Frequency Impedance 10MHz 18.2 + j0.14 240MHz 18.1 + j2.8 450MHz 18.1 + j5.2
One simple approach seems to be:
50 ohm input - ----------- -----------|---------- IN+ ) ( ) ( ___ ) ( ) (- ground --- 6 pF LT5512 ) ( ) ( | ground - ----------- -----------|---------- IN- 3:1 1:1 Coilcraft Coilcraft WBC3-1TL WBC1-1TL
The 3:1 transformer converts the unbalanced 50 ohm input to 16.67 ohms. The 1:1 center tap transformer converts the unbalanced 16.67 ohms to a balanced 16.67 ohms. The center tap also supplies a DC ground for biasing the LT5512 inputs. The 6 pF capactor reactance is:
Frequency Reactance 10MHz 2652.58 240MHz 110.52 450MHz 58.95
placing it in parallel with the LT5512 inputs results in:
Frequency Impedance 10MHz 1/(1/(18.2 + j0.14) + 1/2652.58) = 18.21 240MHz 1/(1/(18.1 + j2.8) + 1/110.52) = 17.58 450MHz 1/(1/(18.1 + j5.2) + 1/58.95) = 16.7
Questions:
1) How off base is my simple analysis?
2) Is there a better way to do this?
3) How significant is the 1.6 ohm mismatch at 10 MHz?
-- John
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