BIOEN 302, Section 3: AC electronics

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1 BIOEN 3, Section 3: AC electronics For this section, you will need to have very present the basics of complex number calculus (see Section for a brief overview) and EE5 s section on phasors.. Representation of sinusoids in the complex plane For a magnitude (say, a voltage) W(t) varying sinusoidally with time, we can write: W(t) M cos(t + θ) hence W(t) Re [M e j(t + θ) ] where M e j(t + θ) W(t) is the complex-plane representation of W(t). W(t) can be thought of as a rotating vector the real part of which represents a sinusoid.. Representation of sinusoids as phasors The phasor representation of a time-varying sinusoid v(t) m cos(t + θ) is exactly the same as its complex representation, most often used in its exponential or polar form (for convenience). The nomenclature is slightly different: j( t m e θ + θ) m where m is the maximum value of the sinusoid, t is suppressed and θ is the phase angle of the sinusoid. Thus, the phasor is a transformed version of a sinusoidal voltage or current waveform and consists of the magnitude and phase angle information of the sinusoid. Equivalently, v(t) is then referred to as the representation of the phasor in the time domain: Phase lead is expressed by a positive angle θ in the phasor notation, while phase lag is expressed by a negative angle θ in the phasor notation. See the diagrams below. v(t) m cos ( t + θ) ( ) [ ] j t+θ v(t) Re m e Important: As we shall see, the phasor concept may be used ONLY when: ) the circuit is linear, ) the steady-state response is sought 3) all independent sources are sinusoidal and have the same frequency. BIOEN 3, 3: AC electronics Instructor: Albert Folch

2 BIOEN 3, Section 3: AC electronics 3. Operations with phasors a) Addition of phasors: just like addition of vectors! (Please do NOT remember by ear!) b) Multiplication of phasors: just like exponentials (NOT vectors)! c) Division of phasors: just like exponentials (NOT vectors)! Exercise : θ arctan θ θ θ Re[ ] + Re[ ] + j{ Im[ ] + Im[ ]} cos + + ( + ) θ cosθ j sin θ sin θ ( cosθ + θ ) + ( θ + θ ) cos sin sin sinθ + sinθ cosθ + cos θ [ ] θ ( θ + θ ) ( θ θ ) θ Given 3 and, give + and / Answer: / -9 Example : Reduce the equation v(t) cos(t) + 5 sin(t + 6 o ) + 5 cos(t + 9 o ) to an equation of the form v(t) m cos(t + θ) This operation is greatly simplified by using complex (phasor) notation. Converting all sinusoids to cosines, we get v(t) cos(t) + 5 cos(t + 6 o 9 o ) + 5 cos(t + 9 o ) All cosines are in turn the real part of complex numbers: v(t) Re[e jt ] + 5 Re[e j(t 3o) ] + 5 Re[e j(t+9o) ] BIOEN 3, 3: AC electronics Instructor: Albert Folch

3 BIOEN 3, Section 3: AC electronics v(t) Re[( + 5 e j3o + 5 e j9o ) e jt ] In polar form, we have v(t) Re[( o o o ) e jt ] Now, we can combine the complex numbers as o o o j + 5j j o 4.54 e j9.9o Thus we obtain v(t) Re[(4.54 e j9.9o ) e jt ] Re [4.54 e j(t+9.9o) ] 4.54 cos(t o ) Answer: m 4.54 and θ 9.9 o Example 3: Exercise 4: Reduce the following expressions using phasors. BIOEN 3, 3: AC electronics 3 Instructor: Albert Folch

4 BIOEN 3, Section 3: AC electronics Exercise 5. Answer: Exercise 6. Exercise 7. Summarizing, when performing operations with phasors, rather than remembering the formulas, think of phasors as if they were arrows described by a length and an angle when you are multiplying or dividing, and think of them as arrows with two rectangular coordinates when you are adding or subtracting it s as easy as that! 4. Complex impedance a) Inductance Suppose the sinusoidal current going through an inductance is given by i L (t) I max sin(t+θ) ; in other words, i L (t) I max cos(t+θ-π/) In complex notation: i L (t) Re[I max e j(t+θ-π/) ] BIOEN 3, 3: AC electronics 4 Instructor: Albert Folch

