ET 304b

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ET 304b
ET 304b
Laboratory 4
RC Circuits: Frequency Response and Rise time
Objective: Observe transient response effects for RC circuits excited with square wave inputs.
Relate RC time constants to the sinusoidal bandwidth of a RC circuit. Compensate oscilloscope
probes to achieve voltage division independent of frequency. Apply the RC circuit concepts of
rise time, frequency response, and source impedance to the oscilloscope measurement problem.
Theoretical Background
The RC combinations in amplifiers and other electronic circuits control the high frequency
response of the system. Complex networks can be reduced to simple RC circuits for analysis if a
dominate time constant exists. The dominate time constant is the slowest response to abrupt
changes in the circuit. It has a value much greater than all other circuit RC combinations.
The simplified model below shows a signal generator with an input resistance, Rg,
connected to an ideal amplifier through the parallel combination of amplifier input resistance Ri
and capacitance, Ci.
Rg
Vg
Vi
Ri
Amp
Ci
Vo
Figure 1. Simplified Time Response Model of an Amplifier.
The voltage divider formed by the components Rg, Ri, and Ci presents a fraction of the generator
voltage to the amplifier input. The capacitor in parallel with the input resistance makes this
voltage divider dependent on frequency. The input voltage in terms of the circuit impedances is:
 R i || X Ci
Vi  Vg 
 R g  R i || X C
i





Substituting in the definition of capacitive reactance gives the amplifier input in terms of the
divider resistance and capacitance.
 Ri
Vi  Vg 
R R
i
 g
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

1


 2  R || R   C  f  j  1 
i
g
i


(1)
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Equation 1 shows that a resistive divider determines the input voltage. The third term of the
product is a lowpass filter characteristic with the cutoff frequency determined by equation 2.
f Hi 
1
Hz
2  R i || R g  C i
(2)
The cutoff frequency fHi, is the frequency where Vg is reduced to a level of 0.707 times the
maximum value determined by the resistive divider network. The frequency, fHi, governs the
high frequency response of the overall circuit.
Equations 1 and 2 define the Figure 1 circuit response to sinusoidal inputs. If Vg is a
square wave or a square pulse, Equation 3 determines the value of Vi. The input and output
voltages are now functions of time.
t
 R i 


 1  e Hi
Vi ( t )  Vg ( t )
 R  R 
i 
 g




(3)
The circuit time constant, Hi is given by  Hi  R i || R g   C i . This is the time constant of the
input circuit that controls the overall circuit response to rapidly changing inputs.
Since the voltage across the input capacitance, Ci, can not change instantaneously, the
input voltage will not follow the abrupt transitions of pulse inputs. After some time, Ci will
charge to a final value. If the period of the pulse wave is too short, then Vi will never reach its
final value. If Hi is much shorter than the period of the pulse, the input to the circuit will reach
its final value. The resistive divider of equation 4 determines the final value.
 Ri
Vi  Vg 
R R
i
 g




