mitsos
Well-known member
- Joined
- May 4, 2007
- Messages
- 2,886
I am trying (planning to) to wind a 10K:10K audio transformer for use as input on my PRR176.
In keeping with the PRR varimu mentality of using what you have lying around, I have EI 625 lams in M4 grain oriented steel (a bit thinner than the usual 29 gauge M6 used in the US), as well as about 3/4" of 49% nickel I got as a sample. I planned to make a 5/8 stack because anything larger won't hit (no low-profile bobbins to be had locally). It has to fit inside a 1U case. I started calculating things yesterday but they don't make sense, see below, and my comparison to Edcor at the end to understand my dilemma!
From Ian's inductance thread, I got the rule of thumb Primary Inductance = 10K / 100 = 100 H. From the reactance formula (XL = 2piFL), if I want to go down to 20Hz, I found I need at least 80H, so anywhere between those two is perfect for me.
I have no idea what the permability of the lams is, so I wound 100 turns on a bobbin and measured the following (interleaved 1x1):
M4 = 21mH
49% = 69mH
M4 mixed with 49% = 41mH
From what I've read online and elsewhere, L varies with the square of turns, like the AL number for inductor cores. So, the 41mH of the mixed core is related to turns squared which is 100x100 = 10000. Using this number, I can set up a proportion to find out the number of turns I need. At least I think that's how it works.
So if I set up the relationship as: (0.041H / 80H) = (10000 / X)
X being the turns squared of the coil that will give me 80H. Rearranging and solving for X gives me 19,512,195.1
Again, this is "turns squared", so, SQRT(19,512,195.1) = 4,417.21 turns.
If this is correct so far (that's a lot of turns), going to the tables in the back of Wolpert's Audio Transformer Design Manual, tells me that to get 4417 x 2 (two windings) = 8834 turns, I need at least 40 AWG wire (max turns for square stack using 39 AWG = 8700, so it would just barely not fit, 40 AWG goes to 10,670 max turns)
OK, so even if I can fit all these turns, that leaves me with a DCR of 3377 (for 10670 turns). Wolpert gives a mean length turn of 3.62", we can use this to estimate and subtract the DCR of the extra wire since I'll only be using 8834 turns:
10670-8834 = 1836 turns (times MLT) x 3.62" => 6646 inches = 553.86 feet.
40AWG has 1079 Ohms nominal DCR. I think this is for 1000 feet(?) so [(553.85/1000) * 1079] gives us about 597 Ohms for the extra 553.86 feet.
OK, so 3377 Ohms (for 10670 turns) - 597 Ohms = 2780 Ohms (for both windings) => 1390 Ohms per winding. hmm, that doesn't seem as bad BUT....
that said, I'm confused. I figured that, by mixing in a bit of nickel, I'd be able to use thicker wire or less turns than Edcor, who use M6 only, but then I don't know how they have only a 262 Ohms DCR, and fairly empty looking bobbins (my turns calculated above would pretty much fill the bobbin I'm sure). Well, looking at their PDF again, they list Pri L as 8H, which I had assumed was a typo (missing a zero), because they list the frequency response as +/-1dB, 20-20KHz. How do they get this frequency response out of a 10K trafo with 8H (8H would give 10K reactance down to 199Hz only)?
Am I completely off with these calculations? The whole winding experience is mostly educational, but I would like to get a decent transformer out of it as well, not just waste my friend's time (he's the one with the simple winder), so if you guys can go over my numbers and maybe suggest changes, I'd appreciate it.
Thank you!
In keeping with the PRR varimu mentality of using what you have lying around, I have EI 625 lams in M4 grain oriented steel (a bit thinner than the usual 29 gauge M6 used in the US), as well as about 3/4" of 49% nickel I got as a sample. I planned to make a 5/8 stack because anything larger won't hit (no low-profile bobbins to be had locally). It has to fit inside a 1U case. I started calculating things yesterday but they don't make sense, see below, and my comparison to Edcor at the end to understand my dilemma!
From Ian's inductance thread, I got the rule of thumb Primary Inductance = 10K / 100 = 100 H. From the reactance formula (XL = 2piFL), if I want to go down to 20Hz, I found I need at least 80H, so anywhere between those two is perfect for me.
I have no idea what the permability of the lams is, so I wound 100 turns on a bobbin and measured the following (interleaved 1x1):
M4 = 21mH
49% = 69mH
M4 mixed with 49% = 41mH
From what I've read online and elsewhere, L varies with the square of turns, like the AL number for inductor cores. So, the 41mH of the mixed core is related to turns squared which is 100x100 = 10000. Using this number, I can set up a proportion to find out the number of turns I need. At least I think that's how it works.
So if I set up the relationship as: (0.041H / 80H) = (10000 / X)
X being the turns squared of the coil that will give me 80H. Rearranging and solving for X gives me 19,512,195.1
Again, this is "turns squared", so, SQRT(19,512,195.1) = 4,417.21 turns.
If this is correct so far (that's a lot of turns), going to the tables in the back of Wolpert's Audio Transformer Design Manual, tells me that to get 4417 x 2 (two windings) = 8834 turns, I need at least 40 AWG wire (max turns for square stack using 39 AWG = 8700, so it would just barely not fit, 40 AWG goes to 10,670 max turns)
OK, so even if I can fit all these turns, that leaves me with a DCR of 3377 (for 10670 turns). Wolpert gives a mean length turn of 3.62", we can use this to estimate and subtract the DCR of the extra wire since I'll only be using 8834 turns:
10670-8834 = 1836 turns (times MLT) x 3.62" => 6646 inches = 553.86 feet.
40AWG has 1079 Ohms nominal DCR. I think this is for 1000 feet(?) so [(553.85/1000) * 1079] gives us about 597 Ohms for the extra 553.86 feet.
OK, so 3377 Ohms (for 10670 turns) - 597 Ohms = 2780 Ohms (for both windings) => 1390 Ohms per winding. hmm, that doesn't seem as bad BUT....
that said, I'm confused. I figured that, by mixing in a bit of nickel, I'd be able to use thicker wire or less turns than Edcor, who use M6 only, but then I don't know how they have only a 262 Ohms DCR, and fairly empty looking bobbins (my turns calculated above would pretty much fill the bobbin I'm sure). Well, looking at their PDF again, they list Pri L as 8H, which I had assumed was a typo (missing a zero), because they list the frequency response as +/-1dB, 20-20KHz. How do they get this frequency response out of a 10K trafo with 8H (8H would give 10K reactance down to 199Hz only)?
Am I completely off with these calculations? The whole winding experience is mostly educational, but I would like to get a decent transformer out of it as well, not just waste my friend's time (he's the one with the simple winder), so if you guys can go over my numbers and maybe suggest changes, I'd appreciate it.
Thank you!