5 BIOEN 3, Section 3: AC electronics The voltage across an inductance is given by v L (t) L di L (t)/dt, hence v L (t) Re[jLI max e j(t+θ-π/) ] Re[LI max e jπ/ e j(t+θ-π/) ] LI max Re[e j(t+θ) ] or v L (t) LI max cos(t+θ) In phasor notation, we have that the phasors for current and voltage are I L I m (θ 9 o ) L LI m θ [Discuss: is this equivalent to writing θ+9 o for L and θ for I L?] It is the current, and not the voltage, that carries an extra 9 o when put into phasor notation, because otherwise it does not satisfy i L (t) Re [I L ]. We subtract 9 o to keep the sign of the real part constant. L (L 9 o ) (I m (θ 9 o )) (L 9 o ) I L L jl I L Remember this!!!! We refer to the term (L 9 o ) jl Z L as the impedance of the inductance. Summarizing, Note: Current lags voltage by 9 o in a pure inductance, as illustrated below BIOEN 3, 3: AC electronics 5 Instructor: Albert Folch

6 BIOEN 3, Section 3: AC electronics b) Capacitance Suppose the voltage across a capacitor is given by v C (t) max sin(t+θ) ; in other words, v C (t) max sin(t+θ-9 o ) In complex notation: v C (t) Re[ max e j(t+θ-π/) ] The current charging this capacitor is given by i C (t) C dv C (t)/dt, hence i C (t) Re[jC max e j(t+θ-π/) ] C max Re [e jπ/ e j(t+θ-π/) ] C max Re[e j(t+θ) ] or i C (t) C max cos(t+θ) In phasor notation, we have that the phasors for current and voltage are C max (θ 9 o ) I C C max θ We refer to the term (/(C)) -9 o /jc Z L as the impedance of the capacitance. C I C / jc I C jc C Remember this!!!! Z C /(jc) /(C) -9 o Summarizing: Note: Current leads voltage by 9 o in a pure capacitance: BIOEN 3, 3: AC electronics 6 Instructor: Albert Folch

7 BIOEN 3, Section 3: AC electronics c) Resistance v(t) R i(t) R R I R I R R / R Z R R o You already knew this Summarizing: Note: R and I R are in phase in a pure resistance: BIOEN 3, 3: AC electronics 7 Instructor: Albert Folch

8 BIOEN 3, Section 3: AC electronics Overall summary: ) Imprint in your memory that Z C /jc and Z L jl because you will be using it a lot. ) Appreciate the simplification provided by the phasor notation: Element Differential equation Phasor notation Impedance L v L di/dt jl I jl C i Cdv/dt I jc /jc In other words, except for the fact that we use complex arithmetic, sinusoidal steady-state analysis of RLC circuits is virtually the same as the analysis of resistive circuits no RLC differential equations! (We pay a price: the phasor notation applies only to steady-state sinusoidal signals.) 3) Rules of conversion from one notation to another: W A+jB W W θ where W (A +B ) ½ and θ arctan (B/A) W W θ W A+jB, where A Wcosθ and B Wsinθ Exercise 8: Exercise 9: BIOEN 3, 3: AC electronics 8 Instructor: Albert Folch