(4)
The rise time of the pulse wave applied to Figure 1 relates to the sinusoidal cutoff
frequency through the relationship:
tr 
0.35
f Hi
(5)
where
tr = the rise time in seconds
fHi = the high frequency cutoff.
This is accurate when the circuit has a dominate cutoff frequency. Rise time is the time it takes
for a pulse input to change from 10% to 90% of its final value. Figure 2 shows the rise time
measurement for a single pulse input. The rise time is defined as tr=t10-t90. This measurement is
made using an oscilloscope.
Assuming that the low frequency limit of the circuit is dc (0 Hz), fHi determines the
bandwidth of the circuit. Equation 5 shows that the rise time-bandwidth product is constant.
Circuits with fast rise times will have wide bandwidths and high upper cutoff frequencies. This
also implies that the circuit has smaller values of Ci since Vi quickly reaches its final value.
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120
t 10
t 90
Input Voltage (% Final Value)
100
V i90
80
V i( t )
60
40
20
V i10
0
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
t
Time (sec)
Figure 2. Definition of Rise time for Pulse Waveforms.
Conversely, Equation 5 relates time response of the circuit to frequency response. The bandwidth
of a circuit can be determined by injecting a square wave signal into it and measuring the rise
time. Computing the fHi using equation 5 should give the same cutoff frequency as applying sine
waves and producing a frequency response plot.
Oscilloscope inputs are examples of how RC networks effect the time and frequency
response of a circuit. The input capacitance of the scope and the probe cable effect the high
frequency response of the instrument. A equivalent resistance, Rs, paralleled with a capacitance,
Cs, models vertical input of scope amplifiers. The input voltage Vi develops across these
components. An signal source, Vg ,with an source resistance ,Rg, connects the circuit under test
to the scope input.
vertical amp.
Vi
Rg
Vg
1M
Rs
Cs
25pF
Figure 3. Circuit Model of Scope with 1x Probe and Signal Source.
Figure 3 models the scope connected to a signal source through a 1x probe. This circuit is
similar to the RC circuit in Figure 1. The input capacitance of the scope increases due to the
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shunt capacitance of the cable. Using a 1x probe significantly reduces the frequency response of
the instrument due to this increase in input capacitance.
The bandwidth limit of the scope with a 1x probe is the high frequency limit defined by
equation 2 above. In terms of the scope parameters it becomes:
f Hi 
1
2  (R g || R s )  C s
(6)
Sine inputs above this frequency will be attenuated. Pulse inputs will have a significant rise time
that can be computed from equation 5. Equation 6 shows that the high frequency cutoff of the
instrument depends on the signal sources resistance, Rg, and the scopes input parameters. Low
impedances sources will exhibit higher cutoff frequencies for constant scope parameters. It may
be possible to accurately measure low resistance sources at moderately high frequencies with a
1x probe, but measuring high resistance sources at high frequency will introduce significant
error.
The effects of input capacitance and instrument loading can be reduced by using an
attenuator probe (10x probe). A 10x attenuator probe is a frequency compensated voltage
divider that increase the scope's input impedance to 10 M and almost eliminates the
capacitance of the cable and scope input. Figure 4 gives the scope/probe circuit model.
Ra
vertical amp.
Vi
Rg
Ca
Vg
Rs
1M
Cs
25pF
Ra=9M
Figure 4. Circuit Model of Scope with 10x Attenuator Probe.
Setting Ra=9 M reduces Vi by a factor of ten. Adjusting Ca such that RsCs=RaCa removes the
capacitive effects of the probe cable and the scope, which flattens the circuit frequency response.
This is called compensating the probe. The probe is compensated by adjusting Ca with a square
wave input. The probe is compensated when the dc portion of the square wave is horizontal.
Figure 5 shows test waveforms for 10x probes that are correctly compensated, over
compensated, and undercompensated.
Overcompensation occurs when Ca is too large. The square wave test signal has
excessive overshoot, which reflects an amplification of high frequencies. This is shown in the
sine trace of Figure 5-b. A high frequency sine wave will display larger that normal with an
overcompensated probe.
Undercompensation occurs when Ca is too small. The test signal will have a rounded
leading edge, which indicates high frequency attenuation. High frequency sine waves will be
smaller than normal when measured with an undercompensated probe. This is shown in the sine
trace of Figure 5-c.
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Figure 5. Effects of Over and Under Compensation of Attenuator Probe.
All scope measurements should be made using a properly compensated 10x probe for maximum
accuracy.
The effect that the scope probe has on the circuit depends on the impedance of the test
circuit and the type of probe used. For low impedance sources such as the signal generators in
the lab, Figure 6 represents the circuit analysis model.
Signal Generator
Vi
Rg
vertical amp.
50 
Rs
Vg
1M
Cs
250pF
Figure 6. 1x Probe Model connected to a Low Impedance Signal Generator.
The parallel combination of Rg and Rs is approximately Rg. The value of Cs reflects the probe
cable capacitance. The value of fHi is:
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f Hi 
1
 1.274 MHz
250 || 1M250pF
that gives a rise time of :
0.35
0.35

 275 nS
f Hi 1.274 MHz
These calculations indicate that the 1x probe will give reasonably accurate measurements up to
1.274 MHz. This is much lower than the frequency limit of the scopes in the lab.
Adding the attenuator probe will increase this limit by increasing the scope's input
resistance and lower the effective value of Cs. Figure 7 shows the circuit analysis model of the
10x probe connected to a low impedance signal generator.
tr 
Signal Generator
Ra=9M
Vi
Rg
vertical amp.
Ca
50
Rs
Vg
1M
Cs
2.5pF
Figure 7. Attenuator Probe and Low Impedance Signal Source.
The computing fHi and tr gives:
1
 1274 MHz
250 || 10M 2.50pF
0.35
0.35
tr 

