# Since any real component also has loss due to the resistive component, the average power dissipated is 2 2R

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2 Quality factor, Q Reactive components such as capacitors and inductors are often described with a figure of merit called Q. While it can be defined in many ways, it s most fundamental description is: Q energy stored = ω average power dissipated Thus, it is a measure of the ratio of stored vs. lost energy per unit time. Note that this definition does not specify what type of system is required. Thus, it is quite general. Recall that an ideal reactive component (capacitor or inductor) stores energy E = 1 2 CV 2 pk or 1 2 I 2 pk Since any real component also has loss due to the resistive component, the average power dissipated is 2 1 V 2 pk Pavg = I pk R = 2 2R If we consider an example of a series resonant circuit. R C At resonance, the reactances cancel out leaving just a peak voltage, Vpk, across the loss resistance, R. Thus, Ipk = Vpk/R is the maximum current which passes through all elements. Then, Q = ω O I I 2 pk 2 pk R 2 2 ωo = R 1 = ω RC O In terms of the series equivalent network for a capacitor shown above, its Q is given by: 1 Q = = ω RC X R

3 where we pretend that the capacitor is resonated with an ideal inductor at frequency ω. X is the capacitive reactance, and R is the series resistance. Since this Q refers only to the capacitor itself, in isolation from the rest of the circuit, it is called unloaded Q or Q U. The higher the unloaded Q, the lower the loss. Notice that the Q decreases with frequency. The unloaded Q of an inductor is given by Q U ωo = R where R is a series resistance as described above. Note that Q is proportional to frequency for an inductor. The Q of an inductor will depend upon the wire diameter, core material (air, powdered iron, ferrite) and whether or not it is in a shielded metal can. It is easy to show that for a parallel resonant circuit, the Q is given by susceptance/conductance: Q = B G where B is the susceptance of the capacitor or inductor and G is the shunt conductance. oaded Q. When a resonant circuit is connected to the outside world, its total losses (let s call them R P or G P ) are combined with the source and load resistances, R S and R. For example, C R P R S R Here is a parallel resonant circuit (C, and R P )connected to the outside. The total Q of this circuit is called the loaded Q or Q and is given by Q = ωo C ( RP RS R ) or

4 Q ωoc ωoc suscep tan ce = = = G + G + G G conduc tan ce S P total The significance of this is that Q can be used to predict the bandwidth of a resonant circuit. We can see that higher Q leads to narrower bandwidth. o BW = ω Q where ω = o 1 C 1. So, large C will increase the loaded Q at a given resonant frequency and reduce bandwidth. 2. Or, we could vary G total. How? A. Unloaded Q (Q U ) is an attribute of the passive and C components and affects G P. This will vary if we change and C. And, G P = G C + G ind, ωoc Capacitor: QU ( capacitor) = G C Inductor: 1 QU ( inductor ) = ω G o ind Using this to set the bandwidth of a resonator is not a good idea. If Q U becomes comparable to Q, then the loss in the resonator becomes very high as we shall see.

5 Insertion oss The resonator can be used as a bandpass filter or matching network. In these applications, insertion loss can be important. G S Vout V S C Gp G Consider this resonant CR circuit. When w = wo, Vout ( jω o) G = S < 1 V G + G + G S S P Typically, G S = G = G. We need to find S 21 to determine insertion loss. Recall that S 21 2 = transducer gain = Power delivered to the load/available power from the source when the source and load are both Zo. First note that Q GP = Q G + 2G U P And, S 21 2Vout Q 2G = = 1 = V Q G + 2G S U P So, insertion loss (db): Q I = 20log 1 Q U Thus, narrow bandwidth where Q and Q U are similar can be very lossy.

6 Here is an example of a simple resonant circuit. The unloaded Q is infinite, since no losses are included in the network. We see that there is no insertion loss in this case. oaded Q varies from 1 to 10 with the given parameter sweep. Q=1 Q=10

7 Now, the circuit is modified to include a 500 ohm resistor (R P ) in parallel with the C network. This resistance represents the parallel equivalent loss due to both the and the C. So, now we have a finite unloaded Q. Note that the insertion loss increases as loaded Q, Q, approaches Q U. Sweeping RS, we see at resonance, the reactances cancel, and we are left with a resistive divider. Vout = Vin [R 1 /(2R 1 +R S )]. R S =200 R S =2000 S21 (db) = 20 log(2vout/vin)

8 So, we can set the Q and bandwidth by adjusting the loading conductances/resistances. BW ωo = Q But, we must make sure that Q U >> Q to avoid excessive losses.

9 So, if really narrow bandwidth is required, the solution generally requires multiple resonators or more complicated bandpass filter approaches or mechanical structures like quartz crystal filters. 3. How can you set G S and G? Aren t these dictated by the generator and load? Not necessarily! We can use tapped C matching circuits to transform source and load impedances to whatever we desire (assuming we don t increase loss too much by approaching Q U ). But, first, we will introduce a convenient approximation which can be used to transform a parallel equivalent circuit into a series equivalent circuit and vice-versa.

