Radio Frequency Electronics - University of Iowas-iihr64.iihr.uiowa.edu/MyWeb/Teaching/ece_55195_2015/Lectures/...A. Kruger Radio Frequency Electronics The University of Iowa 1

A. Kruger 1 Radio Frequency Electronics The University of Iowa

Radio Frequency Electronics

Edwin H. Armstrong

Image from Wikipedia

• Invented • Regenerative circuit while an undergraduate

(1914)

• Superheterodyne receiver (1918)

• Super-regenerative circuit (1922)

• Frequency modulation (FM) radio

• Professor at Columbia University

• Held 42 patents

• Awards

• IRE (now IEEE) Medal of Honor

• French Legion of Honor,

• National Inventors Hall of Fame

Preliminaries I


Calculators

Be sure to set your calculator to display in engineering mode.

Unless stated otherwise, provide answer to 3 significant figures.

Be sure that you know how to do complex number arithmetic on your calculator.

Be sure that you know how to do 2 × 2 and 3 × 3 matrix arithmetic on your calculator.

Consider programming frequently-used formulas into your calculator.

Calculators should be a real calculator. Smartphone apps the simulate calculators are not

allowed.

If your calculator has wireless/network functionality, turn the wireless off during exams.

I can’t teach you how to use your calculator. Read the manual or look for tutorials on

the web. There are some decent tutorial for TI calculators on YouTube.


Matlab, Mathematica, SPICE, …

The use of tools such as Matlab, Mathematica, Maple, SPICE, the

programmable features of your calculator, etc. is strongly

encouraged.

If you solve homework problems using these tools, or check your

answers with these tools, you are eligible for extra credit.

Some homework assignments will require SPICE simulations.

Some homework will involve writing Matlab scripts.


Metric/Engineering Prefixes

Since we are engineers, be sure to

express numbers in formats that other

engineers do.

When in Rome, do as the Romans do.

5.2E − 4

If your calculator displays

set it so it displays the number as

520E − 6

If this were the answer to a voltage

calculation, you would write

Vo = 520 𝜇V

Note “V” is not

in italics

Note space between

number and units

Grayed-out prefixes are generally not used in electrical

engineering, but we have decibel and use phrases such

as 20 dB per decade.


Metric/Engineering Formats

>> format shortEng

>> u = 4*pi*1e-7

You can tell Matlab to format its answer in engineering format

u = 1.2566e-00

Right click Format Custom and then

type in ##0.0E+0

To make Microsoft Excel format cells in engineering format

To increase the number of decimals just add

more zero's i.e., ##0.00E+0

There are of course exceptions. For example, the speed of light is 𝑐 = 3 × 108

m/s, and the permeability of vacuum is 4𝜋 × 10−7 H/m and the charge on an

electron is 1.6 × 10−19 C, and so on.


Complex Number Review

𝑧 = 𝑎 + 𝑗𝑏

𝑧 = 𝑎2 ++𝑏2

Real Axis

Imaginary

Axis

𝜃 = tan−1𝑏

𝑎

𝑎

𝑏

𝜃

Rectangular Polar

𝑧 = 𝑧 ∠𝜃

Exponential

𝑧 = 𝑧 𝑒𝑗𝜃

Euler’s identity 𝑒𝑗𝜃 = cos 𝜃 + 𝑗 sin 𝜃

𝑧1 = 𝑎 + 𝑗𝑏 𝑧2 = 𝑐 + 𝑗𝑑

𝑧1𝑧2𝑧3 = 𝑎 + 𝑗𝑏 𝑐 + 𝑗𝑑 𝑒 + 𝑗𝑓

𝑧1𝑧2𝑧3 = 𝑧1 𝑧2 𝑧3 ∠ 𝜃1 + 𝜃2 + 𝜃3

𝑧 = 𝑧1 𝑧2 𝑧3 𝑒𝑗 𝜃1+𝜃2+𝜃3

𝑧3 = 𝑒 + 𝑗𝑓 Let

Then = 𝑎𝑐 − 𝑏𝑑 + 𝑗𝑏𝑐 + 𝑗𝑎𝑑 𝑒 + 𝑗𝑓 = ⋯

cos 𝜃 =𝑒𝑗𝜃 + 𝑒𝑗𝜃 +

2 sin 𝜃 =

𝑒𝑗𝜃 − 𝑒𝑗𝜃

2𝑗



𝑧1 = 𝑎 + 𝑗𝑏 𝑧2 = 𝑐 + 𝑗𝑑 𝑧3 = 𝑒 + 𝑗𝑓 Let

Then 1

𝑧1=

1

𝑎 + 𝑗𝑏 =

(𝑎 − 𝑗𝑏)

