Bài báo cáo thực hành 1 môn lý thuyết mạch (Circuit Theory)

EE233 lab report 1

The University of Danang

Danang University of Science and Technology

LAB REPORT

Instructor : Nguyen Tri Bang

Group members: Tran Viet Tu

Nguyen Cong Thien

Dinh Ngoc Tien

Le Dinh Hoai Nam

Danang 2017 Danang 2017

Step Response of RC Circuits

specifications………

Step Response of RC Circuits

1. 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

2 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

3 Circuits

 Figure 1 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 2 shows the first-order RC circuit whose step response will be studied in this lab

 Figure 3 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.)

4 Components and Specifications

2 10 KΩ resistorΩ resistor Real value : 9.98 kΩ

1 27 KΩ resistorΩ resistor Real value : 27.30 kΩ

2 0.01 μF capacitorF capacitor Real value : 0.01μF capacitorFcapacitor

-5 Experimental proceduces:

An arbitrary waveform generator

A multimeter

A board and an oscilloscope

1 Build the circuit in Figure 1 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 3 or 4 complete cycles of the signal

3 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 V out 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 1 V, using 100

mV step size (e.g the amplitudes are 400 mV, 500 mV up to 1 V) Repeat step 3 to measure the amplitude of Vout on the scope for each setting Get a hardcopy for the case of 500 mV amplitude only

6 Remove the 50 Ωresistor and replace it with a 27 KΩresistor resistor and replace it with a 27 KΩ resistorΩresistor and replace it with a 27 KΩresistor resistor Repeat the steps 1 through 4 above Observe and explain any difference insignal amplitudes when the loading on the function generator is changed from50 Ωresistor and replace it with a 27 KΩresistor to 27 KΩ resistorΩresistor and replace it with a 27 KΩresistor

Amplitu

- When R1= 27 kOhm, we can see that the value of Vout equals

to the value of Vin

- When the loading on the function generator is changed to 27

k Ω, the inaccuracy of signal amplitudes is very small Thus, the output voltage is approximately the input voltage

- When using the formula in the prelab 1, since R1 >> Rs, 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

7.3 Step response of first-order RC circuits

1 Build the circuit in Figure 2 using R = 10 KΩ resistorΩ and C = 0.01 μF capacitorF 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 0 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 1 to make sure the amplitude is correct

We use this amplitude since it is common in computer systems

c Set both channel 1 and channel 2 to DC coupling

2 Use Channel 2 of the

oscilloscope to display the

output signal waveform Adjust

the timebase to display 2

complete cycles of the signals

Record the maximum and the

minimum values of the output

signal

V out Max=4.98V V out min = 0 V

3.Use the measurement capability of 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

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).

t rise = 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).

t fall = 304 µs

t PLH = 68.0 µs

4 Save a screenshot from the display with both waveforms and the measured values

5 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

6 Save a screenshot from the display with both waveforms and the measured values

7 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 = 0 at the point where the input signal rises from

0 V to 5 V or falls from 5 V to 0 V Record these 10 measurements

Time 16 μF capacitors 20 μF capacitors 22 μF capacitors 36 μF capacitors 42 μF capacitors 48 μF capacitors 108 μF capacitors134 μF capacitors190 μF capacitors240 μF capacitors

Voltage 638mV 790mV 1.1 V 1.4 V 1.6 V 1.8 V 3 V 3.4 V 4 V 4.32 V

7.4 Step response of cascaded RC sections

1 Build the circuit in Figure 3, using 2 identical resistors R = 10 KΩ resistorΩresistor and replace it with a 27 KΩresistor and 2 identical capacitors C =0.01 µF Use the same square input as

in item 1, section 7.3 above and display it on Channel 1

2 Display Vout on Channel 2 and adjust the time base to display 2 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

t PHL = 220 μF capacitors

7.5 Manufacturing test time and test cost considerations

1 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 7 “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

2 The minimum number of data points do 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

0 500

1000

1500

2000

2500

8 Analysis, calculation and results (Tabulated data) and answers to questions:

waveform generator

1 Vout = R1/(R1+Rs) * Vs

R1 = 50 Ωresistor and replace it with a 27 KΩresistor Vout = 50/(50+50) * 500 = 250 (mV)

R1= 27 kΩresistor and replace it with a 27 KΩresistor Vout = 2700/(2700+50) * 500 = 499.07 (mV)

This value doesn’t agree with the recorded data in the lab

2 From the data recorded in section 7.2 : Rs=50Ω

3 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

1 R = 10.013 kΩresistor and replace it with a 27 KΩresistor C=0.01 µF

Vout = Vs (1 - e –t/RC)

Vs = 5 (V)when t = T Voutmin= 0 (mV) when t = 0

Voutmax(measured) = 4.96V < 5V due to the existence of sources internal resistance

2

Calculated value Measured value 0->10%: Δt1 = -RC.lnt1 = -RC.ln

0.9 10.53 μF capacitors 10 μF capacitors

0->50%: Δt1 = -RC.lnt2 = -RC.ln

0.5 69.31 μF capacitors 80 μF capacitors

0->90%: Δt1 = -RC.lnt3 = -RC.ln

0.1 230.25 μF capacitors 280 μF capacitors

Error: 0->10% : 5.03%

0->50% : 15.42%

0->90% : 21.6%

Because the source has internal resistance so Req = Rs + R > R

so t in practical is larger than t in theory

3

tfall= trise = Δt3 - Δt1t3 - Δt3 - Δt1t1 219.72μF capacitors trise=270 μF capacitors, tfall= 260

μF capacitors

tPHL=tPLH= Δt3 - Δt1t2 69.31μF capacitors 80 μF capacitors

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%

4

Time 16 μF capacitors 20 μF capacitors 22μF capacitors 36μF capacitors 42μF capacitors 48μF capacitors 108μF capacitors 134μF capacitors 190μF capacitors 240μF capacitors

Volta

ge 720mV 880mV 1 V 1.4V 1.6V 1.8V 3 V 3.4 V 4 V 4.32V

(Ƭ is time constant)

Ƭ=RC=0.0001s

Ƭ measured:0.000104658s

The difference in percent: 4.658%

5

(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

1 For the measurements in section 7.4 item3, the delay time for

the cascaded circuit in Figure 3 ( of 2 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))

2 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 = 10kΩresistor and replace it with a 27 KΩresistor , 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