Stepdown Transformer - Turns Vs. Ampere

Thread Starter


Joined Mar 18, 2010
In small step down transformer say 500 mA to 1 Amp. @ 14V, I want to be sure about whether the Ampere increases with increase in Turns.
From a Data Table I found that 38 SWG copper wire is having current capacity of 0.0365 @ 200 amp per and also there is 3507 turns per
So, If I wind 4000 turns then I arrive at current 0.0416. Am I right or wrong. Kindly help.
Raw formula being: (0.0365/3507) x 4000 turns. = 0.0416
It means current increases with increase in turns.
Awaiting reply, I remain


Joined Jan 8, 2017
You are totally wrong. 38 SWG wire has a diameter of 0.152 mm (Radius = 0.076 mm) so it's cross sectional area π x R^2
= 3.142 x 0.076 x 0.076 = 0.01815 sq mm so at a current density of 200 amps per square cm (100 sq mm) so the current carrying capacity would be (200 x 0.01815)/100 = 0.0363 amps. (Which is close the the value that you found.) It does not matter how many turns you have the wire can only carry 0.0365 amps. (You may have to reduce that rating for a very large winding to allow for the heat produced at the centre of the winding to escape without causing an excessive rise in temperature.
The 3507 turns per sq cm is the number of turns that will pass though a space of 1 square cm.



Joined Sep 24, 2015
I'm not the "Transformer Guy", but I can tell you what I know - or what I THINK I know. A transformer, as you know, takes one voltage and transforms it into a different voltage (in most cases). In a step down transformer, the number of turns in the primary side is set by the expected primary voltage. The resistance of the wire needs to be considered so as to not build a HOT transformer. By that I mean if you make a 120 V transformer with just 10 turns on the primary side, there's going to be a tremendous amount of current flowing through the primary. So there needs to be enough resistance in that wire so as to not excessively heat up. Now, here's why I say I'm not the transformer guy: There's also inductance to be considered, and I'm 100% NOT qualified to discuss inductance. Nevertheless, assuming you have the right amount of windings AND proper size of wire for the primary, the secondary has to have a turns ratio adequate to achieve the voltage wanted on the secondary side. For instance, a 12 volt transformer with a 120 V input has to have a turns ratio of 10:1 (ten to one). That means for every 10 turns on the primary side you have one turn on the secondary side. That's how you get your transformation from 120 V to 12 V. BUT WAIT! THERE'S MORE.

There's a number of factors going into a transformer. If you have a 120 V primary rated at 100 watts, you have a primary side that draws (in theory - there are inefficiencies I'm going to ignore for the sake of clarity) the primary side is drawing 833 mA of current. When the voltage is transformed on the secondary side you still have 100 watts. Therefore the amount of current at 12 volts is still going to be equal to that 100 watts. (again, ignoring inefficiencies) So at 12 volts, 100 watts should produce 8.3 amps of current. THEREFORE, the size of the secondary wire size needs to be able to handle that amperage. And I know I'm going to get lambasted for being in error, I welcome the corrections to my statements. It's how I learn new things. Nevertheless, it's that input wattage that dictates the output wattage. BUT WAIT! THERE'S STILL MORE.

The type and size of the iron core ALSO plays a part in how much current that transformer is capable of delivering. Different types of materials, ferrous in nature, will play out on how efficient the transformer is going to be. ALSO, in a standard transformer there's also something known as "Shunts". They bypass some of the electromagnetism thus limiting the amount of current the transformer can handle. Now, here's where I say I will tell you what I THINK I know. I'm not 100% on the shunts. I just know that if they're not there, in some cases they can cause the transformer to overload. I've seen that happen many years ago. A company I worked for bought some replacement transformers for units they routinely serviced. Once those transformers were placed into service they would immediately begin to overheat and fail within hours. I heard it had something to do with improper shunting. Again, I may be wrong; someone who knows more about it will most definitely correct my statements.

My goal in even commenting at all is to share in an understanding level comparable to my own understanding is this: Transformers produce voltage based on the turns ratio, not the size of the wire. Current is based on the size of the core of the transformer and the wire's ability to handle that current. And the amount of power coming out of a transformer can not be greater than the power going in. By power I mean wattage. 1 amp in and 10 amps out means at 120 volts, 1 amp in is equal in power (wattage) to 12 volts, 10 amps out. Multiply the voltage by the amperage and you get the wattage. The wattage can never go up. But due to inefficiencies of the transformer, the iron core, the copper wire, the amount of energy lost as heat, that wattage on the output side WILL BE less than the wattage going in. That's what I meant when I said I was going to ignore the inefficiencies for the purpose of clarifying how a transformer works. Again, this is my SIMPLE way of understanding how a transformer works. It may not clear up everything but hopefully it clears up your notion that more windings will increase the amperage on the output. In short, it won't. More windings will increase the voltage out but will reduce the amount of amps out. The wattage will remain basically constant.
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