EE101 Notes 2

January 4, 2017

Measurement Devices

The measurement devices to be studied are oscilloscope, function generator,dc power supply and spectrum analyzer.

DC power supply

It is a device used for generating DC voltages. It has three terminals as +terminal, - terminal and ground. Current limiter puts a limit on the current

Figure 1: DC Power Signal

supplied by the generator. If the LED for the current lights, this indicates a

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current limit. That is more current is demanded than the device can supply.Usually, a short circuit can cause such a case.

The maximum voltage that can be generated is 30V.

Oscilloscope

It is a measurement device that displays voltage vs. time graph.

Figure 2: Oscilloscope display

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Operation Steps:1. Turn on the device.

Figure 3: Oscilloscope display

2. Connect the probe to channel 1.3. Connect the probe to the ”probe compensation” contacts.

Figure 4: Oscilloscope display

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4. Push the ”autoset” to observe the square wave.

Figure 5: Oscilloscope display

5. If the square wave is not observed properly, this indicates a failure in theprobe.

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Measurement Steps for Tektronix TDS 210-220 SeriesOscilloscopes

The following steps are applied for manual measurements (no autoset):1. Click CH1, menu.Adjust the ground level from the ”position dial” to the center.

Figure 6: Oscilloscope display

2. Click the menu button until the coupling setting appears to be DC.DC shows DC + AC signals whereas AC shows only AC signals.3. Rotate the vertical Volts/Div dial to adjust the ”Volts/Div”.4. Rotate the horizontal Sec/Dic dial to adjust the ”Sec/Div”.

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Function generator

It is a electronic device that generates alternating (time varying) voltage sig-nals.

Time varying nature of the voltage signal is sinusoidal, i.e.

f(t) = Asin(ωt+ φ) + VDC (1)

whereA(V ) is the voltage amplitude,ω(rad/sec) is the radian frequency,φ(rad) is the phase angle,VDC(V ) is the dc voltage.

Figure 7: Sine wave

We have the following relations:

ω = 2πf (2)

where f is the Hertz frequency (number of cycles per second),and

T =1

f(3)

is the period (time for 1 cycle). Also,

Vpp = 2A (4)

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Proper selection of values for Volts/Div and Sec/Div in an oscilloscopeare determined based on the available settings. These settings are given inthe following figure:

Figure 8: Oscilloscope display

Also, note that in an oscilloscope display, there are total of 10 divisionshorizontally, and 8 divisions vertically.

The best oscilloscope setting can be considered to be one division greaterthan the minimum required setting for full coverage.The minimum require full coverage is:Horizontally, t=T (Period), andVertically, v=Vpp (Peak to peak Amplitude).

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Examples

Example 1:Oscilloscope displays which of the following graphs ?a. Power vs. frequencyb. Voltage vs. frequencyc. Voltage vs. timed. Current vs. frequency

The correct answer is C.

Example 2:In oscilloscope measurement steps, which of the followings is the correct cou-pling setting ?a. Groundb. Volts/div.c. DCd. Sec./div.

The correct answer is C.

Example 3:Find the best oscilloscope settings (Volts/div, sec./div.) for the followingsignal ?V (t) = sin(2πt),a. 500mV/div, 250ms/divb. 2V/div, 100ms/divc. 200mV/div, 50ms/divd. 1V/div, 100ms/div

Answer:T = 1/f = 1s.Since there are 10 horizontal divisions, The min. sec/div is:sec/div = T

10= 1

10sec/div = 100ms/div

The best sec/div is:sec/div > T

10= 100ms/div

So, from the table (fig.7),sec/div = 250ms/div

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Also, amplitude isV = 1V ,Vpp = 2V Then, the min. volts/div is:

volts/div > Vpp

8= 2

8volts/div = 250mV/div.

Then, from the table (fig.7),volts/div = 500mvolts/divThen, the correct answer is A.

Example 4:Find the best oscilloscope settings (volts/div, sec./div.) for the followingsignal ?V (t) = sin(2000πt),a. 100mV/div, 500us/divb. 500mV/div, 250us/divc. 200mV/div, 50us/divd. 100mV/div, 250us/div

Answer:T = 1/f = 1ms.Since there are 10 horizontal divisions, The min. sec/div is:sec/div = T

10= 1ms

10div= 100us/div

The best sec/div is:sec/div > 100us/divSo, from the table (fig.7),sec/div = 250us/divAlso, amplitude isV = 1V ,Vpp = 2V Then, the min. volts/div is:

volts/div > Vpp

8= 2

8volts/div = 250mV/div.

Then, from the table (fig.7),volts/div = 500mvolts/divThen, the correct answer is B.

Example 5:Find the best oscilloscope settings (volts/div, sec./div.) for the followingsignal ?V (t) = 0.2sin(40000πt),a. 50mV/div, 250us/divb. 100mV/div, 10us/div

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c. 200mV/div, 10us/divd. 100mV/div, 5us/div

Answer:T = 1/f = 50us.Since there are 10 horizontal divisions, The min. sec/div is:sec/div = T

10= 50u

10div= 5us/div

The best sec/div is:sec/div > 5us/divSo, from the table (fig.7),sec/div = 10us/divAlso, amplitude isV = 0.2V ,Vpp = 0.4V Then, the min. volts/div is:

volts/div > Vpp

8= 0.4

8volts/div = 50mV/div.