𝑎 + 𝑗𝑏 𝑎 − 𝑗𝑏 =

(𝑎 − 𝑗𝑏)

𝑎2 + 𝑏2 =

𝑎

𝑎2 + 𝑏2− 𝑗

𝑏

𝑎2 + 𝑏2

1

𝑧1=

1

𝑧1∠ − 𝜃1

1

𝑧1=

1

𝑧1𝑒−𝑗𝜃

1

𝑧1𝑧2𝑧3=

1

𝑎 + 𝑗𝑏 𝑐 + 𝑗d 𝑒 + 𝑗𝑓 =

1

𝑎𝑐 − 𝑏𝑑 + 𝑗𝑏𝑐 + 𝑗𝑎𝑑 𝑒 + 𝑗𝑓 = ⋯

1

𝑧1𝑧2𝑧3=

1

𝑧1 𝑧2 𝑧3 ∠ − (𝜃1+𝜃2 + 𝜃3)

1

𝑧1𝑧2𝑧3=

1

𝑧1 𝑧2 𝑧3 𝑒−𝑗(𝜃1+𝜃2+𝜃3)



𝑧1 = 𝑎 + 𝑗𝑏 𝑧2 = 𝑐 + 𝑗𝑑 𝑧3 = 𝑒 + 𝑗𝑓 Let

Then 𝑧1𝑧2

=(𝑎 + 𝑗𝑏)

𝑐 + 𝑗𝑑 =

(𝑎 + 𝑗𝑏)

𝑐 + 𝑗𝑑

(𝑐 − 𝑗𝑑 )

(𝑐 − 𝑗𝑑)

𝑧1𝑧2

=𝑧1𝑧2

∠ 𝜃1 − 𝜃2

𝑧1𝑧2

=𝑧1𝑧2

𝑒−𝑗 𝜃1−𝜃2

= ⋯

𝑧1𝑛 = 𝑎 + 𝑗𝑏 𝑛 = 𝑎 + 𝑗𝑏 1 𝑎 + 𝑗𝑏 2 … 𝑎 + 𝑗𝑏 𝑛

𝑧10.32 = 𝑎 + 𝑗𝑏 0.32 =?

𝑧1𝑛 = 𝑧 𝑛∠ 𝑛𝜃

𝑧10.32 = 𝑧1

0.32∠ 0.32 × 𝜃

𝑧1𝑛 = 𝑧1

𝑛𝑒𝑗 𝑛𝜃

𝑧10.32 = 𝑧1

0.32𝑒𝑗 0.32×𝜃


Concepts Review – The Decibel (dB)

Decibel is defined as ratio of two powers 𝑃1 and 𝑃2 Ratio (dB) = 10 log10 𝑃1/𝑃2

Note “10” and not “20” (that is why it is called deci-bell)

Consider a signal source that delivers a power 𝑃𝑖 to an amplifier. The amplifier amplifies the signal and delivers a

power 𝑃𝐿 to a load. The amplifier’s power gain is:

𝑅𝑖 𝑅𝐿

𝑃𝑖

𝑉𝑖 𝑉𝐿

𝑃𝐿

𝐺 = 10 log10𝑃𝐿𝑃𝑖

In terms of the terminal voltages and resistances:

𝐺 = 10 log10𝑉𝐿2 𝑅𝑖

𝑉𝑖2 𝑅𝐿

= 10 log10𝑉𝐿2

𝑉𝑖2

𝑅𝐿𝑅𝑖

= 20 log10𝑉𝐿𝑉𝑖

𝑅𝐿𝑅𝑖

In the special case when 𝑅𝐿 = 𝑅𝑖, the power and voltage gain is 𝐺 = 20 log10𝑉𝐿𝑉𝑖

Strictly speaking, even though the “20” should be used only when 𝑅𝑖 = 𝑅𝐿, “20” is used universally when we

express voltage ratios in dB, regardless of the values of 𝑅𝑖 and 𝑅𝐿.

In microelectronics, voltages and currents are normally measured and calculated for, while in RF work, quite

often power is the quantity of interest. Also, in RF work, in many instances 𝑅𝑖 = 𝑅𝐿 = 50 Ω.


