《电路》（英文版）10-1 Parallel resonance(谐振)

§10-1 Parallel resonance(谐振) Let us now define resonance. in a two-terminal electrical network containing at least one inductor and one capacitor, resonance is the condition which exists when the input impedance of the network is purely resistive μR(4 r+iol+ joc 83J +f(C、1 R at0=0 0c 0 or 0C OL L resonant frequence LC

§10-1 Parallel resonance(谐振） Let us now define resonance. In a two-terminal electrical network containing at least one inductor and one capacitor, resonance is the condition which exists when the input impedance of the network is purely resistive. ) 1 ( 1 L j C R  = +  − j C R j L Y   = + + 1 1 resonant frequence LC r = − − 1  L or C L at C r r r r r       1 0 1 = , − = = R L

Let us examine the magnitude of the response |= +j(ac-) R )2+(oC-1 OL ato=o=1/√LC,V=R orv=RI(max imum) IR atO=0,V->0 IR/2 ato=∞,V→>0 o,ar 0 At resonance the parallel rLC combination acts simply as R, the parallel lC combination behaves as an open circuit. 19002 two half power frequencies B=2-m1- band width(带宽)

Let us examine the magnitude of the response. ) 1 ( 1 L j C R I Y I V  +  − = = • • • 1 2 1 2 ) L ) ( C R ( | I | |V |  +  − = • • or V RI(max imum ) at r / LC ,V IR =  =  = 1 = at  = 0, V → 0 at  = , V → 0 V  r IR IR/ 2 1 2 At resonance the parallel RLC combination acts simply as R, the parallel LC combination behaves as an open circuit. 1 2  , --two half power frequencies. B =2 −1 --band width（带宽）

The quality(品质) factor:Q=2x max imum energy stored total energy lost in a period or =2T /w(t)+w(t) i(t)=√2 I cos a,to(t)=√2 RIcos a,t w(t) C(2RI coso, t)2CR212 cos2@,t 2 RI∠0 RI RI L j0nL,L∠90°a,L ∠-900→i(t)=√2-,cos(art-90) L R WL(=li= sinr≠CR" 1 sin a,t 2 w(t)+w,(t)=CR 1---cons tant 2 Q=2兀 CR CR CR 2兀 =2元 R IRT FO RC- R 2兀/a L

The quality(品质) factor: total energy lost i n a period max imum energy stored Q = 2 P T [w (t ) w (t )] or Q R c + L max = 2 Let i t I t r ( ) = 2 cos t RI t  r ( ) = 2 cos 2 2 ( 2 cos ) 2 1 2 1 w (t) C C RI t c =  = r       90 90 0 =  −   = = • • L RI L RI j L V I r r r L t L R I w t Li r r L   2 2 2 2 2 sin 2 1 ( ) = = w (t ) w (t ) CR I constant c + L = − − 2 2 I RT CR I Q 2 2 2 = 2 ( ) 2 cos( 90 )     = t − L RI i t r r L L C R L R RC / CR T CR r r r = = = = =       2 2 2 CR I t  r 2 2 2 = cos CR I t  r 2 2 2 = sin

R:1 0 R Y=-+j(ac-) +j(ac R R ""@R OL OR 十 0-2)=[1+j2( R·R R At the half power frequencies @, and a2 RI √2 R [1+jo( )=±1 R R B m10)a2 )2+1+ 20 20 2 B=O2-0 OO RCRC B

) 1 ( 1 L j C R Y  = +  − ( ) 1 1 Q Q R j R r r     = + − [1 ( )] 1     r r jQ R = + − ) 1 ( 1 R R R L R j C R r r r r      = +  − At the half power frequencies 1 and 2 R Y Y RI I V 2 2 = =  = R jQ R r r 2 [1 ( )] 1 + − =     ( − ) = 1     r r Q Q( ) ( ) r r r       − = + 2  2 2 1 Q( ) ( ) r r r       − = − 1  1 1 1 2 2 ) 1 2 1 ( 2 1 B Q Q r r = r + −  −    2 2 ) 1 2 1 ( 2 2 B Q Q r r = r + +  +    B or Q Q RC RC B r r r r     = − = = = = 1 2 1

Example 2: Find @r, @, B and v at @=O Solution 1 D20∠0°50 5mH 2uFTV √LC √5×103×2X004 Q=a,RC=104×400×2×106=8 10∠0° B =1250 m44005m2uF V=RI,=400×10×10-3∠0° =4∠0° RIs L jOIs=-j80mA JOL jO,l Ic=jO Cv=jo, crls=jo Is =j80m4 o>>l, parallel resonance--current resonance

Example : Find ,Q,B and V at .  r  =  r • 2 Solution: 4 3 6 10 5 10 2 10 1 1 =    = = − − LC r 10 400 2 10 8 4 6 = =    = − Q r RC = = 1250 Q B  r V V R Is   4 0 400 10 10 0 3 =  = =    − • • jQ I j mA j L R I j L V I s r s r L = = = − = − 80 • • • •   I C = j r CV = j r CRIs = jQ Is = j80mA • • • •   Q>>1, parallel resonance--current resonance