eee117L_Lab1.pdf

EEE 117L Network Analysis Laboratory Lab 1

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EEE 117L Network Analysis Laboratory
Lab 1 – Voltage/Current Division and Filters

Lab Overview

The objective of Lab 1 is to familiarize students with a variety of basic applications of
passive R, C devices, and also how to measure the performance of these circuits using
both Spice simulations and the Digilent Analog Discovery 2 on the circuits constructed.

Prelab

Before coming to lab, students need to complete the following items for each of the
circuits studied in this lab :
• Any hand calculations needed to determine the values of components used in the

circuits such as resistors and capacitors, or specifications such as pole frequencies.
• A Spice simulation of each circuit to get familiar with how it works, and determine

what to expect when the circuit is built and its performance is measured.

Making connections on a Breadboard

Breadboards are used to easily construct circuits without the need to solder parts on a
printed circuit board. As seen in Figure 0 they have columns of pins that are connected
together internally, so that all the wires inserted in a column are shorted together. Note
that the columns on top and bottom are not connected together. There are also rows of
pins at the top and bottom that are connected together. These rows are intended for use
as the power supplies, and are typically labeled + and – and color coded red and blue for
the positive and negative power supplies. These rows are not connected in the middle.

Figure 0.

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Circuits to be studied

When choosing resistor and capacitor values use standard values available to you,
and keep all resistor values between 100 W and 100 kW.

1. Voltage and Current Dividers

One of the most commonly used circuits is a voltage divider
like the one shown in Figure 1.a. For example, if a signal is
too large to be input to a voltmeter or oscilloscope it can be
attenuated (reduced in size) using voltage division. The DC
voltage that an AC signal like a sine wave varies around can
also be reduced using this circuit.

For example, if all of the resistors in this circuit are the same
value, and the VS input source provides a DC voltage of 4V,
then the voltages in this circuit will be VA = 4V, VB = 3V,
VC = 2V, and VD = 1V. That is, voltage division will cause the voltage at node B to be
¾ of VS , the voltage at node C to be ½ of VS , and the voltage at node D to be ¼ of VS.
If a sine wave with an amplitude of 1V is then added so that VS = 4 + sin(wt) Volts, then
voltage division will cause the new values of VA , VB , VC and VD to be :

VA = 1.00*VS = 1.00*(4 + sin(wt)) = 4 + 1.00*sin(wt) Volts
VB = 0.75*VS = 0.75*(4 + sin(wt)) = 3 + 0.75*sin(wt) Volts
VC = 0.50*VS = 0.50*(4 + sin(wt)) = 2 + 0.50*sin(wt) Volts
VD = 0.25*VS = 0.25*(4 + sin(wt)) = 1 + 0.25*sin(wt) Volts

In this example both the amplitude of the sine wave and the DC voltage that the sine
wave varies around are reduced by either 25%, 50%, or 75%, depending on whether
the output voltage is measured at VB , VC or VD. It is also possible to measure the
output voltage as the difference between two node voltages, such as :

VBC = VB – VC = (3 + 0.75*sin(wt)) – (2 + 0.50*sin(wt)) = 1 + 0.25*sin(wt) Volts.

In addition there is no requirement that all the resistors be the same value. By carefully
choosing different resistor values it is possible to adjust the amount of signal attenuation
and the DC bias voltage that the signal varies around separately.

Design the circuit in Figure 1.a to both attenuate and level shift the input VS
Choose the resistor values to set VBC = 0.1*VS ± 10%, and the average DC value of VB
and VC equal to (VB + VC)/2 = 2V. Use VS = 4 + sin(wt) Volts at a frequency of 1k Hz.
(Hint: You may want to set RD = 0 W for this.)

Figure 1.a

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Current dividers are also commonly used to reduce the size
of a current signal, similar to how voltage dividers are used
to attenuate the size of a voltage signal. For example, if in
the circuit shown in Figure 1.b both resistors are set to the
same value, then the current flowing in each resistor will be
½ of the input current IS.

