Radio Frequency Electronics. The University of Iowa. Spring 2015.

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Homework Assignment 05

Question 1 (2 points each unless otherwise indicated)(20 points)

1. Estimate the parallel parasitic capacitance of a 1 mH inductor with an SRF of 220 kHz.

Answer: (2𝜋)(220 × 103) = 1 √(1 × 10−3)⁄ , so that 𝐶 = 532 pF

2. Determine the series resistance of a 100 𝜇H inductor at 5 MHz if it has a 𝑄 0f 80.

Answer: 𝑄 = 𝜔𝐿 𝑅⁄ ⇒ 𝑅 = 2𝜋(5 × 106)(100 × 10−6) 80 ≈ 39 Ω⁄

3. A BJT is biased such that 𝑟𝜋 = 4.5K, 𝐶𝜋 = 100 pF, and 𝐶𝜇 = 14 pF. What is 𝑓𝛽?

Answer: 𝜔𝛽 = 1 𝑟𝜋(𝐶𝜋 + 𝐶𝜇)⁄ = 1.95 rad s⁄ = 310 kHz

4. Assume you need to build a parallel resonant circuit and have several capacitors and inductors

available that you can match up to achieve the desired resonance frequency. Assume the cost and

size of the various combinations are the same. Generally-speaking, which would be the better

combination 𝐿𝑠𝑚𝑎𝑙𝑙, 𝐶𝑙𝑎𝑟𝑔𝑒, or 𝐿𝑙𝑎𝑟𝑔𝑒, 𝐶𝑠𝑚𝑎𝑙𝑙? Briefly explain your answer.

Answer: The 𝐿𝑠𝑚𝑎𝑙𝑙, 𝐶𝑙𝑎𝑟𝑔𝑒 combination is preferable. This is because capacitors are generally

better at being capacitors than inductors being inductors. Smaller-value inductors will use less

wire and have few losses than large value inductors. Consequently, the 𝑄 of the resonant circuit

will probably be higher if we use a 𝐿𝑠𝑚𝑎𝑙𝑙, 𝐶𝑙𝑎𝑟𝑔𝑒 combination.

5. Estimate the self-resonance frequency in Hz of a 100-pF, though-hole ceramic capacitor with

leads trimmed to 2.5 mm.

Answer The total lead inductance is (at 1 nH/mm for component leads) is around 5 nH. The self-

resonance frequency

𝑓 = 1 (2𝜋√𝐿𝐶) = 1 (2𝜋√(5 × 10−9)(100 × 10−12)) = 225 MHz⁄⁄

6. Estimate the SRF in Hz of a 470-pF, though-hole ceramic capacitor with leads trimmed to

2.5 mm.

Answer The total lead inductance is (at 1 nH/mm for component leads) is around 5 nH. The self-

resonance frequency

𝑓 = 1 (2𝜋√𝐿𝐶) = 1 (2𝜋√(5 × 10−9)(470 × 10−12)) = 104 MHz⁄⁄

Radio Frequency Electronics. The University of Iowa. Spring 2015.

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7. Calculate the thermal noise generated by a 10K resistor in a 10 kHz bandwidth at room

temperature.

Answer:

𝑣𝑛 = √4𝑘𝑇𝐵𝑅

= √4(1.38 × 10−23)(300)(10 × 103)(10 × 103) = 1.29 𝜇V rms

8. Calculate the Johnson noise that a 1K resistor generates at 𝑇 = 300 K.

Answer:

𝑣𝑛 = 4.07 nV √Hz⁄ (rms)

9. At low frequencies the 𝑄 of an inductor proportional to the operating frequency (𝑄 = 𝜔𝐿 𝑅𝑠⁄ ,

where 𝑅𝑠 is the series resistance). Ignoring capacitive effects, but taking the skin effect into

account, how does the 𝑄 depend on frequency at high frequencies?

