Sorry this is such a big post. I watched a video about bandpass filters and I am trying to figure out the design strategy they used. In particular, how to calculate the Q of this filter. A screenshot of the filter and parts table:
The transformers in the table all have similar turns ratio, around 5:1. That's a 25:1 impedance ratio. So 50Ω is turned into 1250Ω and vice versa at the ends of the filter. The middle capacitors have a reactance matching that 1250Ω at the secondary. For transformers, I know that a higher impedance means a higher Q, but how high? To make an attempt at verifying my guesses at calculating this Q, I put some parts into simsmith for an 11m filter using values similar to those in the table above, but it didn't answer my question:
Since just under an SWR of 6 is about 50% loss, I assume about there is the -3dB points (don't know how to make simsmith show return loss or anything else yet). That shows about 4.4MHz bandwidth. But I didn't put in a core material, I didn't specify wire diameter, nothing. So how can simsmith know the bandwidth???
If I take the secondary resistance from toroids.info with 16 turns on a 37-6 core, it says 5.1Ω resistance (I assume this swamps the effects of wire diameter). Wikipedia states that when a tank only has the resistance of the coil, it is the series equation for Q (but I have both). So if I convert the parallel combination of inductive reactance and the 1250Ω of the secondary to series equivalents so I can add the coil resistance to that, then convert back to parallel equivalents and use those values in the typical equation for Q of a parallel tank (using Xl=133Ω and R=920Ω instead of Xl=132Ω and R=1250Ω), I get a Q of about 7 and a BW of 3.9MHz. Close, but not the same as simsmith shows. What am I doing wrong???
Am I messing this up somehow????
Thanks!
The transformers in the table all have similar turns ratio, around 5:1. That's a 25:1 impedance ratio. So 50Ω is turned into 1250Ω and vice versa at the ends of the filter. The middle capacitors have a reactance matching that 1250Ω at the secondary. For transformers, I know that a higher impedance means a higher Q, but how high? To make an attempt at verifying my guesses at calculating this Q, I put some parts into simsmith for an 11m filter using values similar to those in the table above, but it didn't answer my question:
Since just under an SWR of 6 is about 50% loss, I assume about there is the -3dB points (don't know how to make simsmith show return loss or anything else yet). That shows about 4.4MHz bandwidth. But I didn't put in a core material, I didn't specify wire diameter, nothing. So how can simsmith know the bandwidth???
If I take the secondary resistance from toroids.info with 16 turns on a 37-6 core, it says 5.1Ω resistance (I assume this swamps the effects of wire diameter). Wikipedia states that when a tank only has the resistance of the coil, it is the series equation for Q (but I have both). So if I convert the parallel combination of inductive reactance and the 1250Ω of the secondary to series equivalents so I can add the coil resistance to that, then convert back to parallel equivalents and use those values in the typical equation for Q of a parallel tank (using Xl=133Ω and R=920Ω instead of Xl=132Ω and R=1250Ω), I get a Q of about 7 and a BW of 3.9MHz. Close, but not the same as simsmith shows. What am I doing wrong???
Am I messing this up somehow????
Thanks!
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