I got a small step down transformer from my local market which has a total of 6 individually isolated winding in it (1 primary of 230V and 5 different secondary wingding with 5 different voltage).

My question is that how can I find out the maximum output current for the secondary winding? I was told generally these used to be made for "Small Sound Systems". As this from a local market it has no datasheet. Transformer weights about 500 grams and looks well made. The dimensions of the transformer are near about 6cm X 3cm X 5cm and all the secondary wires are less than 1mm in dia.

This is my first question on stackexchange. Thanks in advance for all help and suggestions!

The winding voltages and resistance measured are as follows:

  1. Primary Winding (Red thick wire) 230V with a resistance of 72.6 ohms
  2. Blue Wire 8.8V with a resistance of 1.4 ohms
  3. Yellow Wire 11.1V with a resistance of 2.0 ohms
  4. Green Wire 12.8V with a resistance of 2.7 ohms
  5. White Wire 23.1V with a resistance of 9.1 ohms
  6. Red wire 3.7V with a resistance of 1.8 ohms

These are all the output voltages when the transformer is given 230VAC as input.

I tried measuring the maximum output current by a short circuit test (Using some high watt ceramic resistance in series with my DMM in amp mode).But unfortunately the DMM is not showing anything, I double checked my DMM for a blown fuse inside but it was working properly when I tested a DC load.

I want to use this transormer for building up a variable power supply circuit, but without getting sure of it's maximum output current of the secondary windings, I can't proceed further.I am not that much experienced of power electronics hence can't figure out what to do next.

Here are some pictures of it Transformer Top View enter image description here enter image description here enter image description here Thanks again for helping me,

Best Regards,


  • \$\begingroup\$ Your DMM shows nothing because transformers output AC. If your DMM is in DC mode, it'll average the positive and negative halves of the AC signal for a net sum of zero. Switching to AC mode will take the absolute value and adjust for the sinusoidal shape of the waveform. \$\endgroup\$
    – Harry Tsai
    Commented Mar 17, 2018 at 1:07
  • \$\begingroup\$ Thanks for your reply, Yes, you were right ! I tried measuring it with a DC DMM in Amp mode that's why it showed nothing....Thanks a lot for pointing out my mistake @Harry Tsai \$\endgroup\$ Commented Mar 17, 2018 at 18:52

6 Answers 6


I got a small step down transformer from my local market which has a total of 6 individually isolated winding in it (1 primary of 230V and 5 different secondary wingding with 5 different voltage).


As this from a local market it has no datasheet.

A transformer that drops AC dangerous voltages to much safer lower voltages is: -


So, if it doesn't have a data sheet and you bought it from a market I have to point out that you are risking your life using it. I would not use this device at all. Go buy one with a data sheet from a reputable dealer.

  • \$\begingroup\$ Thanks for your reply, I know it's not safe to work with this but I searched a lot for this kind of transformers in the market from some reputable shops with a proper data sheet, but unfortunately I ended up getting this transformer without any data sheet (Bad luck of mine, as almost all of them are made in china having no data sheet☹️) \$\endgroup\$ Commented Mar 17, 2018 at 18:47

First of all, a small transformer like that is going to be limited to tens of watts total power, based on the core size alone.

Determining how that power is distributed among 5 separate windings is going to be tricky. You can start by estimating the relative wire gauge by looking at the ratios between voltage and resistance for each winding — from that, you can infer what relative current values the designer had in mind.

This basically boils down to calculating \$V^2/R\$ for each winding and then figuring out what fraction of the total power it represents. In this case, we find that the four highest voltage windings each get about 24% and the low voltage winding gets about 3%.

If we generously assume that the transformer can handle a total of 50W, we get:

  • 8.8V @ 1.4A
  • 11.1V @ 1.1A
  • 12.8V @ 0.94A
  • 23.1V @ 0.52A
  • 3.7V @ 0.41A

To estimate a transformer's throughput, weigh it, find a similar build transformer of the same weight in an online catalogue, and read off its VA rating.