Answer The skin effect causes the resistance of the wire to be proportional to the square root of

frequency (𝑅𝑠 = 𝑅𝑎𝑐 ∝ 𝑅𝑑𝑐√𝜔). Thus, the inductor 𝑄 will be proportional to the square root of

frequency are high frequencies.

10. Consider the following single-turn inductors, made out the same length of wire (4𝑙). Which one

will have the largest inductance? Briefly motivate your answer.

Answer:

Because inductor (a) closes a larger are than inductor

(b), it will have the larger inductance.

Radio Frequency Electronics. The University of Iowa. Spring 2015.

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Question 2 Design a capacitive transformer to match a 50 Ω load to a 600 Ω generator at 30 MHz. The

overall bandwidth must be 3.5 MHz. Assume the 𝑄𝑠 of the inductor and capacitors you will be using are

very high so that you can ignore losses in these components. However, the inductor you will use has an

SRF of 80 MHz. Calculate the inductor’s “built-in” capacitance and absorb that into the transformer

capacitances. (20 points)

Solution The required transformer ratio is

𝑛 =𝐶1

𝐶1 + 𝐶2= √

50

600= 0.2887

The 50 Ω load is transformed to 600 Ω so that at the generator the effective resistance is 𝑅𝑝 = 300 Ω.

Further, the required bandwidth is 3.5 MHz, so that the capacitance of the transformer should be

𝐶 =1

2𝜋𝐵𝑅𝑝=

1

2𝜋(3.5 × 106)(300)= 151.58 pF

From this follows that the required inductance is

𝐿 =1

(2𝜋𝑓)2𝐶=

1

(2𝜋 × 30 × 106)2(151.58 × 10−12)= 185.68 nH

However, the inductor has a parasitic capacitance which we can obtain from the SRF

𝐶𝐿 =1

(2𝜋 × 𝑆𝑅𝐹)2𝐿=

1

(2𝜋 × 80 × 106)2(185.7 × 10−9)= 21.32 pF

Thus, the capacitive transformer need to supply only 151.58 − 21.32 = 130.26 pF, since the inductor’s

parasitic capacitance will supply the balance

𝑛 =𝐶1

𝐶1 + 𝐶2= 0.2887 and 𝐶 =

𝐶1𝐶2

𝐶1 + 𝐶2= 130.26 pF

Solving for 𝐶1 and 𝐶2 yields 𝐶1 = 183.1 pF and 𝐶2 = 451.2 pF. The final circuit is shown below.

Radio Frequency Electronics. The University of Iowa. Spring 2015.

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To check how good the transformer is, calculate 𝑛𝑄𝑇𝑄𝐸:

𝑄𝐸 = 𝜔𝑅(𝐶1 + 𝐶2) = (2𝜋 × 30 × 106)(50)(183.1 × 10−12 + 451.2 × 10−12) = 5.98 ≈ 6

𝑄𝑇 = 𝜔𝐶𝑅 𝑛2⁄ = (2𝜋 × 30 × 106)(151.58 × 10−12)(50) = 17.13

𝑛𝑄𝑇𝑄𝐸 ≈ 30

The model is good since 𝑛𝑄𝑇𝑄𝐸 > 20

SPICE Validation (Not Required)

The impedance the generator sees was calculated using SPICE. The plot below shows that the

transformer resonates at 29.857 MHz and the resistance is 616 Ω. The second plot shows the bandwidth

is 3.44 MHz. These values are within 2% of the design values of 30 MHz, 600 Ω, and 3.5 MHz.

𝑽𝑳𝑶𝑨𝑫

(𝐝𝐁

) 𝑹𝒊

( 𝛀

)

Radio Frequency Electronics. The University of Iowa. Spring 2015.