9 BIOEN 3, Section 3: AC electronics Exercise : Exercise : BIOEN 3, 3: AC electronics 9 Instructor: Albert Folch

10 BIOEN 3, Section 3: AC electronics 5. Circuits with multiple impedances We now know: ) The expressions for the impedances (i.e. phasor notations) of R, L, and C ) The rules for adding R, L, and C in series vs. parallel in non-phasor notation What about the rules for adding impedances? Importantly, impedances add just like resistors (albeit in the complex plane): in series, they sum to yield the total impedance; in parallel, their inverses (also called admittances, denoted by Y) add to yield the inverse of the total impedance. Note: In Z R + jx, R is called resistance and X reactance (both in ohms) In Y /Z G + jb, G is called conductance and B susceptance (both in siemens) Important: G /R!!! It is straightforward to see that both Kirchhoff s laws hold for phasors. Example : A series RLC circuit The exercise will consist of finding the currents and voltages given the following circuit: The given expression for the source voltage v s (t) tells us that the peak voltage is, the angular frequency is 5, and the phase angle is 3 o. The phasor for the voltage source is thus Important: a disadvantage of the polar notation is that the angular frequency is implicit it s up to you to remember its value! The complex impedances of the inductance and the capacitance are, respectively, The circuit in phasor/complex notation is shown below: BIOEN 3, 3: AC electronics Instructor: Albert Folch

11 BIOEN 3, Section 3: AC electronics Note that all three elements are in series, therefore we can find the equivalent impedance of the circuit by adding the impedances of the elements: Z eq R + Z L + Z c. Substituting values, we obtain Z eq + 5j 5j + j. Converting to polar form, Now we can find the phasor current by dividing the phasor voltage by the equivalent impedance: As a function of time, the current is: Therefore, we can express the phasors for the voltage drops across R, L and C as: As a function of time, the voltages are: v R (t) 7.7 cos (5t 5 o ) v L (t) 6. cos (5t + 75 o ) v C (t) 35.4 cos (5t 5 o ) Exercise 3: Consider the circuit below: BIOEN 3, 3: AC electronics Instructor: Albert Folch

12 BIOEN 3, Section 3: AC electronics i (t) i (t) i 3 (t) Find the values and units of s, Z R, Z C, Z R, Z L, and Z L. Find an expression for the total impedance, for v o (t), for the currents through each branch of the circuit (call them i (t), i (t), and i 3 (t)), and for the voltages across each circuit element (call them v R (t), v C (t), v L (t), and v L (t)). Exercise 4. Suppose that the input signal for the circuit shown below is Exercise 5. BIOEN 3, 3: AC electronics Instructor: Albert Folch

13 BIOEN 3, Section 3: AC electronics Exercise 6. Exercise 7. BIOEN 3, 3: AC electronics 3 Instructor: Albert Folch

14 BIOEN 3, Section 3: AC electronics Exercise 8. Exercise 9. BIOEN 3, 3: AC electronics 4 Instructor: Albert Folch

15 BIOEN 3, Section 3: AC electronics 6. Low-pass and high-pass filters Until now we have only talked generically about how we can analyze circuits with a pure (single frequency) sinusoidal input so we didn t have to worry about the response of the system to other frequencies. Let s take a closer look at the equations for impedance and try to think about what will happen when the frequency of the input is varied. L Consider the following simple circuit: I R Note that the current I and the voltage are related as follows: I I Z [ R + ( L) ] Importantly, the function of the circuit depends on what part of the circuit we consider the output : If the output o R IR, then o varies with in the same way as I does. R R R + jl R R [ R + ( L) ] + arctan jl ( L) L R arctan L R If you plot the magnitude of the voltage gain, G v o /, you will see that this is a low-pass filter: As you can see, the response is far from ideal. BIOEN 3, 3: AC electronics 5 Instructor: Albert Folch

16 BIOEN 3, Section 3: AC electronics An approximate idea of what the filter response is like can be obtained by looking at the limits when and : lim I R What happens when R/L? lim I 9 I 45 R The frequency R/L is therefore referred to as the half-power frequency (or also break frequency for reasons we will see later). If the output o L IjR, then I lags o by 9 o. L jl R + jl or also L R j L [ R + ( L) ] L + 9 R L o arctan arctan L R R L If you plot the magnitude, you will see that this is a high-pass filter: Exercise. Repeat the previous exercise but with an RC circuit instead of RL. Consider also both cases, o C (left) and o R (left). BIOEN 3, 3: AC electronics 6 Instructor: Albert Folch

