Filters, Basic Q, Parallel to Series Conversion, and Impedance Matching
RC Filters
RC Low Pass Filter
R
+
Vi
C Vo
Vo 1 H ( ) = = Vi 1 + j RC
c
Cutoff Frequency c is the frequency at which | H ( ) |= 1 / 2 of the peak value For this example c = 1/ RC Note: f c = c / 2 = 1/ 2 RC In dB: 20 log(0.5) = -3.01 dB EE521 pg. 1
Basic Filters, Q, Parallel to Series Conversion, and Impedance Matching RC High Pass Filter
C
Vi
+ R V o
Vo j RC H ( ) = = Vi 1 + j RC
c
Cutoff Frequency c is the frequency at which | H ( ) |= 1 / 2 of the peak value Peak value occurs at Again, for this example c = 1/ RC Note: f c = c / 2 = 1/ 2 RC
EE521
pg. 2
Basic Filters, Q, Parallel to Series Conversion, and Impedance Matching
Series Resonant RLC Filters
C
Vi
L
+ R V o
Vo R H ( ) = = = 1 Vi R + j L + R+ jC
R 1 j L C
=
R R + jX
Observations At low frequency, X is large due to the capacitance At high frequency, X is large due to the inductance When, L = 1/ C , X = 0 When X = 0, H = 1 Resonant Frequency Resonance occurs when the series reactances cancel: 1 1 1 0 L = 1/ 0C 02 = 0 = , or f 0 = LC LC 2 LC Note at resonance, the total reactance = 0. Thus, only real power is delivered by the source. At this frequency, the dual reactive elements perfectly share energy. EE521 pg. 3
Basic Filters, Q, Parallel to Series Conversion, and Impedance Matching Frequency response:
0 is the resonant frequency (in radians/sec) l , u are the upper and lower -3 dB points, or half-power frequencies is the bandwidth (in radians/sec) over which the magnitude is > -3 dB Solving for the half-power frequency points: Given: R H ( ) = 1 R + j L C The half power points occur when: 1 L = R C
EE521 pg. 4
l o u
Basic Filters, Q, Parallel to Series Conversion, and Impedance Matching Define
u L
Let Then
1 = + R, u C
l L
1 = R l C
C = 1/ 02 L
l o R = o l o L These two equations are simultaneously true iff: u = o o l This leads to: o = ul = the geometric mean of the -3 dB frequencies
Substituting the latter relationship into the above equations, leads to: u l R l R R =+ , u = = o o o L o o L o o L This ratio is defined as the fractional bandwidth of the resonator: f R Fractional BW = = = f o o L o
u o R =+ , o u o L
EE521
pg. 5
Basic Filters, Q, Parallel to Series Conversion, and Impedance Matching
Complex Power
Consider the complex power in the series RLC circuit: 1 1 1 1 2 2 2 S = VI * = Z I = R I + j X I 2 2 2 2 = Pave + jPr Pave = the average power Pr = the reactive power 2 2 L I2 CV 2 L I I = Pr = = ( EL EC ) 2 2 2C 2 Note that Pr = 0 at resonance ( = o ) We define the resonator Quality factor (Q) as: 2 ave energy stored per unit cycle EL o L I / 2 o L = = = Q= 2 Pave R ave power dissipated R I /2 Also: 2 EC o I / 2o2C 1 Q= = = 2 o RC Pave R I /2 EE521 pg. 6
Basic Filters, Q, Parallel to Series Conversion, and Impedance Matching It is observed that the quality factor is reciprocally related to the fractional bandwidth: L 1 1 = o = Q= o = R o RC fractional BW Some observations: Limit R 0 The Q , and 0 Spiked response (output only at the resonant frequency) Limit R The Q 0 , and Flat response (0 output) In general, larger Q series RLC circuits have a very narrow bandwidth Low loss circuit Smaller Q series RLC circuits have a broader bandwidth Larger losses in the circuit
EE521
pg. 7
Basic Filters, Q, Parallel to Series Conversion, and Impedance Matching
Parallel Resonant RLC Filters
Io Ii
C
L
+ G
Io G = H ( ) = = 1 I i G + jC + G+ j L
G 1 j C L
Frequency response:
l o u
Summary o = 1/ LC , o = ul C 1 = o Q= o = o LG G pg. 8
EE521
Basic Filters, Q, Parallel to Series Conversion, and Impedance Matching
Parallel and Series Resistance/Reactance Networks (1st order)
Consider the parallel and series resistance/reactance networks:
Rs
Rp
jX p
jX s