The decide how to partition that total VA rating between secondaries, calculate \$\frac{V^2}{R}\$ for each secondary winding, where V is the open circuit voltage, and R is the DC resistance of that winding. This will give you the relative power rating of that secondary winding compared to the other windings. Scale the total of those to the total VA for the transformer. Although it's approximate, it's much better than guesswork, or trying to estimate the wire gauge from the leads.


Technically I believe it is the RMS resistive load current at 85'C winding temp.

This may help get a rough estimate.

Measure Voc ( open circuit) and secondary DC resistance (DCR) then use the max current as the point where voltage drops 10%.

Note that for a DC bridge this is not true as Peak to Average drops it another 40% resulting in 50% drop from no load to full load.

Max power is usually rated as VA (=W+VAR) output which depends on total losses and rated insulation temperature such as 85'C.

But VA =W only for linear loads and must be de-rated for Diode bridge cap loads by some 40% depending if DC ripple is very low <<10%

My quick and dirty estimate. For a rough estimate use 10%Voc/DCR = I max rms.

Another Rule of Thumb

Small power transformers have a total weight is mainly steel core which has certain losses according to grade of core laminates and balance from copper weight and losses.

My Rule of thumb is 36VA ~ 60VA per kg.

How much does yours weight?

  • \$\begingroup\$ Thanks for your answer ! So, according to you for a rough estimate I should do the calculation like 10% of 8.8 by 1.4 ohms = 0.88/1.4 = 0.628mA (Max) ? \$\endgroup\$ Commented Mar 16, 2018 at 15:14
  • \$\begingroup\$ if that is your Voc/DCR yes which implies a VA Rating of 8.0*.628= 5VA where 8.8-10%=8.0 rated load voltage so you can test this rule of thumb with V/I= 12.7 Ohms and verify Voltage but it depends on how each winding adds up to contribute to core loss where the total core VA rating will be less than the max sum of each winding. \$\endgroup\$ Commented Mar 16, 2018 at 15:23
  • \$\begingroup\$ Thanks for your reply, my transformer weights around 580 grams, so as per your guess it should be between 18~30VA, Right ? \$\endgroup\$ Commented Mar 17, 2018 at 18:39
  • \$\begingroup\$ you got the idea now....... \$\endgroup\$ Commented Mar 17, 2018 at 20:43

This looks suspiciously like those marketed for DVD players. I bought one and it was disappointing. The core heated up with no load. I measured the current draw on the one I bought....

color    Vrating   Vopen    current draw at rated voltage (mA)
red      230
white    11.5V     12.0     150
yellow   21.5      20.8     !!!
black    3.5       4.0      150
blue     13.5      13.8     50
green    12        12       50
purple   10        10.1     100

You may or may not have something similar...

From experience, there is only one way to figure out maximum secondary current output. Make an electronic load and measure it. Make a plot of VxI, and VxVA. Linear region of the VxVA plot will show the max current that can be safely drawn. The VxI plot will be absolutely linear (one can possibly use this property to test transformer voltage drop at the market, before buying). Double check with a thermometer after the transformer has warmed up under load. Make sure not to overlook airflow from table/ceiling fans, overly hot transformers might require air cooling. Most enamel coatings are rated at 130C, with a safe limit of max 85C. The copper winding will generally feel cooler than the core.

Also, plugging in the primary and secondary inductances and resistances into ltspice, gives a ballpark figure of the voltage drop under full load. However, it is not too useful. Also, calculating only with secondary resistance is inaccurate at low currents. This is because the current will be limited by the inductance ratio, and not the wire diameter.

PS: Here's my first question on stackexchange How to find step down transformer secondary output current rating?


Perhaps by monitoring for the 10% drop (suggested by a previous post) from the open circuit voltage with a varying load, the current measured at the 10% point might be the best ballpark figure.


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