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Question 3 Consider the common emitter voltage

follower amplifier. The transistor is biased at 30 mA

and 𝐶𝐶 → ∞. (a) Estimate the magnitude of the voltage

gain 𝑣𝑂 𝑣𝑔⁄ and express this in dB. (b) Estimate the

power gain 𝑃𝐿 𝑃𝑔⁄ of the amplifier and express this in

dB. (20 points)

Note: “Estimate” means you may make reasonable

assumptions, but you have to state them.

Solution

Part (a) This is a BJT voltage follower so that the emitter signal voltage is slightly less than the base

signal voltage, say 𝑣𝐸 = 0.95𝑣𝐵. Further,

𝑣𝐵 =𝑅𝑖

𝑅𝑖 + 𝑅𝑔𝑣𝑔 ⇒

𝑣𝑜

𝑣𝑔= 0.95

𝑅𝑖

𝑅𝑖 + 𝑅𝑔,

where 𝑅𝑖 is the follower’ input resistance as indicated. Using BJT scaling, we calculate 𝑅𝑖as follows:

𝑟𝜋 =𝛽

𝑔𝑚=

200

(40)(0.03)= 167 Ω

𝑅𝑖 = 𝑅1‖𝑅2‖(𝑟𝜋 + (1 + 𝛽)𝑅𝐿) = 56K‖100K‖(167 + (201)(50)) ≈ 8K

Since 8K ≫ 50 Ω, we have 𝑣𝑏 ≈ 𝑣𝑔 and the voltage gain 𝑣𝑜 𝑣𝑔⁄ is

𝑣𝑜

𝑣𝑔≈ 0.95 = 20 log(0.95) = −0.45 dB.

Part (b) Since 𝑅𝑖 ≫ 𝑅𝑔 we can ignore the effect of 𝑅𝑔, and the generator and load signal powers are

𝑃𝑔 =𝑉𝑔

2

𝑅𝑖, 𝑃𝐿 =

(0.95Vg)2

𝑅𝐿.

The power gain is

𝐺 =𝑃𝐿

𝑃𝑔= (0.95)2

𝑅𝑖

𝑅𝐿= (0.95)2

8 × 103

50= 144.4.

In dB, the power gain is

10 log(144.4) = 21.6 dB

Radio Frequency Electronics. The University of Iowa. Spring 2015.

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Question 4 A 1.5 GHz source has an internal

resistance 𝑅𝑔 = 12 Ω in series with a 1 nH inductor.

(a) Design an 𝐿-impedance matching network for

maximum power transfer to a 58 Ω load. The load

has a 1.5 pF stray parallel capacitance. Use a low-

pass 𝐿-network and the method of absorbing the load- and generator reactance (b) Now design a high-

pass 𝐿-network and use the technique of resonating out the load-and generator inductances. (15 points)

Solution

Part (a) Ignore the source-and load reactances and match the 58 Ω load to the 12 Ω generator with a

parallel capacitor and series inductor.

𝑄𝑠 = 𝑄𝑝 = √𝑅𝑝

𝑅𝑠− 1 = √

𝑅𝐿

𝑅𝑔− 1 = √

58

12− 1 = 1.958

𝑄𝑝 =𝑅𝑝

𝑋𝑝⇒ 𝑋𝑝 =

𝑅𝑝

𝑄𝑝=

58

1.958= 29.59 Ω

This corresponds to a capacitor with value 𝐶𝑝 = 1 (2𝜋 × 1.5 × 109 × 29.59)⁄ = 3.586 pF. Further

𝑄𝑠 =𝑋𝑠

𝑅𝑠⇒ 𝑋𝑠 = 𝑅𝑠𝑄𝑠 = (12)(1.958) = 23.49 Ω

This corresponds to an inductor with value 𝐿𝑠 = 23.49 (2𝜋 × 1.5 × 109)⁄ = 2.493 nH. The resulting

network is shown in (i) below. Applying the absorption technique, move 1 nH from the 𝐿-match

inductor to the generator and move 1.5 pF from the 𝐿-match network to the load. The resulting network

is shown in (ii) below.