𝑃𝐿 = 100 mW

Signal

Source

𝑅𝐿 = 50 Ω

(a) 𝑃𝐿 = 50 mW

Signal

Source

𝑅𝐿 = 50 Ω 3 dB Attenuator

(b)

Concepts Review – Meaning of 3-dB We often encounter phrases such as “3-dB point”, “3-dB frequency”, “3-dB bandwidth”, “half power point”, etc.,

and there is quite a bit of confusion surrounding these terms.

log10 2 = 0.301 First, the “3” comes from the fact that and 10log10 2 ≈ 3

Thus, when an amplifier has a power gain of 2, the power gain in dB will be 3 dB because:

𝑃𝐿 = 2𝑃𝑖 ⇒ 𝐺 = 10 log𝑃𝐿𝑃𝑖

= 10 log 2 = 3 dB

Consider the attenuator shown

Devices such as this are frequently used in RF work. The 50 Ω indicates the input impedance of the device. In

(a) below a sinusoidal signal source delivers 100 mW to a 50 Ω load. In (b), the attenuator reduces the power by

3 dB, which is a factor 2 so the load now dissipates 50 mW.

Note that the load voltage in (a) is 0.1 50 = 2.236 VRMS and in (b) the load voltage is (0.05)(50) =

1.581 VRMS. While the power was reduced by a factor 2, the voltage was reduced by 1.581 2.236 = 1 2 .


dBm, dBW, dB𝝁, etc.

The decibel is related to the ratio of two powers. It is sometimes convenient to express a power

relative to some reference power. One such reference is 1 mW, and this leads to the dBm:

𝑃dBm ≡ 10 log𝑃

1 mW

Example. Express the following power levels as dBm: 1 mW, 10 mW, 1 W, and 5 𝜇W

Solution 1 mW = 10 log

1 mW

1 mW= 0 dBm

10 mW = 10 log10 mW

1 mW= 10 dBm

1 W = 10 log1,000 mW

1 mW= 30 dBm

5 𝜇W = 10 log5 × 10−6

1 × 10−3= −23 dBm



dBW ≡ 10 log𝑃

1 W

dB𝜇 ≡ 10 log𝑃

1 𝜇W Power, reference is 1 𝜇W.

Power, reference is 1 W.

dBm ≡ 10 log𝑃

1 mW Power, reference is 1 mW.

dBV ≡ 20 log𝑉

1 V𝑅𝑀𝑆 Voltage, reference is 1 VRMS regardless of impedance.

dBmV ≡ 20 log𝑉

1 mV𝑅𝑀𝑆

Voltage, reference is 1 mVRMS across 75 Ω. Used in

cable TV.

dB𝜇V ≡ 20 log𝑉

1 𝜇V𝑅𝑀𝑆

Voltage, reference is 1 𝜇VRMS. Used in radio sensitivity,

amplifier and antenna specifications.

dBZ, dBa, dBi,… Many others, used in radar, sound, etc.



Since log 𝐴 𝐵 = log 𝐴 + log 𝐵 it follows that we know an amplifier’s gain in dB, we can

easily perform power level calculations.

Example. An amplifier with 15 dB power gain amplifies a -10 dBm signal. What is the resulting

output power, expressed in dBm?

Solution

10 dBm is equivalent to 101 1 × 10−3 = 10 mW = 0.01 W. Further, 15 dB gain is equivalent

to a factor 101.5, so that the output power is 𝑃𝑜 = 101.5 0.01 = 316 mW = 0.316 W.

Converting to dBm gives 10 log 0.316 1 × 10−3 = 25 dBm .

However, using the logarithm property log 𝐴 𝐵 = log 𝐴 + log 𝐵 we can write directly

𝑃𝑜 = 𝑃𝑖 + 𝐺 = 10 dBm + 15 = 25 dBm

This works for the other dB units: dB𝜇, dBW, etc.



If an antenna delivers 𝑉𝑔 =10 dB𝜇V to a receiver with input an impedance of 50 Ω, what is the signal

level in dBm?

Solution

𝑃 =𝑉𝑔

2

𝑅

10 = 20log𝑉𝑔

1 × 10−6 VRMS ⇒ 𝑉𝑔= 1 × 10−6 10

1020 = 3.16 × 10−6 VRMS

=3.16 × 10−6 2

50 = 200 × 10−15 W = 10 log

200 × 10−15

1 × 10−3 dBm = −97 dBm

A sine wave has an amplitude of 𝑉𝑔 = 20 V. Express this as dBV.

Solution

𝑉𝑔 = 20log20 2

1

The definition of dBV requires that we use the RMS value of 𝑉𝑔. Thus:

= 23 dB



A spectrum analyzer with 50 Ω input impedance has a label next to the input warning the user to limit

the input power to 10 dBm. What is maximum amplitude of a signal 𝑣 𝑡 = 𝐴 cos 𝜔𝑡 that one can

apply to the spectrum analyzer?

