time The differences between theoretical calculations and experimental data are significant The data collected correlated strongly to the hypotheses, although percent errors reaching high value ( largely >15%) because the internal resisters have influence on measured results Furthermore, the differences between theoretical calculations and experimental data due to the mechanical error ... :18 .33% tPHL and tPLH :15 .42% Time 16 μs 20 μs Voltag 720 e mV 22 36 42 48 10 8 13 4 19 0 240 μs μs μs μs μs μs μs μs 880 1. 4 1. 6 1. 8 4.32 1V V 3.4 V V mV V V V V (Ƭ is time constant) Ƭ=RC=0.0001s... or falls from V to V Record these 10 measurements Time 16 μs 20 μs 22 μs 36 μs 42 μs 48 μs 10 8 μs134 μs190 μs240 μs Voltage 638 mV 790 mV 1. 1 V 1. 4 V 1. 6 V 1. 8 V 3V 3.4 V V 4.32 V 7.4 Step response... 10 .53 μs 0- >10 %: Δt1 = -RC.ln 0.9 0->50%: Δt2 = -RC.ln 69. 31 μs 0.5 0->90%: Δt3 = -RC.ln 230.25 μs 0 .1 Error: 0- >10 % : 5.03% Measured value 10 μs 80 μs 280 μs 0->50% : 15 .42% 0->90% : 21. 6% Because