We define the Q of the impedance as: R EL X s Qs = = , dually: Q p = p Pave Rs Xp Note the impedance Q is not to be confused with the resonant Q, or quality factor Often, for analysis, it is convenient to be able to convert between a series impedance and a parallel impedance network Example:
Ii
jX c
jX s Rs
Ii
jX c
jX p
Rp
EE521
pg. 9
Basic Filters, Q, Parallel to Series Conversion, and Impedance Matching
Parallel to Series Conversion
We can start out with a parallel to series conversion:
Rs
=
jX s
Rp
jX p
For an equality, the impedances must be equated: Zs = Z p We can express: Z s = Rs + jX s 2 2 jX p R p R p X p + jR p X p Z p = R p jX p = = 2 2 R p + jX p Rp + X p Therefore: 2 2 Rp X p Rp X p Rs = 2 , Xs = 2 2 2 Rp + X p Rp + X p
EE521
pg. 10
Basic Filters, Q, Parallel to Series Conversion, and Impedance Matching
Parallel to Series Conversion
We can start out with a parallel to series conversion:
Gs
Gp
jB p
=
jBs
For an equality, the admittances must be equated: Ys = Yp We can express: Yp = Yp + jB p jBsGs Gs Bs2 + jGs2 Bs Ys = = Gs + jBs Gs2 + Bs2 Therefore: Bs2 Gs Gs2 Bs Gp = 2 , Bp = 2 Gs + Bs2 Gs + Bs2
EE521
pg. 11
Basic Filters, Q, Parallel to Series Conversion, and Impedance Matching Gs Bs2 Gs2 Bs Gp = 2 , Bp = 2 2 Gs + Bs Gs + Bs2 This can be related back to reactances as: 2 1 1 1 1 Rs X s Rs2 X s 1 1 = = , 2 2 X p Rp 1 1 1 1 + + 2 2 Rs X s Rs X s Thus: Rs2 + X s2 Rs2 + X s2 Rp = , Xp = Rs Xs
EE521
pg. 12
Basic Filters, Q, Parallel to Series Conversion, and Impedance Matching
Transmit Filter
Transmit filter
50
5pF
C 37
+
Vin
L6
Rs
C 38 100 pF
C 39 Vout
2nd-order Band pass filter (~7 MHz center frequency) Note that Rs is due to series resistance of L6 that cannot be neglected (impacts the filter Q) Analysis: Note that at 7 MHz, Z C 37 = j 4,547 >> 50 Write Norton equivalent for source, and combine parallel capacitance. Vin Vin = I sc = Vin jC37 1 1 50 + jC37 jC37 Approximate the circuit near resonance as:
~ Vin jC37
C 37
5pF
+
L6
Rs
C 38 + C 39 Vout
EE521
pg. 13
Basic Filters, Q, Parallel to Series Conversion, and Impedance Matching Finally, convert the series L6-Rs branch to a parallel branch:
~ Vin jC37
Lp Rp
+
C 38 + C 39 Vout +C 37
This is now explicitly expressed as a parallel RLC resonator Note that the prediction of L p is frequency dependent. The prediction is valid near/at the resonant frequency 2 2 Xp 1 Rs + (o L6 ) s Lp = = L6 (assuming o L6 >> Rs ) o o o L 6 Rs2 + (o L6 ) Rp = Rs Note that the Q is impacted by R p .
2
Note: L6 3.1 H. If Rs ~ 1 , then R p ~ 19 k . Recall, the filter Q = oCR p ~ 136 Without the coupling capacitor C37, what would happen to the Q of the filter?
EE521
pg. 14
Basic Filters, Q, Parallel to Series Conversion, and Impedance Matching
Matching Networks
Often, it is desired to match a complex load to a real source impedance source Maximum power transfer requires the load impedance = conjugate of the source impedance If the source impedance is real, then the load must be equal to the source resistance for maximum power transfer If the load is complex, we can introduce a lossless matching network (purely reactive) that effectively transforms the complex impedance to a purely real impedance that is equal to the source resistance.
Rs Vs
Rs
Matching Network
ZL
Objective: Input impedance of the matching network terminated by Z L is Rs The network is lossless (hence all the power is transferred to the load)
EE521
pg. 15
Basic Filters, Q, Parallel to Series Conversion, and Impedance Matching
L-Section Matching Network
A simple matching network can be achieved with only 2 reactive elements Transforms both the real and the imaginary part of the input impedance A common configuration of the 2 reactive elements is referred to as the L-section matching network (or el-section) Two types:
jX jX
Z in
jB
ZL
Z in jB
ZL
Network #1
Network #2
where, Z L = RL + jX L Network #1 is used when RL > Rs . Network #2 is used when RL < Rs
EE521
pg. 16
Basic Filte...