Homework Assignment 05

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1 Homework Assignment 05 Question (2 points each unless otherwise indicated)(20 points). Estimate the parallel parasitic capacitance of a mh inductor with an SRF of 220 khz. Answer: (2π)( ) = ( 0 3 ), so that C = 532 pf 2. Determine the series resistance of a 00 μh inductor at 5 MHz if it has a Q 0f 80. Answer: Q = ωl R R = 2π(5 0 6 )( ) Ω 3. A BJT is biased such that r π = 4.5K, C π = 00 pf, and C μ = 4 pf. What is f β? Answer: ω β = r π (C π + C μ ) =.95 rad s = 30 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 L small, C large, or L large, C small? Briefly explain your answer. Answer: The L small, C large 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 L small, C large combination. 5. Estimate the self-resonance frequency in Hz of a 00-pF, though-hole ceramic capacitor with leads trimmed to 2.5 mm. Answer The total lead inductance is (at nh/mm for component leads) is around 5 nh. The selfresonance frequency f = (2π LC) = (2π (5 0 9 )( )) = 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 nh/mm for component leads) is around 5 nh. The selfresonance frequency f = (2π LC) = (2π (5 0 9 )( )) = 04 MHz

2 7. Calculate the thermal noise generated by a 0K resistor in a 0 khz bandwidth at room temperature. Answer: v n = 4kTBR = 4( )(300)(0 0 3 )(0 0 3 ) =.29 μv rms 8. Calculate the Johnson noise that a K resistor generates at T = 300 K. Answer: v n = 4.07 nv Hz (rms) 9. At low frequencies the Q of an inductor proportional to the operating frequency (Q = ωl R s, where R s 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 (R s = R ac R dc ω). Thus, the inductor Q will be proportional to the square root of frequency are high frequencies. 0. 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. 2

3 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 = C C + C 2 = = The 50 Ω load is transformed to 600 Ω so that at the generator the effective resistance is R p = 300 Ω. Further, the required bandwidth is 3.5 MHz, so that the capacitance of the transformer should be From this follows that the required inductance is L = C = = 2πBR p 2π( = 5.58 pf )(300) (2πf) 2 C = (2π ) 2 ( = nh ) However, the inductor has a parasitic capacitance which we can obtain from the SRF C L = (2π SRF) 2 L = (2π ) 2 ( = 2.32 pf ) Thus, the capacitive transformer need to supply only = pf, since the inductor s parasitic capacitance will supply the balance n = C C + C 2 = and C = C C 2 C + C 2 = pf Solving for C and C 2 yields C = 83. pf and C 2 = 45.2 pf. The final circuit is shown below. 3

4 V LOAD (db) R i (Ω) Radio Frequency Electronics. The University of Iowa. Spring 205. To check how good the transformer is, calculate nq T Q E : Q E = ωr(c + C 2 ) = (2π )(50)( ) = Q T = ωcr n 2 = (2π )( )(50) = 7.3 nq T Q E 30 The model is good since nq T Q E > 20 SPICE Validation (Not Required) The impedance the generator sees was calculated using SPICE. The plot below shows that the transformer resonates at MHz and the resistance is 66 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. 4

5 Question 3 Consider the common emitter voltage follower amplifier. The transistor is biased at 30 ma and C C. (a) Estimate the magnitude of the voltage gain v O v g and express this in db. (b) Estimate the power gain P L P g 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 v E = 0.95v B. Further, v B = R i v R i + R g v o R i = 0.95, g v g R i + R g where R i is the follower input resistance as indicated. Using BJT scaling, we calculate R i as follows: r π = β 200 = g m (40)(0.03) = 67 Ω R i = R R 2 (r π + ( + β)r L ) = 56K 00K (67 + (20)(50)) 8K Since 8K 50 Ω, we have v b v g and the voltage gain v o v g is v o v g 0.95 = 20 log(0.95) = 0.45 db. Part (b) Since R i R g we can ignore the effect of R g, and the generator and load signal powers are The power gain is In db, the power gain is P g = V g 2, P R L = (0.95V 2 g). i R L G = P L = (0.95) 2 R i = (0.95) = P g R L 50 0 log(44.4) = 2.6 db 5

Question 4 A.5 GHz source has an internal resistance R g = 2 Ω in series with a nh inductor. (a) Design an L-impedance matching network for maximum power transfer to a 58 Ω load. The load has a.

Use a lowpass L-network and the method of absorbing the load- and generator reactance (b) Now design a highpass L-network and use the technique of resonating out the load-and generator inductances.

958 Q p = R p X p X p = R p Q p = 58.958 = 29.59 Ω This corresponds to a capacitor with value C p = (2π.5 0 9 29.59) = 3.586 pf. Further Q s = X s R s X s = R s Q s = (2)(.958) = 23.

