how to generate qr code in vb.net L- NETWORK RS 12 15 GHz SOURCE 315 nH L 75 nH 452 pf CSTRAY 15 pf 58 LOAD RL in Software

Encoder Code 128 Code Set A in Software L- NETWORK RS 12 15 GHz SOURCE 315 nH L 75 nH 452 pf CSTRAY 15 pf 58 LOAD RL

L- NETWORK RS 12 15 GHz SOURCE 315 nH L 75 nH 452 pf CSTRAY 15 pf 58 LOAD RL
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Adding the two inductors
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L-NETWORK RS 12 15 GHz SOURCE 222 nH LOAD 452 pf CSTRAY 15 pf 58 RL
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Combining the two inductors into one for an L network
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L1 L2 , L1 + L
222 nH = LNEW
The 50- source is now perfectly matched to the complex load
PI and T Network Matching
Three-element impedance-matching (PI or ) networks are popular in many narrowband applications The narrowband popularity is due to the higher loaded Q over what the L network possesses, yet PI and T network s also permit almost any Q to be selected Nonetheless, T and PI circuits can never be lower in Q than an L network The Q desired for a particular application may be calculated with the following formula, assuming the utilization of high-Q inductors: Q = fC/( f2 f1) where Q = loaded quality factor of the circuit fC = center frequency of the circuit f2 = upper frequency that we will need to pass with little loss f1 = lower frequency that we will need to pass with little loss Employ the guidelines below to design a PI network capable of matching two different pure resistances (Fig 341) With the following design methodology, consider the PI network as two L networks attached back to back; with a virtual resistor ( R ) in the center, which is used only as an aid in designing these networks and will not be in the final design 1 Find R, the virtual resistance, as shown in Fig 342 In this example, the loaded Q of the PI network is chosen to be 10: "R" = RH 58 = = 057 Q 2 + 1 102 + 1
RH is equal to whichever source or load resistance is larger: RL or RS
Three
RS L
C2 RL
FIGURE 341 A PI matching network between a source and its load
SOURCE SIDE RS 15 GHz 12 XP1 268 XS1 255 XS2 57 "R" VIRTUAL RESISTOR XP2 58 58 RL (Rp)
LOAD SIDE
FIGURE 342 Using a virtual resistor to design a PI network
2 Find XP2 and XS2 by: XP2 = RL 58 = = 58 Q 10 and X S2 = Q " R " = 57 for the load side values
3 Find the value of XP1 and XS1: X P1 = RS Q1 Q1 = RS 12 1 = 448 = "R" 448 12 1 = 268 057
X S1 = Q1 " R " = 448 057 = 255 4 Combine XS1 and XS2 (XS1 + XS2) (Fig 343) 5 One of four different PI matching configurations can be chosen, depending on whether we must get rid of stray reactances, pass or block DC, or filter excess harmonics (Fig 344) 6 Convert the reactances calculated to L and C values by: L= XS 2 f and C= 1 2 fX P
To match two stages with a PI network, while canceling reactances and matching resistances (Fig 345), observe the following procedures Convert the load/source to/ from parallel or series equivalences as required to make it easier to absorb any reactances (These conversion equations are found later in this chapter)
Amplifier Design
RS 825 12 15 GHz 268 58 58 RL (Rp)
FIGURE 343 Reactance values as calculated for PI network
255 268
57 58 268
825 58
255
57
825
268
58
268
268
255
57
315 58
268
58
268
255
57
315 268
268
58
58
FIGURE 344
Different legitimate PI networks before and after combining components
Three
SOURCE RS 50 CSTRAY1 CSTRAY2 75 RL LOAD
FIGURE 345 A PI network used in a resistive and reactive source and load
1 Select a proper network topology that will absorb both stage s reactances In this case, we would choose a PI network with two parallel capacitors (see above on Absorption under Lumped L Matching ) 2 Choose a desired loaded Q and frequency of operation 3 Find R, the virtual resistance: "R" = RH Q2 + 1
NOTE : RH is equal to whichever source or load resistance is larger: RL or RS
4 Find XC2 and XL2 by: XC 2 = RL and X L2 = Q " R " for the load side values Q
5 Find XC1 and XL1 by: XC1 = RS Q1 Q1 = RS 1 "R" and X L1 = Q1 " R "
6 As shown in Fig 346, add XL1 and XL2 to form XL (NEW); combine XC (STRAY1) and XC1; then combine XC (STRAY2) and XC2 (XC1 and XC2 must be smaller than XC (STRAY1) and XC (STRAY2) respectively, since adding two capacitor s reactances in parallel involves: X C X C(STRAY) = XC X C + X C( STRAY ) Thus, if XC(STRAY) < XC1, then XC(TOTAL) will not be able to reach the proper XC value Also, increase XC1 until: X C 1 X C( STRAY1) = X C(TOTAL) = X C 1 X C 1 + X C( STRAY1) or X C 1 X C(STRAY1) = X C(NEW) X C 1 X C(STRAY1)
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