clintrubber
Well-known member
[quote author="rodabod"][quote author="clintrubber"]
As I hope and am reasonably sure the situation is like this:
Connecting them in series gives you a centre tap to which you can or won't be connecting phantom by what was it, a 3k4 ? Ratio will then be 1:N.
By connecting them in parallel the ratio will be 1/2:N, so 1:2N (and the impedance-ratio will scale fourfold).
As I understand bifilar, it's just that two windings are wound at once, so they're kind of twins i.s.o. one after the other (with 9 months delay :wink: ). For the principal functioning of the TX it won't matter I figure.
[/quote]
Hi Peter,
Is having a bifilar twin primary the same as having a split primary?
[/quote]I think the former is always the latter, but the latter not necessarily always the former I think. The difference is in how they're wound.
As you stated it, from 'twin' & 'split' both follows that the primaries are not already connected. If they were the word 'centertapped' would have been used somewhere.
You likely will have had a look already at various possible permutations in the Beyer catalog (the 4th digit I thought).
So, will the ratio not just be the same if you parallel these bifilar primaries? The turns number on the primary would just be the same, no ?
[/quote]Nice explanation, understanding this. Yes, the ratio will stay the same, as CJ said.
But how about impedance ? May get confusing since some 'fixed' relationship between ratio (N) & impedance-scaling (N*N) seems to get broken:
For each doubling of the amount of parallel windings the impedance when looking into the whole thing halves, correct ?
Say you got a 1+1:N and a sec. termination of 100kOhm. Just using 1:N gives an impedance when looking into that one primary of 100k/(N*N).
Add the other primary in parallel to the first primary, we maintain the ratio, but now get 0.5*(100k/(N*N)), correct ?
Yes that was too funny, and that even in one posting. Sigh...
You're right, should be terminated.
As I hope and am reasonably sure the situation is like this:
Connecting them in series gives you a centre tap to which you can or won't be connecting phantom by what was it, a 3k4 ? Ratio will then be 1:N.
By connecting them in parallel the ratio will be 1/2:N, so 1:2N (and the impedance-ratio will scale fourfold).
As I understand bifilar, it's just that two windings are wound at once, so they're kind of twins i.s.o. one after the other (with 9 months delay :wink: ). For the principal functioning of the TX it won't matter I figure.
[/quote]
Hi Peter,
Is having a bifilar twin primary the same as having a split primary?
[/quote]I think the former is always the latter, but the latter not necessarily always the former I think. The difference is in how they're wound.
As you stated it, from 'twin' & 'split' both follows that the primaries are not already connected. If they were the word 'centertapped' would have been used somewhere.
You likely will have had a look already at various possible permutations in the Beyer catalog (the 4th digit I thought).
I asked what would happen when paralleling a split primary (because I thought it would halve the ratio) but this is what CJ said:
[quote author="CJ The Transformer Murderer"]Imagine if you used 100 strand litz wire to wind a primary. How many turns do you have if you wrap the litz wire around the core one time?
1 turn. Even though that turn consists of 100 individual strands of wire.
So if you parallel a primary, you are just doubling the strands in your wire, not doubling the turns.
So the ratio stays the same.
So, will the ratio not just be the same if you parallel these bifilar primaries? The turns number on the primary would just be the same, no ?
[/quote]Nice explanation, understanding this. Yes, the ratio will stay the same, as CJ said.
But how about impedance ? May get confusing since some 'fixed' relationship between ratio (N) & impedance-scaling (N*N) seems to get broken:
For each doubling of the amount of parallel windings the impedance when looking into the whole thing halves, correct ?
Say you got a 1+1:N and a sec. termination of 100kOhm. Just using 1:N gives an impedance when looking into that one primary of 100k/(N*N).
Add the other primary in parallel to the first primary, we maintain the ratio, but now get 0.5*(100k/(N*N)), correct ?
[/quote]Of course, it is easy to double the ratio with the twin primary...
P.S, Peter, those puns are getting worse and worse....
[quote author="Peter The Comedian"]they're kind of twins i.s.o. one after the other (with 9 months delay :wink: )...
Please let us know how they turn(...) out !
Yes that was too funny, and that even in one posting. Sigh...
You're right, should be terminated.