6 Question 4 A.5 GHz source has an internal resistance R g = 2 Ω in series with a nh inductor. (a) Design an L-impedance matching network for maximum power transfer to a 58 Ω load. The load has a.5 pf stray parallel capacitance. Use a lowpass L-network and the method of absorbing the load- and generator reactance (b) Now design a highpass L-network and use the technique of resonating out the load-and generator inductances. (5 points) Solution Part (a) Ignore the source-and load reactances and match the 58 Ω load to the 2 Ω generator with a parallel capacitor and series inductor. Q s = Q p = R p R s = R L R g = 58 2 =.958 Q p = R p X p X p = R p Q p = = Ω This corresponds to a capacitor with value C p = (2π ) = pf. Further Q s = X s R s X s = R s Q s = (2)(.958) = Ω This corresponds to an inductor with value L s = (2π ) = nh. The resulting network is shown in (i) below. Applying the absorption technique, move nh from the L-match inductor to the generator and move.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 6

We can cancel these elements by adding a series capacitance at the generator and a parallel inductance at the load. At the load, the.5 pf capacitor has a reactance j70.7 Ω at.5 GHz. An inductor L = 7.

7 Part (b) Note below we show how to resonate out the load-and generator inductances for the lowpass 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.5 pf capacitor has a reactance j70.7 Ω at.5 GHz. An inductor L = 7.5 nh has a reactance j70.7 Ω at.5 GHz. Similarly, at the generator, a nh inductor at.5 GHz has a reactance j9.43 Ω which one can cancel with a series capacitor with value.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. 7

8 Question 5 Design a common-base amplifier using a 2N2222 BJT. Bias the BJT at I C = 2 ma, and use a power supply of V CC = 9 V. The amplifier will be used at 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 R C = 4.5 V (2 ma) = 2.25 K. For bias-point stability, set V RE = 0.5 V CC =.35 V. Then R E = V RE 2 ma = 675 Ω. Use 680 Ω, the closest standard value. The base voltage is 0.65 V higher than V E so that V B = 2 V. Choose the current that flows through R and R 2 to be 0.I C = 0.2 ma. This will ensure we can ignore the base current and makes the design of R and R 2 easier. R 2 = 2 V 0.2 ma = 0K, and R = (9 V 2 V) 0.2 ma = 35K. This completes the bias design. Assume that β = 200, then g m = 40I C = 80 ms, and r π = β g m = 2.5K. Next, ground the base for ac signals by adding a capacitor C B with reactance much less than r π R R 2 = 2K at MHz. Choose C B = 0. μf which has a reactance of.6 Ω at MHz. The CB amplifier s input is at the emitter and is coupled into the amplifier through C E. The input impedance of the CB amplifier is R i g m = 2.5 Ω. Pick C E such that its reactance is 20 times smaller than this = (2π)( 0 6 )(C E ) C E = μf Use 0.27 μf, the closest standard value. For the output coupling capacitor, use 0. μf and decouple the power supply with a 0. μf capacitor. The voltage gain is A v g m R C = (0.08)( ) = +80 = 45 db The power generated by the load and dissipated by R C are The power gain is P g = v g 2 R i = v g 2 g m, P L = v 2 L = (g mr C v g ) = R R C R C v 2 2 g g m C G = P L = R Cv 2 2 g g m P g g m v2 = g m R C = 80 = 22.5 db g 2 8

9 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 I C = ma, and V CE = 5 V. Assume T = 25 and take V bi = 0.75 V for both pn junctions. Find β 0, r ο, I s, r ο, C μ, f T, C π, r π, and f β. 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 β 0 h FE = 390, V BE = 0.69 V, and r ο = 80K, at V CE = 5 V, I C = ma. V BE I C = I S [e V T ] 0 3 = I S [e ] I S = 2.98 fa From the datasheet C CB C μ = 0.46 pf at V CB = 3 V. From this we calculate C CB0 the C-B junction capacitance at zero bias voltage: C CB = C CB0 ( + V MJC R V ) bi 0.46 pf = C CB0 ( C CB0 =.03 pf 0.75 ) Since we don t have information on MJC, we assumed MJC = 0.5. At V CE = 5 V, V CB = = 4.3 V, so that the C-B junction capacitance at V CE = 5 V is C μ = C CB =.03 pf ( C μ = C CB = pf 0.75 ) For a quick estimate, which is probably good enough for most cases: C CB2 C CB V R V R2 = = pf From the data sheet f T = 280 MHz at V CE = 5 and I C = ma. Thus C π = g m ω T C μ = (40)( 0 3 ) (2π)( ) = pf pf = 22.3 pf r π = β = g m (40)( 0 3 ) = 9.75K The beta-cutoff frequency is f β = 2π r π (C π + C μ ) = 2π Alternatively, f β = f T β 0 = = 78 khz ( )( = 78 khz ) 9

Homework Assignment 12

Homework Assignment 12 Question 1 Shown the is Bode plot of the magnitude of the gain transfer function of a constant GBP amplifier. By how much will the amplifier delay a sine wave with the following