Ee3101 Lab Report Exp

EE3101 LAB REPORT EXP#3 26 OCT 2006 -1-

Nonideal Behavior of Electronic Components at High Frequencies and Associated Measurement Problems Matt xxxxx Student ID : xxxxxxxx 10/2/06 – 10/16/06 Abstract Throughout this experiment we take the input and output measurements of given circuits at various frequency rates. This is to demonstrate the frequency response of these circuits. In other words, the circuits behave differently at different frequencies. At high frequencies we can see the effects of shunt capacitance of the measurement terminals and interconnection cables, resonance of the circuits, and the nonidealistic frequency behavior of passive components. This experiment is designed to explore the response of circuits at high frequencies and to modify the circuits to have the proper responses that are would be required in electrical circuit design.

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Introduction AS opposed to dc or low frequencies, the characteristic of the circuits changes with each component. This experiment focuses on the aspects of high frequency measurement divided into three basic parts. We start with experiments to understand the characteristic of shunt capacitance of the interconnection cables and the measuring instruments. Next we study about the resonance in RLC circuits. Finally, we look at the nonideal frequency behavior of passive circuit components. All of the measurements throughout these three basic parts use similar procedures. We vary the input frequency from low to high to gather the most important value, either the resonant frequency or the break frequency of the circuit. These two characters are results of the capacitance in the circuit that works as a short circuit at high frequencies. RC

f B 2p 1 =, LC f R 2p 1 = The break frequency is when the output is the 3dB of the max value, and the resonant frequency is the when the phase shift of the input to the output signal is zero. Throughout the report you are able to see other characters as Q – factors. Through the following 7 experiments we are able to gather a broad understanding on high frequency responses Body Part 1 – Shunt Capacitance and the RC Compensator Experiment 3.1 In the first experiment we measure the transfer function, which would be the gain, or | Vo / Vi | of the circuit. We take our measurements from low to high frequencies. We do this to see the effects of the shunt capacitance that the oscilloscope and the interconnecting cables that occur. We start at the low frequency of 1 kHz and go up to 1MHz. We construct circuit Figure 1 to measure the shunt capacitance. We could expect that, because of the shunt capacitance, Vout will have a smaller value as the input voltage frequency increases.

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Fig 3.1.1

1. Data & Results f(hz) Vin Vout | Vo / Vi | 1000 5.000 2.380 0.476 5000 5.000 2.380 0.476 10000 4.940 2.280 0.462 28160 5.000 1.767 0.353 50000 5.000 1.300 0.260 100000 5.000 0.750 0.150 500000 4.820 0.167 0.035 1000000 4.820 0.099 0.021

Fig 3.1.2

Gain vs Frequency 0.000 0.050 0.100 0.150 0.200 0.250 0.300 0.350 0.400 0.450

0.500 0 200000 400000 600000 800000 1000000 1200000 Frequency (Hz) Gain (Vout/Vin) Gain vs Frequency

Fig 3.1.3

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As expected, we can see a decrease of gain that is result of the decrease in the Vout. This is the effect of the shunt capacitance as it works as a short circuit. Now we calculate the exact value of the shunt capacitance we need further calculations. 2. Shunt Capacitance In order to calculate the shunt capacitance we look for the break frequency, which has the information of the capacitor in the circuit: RC f B 2p 1 = , where R = R1||R2 = 50KΩ VV V RR VR out out in

5 0. 7 07 1. 787 2 1 ( 70. 7 %) 12 2

=´´= ´´ + = Then we pick the closest value from our measurements, which would be: pF Rf C f KHz B B

1 .1 3 10 113 2 1 28. 16 \ = = ´ 10 = \@ -

p So the shunt capacitance of the oscilloscope and the interconnecting wires are 113pF by

measurement. As you can see the shunt capacitance has a very small value. This is why we can see the effects of the shunt capacitance only at the high frequency rates that make the shunt capacitance work as a short circuit. This effect needs to be considered when we are making measurements of circuits at high frequencies as mentioned. Part 2 – Resonance in RLC Circuits Experiment 3.2

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We now build our second circuit (Fig 3.2.1) to see the effects of the resonance in the RLC circuits that has its resonant frequency at 2 kHz. Fig 3.2.1 Our two goals here are to: 1. Determine the resonant frequency and Qfactor. 2. Determine the impedance Z in ( jw o ) at the resonant frequency. We design our circuit to have an R relatively very small R<<1kΩ in order to make the circuit dependent more on the inductor and capacitor. So we put our R as 10Ω. For L we use a 100mH component from our circuit kit. To determine our capacitor value, we need further calculations to make the resonant frequency at approximately 2 kHz: LC f R 2p 1 = LC f R2 2

