Radio Frequency Electronics

Transcription

1 Radio Frequency Electronics Frederick Emmons Terman Transformers Masters degree from Stanford and Ph.D. from MIT Later a professor at Stanford His students include William Hewlett and David Packard Wrote a very influential book Electronic and Radio Engineering, that is still referenced 60 years later During WWI, directed the Radio Research Lab at Harvard, directing a staff of more than 850 Became Dean of Engineering at Stanford Founding member of the National Academy of Engineering Dies in 1982 Image from Wikipedia 1

2 RC Parallel Series RC Summary Derived in class, make sure you can derive the corresponding series parallel transformation 2

3 LC Parallel Series LC Summary Derived in class, make sure you can derive the corresponding series parallel transformation 3

4 Summary of LC, RC Transformations 4

5 Example Example. Compute the Q of a 50 nh inductor with series resistance R s = 10 Ω at 100 MHz. Transform the circuit to an equivalent RL network. Solution. The Q of the inductor (and the network) is Q s = ωl = 2π R s 10 = 3.14 Since Q s is low, we will not use the approximation R p Q s 2 R s. Rather R p = 1 + Q s 2 R s = = 1,08.7 Ω L p = L s 1 + Q s 2 Q s 2 = = 55.1 nh Note how the reactive element transformed the100 Ω resistor to a Ω resistor. 5

6 Parallel LC circuit with Series Loss (C) Similar to the case of a parallel LC circuit with loss in series with L, for the case where the loss is in series with C, one can transform the circuit to a parallel RLC circuit. In this case, one will use a series parallel RC network transformation. Loss in series with C Equivalent RLC Transformation used Start with computing Q s Q s = 1 ωr s C s Then use Q s to compute C p and R p C p = C s Q s Q s 2 R p = R s 1 + Q s 2 6

7 Parallel LC with loss in both L and C 7

8 Impedance Matching In RF work, the input- and output impedances of the various building blocks are often a very important consideration. Receiver Antenna Amplifier Transmitter Antenna Z i Z o Z i Z o Very often the goal is to maximize the power transfer from one stage to another, and RF engineers concern them more with power gain than with voltage- or current gain. Z g V g P g Z i Z o Z L P L 8

9 Impedance Matching For maximum power transfer from a generator to a load, the load s input impedance must be the complex conjugate of the generator. Z g V g P g Z i Z o Z L P L For the block diagram above, to maximize P L P g, we should have Z g = Z i Z o = Z L To accomplish this, we insert impedance-matching blocks between stages Z g V g P g Z i Z o Z L P L Impedance matching network makes Z i = Z g Impedance matching network makes Z o = Z L 9

10 Impedance Matching There are several method for impedance matching. If the load is purely resistive, then a traditional transformer can be used. This technique is popular in audio applications. One can model a moving-coil speakers as a small-value resistor (4 Ω, 8Ω). The output stage of the amplifier may want to see a large resistor. A transformer with a n: 1 turn ratio will transform the load resistance by a factor n 2. V cc n:1 Turns ratio 10 mm Dot convention R L Resistive Load Output transistor of power amplifier Small audio transformer A typical requirement would be transform an 8 Ω load to 600 Ω. This would require a turns ratio of = 8.66 or 8.66: 1. 10

11 Transformers for Impedance Matching Transformer Transformer Auto-transformer or tapped inductor High-pass L High-pass L Low-pass L Low-pass L Resonant capacitive transformer Capacitive transformer π Network Sections of transmission lines 11

12 RF Transformers RF transformers are often tuned. That is, a capacitor is added that will resonate with the inductor. The were once widely-used in radios as part of the intermediate frequency (IF) amplifiers. Companies manufactured IF transformers and house them in small metal enclosures, which leads to the term IF can. Note the tapped inductor, which provides additional flexibility for impedance transformations Often these transformers have a ferrite slug that one can screw in or out to change the coupling between the windings Aluminum can provides electrical shielding as well as mechanical stability. 12

13 RF Transformers The intermediate frequency for AM radio normally 455 khz and companies developed a range of transformers for the frequency. Color coding indicate different transfer properties (transformer ratio, etc.) Windings Tuning slug IF Transformer Use of such transformers waning fast 13