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15 Example. Design bandpass matching network to transform 50 ohms to the input of an SA602 mixer. O The mixer input impedance is given on the data sheet as Z IN = 1.5k 3 pf for the SA602 IC. Choose a convenient frequency for this exercise: 159 MHz = 1 x 10 9 rad/sec. Use the tapped C procedure to transform 50 ohms to 1500 ohms. Include the 3 pf into the equivalent series capacitance of C1 and C2. The equivalent parallel resistance, R P, due to the unloaded Q of the components should also be taken into account in calculating the loaded Q and the transformation ratio needed to match the source and load ohms RS = 50 C ohms 3 pf Gp C 2 We need to specify a bandwidth to begin the process. et s choose 10 MHz. 159 Q = = Q U is generally limited by the inductor. The manufacturer s data generally specifies an unloaded Q at a certain frequency. For a T powdered iron core material, Q U of 120

16 is typical at this frequency. The external capacitors normally have a higher unloaded Q which can be neglected. Our first design equations: Q 1 QU = 120 ωog = P B 1/ ωo = = = 16 G G + G + G total S P Assume that G S = G = 1/1.5k = 6.7 x That is, design the tapped C network to provide a transformed impedance of 1.5k. This will get close to the solution needed. So, now we have two equations with 2 unknowns. We can solve for and for G P. G P = 2.04 x 10-4 S or R P = 4900 ohms. B = Q U G P = 2.44 x 10-2 S 1 = 41nH ω B = o This is a rather small value and will require some care in layout to implement accurately on a PC board. Q Check insertion loss: I= 20 log 1 = 1.2 db QU If we required less loss, then a wider BW or higher Q U is necessary.

17 Next, design the tapped C network to give 1.5k R P = 1150 ohms ohms RS = 50 C 1 3pF C 2 Because is known, we can calculate the required C total to resonate at 159 MHz. C total 1 = = 24 pf ω 2 o Now, deduct the 3 pf from C total so that is is absorbed into the resonator. Design C1 and C2 for 21, not 24 pf. The transformation ratio relates C 1 and C 2. It should transform 50 ohm source into the load impedance, 1500 RP = 1150 ohms. C1+ C n = = = 4.8 C 50 1 The series combination of C 1 and C 2 CC pf C + C = = equations; 2 unknowns. Solve for C 1 and C 2. C 1 = 27 pf; C 2 = 103 pf.

18 Check the result with ADS: A small signal AC simulation is performed which includes the RP and excess C of the load. If the load is matched correctly to the source, we should see half of the source voltage at node Vin, 0.5V (available power). Vout should be 4.8 x 0.5 = 2.4V. Everything looks as expected except for the bandwidth, 11.4 MHz rather than 10 MHz. This is due to the effect of R P. If it mattered, we could go back through again and include this effect and get a 10 MHz BW, but most applications as this one are not so critical. The important thing is to get the correct impedance transformation. Of course, a network, PI or T network could also have been used here with somewhat less flexibility in choosing loaded Q.

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20 Example of resonator used as bandpass filter: 50 ohms to 50 ohms. 100 MHz. The tapped C transforms up to 1000 ohms for this example. You can work through the design similar to the earlier example. Tapped Capacitor Resonator

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23 Note the improvement in the stopband rejection provided by the coupled resonator compared with the single resonant tapped C circuit.

24 Temperature Compensation of Resonant Circuits Oscillators are frequently used to set the transmit or receive frequency in a communication system. While many applications use a phase locked loop technique to correct for frequency drift, it is good practice to build oscillators with some attempt to minimize such drift by selecting appropriate components. Inductors and capacitors often drift in value with temperature. Permeability of core materials or thermal expansion of wire causes inductance drift. Variations in dielectric constant with temperature in capacitors is the main source of drift for these components. Temperature drift is expressed as a temperature coefficient in ppm/ o C or %/temp range. Capacitors The 3 most common types of dielectrics for RF capacitors are: Dielectric type Temp coefficient (TC) Temp range C0G (or NP0) +/- 30 ppm/ o C -55 to +125C X7R (BX) (+15% to -15%) -55 to +125 Z5U (+22% to -56%) 10 to 85 Clearly, the Z5U is not much good for a tuned circuit and should be used for bypass and AC coupling (DC block) applications where the value is not extremely critical. At lower radio frequencies, polystyrene capacitors can be used. These have a 150 ppm/c TC. The C0G and X7R can be used in tuned circuits if their values are selected to compensate for the inductor drift. CK05 BX 330K 330K NP0 The two leaded capacitors above illustrate the labels found on typical capacitors of the X7R and NP0 types. The value is given by the numerals: 330. In this example, this is 33 pf. It goes 1 st significant digit (3), 2 nd significant digit (3), and multiplier (10 0 ). The letter K is the tolerance, which is +/- 10%. As always, the parasitic inductance and self resonance of any capacitor must be considered for RF applications.

25 Inductors There are many types of inductor core materials which are intended for different frequency ranges, permeability, and TC. Powdered Iron and Ferrites are the two categories of these materials. For example, the material you will have available for the VCO lab is powdered iron, Type 12 (green/white). This is useful from 50 to 200 MHz and gives Qu in the range. µ/µ 0 = 4. Manufacturer s data sheets can be found on the web that specify TCs for the many materials. This one has a weird TC vs temperature behavior, but we are mainly interested in the 25 to 50C range for this project. Temperature range TC 25 50C +50 ppm/c So, how can you compensate for component drift? NP0 X7R C 1 C 2 The equation below shows how the TCs of individual components combine 1. Suppose that the inductor was resonated with a drift free capacitor (NP0). The frequency drift will be 25 ppm/c. If the design frequency is 100 MHz, this corresponds to a drift of 2.5 khz/c. But, the equation shows that you can set the total frequency TC (TCF) of a circuit to zero by combining capacitors with different TCs. f 1 C1 C2 TCF = = TC + TCC1 + TCC2 f0 2 CTOTA CTOTA Thus, if the inductor has a positive TC, you can correct for temperature drift with the right combination of non drift and drifty capacitors. In this case, we want the total capacitance of C1 and C2 to have a net TC of 50 ppm/c. The best oscillators will be designed with components with low intrinsic TCs so that you do not have to compensate them with different components having large and possibly unreliable TCs. 1 W. Hayward, R. Campbell, and B. arkin, Experimental Methods and RF Design, ARR Press, 2003.

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