# How to calculate the output amperage of transformer

I am going to attempt to wind my own transformer, but I need to know the how to simply calculate the output amperage. I know how to calculate the voltage:

Np/Ns = Vp/Vs

Power = Voltage x Current

Vp = 240
Np = 100
Ns = 2

100/2 = 240/X
100/2 = 50
240/X = 50
240 x 50 = X
X = 12000
240/12000 = 0.02

voltageSecondary = 0.02V


But I don't know the amperage calculations. Keep in mind I am very young. I cannot find it anywhere on the internet. Thanks!

• I'm surprised you could not find: en.wikibooks.org/wiki/Electronics/Transformer_Design and there are many more. Perhaps this looks complicated, well that's because it is not so easy to design a transformer. Aug 30 '16 at 7:30
• Popular hobby items are scrap Microwave Oven transformer (MOT) then users strip the thin seondary High V windings that step up to high voltage and replace with heavy few turns for high current low voltage output based on turns ratio. Many videos and sites use this with step by step instructions. Remember safety first. Aug 30 '16 at 7:42
• How much iron have you got in the core? You estimate the total power throughput as a function of the weight of iron, or vice versa if you're starting with a power requirement and choosing a core. Aug 30 '16 at 7:58
• @tony Stewart - I tried this, but stripping the secondary coil with a hacksaw and drill took me 3 and a half hours, and even then I was only half way through. Sep 3 '16 at 22:58
• Oct 18 '18 at 3:12

You start by specifying what power you want through the transformer. A 1kW transformer needs much more iron than a 6 watt.

Given the power, you then choose a suitable core. Broadly, power goes as weight of core, over quite a wide range. Core manufacturers have tables that shows you core size required for any given power rating.

Only after you have the core size, then you can work out the primary turns. Too few and the transformer won't work, it will blow fuses or burn. Too many, and it will be inefficient. Calculate the primary turns based on your input voltage, and swinging the core over a 'reasonable' range of flux, typically +/- 1.5T for iron. Size the primary wire to fill half of the winding window.

Now you can calculate your secondary turns, as (Vout/Vin) * primary_turns. Size the secondary wire to fill the other half of the winding window.

If you have started with the right size iron core, then filling the winding window with copper (allowing for insulation, the bobbin, and a few gaps) will 'just work' to give you enough current handling capacity to meet your power specification.

• +1 for explaining how to actually design a transformer (instead of just analyzing the proposed design). Aug 30 '16 at 8:19

Amperage depends on load and thickness of secondary wiring. You have to specify amperage yourself and then use wire suitable to handle that current.

Power going to primary is the maximum power you can get from secondary (minus loses). So in short Vp*Ip=Vs*Is, then Is=Vp*Ip/Vs

• Could you supply a calculation showing this? Aug 30 '16 at 7:23

Ideally, if the turns ratio is 100:2 then the voltage step down ratio is also 100:2 hence 240 V RMS on primary becomes 4.8 volts RMS on secondary. If there is a load on the secondary of 4.8 ohms then the current in that load (due to ohms law) is 1 A RMS and this will be seen as a much reduced current on the primary of 20 mA i.e. when you step down voltage you use the step down turns ratio and this ratio is also used for determining the primary current for a current in the secondary.

In your calculation you got the algebra wrong i.e. Vs should be 240/50

Your bigger problem will be the 100 turns on the primary. Almost certainly this is too small for most modest home-appliance type transformers and will cause transformer core saturation. Probably a figure of about 1000 turns is more realistic.

For ratio = a:1

• a=Vp/Vs
• a=Ip/Is such that for ideal lossless VpIp=IsVs
• L inductance of primary must be high enough to only conduct 10% of rated current. This is the current used to couple primary and secondary through a steel laminate core.
• rated current is mainly from resistive loss, core saturation limits and temperature rise
• Vs*Is=VA rating of output which may be 10% less than input due to excitation current which occurs with no load.
• side with the most turns uses the smaller gauge wire.
• V*I or VA ratings depend on linear R load, but if using diode bridge and cap, it must be derated at least 30% depending on load and peak ripple current.
• isn't it a Vp/Vs = a = Is/Ip? You have a = Ip/Is. May 27 '18 at 18:30