Then, from the table (fig.7),volts/div = 100mvolts/divThen, the correct answer is B.

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Spectrum Analyzer

It is a measurement device that displays power vs. frequency graph.

Figure 9: Spectrum Analyzer display

Power Analysis:Power is the time ratio of energy, i.e.

P =W

t(Joules/sec) (5)

In electrical engineering, the power consumed by a resistor R is:

P = V I = V 2/R = I2R(W ) (6)

Many times, we end up having very large or very small values for thepower. Thus, we often use decibel as the unit for power.

Decibel (dB):It is a logarithmic unit to measure power ratios:By definition:

PdB = 10logP

Pref

10 (7)

is called the ”decibel”. The term

PdB = logP

Pref

10 (8)

is called the ”bell”.

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If not specified, we take Pref = 1W .Millidecibel (dBm):

It is defined as

PdBm = 10logP

1mW10 (9)

Then,PdBm = 10logP.10

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10 (10)

orPdBm = 10logP10 + 10log10

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10 (11)

orPdBm = PdB + 30 (12)

AC Power

Consider the voltage signal

V = Asin(ωt+ φ) (V ) (13)

applied across a resistor R as shown in the following figure. The power

Figure 10: AC power

consumed by the resistor R is given by

P =V 2

R=A2sin2(ωt+ φ)

R(W ) (14)

or

P =A2

R

1

2[1− cos(2ωt+ 2φ)] (W ) (15)

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If we plot this power expression in (15), we obtain the following graph

Figure 11: AC power

Since the power is not constant, we define a concept of average power definedas

Pavg =A2

2R(W ) (16)

where A is the amplitude of the voltage. If A is the amplitude of the current,then we have the average power as

Pavg =A2

2R (W ) (17)

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Physical Operation of the Spectrum Analyzer

The reference level can be selected by clicking the amplitude button and themenu button as shown in the figure below. Either the keypad or the dial can

Figure 12: Spectrum Analyzer

be used to set the ref level in dbm. Once it is set, each division correspondsto 10dB drop (-10dB) in power level towards the bottom of the screen. Thereis also attenuation level which corresponds to the decrease in power level.

The setting along the horizontal axis (frequency) is carried out by clickingthe Frequency and the Span buttons and using the keypad or the dial toadjust the ”center frequency” and the ”span” values.

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Examples

Example 1:Find the power in dB for P=1000W ? (Pref=1W)a. 0dBb. 10dBc. 20dBd. 30dB

Answer:

PdB = 10logP

Pref

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orPdB = 10log100010 = 30dB. The correct answer is D.

Example 2:Find the power in dB for P=1W ? (Pref=1W)a. 0dBb. 10dBc. 20dBd. 30dB

Answer:PdB = 10log110 = 0dB. The correct answer is A.

Example 3:Find the power in dB for P=1nW ? (Pref=1W)a. 0dBb. -60dBc. -90dBd. -120dB

Answer:PdB = 10log10

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10 = −90dB. The correct answer is C.

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Example 4:Find the power in dBm for P=1nW ? (Pref=1W)a. 0dBb. -60dBc. -90dBd. -120dB

Answer:PdBm = PdB + 30 = −90 + 30 = −60dB. The correct answer is B.

Example 5:What settings are set for the horizontal axis of the spectrum analyzer ?a. ref levelb. Span and frequencyc. Attenuationd. Average powerThe correct answer is B.

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Electromagnetic Waves and Radiation

-Electromagnetic wave radiation is a propagation of energy due to time vary-ing electric and magnetic fields.

Electric Field

-Electric field: The force field acting on charges (Force per charge).

E =F

Q(N/C) (18)

- Simple way to obtain electric field: Connecting voltage source to parallelconducting plates:

Figure 13: Static Electric Field

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- If we extend the distance:

Figure 14: Static Electric Field

Electric field is weaker due to increased distance.Consider the force between two point charges Q1 and Q2:

F =kQ1Q2

d2(N/C) (19)

where k is a constant.Then,

E ∝ 1

d2, d = distance (20)

If we connect an alternating voltage source:We obtain electromagnetic waves:

Figure 15: Electric Field as wave

The electromagnetic waves travel with the speed of light (3× 108 m/s)

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They carry energy in free space (vacuum). The energy is contained insidethe electric field.

There is an attenuation of waves as they move farther away from thesource due to spreading effect.

The electromagnetic waves are used mostly in wireless communication.

The distance that 1 cycle of a wave travels is called the ”wavelength” anddesignated by λ.

The relation between the wavelength and frequency is given by:

λ =c

f, c = speedoflight. (21)

The energy as electromagnetic wave is radiated more efficiently as thelength of the conductor approaches the wavelength. Practically the limit fora radiation from a linear antenna is

D = dimension of the conductor ≥ λ

4(22)

Thus, in order to radiate electromagnetic energy as wave at f=20MHz,what should be the min. practical wire antenna dimension ?

Ans:λ =

c

f(23)

λ =3× 108

20× 106=

3× 108

2× 107(24)

orλ = 15m (25)

Therefore,

D =15

4= 3m 75cm ≈ 3.5m (26)

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LED vs Photodiode

A diode is semiconductor electronic device that acts as a short circuit whena positive voltage is applied and an open circuit when a negative voltage isapplied.

Figure 16: Diode

LED (Light Emitting Diode) is a diode that lights when the diode con-ducts (passing current).

Photodiode is a diode that conducts as light falls into the device.

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