# 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 Spring 2002 1 2 R || R C f j 1 i g i (1) exp404b.doc 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. Spring 2002 2 exp404b.doc 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 Spring 2002 3 exp404b.doc 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. Spring 2002 4 exp404b.doc 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: Spring 2002 5 exp404b.doc f Hi 1 1.274 MHz 250 || 1M250pF 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 250 || 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. Spring 2002 6 exp404b.doc 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. Spring 2002 7 exp404b.doc 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. Spring 2002 8 exp404b.doc 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? Spring 2002 9 exp404b.doc 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 Spring 2002 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) 10 exp404b.doc Channel 1 Volts/div ______ Channel 2 Volts/div _______ Time/div __________ Channel 1 Volts/div ______ Channel 2 Volts/div _____ Time/div __________ Spring 2002 11 exp404b.doc Channel 1 Volts/div ______ Channel 2 Volts/div _______ Time/div __________ Channel 1 Volts/div ______ Channel 2 Volts/div _____ Time/div __________ Spring 2002 12 exp404b.doc