17 BIOEN 3, Section 3: AC electronics What type of filter are these? Sketch a plot for the voltage gain G v o / in both cases. 7. The Transfer Function We have started talking about voltage gain, G v output / input. The voltage gain is just one of several transfer functions, which are defined as the ratio of the output phasor (of interest) to the input phasor (defined by the source). They are expressed usually as a function of frequency, H(f), where f /π is measured in Hertz. Let s find the transfer function of the following circuit: Z T R + j C The phasor current is the input voltage in divided by the complex impedance Z T of the circuit: Since the output is defined across the capacitor, the phasor for the output voltage is the product of the phasor current and the impedance of the capacitance, /jc: Substituting for I, we have The transfer function is thus: If is useful to define the parameter f B, Then, the transfer function can be written as: We see that the transfer function is a complex quantity having a magnitude and a phase: BIOEN 3, 3: AC electronics 7 Instructor: Albert Folch

18 BIOEN 3, Section 3: AC electronics Plots of the magnitude and phase are shown below: This is a low-pass filter. Also, f B is the half-power frequency and the frequency at which the output lag is half of 9 o. It is important to note that the low-pass filter built with R and L ( L as the output, see above) would have the same transfer function, even though the definition of f B would be different. This makes the concept of f B all the more useful it characterizes the filter, regardless of how it is physically built. The same applies to the two high-pass filters: ) series RL circuit (output taken across L), and ) series RC circuit (output taken across R). 8. Bode Plots Most of the times, it is easier to visualize the ranges of H(f) and of f in a logarithmic scale, in what is called a Bode plot. In a Bode plot, we choose to represent log H(f) as a function of log instead of G() as a function of. In this representation, a change in db (decibels) represents a change in an order of magnitude in H(f). A change in represents 3 db. Let s see an example, taking again the transfer function of the low-pass filter: In decibels, Thus, Using the properties of the logarithm, we obtain BIOEN 3, 3: AC electronics 8 Instructor: Albert Folch

19 BIOEN 3, Section 3: AC electronics Since the logarithm of is zero and the logarithm of x ½ is ½ log(x), The magnitude Bode plot for the first-order low-pass filter thus looks like: Note the following: For very small f, f << f B, H(f) db (a horizontal line) For very large f, f >> f B, H(f) db log (f / f B ) (a straight line in the log(f) scale that intersects the frequency axis at f B ) The high- and low-frequency asymptotes appear to intersect (or break ) at f B, hence the name break frequency. A table of useful values for the f >> f B approximation is shown below: Often in the context of decibels and Bode plots, you will hear the terms decade (a frequency range spanning one order of magnitude, i.e. a factor of ) and octave (spanning a factor of ). In the logarithmic scale: BIOEN 3, 3: AC electronics 9 Instructor: Albert Folch

20 BIOEN 3, Section 3: AC electronics An important feature of the Bode plots and their decibel representation is that a product of two transfer functions appears as a sum of the two: This is important when one tries to figure out the transfer function corresponding to a given, more complex Bode plot with multiple breaks. For example, a band-pass filter can be straightforwardly seen as the addition of the Bode plots of the high-pass and the lowpass filters, but in reality the two filters can be assembled in two stages (e.g. the input of the high-pass stage is the output of the low-pass stage, so the total transfer function is the product of the transfer functions of each stage). Exercise. Exercise. Exercise 3. Exercise 4. BIOEN 3, 3: AC electronics Instructor: Albert Folch

21 BIOEN 3, Section 3: AC electronics Example 5. Example 6. BIOEN 3, 3: AC electronics Instructor: Albert Folch

22 BIOEN 3, Section 3: AC electronics BIOEN 3, 3: AC electronics Instructor: Albert Folch

23 BIOEN 3, Section 3: AC electronics Example 7. BIOEN 3, 3: AC electronics 3 Instructor: Albert Folch

24 BIOEN 3, Section 3: AC electronics 9. Resonance in RLC circuits Suppose we connect a series RLC circuit to a voltage supply as indicated. The impedance of the circuit is: BIOEN 3, 3: AC electronics 4 Instructor: Albert Folch

