I have an old power transformer that I'm retrofitting into a new valve amplifier. Having measured the voltages on the secondaries when 230V is applied to the primary and got the following results:

P: 230 V
S1: 356.2 V
S2: 34.56 V
S3: 6.453 V

The turns ratio of each winding is therefore:

S1: 356.2 / 230 = 1.549
S2: 34.56 / 230 = 0.150
S3: 6.453 / 230 = 0.028

The current flowing in primary at 230V was 0.07A. Using the turns ratio I we can calculate the current flow in each secondary:

S1: 0.07 * 1.549 = 45 mA
S2: 0.07 * 0.150 = 460 mA
S3: 0.07 * 0.028 = 2.495 A

This seems correct at first, but using these figures the combined power output of the secondaries would be three times that of the primary which is obviously incorrect as Vp * Vp = Vs * Is.

Two of the questions to this answer suggest that we can calculate the VA value as V^2 / R, sum the three VA values and then us the result to calculate how the VA of the primary is distributed between the secondaries. In this case the resistance would be DC resistance of the primary multiplied by the square of the turns ratio, plus the DC resistance of the secondary.

The measured DC resistance of each winding is:

P: 7.5 R
S1: 28.60 R
S2: 3.10 R
S3: 0.10 R

Therefore the total resistance on each secondary is:

S1: ( 7.5 * (1.549 ^ 2) ) + 28.60 = 46.59 R
S2: ( 7.5 * (1.150 ^ 2) ) + 3.10 = 3.27 R
S3: ( 7.5 * (0.028 ^ 2) ) + 0.10 = 0.11 R

Therefore the VA rating of each winding is

S1: ( 356.2 ^ 2 ) / 46.69 = 2723
S2: ( 34.56 ^ 2 ) / 3.27 = 365
S3: ( 6.453 ^ 2 ) / 0.11 = 393

Therefore the total VA rating is 3482 and the power distribution between each secondary is:

2723 / 3482 = 78.2%
365 / 3482 = 10.5%
393 / 3482 = 11.3%

So now we can revise the current values we calculated earlier:

S1 = ( 0.07 * 0.782) * 1.549 = 0.035 A
S2 = ( 0.07 * 0.105) * 0.150 = 0.049 A
S3 = ( 0.07 * 0.113) * 0.028 = 0.282 A

This balances the power on each side of the transistor however it doesn't really correlate to reality. The third winding is for the valve heaters and given that his amp contains 4x 12AX7 and 4x EL34 the current required is going to be around 5.4 A however that's clearly not the case. The first secondary should also be much higher, in the range of 500 mA.

Where have I gone wrong? Is my original current measurement on the primary wrong? Do I need to be considering the load into which each secondary will be delivering the current?

Thanks for any help.

  • 3
    \$\begingroup\$ Did you actually have any loads connected to any of the secondaries when you took this measurement? \$\endgroup\$
    – brhans
    Commented May 29, 2018 at 21:20
  • \$\begingroup\$ what's the weight of your transformer? Is it EI, or toroidal? \$\endgroup\$
    – Neil_UK
    Commented May 30, 2018 at 14:32

2 Answers 2


The current flowing in primary at 230V was 0.07A. Using the turns ratio I we can calculate the current flow in each secondary:

S1: 0.07 * 1.549 = 45 mA
S2: 0.07 * 0.150 = 460 mA
S3: 0.07 * 0.028 = 2.495 A

I didn't read through the whole post but I suspect that your secondaries were unloaded (open-circuit) and the actual secondary currents therefore were zero. Unfortunately this renders the rest of your post irrelevant!#

enter image description here

Figure 1. A 50 / 60 Hz RS 100 VA transformer. Dimensions: 89 x 68 x 75 mm. Weight: 1.6 kg. Source: RS-online.

You don't give any indication of the dimensions of your transformer. With 230 V x 0.07 A you have a load of 16 VA which might be reasonable for a > 100 VA transformer. (A 100 VA transformer would weigh about 1.6 kg.)

Your primary current is just magnetising losses on the primary.

  • \$\begingroup\$ I just measured it to be 115 mm x 50 mm x 96 mm. How can I use that to help estimate the VA rating? \$\endgroup\$ Commented May 29, 2018 at 21:34
  • \$\begingroup\$ With these measures and given it's 50/60Hz, it's roughly a 60VA transformer. \$\endgroup\$
    – Janka
    Commented May 29, 2018 at 21:40
  • 1
    \$\begingroup\$ How do you know that? \$\endgroup\$ Commented May 29, 2018 at 21:41
  • \$\begingroup\$ From the mechanical size and the frequency. There are slight size differences between EI cores and (split) toroid cores of the same power rating but in general, the size of the core is a good measure for the power a 50/60Hz transformer can transfer. For higher frequencies, the core shrinks but the material has to allow it. \$\endgroup\$
    – Janka
    Commented May 29, 2018 at 21:50
  • \$\begingroup\$ Any chance you could link me to somewhere I can read up on that? And is there any means of accurately calculating the VA rating or is it just a case of estimating if you're without a datasheet? \$\endgroup\$ Commented May 29, 2018 at 22:02

The steps for estimating the parameters of an unknown mains transformer, where you already know the primary voltage (Vpri), are as follows.

