Ee3101 Experiment 3

EE3101 LAB REPORT

EXP#3

26 OCT 2006

Non-ideal 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 non-idealistic 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 non-ideal 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. 1 1 f = fB = R 2πRC , 2π LC 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 5000 10000 28160 50000 100000 500000 1000000

5.000 5.000 4.940 5.000 5.000 5.000 4.820 4.820

2.380 2.380 2.280 1.767 1.300 0.750 0.167 0.099

0.476 0.476 0.462 0.353 0.260 0.150 0.035 0.021

Fig 3.1.2 Gain vs Frequency 0.500 0.450 Gain (Vout/Vin)

0.400 0.350 0.300 0.250

Gain vs Frequency

0.200 0.150 0.100 0.050 0.000 0

200000

400000

600000

800000 1000000 1200000

Frequency (Hz)

Fig 3.1.3

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EXP#3

26 OCT 2006

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: 1 fB = , where R = R1||R2 = 50KΩ 2πRC Vout =

R2 × Vin × (70.7%) R1 + R2

Vout =

1 × 5 × 0.707 = 1.787V 2

Then we pick the closest value from our measurements, which would be: ∴ f B ≅ 28.16 KHz ∴C =

1 = 1.13 × 10 −10 = 113 pF 2πRf B

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 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. 4

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Fig 3.2.1 Our two goals here are to: 1. Determine the resonant frequency and Q-factor. 2. Determine the impedance Z in ( jω 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: fR = 2

1 2π LC

fR = C=

1 4π 2 LC 1

4π 2 Lf R

2

For our resonant frequency fr = 2 kHz, and L = 100mH, 1 C= = 63.33nF ≈ 100nF = 0.1µF 2 2 4π Lf R However, in our lab kit the closest value of capacitor we have was 100nF. With a modified capacitance: C = 100nF = 0.1µF (104Z ) f R = 1591.5Hz ≅ 1.6kHz If we would have used a series of 100nF capacitance we could have gathered a resonant 5

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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 1796 1670

-45° 45° 0°

Fig 3.2.2 * Determine the resonant frequency and Q-factor. 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 = 1.67 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 Q-factor: Q − factor =

fo 1.67 k = = 6.844 ∆f 1.79k − 1.552k

* Determine the impedance Z in ( jω o ) at the resonant frequency. To determine the impedance of our circuit (Figure 3.2.1), we can use another formula for the Q-factor and solve for Zin. Q − factor =

2πf R L 2π × 1.67 k × 100mH = = 6.844 R Zin

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26 OCT 2006

Zin = 153.316Ω

Since our Zin is R plus the parasitic resistances of the inductor and capacitor we can calculate what the parasitic resistance is by subtracting 10 Ω from it. The calculation will be the following: Z in ( jω o ) = Z parasitic − R = 153.316 − 10 = 142.316Ω

Experiment 3.3 We now modify our circuit (Figure 3.2.1) to have a Q-factor = 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 Q-factor at 5: f 2πf R L Q − factor = o = ∆f R R=

2πf R L Q with Q = 5, fr = 1.6kHz, and L = 100mH R=

2π × 1.670 × 0.1 = 209Ω 5

If we look back to our circuit we see the parasitic resistance of the circuit and the original R. R = 209Ω = 143.316Ω + R ′

So we should put a resistance of value of R ′ = 65.68Ω We can do this by putting two 100 Ω resistors in parallel to give 50 Ω and then put a 10 Ω resistor in series to make the equivalent resistance 60 Ω R ' = 100Ω || 100Ω + 10Ω = 60Ω ≈ 65.68Ω 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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fr = C(uf) 0.001 0.01 0.1

26 OCT 2006

1 LC 2π

fr calculated (hz) 15916 5033 1592

fr measured (hz) 16130 5066 1776

% error 1.34% 0.66% 11.56%

Fig 3.4.1

Frequency vs Capacitance 18000 16000

Frequency (Hz)

14000 12000 10000

fr calculated (hz)

8000

fr measured (hz)

6000 4000 2000 0 0

0.02

0.04

0.06

0.08

0.1

0.12

Capacitance (uF)

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: 1 fr ∝ C 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. 4. Measure the modified circuit and gather the resonant frequency and Q of the circuit 10

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26 OCT 2006

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. Is = Vs / Rs. We will use 100 kΩ.

