Chapter 4 Resonant Circuits

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Chapter 4 Resonant circuits 4.1 Series resonant circuits steady state response, Q factor, half power frequencies, input impedance, 4.2 Parallel resonant circuits steady state response, input impedance, loaded Q, external Q 4.3 Transformer coupled circuits equivalent resonator circuit 4.4 Transmission line resonant circuits short-circuited λ/2 and λ/4 lines, open-circuited λ/2 and λ/4 lines 4.5 Microwave resonators rectangular cavity, circular cylindrical cavity, dielectric resonator

4-1 微波工程講義 4.1 Series resonant circuits Basics 1. Resonator applications: filter, frequency selective components in amplifier and oscillator, impedance transformer, matching element. 2. Steady state response L C V ( jw) A( jw) = o Vin ( jw) v (t) v (t) = R = 1 in R o 1 jwL 1 jwL + + R + +1 jwC R jwRC = 1 = 1 = 1 = 1 j − 1 + j w − 1 + + j w − wo + w − wo (LCw ) 1 ( 2 ) 1 1 ( ) 1 jQ( ) RC w RC wo w wo RC wo w wo w 1 resonant frequency w ≡ o LC + ≡ average stored energy = Wm We = wo L = 1 quality factor Q wo wo power loss Ploss R wo RC

1 2 1 2 1 2 1 2 1 P = I R,W = I L,W = V C = I loss 2 m 4 e 4 c 4 ω 2C 4-2 微波工程講義 w 3. Half power frequencies Q = o − w2 w1 4. Input impedance near resonance 1.0 A Q=1 1 0.707 Z ( jw) = R + jwL + in jwC Q=10 w 2 (w + w )(w − w ) Q=100 = R + jwL (1− o ) = R + jwL o o δ w 2 w 2

≈ R + j2 wL wo j2RQδw ∠A (1) = R + 180o w Q=100 o Q=10 L w Q=1 (2) = w + j2(w − w )L = j2(w − w + o )L 0 o Q o o j2Q 1 -180o = j2[w − w (1+ j )]L o 2Q wo Lossy resonator Z in 1 = lossless resonator Zin by letting w → w (1+ j ) o o 2Q

4-3 微波工程講義 Discussion w 1. Derivation of half power frequencies equation Q = o − w2 w1 V ( jw) 1 A( jw) = o = V ( jw) w w in 1+ jQ( − o ) wo w 1 1 w w w w = → 2 = 1+ Q 2 ( − o ) 2 → Q( − o ) = ±1...(1) 2 2 w w 2 w w w w 1+ Q ( − o ) o o wo w 2 2 → w2 − wo = − w1 − wo → − wo = − + wo → + = 2 1 + 1 ( ) w2 w1 w1 w2 wo ( ) wo w2 wo w1 w2 w1 w1 w2 → = 2 w1w2 wo ...(2) w w 1 w 2 w (2) w w (1) → 1 − o = − → w − o = − o → w − w = − o → Q = o 1 1 2 − wo w1 Q w1 Q Q w2 w1

4-4 微波工程講義 2. Zin(jw)=R+jX, and R=X at half power frequencies 2 − 2 = + w1 wo Z in ( jw1 ) R jw1L 2 w1 ω Zin ( ) 2 − 2 1 = w1 wo Q wo w1 w L 3dB BW → Z ( jw ) = R + j o in 1 Q 2 R R = L Q wo R ω ω → R + jR 1 0

3. Ex.4.1 A series LC circuit passes signals from 9MHz to 11MHz, and

is connected to a communication system with Zin=50Ω. →L, C = = × = f o f1 f 2 9 11 9.5MHz w w L R 50 Q = o = o → L = = = 4uH − − π × − × 6 w2 w1 R w2 w1 2 (11 9) 10 = 1 = C 2 64.3 pF Lwo 4-5 微波工程講義 4.2 Parallel resonant circuits Basics 1. Duality with series resonator: V→I, R→G, L→C, C→L 2. Steady state response iin (t) io (t) I ( jw) A( jw) = o I in ( jw) R = 1/ R = 1 L C 1 1 R jwC + + jwRC + +1 jwL R jwL = 1 = 1 = 1 = 1 jR − 1 + jR w − 1 + + jR w − wo + w − wo (LCw ) 1 ( 2 ) 1 1 ( ) 1 jQ( ) L w L wo w wo L wo w wo w 1 resonant frequency w ≡ o LC + ≡ Wm We = R = quality factor Q wo wo RC Ploss wo L 2 2 1 V 1 2 1 V 1 2 P = ,W = I L = ,W = V C loss 2 R m 4 L 4 ω 2 L e 4 4-6 微波工程講義 3. Input impedance near resonance