A common application of a current divider is to sense the
amount of current flowing in a load resistor by placing a
second resistor in parallel with it, and then measuring the
current flowing in that resistor instead. This is particularly
useful when a large current is flowing in the load resistor,
which makes it difficult to measure that current directly.
For example, if in Figure 1.b the resistor RB has a value of
10 W and the resistor RA has a value of 10 kW, then when
there is 1A flowing in RB there is only 1mA flowing in RA ,
which is much easier to measure. And since the ratio
between RA and RB is known to be 1000:1, we know by
current division that a current of 1mA in RA means there
must be 1A flowing in RB.

Sometimes it’s not possible to measure a current using an ammeter (e.g., the Digilent
Analog Discovery 2 doesn’t have an ammeter), so an alternate approach that can be
used is to measure the voltage across a known resistor, and then use Ohm’s Law to
convert this measured voltage to a current value. For example, in Figure 1.c an extra
resistor RAi is added in series with the RA resistor to measure the current flowing in RA
by measuring the voltage across the known resistor RAi and then using IA = VAi/RAi.

Design the circuit in Figure 1.c to measure the current in RB by measuring VAi
Set RB = 100 W and choose the values of RA and RAi to be able to measure the current
flowing in RB within ± 10% as VS is varied by measuring VAi. Be sure to make RA >> RB
so that the current flowing in RA is much smaller than the current flowing in RB.

2. Low-pass Filter

Another type of circuit that is often needed is a filter that allows
only a certain range of input frequencies to pass through it and
get to the output. The first one of these we’re going to consider
is the low pass filter shown in Figure 2. This circuit uses the fact
that the impedance of a capacitor goes down as frequency goes up
to create a voltage divider whose gain decreases as the input signal
frequency increases. A typical application of a low pass filter is to be
able to adjust the amount of bass heard in audio signals like music.

Figure 1.b

Figure 1.c

Figure 2

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The transfer function for this low-pass filter is given by :

𝐴” =
%&
%’
= (

)

)* +, ,-.
/ where the pole frequency is given by : 𝜔1 =

)
2′ 34

which gives the magnitude and phase of the transfer function as frequency varies to be :

|𝐴”| = 6
)

7)* 8, ,-. 9
:
; and the phase = 0° − 𝑎𝑟𝑐𝑡𝑎𝑛 8𝜔 𝜔1. 9

Which shows that at frequencies << wp where the capacitor looks like an open circuit, the voltage gain is approximately 1 with a phase of 0°. And at frequencies >> wp the
capacitor looks like a short circuit, so the voltage gain approaches 0 with a phase of -90°.

Key concept: Capacitors look like short circuits at high frequencies and open circuits at
low frequencies. So a capacitor in parallel shorts out the signal at high frequencies, and
a capacitor in series blocks the signal at low frequencies.

To be more exact, the magnitude of the voltage gain drops off at -20dB/decade as the
frequency increases above wp. And the phase decreases with a slope of -45°/decade
from 1 decade below wp to 1 decade above wp , until it reaches a maximum of -90°.
At exactly the pole frequency w = wp the magnitude of the voltage gain is reduced by
)
√E

= 0.707 or -3dB, and the phase of the voltage gain is equal to -45°.

Note: A decade means a factor of 10 in frequency. So 1 kHz is 1 decade below 10 kHz,
and 100 kHz is 1 decade above 10 kHz.

Design the circuit in Figure 2 to have a pole frequency of 10 kHz ± 10%.
Before constructing the circuit use an AC analysis in Spice to create a Bode plot for
this circuit, and verify that the pole frequency is as expected with the R and C values you
selected. Then construct the circuit and measure the actual pole frequency you achieved.
Also measure both the magnitude and phase of the voltage gain at several frequencies
much lower than the pole frequency (at least 10 to 100 times smaller), and at several
frequencies much higher than the pole frequency (at least 10 to 100 times larger).

Tip: To measure the magnitude of the voltage gain, input a sine wave to the circuit at
the desired frequency, and compare the peak-to-peak amplitudes of the input and output
signals. To measure the phase of the voltage gain, measure the phase shift between the
input and output signals. At the pole frequency the phase shift is -45°. Also note that the
Analog Discovery 2 can measure a Bode plot using the Network Analyzer feature.