25 BIOEN 3, Section 3: AC electronics Note we can also write L Z R + j L R + L arctan C C R CR which has a minimum in magnitude when L /C, i.e. at (LC) ½. Most important, at this frequency the phase becomes zero. By definition, the frequency at which the output (either current or voltage, or both) is in phase ( in resonance ) with the input (a condition termed resonance ) is called the resonant frequency, which is often (but not always) characterized by yielding a maximal output. The circuits themselves are said to be in resonance at this frequency. Remember this value because it will come back again and again! For a parallel RLC circuit we would write the total admittance (i.e. the inverse of the total impedance) of the circuit as: Y(j ) + jc + R jl where Y(j) /Z(j). It is straightforward to see that a plot of the magnitude of Y(j), Y(j), for the parallel RLC circuit is essentially the same as the series-rlc Z(j) impedance plot, with a minimum at (LC) ½. Summarizing for the series as well as parallel RLC circuits: (LC) ½ At resonance (by definition) Z( ) R o (no imaginary component, i.e. no phase either) At low frequencies, Z is dominated by the capacitive term for the series circuit and by the inductive term for the parallel circuit At high frequencies, Z is dominated by the inductive term for the series circuit and by the capacitive term for the parallel circuit As a general rule (for any circuit), BIOEN 3, 3: AC electronics 5 Instructor: Albert Folch

26 BIOEN 3, Section 3: AC electronics Resonance conditions are defined as those that satisfy Im(Z) or Im(Y), not necessarily (LC) ½ At resonance current and voltage are by definition in phase because the impedance won t change their phase. (Recall the phase is introduced by the imaginary component.) Example of a more general case: consider a circuit of the type RL//C (R in series with L, C in parallel with them), then R L Y jc + + j C R + jl R + L R + L The frequency that satisfies Im(Y) is R LC L which, incidentally, is not a maximum or minimum of Y. In other words, in general the output is not necessarily maximum at resonance. Resonance is a term more related with the phase of the output than with its magnitude. Example 8. Consider the voltage drop across the resistor in a series RLC circuit: In phasor notation, the input voltage s divided by the total impedance Z gives us the current going through the circuit, I s /Z. The voltage drop across the resistor R will thus be o I R (R/Z) s. Hence we obtain: [Discuss: what is the expression for the voltage across the inductance? And across the capacitance?] Rearranging, and substituting values, The plots of the magnitude and phase of o look as follows. Note the resonance peak in o at low frequencies. BIOEN 3, 3: AC electronics 6 Instructor: Albert Folch

27 BIOEN 3, Section 3: AC electronics Because of their steep slopes, magnitude and phase plots are usually shown with frequency on a log axis: Let s take a closer look at the phases for each particular phasor in the circuit. In the case of the series RLC circuit, we consider and plot the phasors c, L and R versus I because I is what is conserved along the circuit. (Here represents s in the above circuit.) [Discuss the following plots.] Note what happens at resonance: the voltage across the resistor and the current through the resistor are in phase. That is precisely because Z( ) R, hence there is no change in phase when we compute I s /Z, and both Z L Z C at resonance. [Discuss the implications on energy storage.] BIOEN 3, 3: AC electronics 7 Instructor: Albert Folch

28 BIOEN 3, Section 3: AC electronics BIOEN 3, 3: AC electronics 8 Instructor: Albert Folch. Filters, Bandwith, and Quality factor Let s take a closer look at the expression for the gain in the series RLC circuit as defined above (i.e. output voltage is defined as the voltage across the resistor): Thus, RC R L Q with, jq s o + G (recall (LC) -½ ) Writing the phasor G in terms of magnitude and phase, G M() e jφ(), we have: Q ) M( + and φ Q tan ) ( An approximate log graph of these two functions is shown here: We define bandwidth (BW in the graph) as the difference between the two half-power frequencies or, equivalently (since power is the square of magnitude), as the range of frequencies that satisfy M() > /. Looking at the above relationship for M(), that condition is satisfied when Q < + which in turn is satisfied when s o jq C L j R R C L j R R