2 1 1 − 1 w − 1 (w + w )(w − w ) − Z ( jw) = ( + jwC + ) 1 = [ + jwC (1 − o )] 1 = [ + jwC o o ] 1 in R jwL R w 2 R w 2 δ 1 − ≈ ( + j2 wC ) 1 R

1 j2Qδw − (1) = ( + ) 1 R Rwo

w C − C − (2) = [ o + j2(w − w )C ] 1 = [ j2(w − w + )C] 1 Q o o j2Q

1 − = { j2[w − w (1+ j )]C} 1 o 2Q

→ + 1 Lossy resonator Z = lossless resonator Z by letting wo wo (1 j ) in in 2Q

4-7 微波工程講義 Discussion

1. Yin(jw)=G+jB, and G=B at half power frequencies 2 2 1 w − w − Z ( jw ) = ( + jw C 1 o ) 1 Z (ω ) in 1 R 1 w 2 in 1 R 2 − 2 1 = w1 wo Q w w o 1 1 w C − R 2 → Z ( jw ) = ( + j o ) 1 in 1 R Q 3dB BW Q =w RC →o = R Z in ( jw1 ) ω ω 1+ j 1 0

2. Unloaded Q, Qu, loaded Q, QL, external Q, Qe 1 1 1 = + QL Qe QU ω  L resonator  o for series resonant circuit RL  R Qu Q =  L e R  L for parallel resonant circuit ω  o L

4-8 微波工程講義 3. Resonator relations Series resonator Parallel resonator

1 1 1 − 1 − R + jwL + ≈ R + j2 wL ( + jwC + ) 1 ≈ ( + j2 wC) 1 Z jwC δ R jwL R in δ 2RQδw 1 2Qδw − ≈ R + j ≈ ( + j ) 1 wo R Rwo 1 wo LC w L 1 R o = w RC = Q o u R wo RC wo L

wo L RL Q e RL wo L 1 = 1 + 1 Q L QL Qe QU

4-9 微波工程講義 4. Ex.4.2 A parallel resonator with R=10kΩ, L=10uH, C=10pF and

its load RL=100kΩ→wo, Qu, QL = 1 = 1 = 8 wo 10 rad / sec LC 10−5 ×10−11 R 105 R Q = = =100,Q = L =100,Q = 50 u 8 × −5 e L wo L 10 10 wo L

4-10 微波工程講義 4.3 Transformer-coupled circuits Basics I 1. Equivalent circuit I1 2

I V = nV , I = − 2 Z S 1 2 1 n V1 L V2 Z V = jwL I + jwMI L1 2 L  1 1 1 2 Vs = + V2 jwMI 1 jwL 2 I 2 n :1  = − + + I 2 = + I 2 V1 ξ jw(1 )L1 I1 jw L1 (I1 ) jwL1I1 jw L1 n n −ξ  (1 )L1 I I / n I ξ 1 I 1 2 2  V = jw L (I + 2 )  2 n 1 1 n ξ L L Z S → 1 =ξM , 1 = L ξ ξ n n 2 2 V ξ V 1 L1 2 Z L M V → =ξ L1 = nM = s n , n :1 L2ξ L1 L1L2 M ξ → ξ = : coefficien t of coupling L1L2

4-11 微波工程講義 Discussion 1. Ex.4.3 A tightly coupled transformer circuit → equivalent resonator ξ ξ circuit, wo, Q 320nH 20nH L 320 6pF 30.7pF 62.5kΩ ≈1,n = 1 = = 4 L2 20 Is 1 1 Z = n2Z →Y = ( + jw×30.7 p) 1 2 1 n2 62.5k R =16×62.5×103 =1MΩ 30.7 C = 6 + = 7.9pF C R L 16 I s L = 320nH n:1 1 f =π = 99.97MHz o 2 LC R 106 Q = = = 4975 π × × 6 × × −9 woL 2 99.7 10 320 10

4-12 微波工程講義 2. Ex.4.4 A double-tuned transformer-coupled circuit → equivalent ξ circuit, (a) wo and Zin as R1, C1 removed (b) Io/Is with R1, C1

(a) without R1,C1

1 2 = = I1 I 2 wo ,Zin ( jwo ) n R2 L C Io 2 2 C R1 C1 L1 L2 2 2 R2 jwn L2 I V2 = ξ − + s V1 Zin ( jw) jwL1(1 ) − 2 + L2 + w L2C2 jw 1 R2 I1 I / n I (b) with R1,C1 2 2 (1−ξ) I  L1 o R1 C L C2 = + I2 1 2  nV jw L (I ) ξ R2 2 1 1 I V L nV V2  nξ s 1 1 2  R  V = −I 2 = I R n:1 2 2 + o 2  1 jwR2C2  = + 1 + + − ξ Is I1 ( jwC1)[nV2 jw(1 )L1I1]  R1