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Another way to measure the pole frequency of this filter is to input a square wave and
measure either the time constant for the circuit, or the 10% – 90% rise/fall times. This
uses the fact that for a single pole low-pass filter the time response to a step is given by:

𝑉G(𝑡) = 𝑉G(𝑖𝑛𝑖𝑡𝑖𝑎𝑙) + ∆𝑉G 81 − 𝑒P

Q R. 9 where the time constant t is : 𝜏 = 𝑅U 𝐶W

Here VO(initial) is the voltage just before the step is taken, and DVO is the size of the step
taken = VO(final) – VO(initial). The value of the time constant can be found by measuring
how long it takes the output voltage to take 63.2% of the step, since at t = t the output
voltage is :

𝑉G(𝑡 = 𝜏) = 𝑉G(𝑖𝑛𝑖𝑡𝑖𝑎𝑙) + ∆𝑉G(1 − 𝑒P)) = 𝑉G(𝑖𝑛𝑖𝑡𝑖𝑎𝑙) + ∆𝑉G(0.632)

Once t is known, then the pole frequency can be found using : 𝜔1 = 1 𝜏.

We can also use the pole frequency to find how long it takes for the output voltage to go
from 10% of the step to 90% of the step, which is called the “10% to 90% rise time”.
(This is also the same time it takes for the step to fall from 90% back down to 10%,
which is called the “10% to 90% fall time”.) This 10% – 90% rise/fall time is given by :

𝑡]^_ = 𝑡 abb = 0.35 𝑓1.

where fp is the pole frequency in Hertz : 𝑓1 =
𝜔1

2𝜋.

Measure the pole frequency from the step response to a square wave.
Input a 1 Vpeak-to-peak square wave to your low-pass filter, and find the pole frequency
both by measuring the time constant, and also by measuring the 10% to 90% rise time.
Compare the values you get to your previous value found when the phase shift = -45°.

Interview question: If the resistors used to construct the low-pass filter in Figure 2 have
a tolerance of ± 5% , and the capacitors have a tolerance of ± 10% , then how much
variation will there be in the pole frequency when millions of these are manufactured?

3. High-pass Filter

The second filter that we’re going to consider is the high pass
filter shown in Figure 3. This is similar to the low pass filter
just examined, but with the resistor and capacitor swapped.
This circuit uses the fact that the impedance of a capacitor
goes down as frequency goes up to create a voltage divider
whose gain increases as the input signal frequency increases.
A typical application of a high pass filter is to be able to adjust
the amount of treble heard in audio signals like music.

Figure 3

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The transfer function for this high-pass filter is given by :

𝐴” =
%&
%f
= (

+,
,-.

)* +, ,-.
/ where the pole frequency is given by : 𝜔1 =

)
24 3′

which gives the magnitude and phase of the transfer function as frequency varies to be :

|𝐴”| = 6
, ,-.

