A
Anthony William Sloman
Guest
On Saturday, May 13, 2023 at 4:38:21â¯AM UTC+10, Lamont Cranston wrote:
But their arrangement can make a lot of difference. A secondary tat is wound as a series of layers, one on top of another, has a higher interwinding capacitance than one that is banked, wound as series of multilayer coils, each one stacked on top of one another electrically speaking, while physically side by side.
> Well, I thought I knew what I was doing, I used C1-4C2 / 3.
But what do you think you are doing?
Every last bit of capacitance inside the transformer has got to get charged up when you apply a voltage across the transfomer. You\'ve thrown a few diodes into the system and it looks as if you are charging some of those capacitances with DC and revers biasing the diodes inot thier non-conducting state for most of the time.
You need to think harder about what you are looking at.
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Bill Sloman, Sydney
On Friday, May 12, 2023 at 9:42:37â¯AM UTC-5, Anthony William Sloman wrote:
Mikek\'s experiments with his neon sign transformer produce strange results - the fact that he has to load the output with 2.5M of resistance before he can see a half wave suggests that there are capacitors embedded in his set-up. The fact that 18pF killed it off again says that it might just be the interwinding capacitance in the transformer - easy enough to measure if you know what you are doing.
I don\'t doubt there is a large self capacitance in the secondary, there are lots of turns.
But their arrangement can make a lot of difference. A secondary tat is wound as a series of layers, one on top of another, has a higher interwinding capacitance than one that is banked, wound as series of multilayer coils, each one stacked on top of one another electrically speaking, while physically side by side.
> Well, I thought I knew what I was doing, I used C1-4C2 / 3.
But what do you think you are doing?
But C1 = 3,208pf @ 75Hz and C2 = 945pf @ 150Hz.
3208 - (4x 945) / 3 = 3208 -3780 / 3
That doesn\'t compute, I did it twice and got very similar results.
Any help?
Every last bit of capacitance inside the transformer has got to get charged up when you apply a voltage across the transfomer. You\'ve thrown a few diodes into the system and it looks as if you are charging some of those capacitances with DC and revers biasing the diodes inot thier non-conducting state for most of the time.
You need to think harder about what you are looking at.
--
Bill Sloman, Sydney