ξjw L1 I nR → o = 2 : four poles L 1 1 Is − 2 + 2 + + − + + ξ + ( w L2C2 jw 1)[1 jw(1 )L1( jwC1)] jw L1( jwC1) R2 R1 R1 4-13 微波工程講義 as ξ → 0 ξ jw L1 I o ≈ nR2 L 1 1 Is − 2 + 2 + + + + + ( w L2C2 jw 1)[1 jwL1( jwC1)] jw L1( jwC1) R2 R1 R1 = = = = ξ for w1 w2 wo ,Q1 Q2 Qo 1 R R L C = L C = , 1 = 2 = w Q 1 1 2 2 w2 L L o o o 1 ξ 2 jw L1 I → o ≈ nR2 2 2 Is − w + jw + 2 +ξ − w + jw ( 2 1) ( 2 ) wo woQ wo woQ

4-14 微波工程講義 γ 4.4 Transmissionγ line resonant circuits Basicsα γ 1. Short-circuitedβ line l α Z + Z tanh l Z = Z αL o = Zβ tanh l in o + α α o Z o Z L tanh l = Zo, α, β β Z L 0 β tanh l + j tan l = Z tanh( δ+ j )l = Z o o 1+ j tanh l tan l β δ πδ lowloss line tanh l ≈ l π λ π  w πδ+ l = β wl (w π+ wπδ)l wl  w 2 l = = o = l + =  o λ v v r πδv w p p p  + l = π πδ πδ  2 2wo 4 λ  w πδw w  tan( + ) = tan ≈ l =  w w w 2 tan l =  o o o w w 2πδw λ tan( + ) = − cot ≈ − o l =   2 2wo 2wo w 4

4-15 微波工程講義 α β α α πδ β

α πδ 2. λ/2 short-circuited line→series resonatorα πδ w δ l + j α tanh l + j tan l w Z w Z = Z α λ ≈ Z o ≈ Z l + j o in o 1+ j tanh l tanπ l o w o w 1+ j l o wo for series resonator Z ≈ R + j2 wL in π → equivalent circuit parameters π α Z 1 2 w L β R ≈ Z l = Z r , L ≈ o ,C = ≈ ,Q = o ≈ α = r o o 2 β 2w w2 L w Z R 2 l 2α α o α o o o β 3. λ/4 short-circuited line→parallelπδ resonator 2w 2w δwπ Z l −αj o − j o j o tanh l + j tan l πδ 2w Z = Z ≈ Zδ πδw ≈ Z wδ × o = l in o + o 2w o α 2w π w α w 1 j tanh l tan l 1− j l o 1− j l o j 1+ j α πδ πδ αλ π w w 2wo l2wo δ α 1 − R for parallel resonator Z ≈ ( + j2 wC ) 1 = in R 1+ j2 wRC π → equivalent circuit parameters π Z 4Z 1 4Z R β R ≈ o = o ,C ≈ , L = ≈ o ,Q = ≈ α = r 2 α i r 4wo Z o wo L wo wo L 4 l 2

4-16 微波工程講義 4. Open-circuited line γ Z + Z tanh l Z Z 1+ j tanh l tan l Z = Z L o γ = o = o = Z in o + + o + Z o Z L tanh l =∞ tanh l tanh( j )l tanh l j tan l α Z L γ πδ α≈ lowloss line tanh l l α β λ l w β ° l = α ° w 2 α β tan l ≈ ® o πδ λ °− 2wo = Zo, α, β β ° l ¯ w 4 π 5. λ/2 open-circuited αline→parallel resonator α β α δw 1πδ+ jαl 1+ j tanh l tan βl w Z Z = Z ≈ Z αo ≈ πδ o in o tanh l + j tan l o w w δ l + j l + j wo wo

α 1 − R αλ π ≈ + 1 = for parallel resonator Z in ( j2 wC ) δ R 1+ j2 wRC → equivalent circuit parameters π π Z 2Z 2Z β R ≈ o = o ,C ≈ , L ≈ o ,Q = α = r α l r 2wo Z o wo 2 l 2 4-17 微波工程講義 α α β β α 6. λ/4 open-circuited line→seriesπδ resonator α 2w α 2w 1− j l o 1− j l o 1+ j tanh l tan l πδ πδ w Z = Z ≈δZ w ≈ Z w = Z ( l + j ) in αo o o o tanh l + j tan l 2wo 2wo α πδ 2w α λ l − j πδ− j o π w w ≈ + for series resonator Z in R j2 wL → equivalent circuit parameters π π Z 4 β R ≈ Z l = Z r , L ≈ o ,C ≈ ,Q ≈ α = r o o α 4 4wo wo Z o 4 l 2