1) Estimate the approximate total VA (VAtot) from its weight (as per this link you identified).

2) Check the transformer is 'OK' by measuring the primary current when all the secondaries are unloaded. This is the magnetising current (Im), and should be well below the running current (Ipri_full_load) you'd estimate from the ratio VAtot/Vpri.

On a big toroidal it may be a few percent of this, but on a tiny transformer may be an appreciable fraction. If it's too high, the transformer will get hot, even on no load. If it's very high, you may have shorted turns. If it's very very high, you may have Vpri wrong.

3) Measure the off-load voltage on each secondary to calculate the turns ratios.

4) Measure the resistance of each secondary and use the formula V^2/R for each secondary to apportion the estimated VAtot (not your measured Im*Vpri) to individual nominal VAs (VAsec) for each secondary.

This step does not tell you the current flowing through the secondaries.

This step tells you, for a fully loaded transformer with all secondaries in use, what load currents should not be exceeded to avoid overloading any particular secondary. It's up to you to make sure the loads don't draw more than that current. The transformer will attempt to supply whatever current the load tries to draw, at that secondary voltage.

The main limitation on a transformer is thermal, not current density. If you are to use only one secondary, then you can allow the load on that secondary to draw a bit more than its estimated VAsec. How much more? You will need to measure the temperature rise of that secondary, by measuring its hot resistance after coming to thermal equilibrium at your desired current. The resistance of a copper winding increases by roughly 10% for every 25C increase in temperature. I normally limit the temperature in a unknown transformer to 70C, you may feel more, or less, ambitious. Start with a load of VAsec, and increase from there. Temperature rise will be more or less proportional to load^2 (as copper heating is given by \$I^2R\$), so there's not much scope for a large increase.

In your case, Janka estimates your VAtotal as 60VA from the dimensions, though Transistor guessed at 100VA from the magnetising current. If you had provided a weight, that would have been better. Let's use 100VA as a round number. Your V^2/R calculation has already split the VAs as 80/10/10 (note the number of significant figures I'm using for the estimates), so the 350v secondary is allocated 80VA, the 35v secondary 10VA and the 6.3v secondary 10VA . Your measured 70mA for Im is quite reasonable, being 16% of VAtotal. When you load your transformer, keep the currents below 80/350 = 230mA on 350v, 10/35 = 300mA on 35v, and 10/6.3 = 1.6A on the heater winding. Measure your winding temperatures on load using the resistance method for better peace of mind. If you do need 5.4A for your heaters, then this is not the transformer for you.

  • \$\begingroup\$ +1. A minor correction: there was no weight or dimensional information in the question when I answered. My 100 VA was just a guess based on the OP's 16 VA losses. \$\endgroup\$
    – Transistor
    Commented May 30, 2018 at 9:13
  • \$\begingroup\$ Neil, thanks for your help. That all makes sense except for the fact that the transformer has been pulled out of an amplifier that would require 7.2A for the heater winding! You check out the schematic in the link below, but for brevity it's come from a Carlsbro TC100 amp that contains 3x 12AX7, 1x 12AU7 and 4x EL34. chambonino.com/carlsbro/carl100tc.html \$\endgroup\$ Commented May 30, 2018 at 17:24
  • \$\begingroup\$ What's the weight of the transformer, which might yeild a better VA guess. The final arbiter of transformer throughput is the winding temperature at a given throughput, all this stuff above is just estimates. Load the 6.3v secondary with 5A, and measure its temperature after a few minutes. If it's still not too hot, let it run longer, should be stable within the hour. If it doesn't get too hot, you can run it at that. \$\endgroup\$
    – Neil_UK
    Commented May 30, 2018 at 18:56
  • \$\begingroup\$ Well I just hooked it back into the amp with all the tubes loaded and it pulled 6.5A at 6.03V. Didn't warm up at all so it seems like it can handled that much current. Not sure why it's 6.5A and not 7.2A as I would expect though. \$\endgroup\$ Commented May 30, 2018 at 20:15
  • \$\begingroup\$ What's the weight of the transformer? \$\endgroup\$
    – Neil_UK
    Commented May 30, 2018 at 20:40

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