4 1 p = 4 22 1 R Lf C

p = For our resonant frequency fr = 2 kHz, and L = 100mH, nF nF F Lf C R

m p

63. 33 100 0. 1 4 1 22= = » = However, in our lab kit the closest value of capacitor we have was 100nF. With a modified capacitance: f Hz kHz C nF F Z R 1591. 5 1. 6 100 0. 1 (1 04 ) =@ ==m

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If we would have used a series of 100nF capacitance we could have gathered a resonant frequency closer to 2 kHz. However with the resonant frequency at around 1.6 kHz, we are still able to see the effect of the resonant frequency at high rates frequencies. In this experiment we start with the low frequency at the input and increase the frequency to see the effects. We cannot rely on correct gain values in this situation because of the parasitic resistance though the inductor and capacitor. Instead, we will look at the phase shift between Vin and Vout. In this case, we know that out break frequencies will occur at phase shifts of +45° and 45° and our resonant frequency will occur at a phase shift of 0°. 1. Data & Results f(hz) Phase Vo > Vin 1552 45° 1796 45° 1670 0°

Fig 3.2.2 * Determine the resonant frequency and Qfactor. By measurements circuit Figure 3.2.2 has its resonant frequency at 1.67 kHz because that is where we have a phase shift of 0°. f R = 1 .6 7 kHz If we compare our value with our theoretical value 1591.5 Hz, we can see we have evaluated a reasonable result. To determine the circuit’s Qfactor: 6 .8 44 1. 79 1. 552

1 . 67 = = D -= kk k f f Q factor o * Determine the impedance Z in ( jw o ) at the resonant frequency. To determine the impedance of our circuit (Figure 3.2.1), we can use another formula for the Qfactor and solve for Zin. 2 2 1 .6 7 100 6. 844 = ´´ -== Zin k mH R fL Q factor R p p

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Zin = 153. 316W Since our Zin is R plus the parasitic resistances of the inductor and capacitor we can calculate what the parasitic resistance is by subtracting 10W from it. The calculation will be the following: Z ( j ) = Z - R = 153. 3 16 - 10 = 142. 316W in o parasitic w Experiment 3.3 We now modify our circuit (Figure 3.2.1) to have a Qfactor = 5. Then we drive the resultant circuit with a 2 kHz square wave to compare the spectra of the input voltage and the current waveforms. 1. To modify the circuit to have a Qfactor at 5: R fL f f Q factor o R 2p = D -=

Q R f R L 2p = with Q = 5, fr = 1.6kHz, and L = 100mH =W ´´ = 209 5 R 2p 1 . 670 0. 1 If we look back to our circuit we see the parasitic resistance of the circuit and the original R. R = 209W = 143. 3 16W + R ¢ So we should put a resistance of value of R ¢ = 65. 6 8W We can do this by putting two 100W resistors in parallel to give 50W and then put a 10W resistor in series to make the equivalent resistance 60 W R ' = 100W || 100W + 10 W = 60W » 65. 6 8W We now modify the circuit to have a Q factor of 5 by making our circuit look like the following:

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Fig 3.3.1 2. Gather spectra of the input voltage and current waveforms. Figure 3.3.2

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Figure 3.3.3 Figure 3.3.2 would be the FFT of the input voltage, and Figure 3.3.3 would be the FFT of the output voltage. Because the output voltage is a part of the input current we used the output voltage of the circuit. If we compare the differences of the two spectra of the waveforms we could see that while the input voltage (Figure 3.3.2) is a combination of several harmonics, the current (Figure 3.3.3) has only one harmonic as an effect. The other harmonics could be assumed to be cut off by the circuit. The combination of the input harmonics makes the input to be a square wave, and the only harmonic on the output makes the wave to look as a sinusoidal waveform. Note: In order to gather a clear waveform, we turned on the Noise Rej., and put the center of the FFT at 2 kHz with a Span of 50 kHz. Experiment 3.4 We now determine the resonant frequencies of the circuit of Figure 3.3.1 for C values from

0.0001uF to 0.1uF to see the relationship between the capacitance of the circuit with its resonant frequency. The circuit used in this experiment is identical to Experiment 3.2. To make the experiment easy, I observed the circuit phase shift and took the 0° frequency with different values of C. Where fr calculated is:

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2p 1 LC f r=

C(uf) fr calculated (hz) fr measured (hz) % error 0.001 15916 16130 1.34% 0.01 5033 5066 0.66% 0.1 1592 1776 11.56%

Fig 3.4.1

Frequency vs Capacitance 0 2000 4000 6000 8000 10000 12000 14000 16000 18000 0 0.02 0.04 0.06 0.08 0.1 0.12 Capacitance (uF) Frequency (Hz) fr calculated (hz) fr measured (hz)

Fig 3.4.2 From the results in Fig 3.4.1, Fig 3.4.2, we can see that the resonant frequency is inversely proportional to the square of the capacitance: C fr 1 µ Experiment 3.5 Goals for experiment 5. 1. Design a parallel resonant circuit (with R = infinity) with a resonant frequency of 2 kHz. 2. Determine the Q of the circuit. 3. Modify circuit to make Q = 5.