4-18 微波工程講義 Discussion 1. Ex.4.5 A coaxial λ/2 short-circuited line has inner conductor radius a=0.455mm, outer conductor radius b=1.499mm→Q at

5GHz for air dielectric and Teflon dielectric (εr=2.08, tanδ=0.0004)

µ w 2 × 5×10 9 × 4 ×10 −7 copper R = π o = = 0.0184Ω α σ s 2 2× 5.813 ×10 7 R 1 1 π 0.0588 Np / m air µ = s ( + ) =  c b a b 0.085 Np/m Teflon εε 2 o ln a α r o  0 air =µw = d ε ε o o r tan  2 δ 0.03Np/m Teflon β π 2 εf 104.7 rad/m air = r =  c 151.03 rad/m Teflon β 891 air Q = =  2α 657 Teflon

4-19 微波工程講義 2. Ex.4.6 A 50Ω microstrip λ/2 short-circuited line has Teflon

substrate (εr=2.08, tanδ=0.0004) with thickness h=0.159cm, ε t=0.159um →Q and l at 2GHz  A2.29,30 = = Ω = → w = 0.51cm r 2.08, Z o 50 , h 0.159cm A2.21,22 λ  → =  re 1.79 c l = = = 5.6cm ε ε 2 2 f re ε w = 0.51cm, h = 0.159cm,t = 0.159um  A2.26,27  → = 0.0428 Np / m =σ = × 7 = c r 2.08, 5.813 10 S / m, f 2GHz  w = 0.51cm, h = 0.159cm, = 2.08 A2.28 δ r  → = 0.0095 Np / m = = d tan 0.0004, f 2εGHz  α β π 2 π =λ = = 56rad / m l α β 56 Q = = = 535.4 2α 2× (0.0428 + 0.0095)

4-20 微波工程講義 4.5 Microwave resonators Basics 1. Rectangular cavities dimension a, b,cπ π π ×µε 8 = 3 10 m 2 +π n 2 + p 2 TE mnp ,TM mnp mode resonant frequency f r µε ( ) ( ) ( ) 2 a b π c 2W 60b(acw )3 µ TE mode Q = e = r 10 p c 2 3 + 3 + 3 + 3 ε Pl Rs (2 p a b 2bc pa c ac ) r = 1 Qd χ taπn δµε π 2. Circular cylindrical cavities  p TM mode = c nm 2 + p 2 = nm nmp f r ( ) ( ) , p  2 r h  p'nm TE nmp mode

pnm : mth zero of the Bessel function J n (x) of the first kind and order n

p'nm : mth zero of J'n (x) : Table 4.3  4.5.10 TE nmp mode 1 Q =  , Q = c d δ 4.5.11 TM nmp mode tan 4-21 微波工程講義 3. Dielectric resonator (DR): high ε, low tanδ, 2r

low temperature coefficient, TE01δ mode 34 r f = (3.45 + )GHz TE mode h r ε 01δ r r h Discussion 1. Ex.4.7 A rectangular cavity made of WR-90 resonates at 9.379GHz

in TE101 mode → length l and Q − = = = = WR 90waveguide a 0.9in 2.286cm,π b 0.4in 1.016cm π µε 3×108 π TE mode 9.379 ×10 9 = ( )2 + ( )2 → l = 2.238cm 101 2 2.286 l = (4.5.2)Qc 7858 2. Ex.4.8 A rectangular cavity with a=1.6cm, b=0.71cm, c=1.56cm is filledπ with Teflonπ → TE mode f and Q µε π 101 r 3×108 f = ( )2 + ( )2 = 9.375GHz r 2 1.6 1.56l 1 1 1 1 (4.5.2)Q = 5489,Q = = 3417, = + → Q = 2106 c d δ tan Q Qc Qd 4-22 微波工程講義 3. Ex.4.9 A cylindrical cavity is operated at 5GHz in TE011mode → Q and height h (h=2r) χ = = 01 p'01 3.832 π µε c 3.831 π 5×109 = ( )2 + ( )2 → r = 0.0395cm, h = 0.079cm 2 r 2r Q = 39984 .6 4. DR selection in the operation with microstrip line ε 1.2892 ×108 1.2892 ×108 > r > ε d f r s f r r → ε DR (4.5.17) ~ (4.5.23) h r h

5. Ex.4.10 DR er=36, substrate es=9.9, ε 2r s t t=0.25mm→DR dimensions for 35GHz 1.17mm > r > 6.139, → r = 0.835mm Hw #3(due 2 weeks) 5, 7, (4.5.17) ~ (4.5.23) → h = 0.668mm 12, 15, 18

3-22 微波工程講義