29 BIOEN 3, Section 3: AC electronics Q > or when Q < i.e. /Q > or else + /Q < Each equation has roots, but only one is positive; discarding these negative frequencies (no physical meaning), we only have one value per equation. We call the lower value LO and the larger one HI. In other words, M() > / in the interval LO < < HI, where + LO + and Q Q + HI + Q Q The difference between HI and LO gives us (by definition) the bandwidth: BW HI LO /Q i.e. BW R/L for the series RLC circuit. The latter is generally used as the general definition of Q, i.e. Q /BW. Hence circuits with high Q have narrow bandwidths. In addition, LO HI, i.e. the resonance frequency is the geometric mean of the two half-power frequencies. Bode plots of the transfer function vs. frequency for various values of Q are shown below. As we can see, the series RLC circuit (with output taken across the resistor) is a (second-order) bandpass filter. Exercise 9. Show that for a series RLC circuit the ratio of impedance at resonance, Z Z( ), to impedance at any radian frequency, Z, is: Z Y Z Y + jq L where Q BIOEN 3, 3: AC electronics 9 R Instructor: Albert Folch

30 BIOEN 3, Section 3: AC electronics Show also that for a parallel RLC circuit the expression is the same provided we invert the definition of Q: s R L C Z Z where Y Y Q + jq R L Summarizing: The expression of Q for a series RLC circuit is the inverse of the expression of Q for a parallel RLC circuit. For example, Q decreases with increasing R in the series circuit, but it increases with increasing R in the parallel circuit. Exercise 3. Exercise 3. Exercise 3.. Other filters using RLC series circuits We have only been considering the series RLC circuit with the output defined as the voltage across the resistor. What would happen if we chose the voltage across the capacitor C (or across the inductance L ) as the output of the circuit, such as depicted below? Output Capacitor BIOEN 3, 3: AC electronics 3 Instructor: Albert Folch

31 BIOEN 3, Section 3: AC electronics We have: I s /Z, L jli and C I /(jc) where Z R + jl + /(jc) NOTE that the outputs C and L can NEER be in phase with the input s, but the current at the output (L, C or R) CAN be in phase with s. In such cases, we still speak of resonance when the current at the output is in phase with the input voltage/current. In L or C, we can t have current and voltage in phase! Thus, C jc R + jl + jc s s LC + jcr Note that I I C is in phase with s for (LC) -½, but note also that at this frequency the phase of C is -9 o (with respect to s ). The transfer function H(f) out / in c / s, can be written as: with Q s L/R /( RC) and f /(π) (π) - (LC) -½ The magnitude of the transfer function is plotted below. This is a second-order low-pass filter circuit, similar to the RC ( first-order ) one, compared here side-by-side: BIOEN 3, 3: AC electronics 3 Instructor: Albert Folch

32 BIOEN 3, Section 3: AC electronics Output Inductor When the output of a series RLC circuit is defined as the voltage across the inductor, we have: L jl R + jl + jc s s R j LC L The magnitude of the transfer function is plotted below: where with Q s L/R /( RC) and f /(π) (π) - (LC) -½ Therefore, this is a (second-order) high-pass filter. Exercise 33. Suppose we need a filter that passes components higher in frequency than khz and rejects components lower than khz. As we know, the series-rcl second-order circuit configuration constitutes a high-pass filter when the output is taken across the inductor. Determine the values or R and C, keeping in mind that we are interested in a filter that has a transfer function that is approximately constant in the pass band (see plot above). Solution: C.57 µf, R 34. Ω. Now, let s see what happens at resonance at both L and C. At resonance, since Z( ) R, we have I( ) s /R and R s Does that mean there is no voltage drop across L or C? No!! We have L ( ) j L I( ) and ( ) I( )/(j C) BIOEN 3, 3: AC electronics 3 Instructor: Albert Folch