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4. Measure the modified circuit and gather the resonant frequency and Q of the circuit 5. Determine the admittance of the circuit at the resonant frequency.

The procedures of this experiment are similar to the earlier experiments. Fig 3.5.1 Rs is used to derive a current source for the circuit. (L=100mH, without R) Is = Vs / Rs. We will use 100 kΩ. 1. To make resonant frequency at 2 kHz with L = 100 mH. LC f R 2p 1 = LC f R2 2

4 1 p = 4 22 1 R Lf C p = nF nF Lf C R

63. 33 50 4 1 22= = » p However, in our lab kit the closest value of capacitor we have was 100nF so we will use two 100nF in series to create an equivalent capacitance of 50nF. With a modified capacitance: f Hz kHz C nF nF nF F Z Z R 2250. 79 2 . 3 100 || 100 50 0. 05 (1 04 || 104 ) =@ ===m

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Results (by using the same experimental procedure as outlined in experiment 3.2): f(hz) Phase Vo > Vin 2205 45° 2503 45° 2361 0°

Fig 3.5.2 2. So the cutoff frequencies would be at 2.205kHz and 2.503 kHz. 7. 923 2. 503 2. 205 2 .3 61 = = D -= kk k f f Q factor o 3. Now to modify Q to be 5. We know: ==W -= R f LQ K fL R Q factor eff R m R eff

2 11. 75 2 p p with the measured Q = 7.923 @W»+W = ´´ + ´ = + ´ -= RKKK Hz mH KR KR fL RR RR fL RR Q factor extra extra extra

R eff extra eff extra R eff extra

20. 13 10 10 5 2 2361 100 11. 75 11. 75 2 2 || pp p So we connect an additional Rextra as 20KW that is a series of two 10KW resistors. We repeat the experiment to do the measurements to see if the modifying worked out. 4. Now, using the same process as outlined previously, the results show:

EE3101 LAB REPORT EXP#3 26 OCT 2006 - 13 f(hz) Phase Vo > Vin 2123 45° 2593 45° 2356 0°

Fig 3.5.3 Analysis of Figure 3.5.3: 5. 013 2. 593 2. 123 2 . 356 = = D -= kk k f f Q factor o This could be considered as a good result as it is very close to 5. 5. Determine the admittance of the circuit at the resonant frequency. To do this, we calculate the equivalent resistance of the resistors in parallel, and then put them in parallel with the equivalent impedance of the capacitor and inductor. Then we can put that impedance in series with Ri. Once we know the total equivalent impedance of the circuit, we can take the reciprocal of that to find the admittance. In phasor form, Zinductor = jwL, and Zcap = 1/(

jwC)

[ ] ( [ ]) 100 0. 0042 || 1 || || 235. 6 || 8488. 9 6 || 20 || 11. 7 5 100 Kj R R Ri j j K K K jwC Zin jwL extra eq =÷÷ + = + ø ö çç è æ= j Zin K j Yin . 00001 .0 04244 100 0 .0 042 11 === This answer makes sense because resonance is where the impedances of the circuit components all cancel out From part 2 we gathered the relationship between L, C, and R components of circuits at high frequencies. We take our experiment further in part 3. Part 3 – Nonideal Frequency Behavior of Passive Components Experiment 3.6 We repeat the measurement of experiment 3.1, however as R1 = 1M Ω and for R2 = 5 KΩ to see the effect of the shunt capacitance at R1 at high frequencies. Note that we are actually

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using two 10KΩ resistors in parallel to represent the 5KΩ R2. Fig 3.6.1 Note: To minimize the shunt capacitance of the oscilloscope we must use a 10x probe. Since the circuit acts as a highpass voltage divider, we need to normalize the gain relative to the high passband gain (~.075) so we will multiply our gains by (1/.075) to normalize them. The .707 normalized gain will be our 3dB

point. Aside from that, the process is the same as stated in experiment 1. The results were as follows. f(hz) Vin Vout | Vo / Vi | Normalized Gain 10 10.130 0.058 0.006 0.0765 50 10.190 0.059 0.006 0.0768 100 10.190 0.058 0.006 0.0760 500 10.190 0.058 0.006 0.0760 1000 10.190 0.058 0.006 0.0760 5000 10.190 0.058 0.006 0.0760 10000 10.190 0.056 0.005 0.0731 50000 10.190 0.058 0.006 0.0760 100000 10.190 0.069 0.007 0.0900 500000 10.000 0.191 0.019 0.2547 1000000 10.060 0.338 0.034 0.4480