Solution

𝑃 =𝐴𝑅𝑀𝑆

2

𝑅

10 dBm = 10 log𝑃

1 × 10−3

⇒ 10 × 10−3 =𝐴RMS

2

50

1 ⇒ 𝑃 = 1 × 10−3 101010

= 10 mW

⇒ 𝐴RMS = 0.707 V

⇒ 𝐴 = 2 𝐴RMS

⇒ 𝐴 = 1 V



The (sine wave) voltage across a 50 Ω resistor is increased from 5 V to 7 V. (a) What power increase

(in dB) does this represent? (b) Repeat but now for a 75 Ω resistor.

Solution

P2𝑃1 (𝑑𝐵) 10 log10

𝑉22 50

𝑉12 50

= 10 log1072

52 = 2.923 dB

Part (a)

Part (b) Since the same resistor is used in the calculation is used, it does not matter whether it

is 50 Ω or 75 Ω. The values cancel and the increase in power is still 2.923 dB



If at a certain frequency, a cable has a loss of 5 dB per 100 feet, how much power will be delivered to

an antenna from a transmitter that puts out 10 W? The length of cable between transmitter and cable

is 40 feet. Assume the transmitter, cable, and antenna are impedance-matched.

Solution

=40

100× 5 dB

10210 = 1.585

The cable attenuation is = 2 dB

This is a factor:

Thus, the power delivered is 10 1.585 = 6.31 W

Alternative Solution

=40

100× 5 dB

10 dBW

The cable attenuation is = 2 dB

10 W is equivalent to

Thus, the power delivered is 10 − 2 = 8 dBW = 10810 = 6.31 W


Decoupling, Coupling, Bias Tees, etc.

Coupling

Capacitor Coupling

Capacitor

Decoupling

Capacitor

Decoupling and bypass capacitors provide capacitors provide a reliable signal or ac ground right at the

point where it is needed in a circuit while blocking dc currents from flowing.

The magnitude of the reactance of decoupling and bypass capacitors’ reactance magnitude , i.e.,

1 𝜔𝐶 , must be small at the working frequency and they must low inductance and ESR.

A radio frequency choke (RFC) provides a short circuit to dc currents but kills (choke to death ) the

ac/signal. RFCs are inductors with large reactance 𝜔𝐿 at the working frequency. They should have low

capacitance and low ESR.

Varactor provides voltage-

variable capacitance

Equivalent ac circuit. 𝐿 and 𝐶 are in

parallel and form a resonant circuit


Decoupling, Coupling, Bias Tees, etc.

A sub-circuit that frequently appears in RF circuits is a

so-called Bias T.

The other branch is a short to both dc.

The last branch is a open to dc and a short to RF.

One branch is a short to dc but an open to RF.

Bias Tee

In the circuit shown there is a bias tee as

highlighted.

However, because of parasitics it is often

non-trivial to get a good inductor/capacitor

are very high frequencies.

Often one can design the bias tee into the

circuit, or make one on a small board.

Image from Wikipedia


Side Bar: Bias Tee

Commercial ($$) Bias Tee. Self-Made Bias Tee

However, because of parasitics it is often non-trivial to get a good inductor/capacitor at

very high frequencies, so some companies market bias tees that will do the job.


American Wire Gauge (AWG)

1 mil = 1/1,000 of an

inch.

40.3 mil = 0.043 inch

AWG numbers run

from 0000 to 40.

In the electronics

industry, odd AWG

numbers are

generally not used.

Shown is a partial

table relevant to this

course.


AWG

The wire used on most solderless bread board is 22 AWG,

sometimes 24.

Thin part of IC (such as 555 timer) and

an LED indicator lamp is ~0.55 mm,

which corresponds to ~22 AWG

diameter wire.

The leads on ¼ W carbon film

resistor commonly used for

bread boarding is 22 AWG


More on Wire

The AWG refers to the conductor and does not include the insulation of the wire.

7/32 7 strands, each 32 AWG

10/30 10 strands, each 30 AWG

More strands means more flexible

Stranded wire have designations such as 7/32, 10/30, …

This means

Strands


Documents

Radio Frequency Electronics - University of Iowas-iihr64.iihr.uiowa.edu/MyWeb/Teaching/ece_55195_2015/Lectures/...A. Kruger Radio Frequency Electronics The University of Iowa 1