(L=100mH, without R)

1. To make resonant frequency at 2 kHz with L = 100 mH. 1 fR = 2π LC 2

fR = C= C=

1 4π 2 LC 1

4π 2 Lf R 1

2

4π 2 Lf R

2

= 63.33nF ≈ 50nF

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: C = 100nF || 100nF = 50nF = 0.05µF (104Z || 104 Z ) f R = 2250.79 Hz ≅ 2.3kHz

Results (by using the same experimental procedure as outlined in experiment 3.2): 11

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f(hz)

26 OCT 2006

Phase Vo -> Vin 2205 2503 2361

-45° 45° 0°

Fig 3.5.2 2. So the cutoff frequencies would be at 2.205kHz and 2.503 kHz. f 2.361k Q − factor = o = = 7.923 ∆f 2.503k − 2.205k

3. Now to modify Q to be 5. We know: Q − factor =

Reff 2πf R L

Reff = 2πf R LQm = 11.75 KΩ

Q − factor = Reff × Rextra Reff + Rextra 2πf R L

with the measured Q = 7.923

Reff || Rextra 2πf R L

11.75K × Rextra 11.75 K + Rextra = =5 2π × 2361Hz × 100mH

Rextra ≅ 20.13KΩ ≈ 10 K + 10 KΩ So we connect an additional Rextra as 20K Ω that is a series of two 10K Ω 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:

f(hz)

Phase Vo -> Vin

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2123 2593 2356

26 OCT 2006

-45° 45° 0°

Fig 3.5.3 Analysis of Figure 3.5.3: f 2.356k Q − factor = o = = 5.013 ∆f 2.593k − 2.123k 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)   −1 Zin =  jwL || || Rextra || Req  + Ri = ( j 235.6 || j8488.96 || [ 20 K || 11.75 K ] ) + 100 K jwC   = 100 K − j 0.0042 1 1 Yin = = = .00001 − .004244 j Zin 100 K − j 0.0042

[

]

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 – Non-ideal 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 using two 10KΩ resistors in parallel to represent the 5KΩ R2. 13

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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 high-pass voltage divider, we need to normalize the gain relative to the high pass-band 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) 10 50 100 500 1000 5000 10000 50000 100000 500000 1000000 1897000 5000000 10000000 20000000

Vin

Vout

| Vo / Vi |

10.130 10.190 10.190 10.190 10.190 10.190 10.190 10.190 10.190 10.000 10.060 10.000 11.300 13.800 17.500

0.058 0.059 0.058 0.058 0.058 0.058 0.056 0.058 0.069 0.191 0.338 0.530 0.830 1.030 1.250

Fig 3.6.2

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0.006 0.006 0.006 0.006 0.006 0.006 0.005 0.006 0.007 0.019 0.034 0.053 0.073 0.075 0.071

Normalized Gain 0.0765 0.0768 0.0760 0.0760 0.0760 0.0760 0.0731 0.0760 0.0900 0.2547 0.4480 0.7067 0.9794 0.9952 0.9524

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Gain vs Frequency 1.2000

Normalized Gain

1.0000 0.8000 0.6000

Gain vs Frequency

0.4000 0.2000 0.0000 0

500000 1E+07 1.5E+0 2E+07 2.5E+0 0 7 7 Frequency (Hz)

Fig 3.6.3

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

fB =

1 2πR ′C ′ Where

R ′ = R || R L = 4.98 KΩ C ′ = C probe + C resistor 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. 15

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fB =

EXP#3

26 OCT 2006

1 1 = = 1.897 MHz 2π ( R || R L ) ( C probe + C resistor ) 2π (1M || 5 K )(15 pF + C resistor ) C resistor = 1.86 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: f(hz)

Vin 100 500 1000 5000 10000 50000 100000

Vout 10.000 10.000 10.000 10.300 10.300 10.300 10.300

9.200 8.800 8.000 3.400 1.900 0.340 0.138

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| Vo / Vi | 0.920 0.880 0.800 0.330 0.184 0.033 0.013

Phase (°) 5 180 30 74 90 89 90

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136000 500000 1000000

EXP#3

10.300 10.300 10.300

0.006 0.425 0.920

26 OCT 2006

0.001 0.041 0.089

unreadable -85 -76

Fig 3.7.4

Gain vs Frequency 1.000 0.800

Gain

0.600 Gain vs Frequency

0.400 0.200 0.000 0

200000

400000

600000

800000

1000000 1200000

-0.200 Frequency (Hz)

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). 1 f R = 136kHz = 2π LC where L = 100mH, fr = 136 KHz ∴ Cω = 13.7 pF 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.

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Z ( jw) = R + jwL +

EXP#3

1 1   = R + j  wL −  = 98 + jwC wC  

26 OCT 2006

 1 j  w( .1) − w 13.7 × 10 −12 

(

)

  

Conclusion 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 Ω ), 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 98.89 Ω 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 designing projects to gather more accurate results.

References 1) Sedra/Smith, “Microelectronic Circuits”, Fifth Edition

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