14 Capacitive Voltage Divider v o s = v i s 1 sc 2 1 sc sc 1 = v i s C 1 C 1 + C 2 v o s = v i s n where n = C 1 C 1 + C 2 In instances where there is a load in parallel with C 2 and the load is light (i.e., high- Q), then we can approximate the voltage division ratio v o s v i s n where n = C 1 C 1 + C 2 v i s v o s /n where n = C 1 C 1 + C 2 14

15 The Capacitive Transformer A very useful sub circuit is a capacitive transformer. It consist of a capacitive voltage divider with a load across one of the capacitors as shown in (a) below. Assuming Q is large, then one can transform the circuit using successive RC transformations to the circuit in (c). 15

16 The Capacitive Transformer On the previous slide we showed the following equivalency v i v 0 The equations suggest we can model the network as a transformer v i v o /n v i v i v 0 v 0 16

17 The Capacitive Transformer We need to quantify by what we mean by large Q. Define Q E as the Q of the RC network which results network with the input shorted so that R sees C 1 and C 2 in parallel. That is, Q E = ωr C 1 + C 2. The network Q (i.e., Q T ) of the equivalent parallel network is Q T = ωr p C p. One can show that if nq T Q E > 20 then the approximation is within 5% and if nq T Q E > 100, the approximation very good. Model is good when nq T Q E > 20 Q E = ωr C 1 + C 2 Model is very good when nq T Q E >

18 The Capacitive Transformer Example. Consider the capacitive voltage divider below and determine the parameters for the equivalent network shown in (b). Check the model assumptions at f = 100 khz. Solution (a) (b) n = C 1 C 1 + C 2 = = 2 3 R p = 1K n 2 = 2.25K C p = C 1C 2 C 1 + C 2 = = 3.33 nf Check assumptions Q E = ω C 1 + C 2 R L = 2π = 9.42 Q T = ωc p R P = 2π = 4.7 nq T Q E = = 29.6 > 20 Since nq T Q E > 20 the model is good, and R p is within 5% 18

19 Example. For the circuit below, determine L, C 1, and C 1 so that the circuit resonates at f 0 = 16 MHz with B = 1.06 MHz and achieves maximum power transfer to R L. L, C 1, C 2 are lossless. Solution We will assume that nq E Q T is large and replace the capacitive voltage divider with the ideal transformer model. At resonance the circuit is purely resistance. For maximum power transfer we want R s to equal the transformed R L. This determines the turns ratio (n) of our model: n = R L R s = 1K 9K = 1 3 = C 1 C 1 + C 2 The total input resistance the source sees is 9K 2 = 4.5K. Further, B = 1 RLC circuit, from with it follows that C = 1 B R and RC for a parallel C = C 1C 2 C 1 + C 2 = 1 (B)(R) = 1 2π = pf Since C 1 C 1 + C 2 = 1 3 we can solve for C 1 = 50 pf and C 2 = 100 pf. Further, since ω 0 = 1 LC, we can solve for L = 1 ω 2 0 C = 30 μh. Self-study: check that the model assumptions are valid. 19

20 Example. Design a resonant LC network so that BW is 10 MHz around 100 MHz. Both the sourceand load resistances are 1K. Assume inductors with Q = 85 and lossless capacitors are available. (a) Solution. The network in (a) transforms to the network in (b) above. The Q T for (b) must be Q T = (100 MHz) 10 MHz = 10. The Qs in the circuit is high, so we will use simplified formulas: R p = Q s 2 R s = 85 2 R s and L p = L s. Further, ω 0 L = 85R s. For the network in (b) we have Q T = 10 = R ω 0 L R = 10 ω 0L Here R is the parallel combination of the source- and load resistances, and R p, namely (b) R = 10 ω 0 L = 500 R p R p 10 85R s = R s R s R s = Ω From ω 0 L = 85R s it follows that L = nh, and from ω 0 = 1 LC it follows that C = 36 pf 20

21 More Transformers Capacitive transformer Inductive transformer Autotransformer Normal transformer 21

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