Would 2 different wire sizes work in a primary winding (single phase transformer) for example 50 turns of 16awg- 100 turns of 22awg- 50 turns of 16awg? All connected End to end sharing the same ac input. In theory (my theory) is it would leave more space for secondary winding and help with heat losses.
\$\begingroup\$
\$\endgroup\$
2
-
1\$\begingroup\$ The smaller wire will surely help increasing the power loss due to resistive heating, but why do you want to do it? \$\endgroup\$– PlasmaHHCommented Dec 7, 2016 at 21:40
-
\$\begingroup\$ If they are all connected end to end they don't share the same ac input, \$\endgroup\$– Andy akaCommented Dec 8, 2016 at 9:28
Add a comment
|
3 Answers
\$\begingroup\$
\$\endgroup\$
Usual practice for a single (not tapped) winding is to use the largest single gauge that will fit nicely.
In this case, very roughly without getting into differing circumferences or packing ratios, the primary copper losses of the suggested arrangement would be 20% worse than just using common AWG20 wire, or 50% worse than using less common AWG 19 wire.
You can easily do this calculation from the cross-sectional areas and resistances in the above-linked table.
If the winding was tapped, and could optionally accept a lower voltage at a higher current then the optimal transformer might have heavier wire for part of the winding.
answered Dec 7, 2016 at 21:50
\$\begingroup\$
\$\endgroup\$
1
One increment in wire gauge is (approximately) a 20% decrease in cross-section area, and a 25% increase in (dc) resistance.
I'm making some simplifying assumptions here, like the lengths and packing factors of the #16 and #22 portions are the same, skin effect and coating thickness are negligible, and so on
In your example, you have 100 turns of #22 wire in series with 100 turns of #16 for a total of 200 turns. #16 wire is 1.95 times the diameter of #22, so it takes up 79% of the winding cross-section area. The resistance of the #22 portion is 3.8 times the resistance of the #16 portion, so if the resistance of 100 turns of #16 is R, then the total resistance of the dual-gauge winding is 4.8 R.
For comparison, #18 wire has an area 64% of #16 wire, so 200 turns of #18 would have about the same winding cross-section as 100 turns of #16 plus 100 turns of #22. Its resistance would be 1.56 R for 100 turns, or 3.12 R for 200 turns. So using #18 for the entire winding would have 35% lower resistance.
-
\$\begingroup\$ I'm quite new at learning this stuff obviously, thank you for answering my question! \$\endgroup\$ Commented Dec 9, 2016 at 16:15
\$\begingroup\$
\$\endgroup\$
1
The maximum current you can put through the primary will be governed by the smallest diameter wire. The sections with larger wire will consume more space than if wound with the same gauge.
-
\$\begingroup\$ Makes sense now, much appreciated! \$\endgroup\$ Commented Dec 9, 2016 at 16:16