EE3101 LAB REPORT EXP#3 26 OCT 2006 - 15 1897000 10.000 0.530 0.053 0.7067 5000000 11.300 0.830 0.073 0.9794 10000000 13.800 1.030 0.075 0.9952 20000000 17.500 1.250 0.071 0.9524

Fig 3.6.2

Gain vs Frequency 0.0000 0.2000 0.4000 0.6000 0.8000 1.0000 1.2000 0 500000 0 1E+07 1.5E+0 7 2E+07 2.5E+0 7 Frequency (Hz) Normalized Gain Gain vs Frequency

Fig 3.6.3 We see from our table (Figure 3.6.2) that the fB is approximately 1.897MHz fb = 1.897MHz

RC f B¢ ¢ = 2p 1 Where R ¢ = R || R L = 4 . 98 K W C ¢ = C probe + C resistor

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Note: We treat our resistors in parallel because they are seen in parallel by the probes. The capacitors are treated in parallel too for the same reason.

Since we know Cprobe = 15pF (as stated on the device), we can plug in all of our values and solve for Cresistor.

( )( ) ( )( )

MHz R R C C M K pF C f L probe resistor resistor

1. 897 2 1 || 5 15 1 2 || 1 = + = + = pp C resistor = 1 .8 6 pF Experiment 3.7 Use the circuit shown below to determine |Z(jw)| of the impedance of the inductor from 100 Hz to 1 MHz. Fig 3.7.1 For the inductor, (Figure 3.7.2) with the parasitic capacitance and resistance (Figure 3.7.3) => Fig 3.7.2 Fig 3.7.3 To obtain the Rw we use dc voltage across the inductor from Figure 3.7.3. Rw = 97.89Ω Now we do another measurement to get the impedance of the circuit: B

f(hz) Vin Vout | Vo / Vi | Phase (°) 100 10.000 9.200 0.920 5

EE3101 LAB REPORT EXP#3 26 OCT 2006 - 17 500 10.000 8.800 0.880 180 1000 10.000 8.000 0.800 30 5000 10.300 3.400 0.330 74 10000 10.300 1.900 0.184 90 50000 10.300 0.340 0.033 89 100000 10.300 0.138 0.013 90 136000 10.300 0.006 0.001 unreadable 500000 10.300 0.425 0.041 85 1000000 10.300 0.920 0.089 76

Fig 3.7.4 Gain vs Frequency 0.200 0.000 0.200 0.400 0.600 0.800

1.000 0 200000 400000 600000 800000 1000000 1200000 Frequency (Hz) Gain Gain vs Frequency

Fig 3.7.4 We could see from the measurements that the resonant frequency is where gain is its minimum value. That is 136 KHz (Figure 3.7.4 & 3.7.5).

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C pF LC f kHz R 13. 7 2 136 1 \= == w p where L = 100mH, fr = 136 KHz Now we could characterize the 100mH inductor that we used through out the experiments that has 97.89Ω parasitic resistance and 13.7pF parasitic capacitance (Figure 3.7.3). We can calculate |Z(jw)| by plugging these values into a phasor form equivalent equation.

()()

()

÷÷

ø ö çç è æ ´ ÷=+ø ö ç è = + + = + æ - 13. 7 10 1 1 98 .1 1 w jw wC R j wL jwC Z jw R jwL Conclusion

-12

First we looked at the effects of the shunt capacitance of the interconnecting cables and oscilloscope. By measuring the transfer function of a basic series connection of resistance we were able to obtain a shunt capacitance of 113pF, which would be able to be a reasonable value of the cables. The break frequency was gathered at the 3dB (70.7%) point of the circuit. Next we checked the resonance of the RLC circuits and we were able to see that the resonant frequency is inversely proportional to the square root of C. We also saw the input and output spectra differences from the RLC circuit by using the FFT math function on the oscilloscope. We found out that through the circuit we are only able to gather one harmonic that results a sinusoidal waveform at the output. Further we made an understanding that in order to modify a circuit to give an expected or proper response at high frequencies, we need to consider the more specific components of the circuit, such as the resistance of the function generator or the impedance of the components, and so forth. Finally, we obtained the knowledge on how to gather the exact characteristics of passive circuit components that are used for circuit designing. Typically a resistor has extra shunt capacitance. For the component we used (R=1M W ), we were able to see a parasitic capacitance of 1.86pF. An inductor has extra parasitic capacitance and resistance contained within the component. And for the component given in our lab kit (L=100mH) we were able to see the 98.89W resistance and the 13.7pF capacitance. In this experiment, we have successfully practiced the process of interpreting the high frequency response of a circuit. This information should be used on further circuit analysis or

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designing projects to gather more accurate results. References 1) Sedra/Smith, “Microelectronic Circuits”, Fifth Edition