 .275 nS
f Hi 1274 MHz
The compensated probe reduces the effects of the capacitance by a factor of 100 and increases
the bandwidth of the scope to over 1 GHz. In practice this bandwidth will be lower due to the
response of other circuit time constants.
f Hi 
Procedure
Part 1-Oscilloscope Attenuator Probe Compensation
1.)
2.)
Select two attenuator (10x) probes and connected them to the scope.
Attach channel 1 to the 1 kHz square wave test signal located on the scope face. Locate
the compensation capacitor. It should be on the probe body or on the BNC connector at
the scope end of the probe. Adjust the compensation capacitor with a small screw driver
until a flat topped square wave results. Repeat this for channel 2.
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3.)
4.)
5.)
6.)
Using the compensated 10x probes, set the output of the function generator for a 2 V
peak ac sine wave at 500 kHz.
With the signal from (3) displayed, adjust the compensation capacitor until the voltage
increases. Remove the function generator and re-attach the probe to the calibration
source. Sketch the resulting waveform and comment on its shape in the report
discussion.
Reconnect the 500 kHz sine source and adjust the compensation screw until the displayed
voltage decreases. Reconnect the probe to the calibration square wave and sketch the
resulting waveform. Discuss the shape in the report.
Recompensate the probe for flat response on the square wave.
Part 2-Effects of Source Resistance
7.)
8.)
Change the function generator output to a 2 V peak square wave at 500 kHz.
Add external source resistance to the 50 ohm function generator output as shown in the
schematic of Figure 8. Measure the rise time of the function generator signal with the
scope probe in the 1x position for the values of Rg given in Table 1.
Signal Generator
R'g
50
Vi
vertical amp.
Rg
Rs
Vg
f=500kHz
V= 2 Vp
1M
Cs
250pF
Figure 8. 1x Probe Connected to Function Generator with Increased Source Resistance.
9.)
Compute the value of fHi from the rise time measurement and enter it in the column
labeled fHi measured. These values will include the actual capacitance values of the probe
and scope input. Compute the values of fHi and tr using equations 5 and 6 and the
estimated values Cs from figure 8. Enter these values in the columns labeled fHi est. and tr
est. From equation 6 and the values of fHi measured, compute the values of Cactual. This
is a better estimate of the total capacitance in the circuit. Enter these values in the Table
also.
Repeat the procedure in step 8 with the scope in the 10x position. Figure 9 shows the
circuit model used for this analysis. Enter the measurements in Table 2. The total source
resistance now will be the sum of Ra, R'g and the external Rg.
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Signal Generator
Ra=9M
Vi
Rg
vertical amp.
Ca
Rg
50
Rs
Vg
1M
Cs
2.5pF
Figure 9. High Source Resistance with 10x Probe.
Part 3 – Effects of Shunt Capacitance
10.)
Construct the circuit shown in Figure 10. Set the function generator for a square wave
output with an amplitude of 2 V peak at a frequency of 5.0 kHz.
R1
1k
Vs= 2 Vp
R2
4.7k
C
5.0 kHz
Figure 10. Circuit for Part 3.
11.)
12.)
13.)
14.)
View the output voltage with an attenuator scope probe and measure the rise time of the
voltage for each of the capacitor values given in Table 3.
Using the measured rise time value in Table 3, compute fHi using equation 5 for all
capacitors. Enter the results in the table under the measured fHi heading.
Compute the theoretical values of fHi and tr using equations 6 and 5 for all C values is
Table 3. Enter these values under the theoretical headings of the table.
Compute the percentage error between the theoretical values and the measured values of
tr and fHi. Place these percentages in the lab report and comment on the agreement
between the values.
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Discussion Points
What effect does changing source resistance have on fHi and tr? What effect does an
increase in C have on the bandwidth and rise time of the circuits presented? What impact does
using a 10x probe have on the scopes input resistance and input capacitance? What effect does
using a 10x probe have on bandwidth of the combined probe and scope network? What is the
procedure for compensating attenuator scope probes?
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Rg (k)
0.0
1.0
2.2
3.9
Rg (k)
0.0
1.0
2.2
3.9
C
0.02 F
0.01 F
0.005 F
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Table 1 High Source Resistance Measurements: 1x Probe
fHi (measured)
fHi (est.)
tr (est)
tr (S)
Cactual
Table 2 High Source Resistance Measurements: 10x Probe
fHi (measured)
fHi (est.)
tr (est)
tr (S)
Cactual
Table 3 Effects of Shunt Capacitance f=5.0 kHz
Measured Values
Theoretical Values
f
fHi
tr (S)
Hi
tr (S)
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Channel 1 Volts/div ______ Channel 2 Volts/div _______ Time/div __________
Channel 1 Volts/div ______ Channel 2 Volts/div _____ Time/div __________
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Channel 1 Volts/div ______ Channel 2 Volts/div _______ Time/div __________
Channel 1 Volts/div ______ Channel 2 Volts/div _____ Time/div __________
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