33 BIOEN 3, Section 3: AC electronics Substituting I( ), we get L ( ) j L s /R j Q s and C ( ) s /(j CR) j Q s In other words, at resonance the voltage across the capacitor and across the inductor have the same magnitude, the gain is Q, and they are out of phase by 8 o, i.e. they add to. However, Q s is not the maximum value of the magnitudes of L and C. For example, C jc R + jl + jc s s LC + jcr d C It is straightforward to show that the frequency that satisfies d is max LC R L Q Q At this frequency, we have c s max Q 4Q Although max, for large Q (Q>>), max and the gain at max is c / s Q. Exercise 34. Derive the transfer function for a series RLC circuit whose output is taken across L and C combined (a band-reject filter): BIOEN 3, 3: AC electronics 33 Instructor: Albert Folch

34 BIOEN 3, Section 3: AC electronics Note, once more, that our choice of output determines what type of filter we have: G v R R + j L C G v (see Exercise 35 just above). Resonance in the parallel RLC circuit From what you know of phasors and resonance in RLC series circuit (and some examples with parallel circuits above), you should be able to comfortably analyze the parallel RLC circuit. Let s take a quick look at the expression for the gain G out /I in in the parallel RLC circuit: It is straightforward to show that the magnitude of the gain is: Exercise 35. Find the total impedance Z s /I, the resonance frequency, and the currents through each element, i.e. I R, I C, and I L in the parallel RLC circuit below. Compare (in arrow plots) the magnitudes and phases of the resonance values I R ( ), I C ( ), and I L ( ). I I R I L I C s R L C BIOEN 3, 3: AC electronics 34 Instructor: Albert Folch

35 BIOEN 3, Section 3: AC electronics Hint: We can now plot the phasors I c, I L and I G versus s ( below) because s is what is conserved along the circuit. Similarly to what we did for the series RLC circuit, we have: Looking at the magnitude of the gain: We note first that the magnitude of the gain has a maximum at the resonance frequency, (LC) ½, and that at this frequency G max R. (Note that the same conclusion is reached by observing that by definition Z at resonance must have Im(Z), and when Z is maximum, out and G are maximum.) If we plotted this curve, we would see that it also has one peak (it looks very similar to the one for the series RLC circuit), and that therefore the parallel RLC circuit acts as a bandpass filter. To find the bandwidth of the circuit, we find the HI and LO that satisfy that the magnitude of the gain is reduced by (i.e. the power is reduced by ): Solving the equation and taking only the positive values of we obtain: Thus the bandwidth (recall that in the series RLC circuit BW R/L) is: BW HI LO /RC and the quality factor Q is: Q /BW RC/(LC) ½ R (L/C) ½ BIOEN 3, 3: AC electronics 35 Instructor: Albert Folch

36 BIOEN 3, Section 3: AC electronics which we already knew from our previous study of the parallel RLC circuit. [To visualize the meaning of the Q factor, recall from your homework the values for the currents through the capacitor and inductance.] Using this value of Q, we can re-write the expressions for HI and LO as: + LO + and Q Q + HI + Q Q which are the same expressions as the ones for the series RLC circuit! Hence the usefulness of the Q factor concept. Exercise 36. Consider the parallel RLC circuit, with L mh, C µf, and R kω. Determine a), b) HI, c) LO, d) BW, and e) Q. Answer: a) rad/s, b) HI 5 rad/s, c) LO 995 rad/s, d) BW rad/s, and e) Q. BIOEN 3, 3: AC electronics 36 Instructor: Albert Folch

37 BIOEN 3, Section 3: AC electronics Exercise 37. Exercise 38. BIOEN 3, 3: AC electronics 37 Instructor: Albert Folch