7)* 8, ,-. 9
:
; and the phase = 90° − 𝑎𝑟𝑐𝑡𝑎𝑛 8𝜔 𝜔1. 9

Which shows that at frequencies >> wp where the capacitor looks like a short circuit,
the voltage gain is approximately 1 with a phase of 0°. And at frequencies << wp the capacitor looks like an open circuit so the voltage gain approaches 0 with a phase of 90°. Key concept: Capacitors look like short circuits at high frequencies and open circuits at low frequencies. So a capacitor in parallel shorts out the signal at high frequencies, and a capacitor in series blocks the signal at low frequencies. To be more exact, the magnitude of the voltage gain increases at +20dB/decade as the frequency increases from << wp up to wp. And the phase decreases from +90° at low frequencies with a slope of -45°/decade from 1 decade below wp to 1 decade above wp , until it reaches a minimum of 0°. At exactly the pole frequency w = wp the magnitude of the voltage gain is reduced by ) √E = 0.707 or -3dB, and the phase of the voltage gain is equal to +45°. Note: A decade means a factor of 10 in frequency. So 1 kHz is 1 decade below 10 kHz, and 100 kHz is 1 decade above 10 kHz. Design the circuit in Figure 3 to have a pole frequency of 100 Hz ± 10%. Before constructing the circuit use an AC analysis in Spice to create a Bode plot for this circuit, and verify the pole frequency is as expected with the R and C values you selected. Then construct the circuit and measure the actual pole frequency you achieved. Also measure both the magnitude and phase of the voltage gain at several frequencies much lower than the pole frequency (at least 10 to 100 times smaller), and at several frequencies much higher than the pole frequency (at least 10 to 100 times larger). Tip: To measure the magnitude of the voltage gain, input a sine wave to the circuit at the desired frequency, and compare the peak-to-peak amplitudes of the input and output signals. To measure the phase of the voltage gain, measure the phase shift between the input and output signals. At the pole frequency the phase shift is 45°. Also note that the Analog Discovery 2 can measure a Bode plot using the Network Analyzer feature. EEE 117L Network Analysis Laboratory Lab 1 7 Another way to measure the pole frequency of this filter is to input a square wave and measure either the time constant for the circuit, or the 10% - 90% rise/fall times. This uses the fact that for a single pole high-pass filter the time response to a step is : 𝑉G(𝑡) = 𝑉G(𝑖𝑛𝑖𝑡𝑖𝑎𝑙) + ∆𝑉G 8𝑒P Q R. 9 where the time constant t is : 𝜏 = 𝑅W 𝐶U Here VO(initial) is the voltage just before the step is taken and DVO is the size of the step. The value of the time constant can be found by measuring how long it takes for 63.2% of the step to decay (how long until only 36.8% of the step is still seen at the output), since at t = t the output voltage is : 𝑉G(𝑡 = 𝜏) = 𝑉G(𝑖𝑛𝑖𝑡𝑖𝑎𝑙) + ∆𝑉G(𝑒P)) = 𝑉G(𝑖𝑛𝑖𝑡𝑖𝑎𝑙) + ∆𝑉G(0.368) Once t is known, then the pole frequency can be found using : 𝜔1 = 1 𝜏. We can also use the pole frequency to find how long it takes for the output voltage to go from 10% of the step to 90% of the step, which is called the “10% to 90% rise time”. (This is also the same time it takes for the step to fall from 90% back down to 10%, which is called the “10% to 90% fall time”.) This 10% - 90% rise/fall time is given by : 𝑡]^_ = 𝑡 abb = 0.35 𝑓1. where fp is the pole frequency in Hertz : 𝑓1 = 𝜔1 2𝜋. Measure the pole frequency from the step response to a square wave. Input a 1 Vpp (peak-to-peak) square wave to your high-pass filter, and find the pole frequency both by measuring the time constant, and also by measuring the 10% to 90% fall time. Compare the values you get to your previous value found when the phase shift = 45°. Interview question: An ideal high-pass filter has a non-zero voltage gain at high frequencies all the way up to w = ¥ . Is that true for an actual high-pass filter built using a real resistors and capacitors? Why? 4. Band-pass filter The low-pass and high-pass filters just considered can be combined to create a band-pass filter, which only allows signals in a limited range of frequencies to pass from the input to the output. A typical application of a bandpass pass filter is to be able to adjust the amount of midrange frequencies heard in audio signals like music. Figure 4 EEE 117L Network Analysis Laboratory Lab 1 8 Combine the low-pass and high-pass filters previously designed as shown in Figure 4 to create a band-pass filter. Before constructing the circuit use an AC analysis in Spice to create a Bode plot for this circuit, and verify that both the lower and upper pole frequencies are as expected from your previous designs for the low-pass and high-pass filters. Then construct the circuit and measure the actual low and high frequency poles that you achieved. Also measure both the magnitude and phase of the voltage gain in the midband frequency range between the two poles, as well as at several frequencies much lower than the low frequency pole (at least 10 to 100 times smaller), and at several frequencies much higher than the high frequency pole (at least 10 to 100 times larger). Tip: To measure the magnitude of the voltage gain, input a sine wave to the circuit at the desired frequency, and compare the peak-to-peak amplitudes of the input and output signals. To measure the phase of the voltage gain, measure the phase shift between the input and output signals. At the pole frequency the phase shift is 45°. Also note that the Analog Discovery 2 can measure a Bode plot using the Network Analyzer feature. Interview question: How could the outputs from a low-pass filter and a high-pass filter be combined to create a filter that attenuates signals in a narrow range of frequencies? This is called a band-reject or “notch” filter.