0
\$\begingroup\$

I am really new to Electronics. I spend time on it as a hobby. I have recently grown my interest on inductors but most of the calculation seems pretty hard to me. Little unsure if what I made can stand 220V AC Mains Current. I have made a Coil of 1000 turns of 32 AWG Annealed Copper Wire whose DC resistance measures 25.5 Ohm. My simple understanding suggests 220V across 25.5 Ohm, approx 8.627 Amps will flow, which seems way too high for this poor coil (I know, since it's inductor, it's not simple a resistive load). I calculated the length of the conductor to be 247.093 foot.

I am basically trying to make a transformer with it. Can I use it as Primary side of my coil?

If yes, then I am expecting the output to be 6V. I have used a 2" X 1" plastic former to wind it. If I expect the transformer to supply 4 Amps at 6V from secondary, which AWG wire (37 Turns Approx) will be suitable?

Please pardon me if the question is too simple. I am kind of stuck, because I still have lot of doubts on these calculations

\$\endgroup\$
10
  • \$\begingroup\$ what type of insulation? its not the gauge that matters \$\endgroup\$
    – JonRB
    Jan 22 '17 at 20:53
  • \$\begingroup\$ What is the coil diameter and length? You should be able to calculate its inductance using diameter, length and number of turns. There are online calculators. Then you can estimate the current at 220V at 50Hz or 60Hz (depending on where you are in the world). If the wire is not insulated, then the coil will probably short out unless you somehow wound it with no contact between adjacent turns. \$\endgroup\$
    – mkeith
    Jan 22 '17 at 20:58
  • 2
    \$\begingroup\$ Unless the magnetic flux has a complete path in the iron through the center of the coil, around the outside and back in the opposite side of the coil, the iron will not be of much use. A complete, detailed transformer design is more than you can expect as an answer here. You need to find a site or a book that will teach you most of what you need to know. Specific sites, books and products are not recommended here. \$\endgroup\$ Jan 22 '17 at 21:39
  • 2
    \$\begingroup\$ Wire guage is irrelevant, air core is silly, iron core must be continuous with no gaps, or you will make a lot of smoke, not a transformer. Look up transformer on wikipedia, plenty of pictures showing suitable continuous core configurations, none showing what you are talking about, go figure. \$\endgroup\$
    – Neil_UK
    Jan 22 '17 at 22:05
  • 1
    \$\begingroup\$ For a 220:6 V transformer with a 4 amp secondary, 32 AWG will be fine for the primary, but you need a proper transformer design. With nothing close to a proper transformer design, wire gauge is irrelevant. \$\endgroup\$ Jan 22 '17 at 22:24
2
\$\begingroup\$

General formula for inductance of a coil of many turns: -

enter image description here

1000 turns on a 50 mm diameter coil with maybe 10 mm length of winding in multiple layers will result in an inductance of between 50 and 250 mH depending on how you stack the turns up.

What does this mean at 230 V AC, 50 Hz and 100 mH? You have a current of 7.32 amps and that is too high - you cannot rely on the coil's resistance for making a good transformer - you have to rely on the inductance to keep primary magnetization current respectably low.

You need several henries inductance to connect to AC power lines and that is why folk use a laminated magnetic iron/steel cores. The iron/steel amplifies the inductance per turn massively so, fewer turns are needed to achieve the required inductance.

\$\endgroup\$
3
  • \$\begingroup\$ Thanks Andy. I had doubt with the resistance and inductance. You response clarifies how inductance gets presedence over resistance to stand AC Mains current. But what is the recommended range of inductance of primary coil to deliver 4amps at 6 volts from secondary? \$\endgroup\$
    – sribasu
    Jan 24 '17 at 10:07
  • 1
    \$\begingroup\$ @sribasu normally primary inductance is chosen so that the basic off-load current into the primary is not too great. Anywhere upwards of 5 henry is aimed for as a bit of a hand wavey approximation. This puts you miles off of course and that is why folk use iron/steel laminates (apart from the reason that with a good lump of high permeability material all the turns couple with each other +90% and inductance pretty much is proportional to turns squared). \$\endgroup\$
    – Andy aka
    Jan 24 '17 at 10:29
  • \$\begingroup\$ I've chosen... :) \$\endgroup\$
    – sribasu
    May 9 '17 at 11:11
1
\$\begingroup\$

The bottom line is that the coil will withstand the voltage only if it has enough inductance to limit the current to a safe level. The inductance is basically a function of how much iron you wrap the wire around. An air core will NOT provide the necessary inductance.

AWG34 wire can safely handle about 57 mA when tightly wound into a coil or transformer (using the rule of thumb of 700 circular mils per amp). That means that you need an inductive reactance of

$$Z=\frac{220 V}{57 mA}= 3860 \Omega$$

At 50 Hz(?), this would require an inductance of

$$L= \frac{Z}{2\pi f} = \frac{3860\Omega}{2\pi\cdot 50 Hz} = 12.3 H$$

Or about 10 H at 60 Hz.

\$\endgroup\$
1
  • \$\begingroup\$ Neat! Clarifies my Primary coil. Any help on the secondary one? \$\endgroup\$
    – sribasu
    Jan 22 '17 at 22:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.