(i) L-match network when ignoring generator- and

load recatances (ii) Generator and load aborbs parts of 𝐿-match

network

Radio Frequency Electronics. The University of Iowa. Spring 2015.

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Part (b) Note below we show how to “resonate out” the load-and generator inductances for the low-

pass network design above.

Start with the network in (iii) below, which shows the 𝐿-match network previously computer and the

generator- and load reactive elements. We can cancel these elements by adding a series capacitance at the

generator and a parallel inductance at the load. At the load, the 1.5 pF capacitor has a reactance −𝑗70.7 Ω

at 1.5 GHz. An inductor 𝐿 = 7.51 nH has a reactance 𝑗70.7 Ω at 1.5 GHz. Similarly, at the generator, a 1

nH inductor at 1.5 GHz has a reactance 𝑗9.43 Ω which one can cancel with a series capacitor with value

11.3 pF.

(iii) 𝐿-match network when disregarding generator- and load reactances

The final match network is shown in (iv) below

(iv) 𝐿-match network with reactive elements canceling generator- and load reactances.

Radio Frequency Electronics. The University of Iowa. Spring 2015.

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Question 5 Design a common-base amplifier using a 2N2222 BJT. Bias the BJT at 𝐼𝐶 = 2 mA, and use

a power supply of 𝑉𝐶𝐶 = 9 V. The amplifier will be used at 1 MHz. Calculate the amplifier input

impedance, voltage- and power gains. (20 points)

Solution

The first step is to design a CE amplifier. We will the 4-resistor bias method. To maximize output

voltage swing, set RC = 4.5 V (2 mA) =⁄ 2.25 K. For bias-point stability, set VRE = 0.15 VCC = 1.35 V.

Then RE = VRE 2 mA⁄ = 675 Ω. Use 680 Ω, the closest standard value. The base voltage is 0.65 V

higher than VE so that VB = 2 V. Choose the current that flows through R1and R2 to be 0.1IC = 0.2 mA.

This will ensure we can ignore the base current and makes the design of R1 and R2 easier. R2 =

2 V 0.2 mA = 10K⁄ , and R1 = (9 𝑉 − 2 V) 0.2 mA⁄ = 35K. This completes the bias design.

Assume that β = 200, then gm = 40IC = 80 mS, and rπ = β gm = 2.5K⁄ .

Next, ground the base for ac signals by adding a

capacitor CB with reactance much less than

rπ‖R1‖R2 = 2K at 1 MHz. Choose CB = 0.1 μF

which has a reactance of 1.6 Ω at 1 MHz. The CB

amplifier’s input is at the emitter and is coupled into

the amplifier through CE. The input impedance of the

CB amplifier is Ri ≈ 1 gm = 12.5 Ω⁄ . Pick CE such

that its reactance is 20 times smaller than this

12.5 20⁄ =1

(2π)(1 × 106)(CE)⇒ CE = 0.254 μF

Use 0.27 μF, the closest standard value. For the output coupling capacitor, use 0.1 μF and decouple the

power supply with a 0.1 μF capacitor.

The voltage gain is 𝐴𝑣 ≈ 𝑔𝑚𝑅𝐶 = (0.08)(2.25 × 10−3) = +180 = 45 dB

The power generated by the load and dissipated by 𝑅𝐶 are

𝑃𝑔 =𝑣𝑔

2

𝑅𝑖= 𝑣𝑔

2𝑔𝑚, 𝑃𝐿 =𝑣𝐿

2

𝑅𝐶=

(𝑔𝑚𝑅𝐶𝑣𝑔)2

𝑅𝐶= 𝑅𝐶𝑣𝑔

2𝑔𝑚2

The power gain is

𝐺 =𝑃𝐿

𝑃𝑔=

𝑅𝐶𝑣𝑔2𝑔𝑚

2

𝑔𝑚𝑣𝑔2 = 𝑔𝑚𝑅𝐶 = 180 = 22.5 dB

Radio Frequency Electronics. The University of Iowa. Spring 2015.

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Question 6 The CA3096 IC consists of several npn- and pnp BJTs on the

same substrate. The objective of this question is to provide you with

experience extracting information relevant to the high-frequency

performance of the npn transistors. Assume one of the npn BJTs is biased

at 𝐼𝐶 = 1 mA, and 𝑉𝐶𝐸 = 5 V. Assume 𝑇 = 25° and take 𝑉𝑏𝑖 = 0.75 V for

both pn junctions. Find 𝛽0, 𝑟𝜊, 𝐼𝑠, 𝑟𝜊, 𝐶𝜇 , 𝑓𝑇 , 𝐶𝜋, 𝑟𝜋, and 𝑓𝛽. If you are

looking for the datasheet for the BJT, consider using an amazing new

invention called Google. (25 points)

Solution

From the data sheet 𝛽0 ≈ ℎ𝐹𝐸 = 390, 𝑉𝐵𝐸 = 0.69 V, and 𝑟𝜊 = 80K, at 𝑉𝐶𝐸 = 5 V, 𝐼𝐶 = 1 mA.

𝐼𝐶 = 𝐼𝑆 [𝑒𝑉𝐵𝐸𝑉𝑇 − 1] ⇒ 1 × 10−3 = 𝐼𝑆 [𝑒

0.690.026 − 1] ⇒ 𝐼𝑆 = 2.98 fA

From the datasheet 𝐶𝐶𝐵 ≡ 𝐶𝜇 = 0.46 pF at 𝑉𝐶𝐵 = 3 V. From this we calculate 𝐶𝐶𝐵0 the C-B junction

capacitance at zero bias voltage:

𝐶𝐶𝐵 =𝐶𝐶𝐵0

(1 +𝑉𝑅𝑉𝑏𝑖

)𝑀𝐽𝐶 ⇒ 0.46 pF =

𝐶𝐶𝐵0

(1 +3

0.75)

1 2⁄⇒ 𝐶𝐶𝐵0 = 1.03 pF

Since we don’t have information on 𝑀𝐽𝐶, we assumed 𝑀𝐽𝐶 = 0.5. At 𝑉𝐶𝐸 = 5 V, 𝑉𝐶𝐵 = 5 − 0.69 =4.3 V, so that the C-B junction capacitance at 𝑉𝐶𝐸 = 5 V is

𝐶𝜇 = 𝐶𝐶𝐵 = 1.03 pF

(1 +4.310.75

)1 2⁄

⇒ 𝐶𝜇 = 𝐶𝐶𝐵 = 0.397 pF

For a quick estimate, which is probably good enough for most cases:

𝐶𝐶𝐵2 ≈ 𝐶𝐶𝐵1 √𝑉𝑅1 𝑉𝑅2⁄ = 0.46√3 4.3⁄ = 0.384 pF

From the data sheet 𝑓𝑇 = 280 MHz at 𝑉𝐶𝐸 = 5 and 𝐼𝐶 = 1 mA. Thus

𝐶𝜋 =𝑔𝑚

𝜔𝑇− 𝐶𝜇 =

(40)(1 × 10−3)

(2𝜋)(280 × 106)− 0.397 × 10−12 = 22.74 pF − 0.397 pF = 22.3 pF

𝑟𝜋 =𝛽0

𝑔𝑚=

390

(40)(1 × 10−3)= 9.75K

The beta-cutoff frequency is

𝑓𝛽 =1

2𝜋

1

𝑟𝜋(𝐶𝜋 + 𝐶𝜇)=

1

2𝜋

1

(9.75 × 103)(22.74 × 10−12)= 718 kHz

Alternatively, 𝑓𝛽 = 𝑓𝑇 𝛽0 = 280 × 106 390⁄ = 718 kHz⁄