# Transformer: current loss

1. ## Transformer: current loss

Hi friends.

I have now two transformers which both suɔʞ at what they do, so I have named the left as "sucky-I" and the right as "sucky-II" (the black tape is self-vulcanizing rubber, excellent at keeping high pressure under control) , both have a turn ratio of about 10:1.

My premise is the one of the two transformers suɔʞ less than the other, so one will be better at transforming than the other, and that difference must visible in the measurements and give a hint of why the current on the secondary side more or less is the same as primary, but voltage gets stepped down alright.

This is the test environment for both transformers

This is "sucky-I"

x-axis is time in nano-seconds

This is "sucky-II"

x-axis is time in nano-seconds.
I guess the trace would be prettier with 30-40% less turns.

Where is the current? Current should go up when voltage is being stepped down.

Then I tried to separate the lower legs of each winding so that primary and secondary are galvaniacally separated, which meant I could only measure one side of the transformer at a time due to lack of diff-probe. But what happened was that then the current on the secondary decreased by more than 40%, now why is that?

Here are the scope traces for "sucky-I" (but the same happens for "sucky-II"), the white trace is ch2 integrated and is the actual current but you have to scale it with -3,552560e6 to get the correct magnitude. ch4 is the voltage which needs to be scaled by a factor of 100.

The integral scale is 5µV/DIV as can be seen in the text on the center axis.

The integral scale is 2.5µV/DIV as can be seen in the text on the center axis.

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2. ## Re: Transformer: current loss

Hi

same current does not mean same power...
And with AC: V_RMS x I_RMS of primary is not V_RMS x I_RMS in secondary. There will be phase shift.

But yes, there will be loss. But it depends on a lot of parameters like frequency and waveform and current...
and maybe the one saturates earlier than the other...

Klaus

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3. ## Re: Transformer: current loss

so they are mains transformers (ie 50Hz)?
What is the coil resistance of each.
Also, cheap mains transformers have loads of turns stuffed in a tight space, and resultant large interwinding capacitance can mess up results and make them very much less than ideal....especially with high rise time pulses, which will just see the stray capacitance as a short.
Also, there may be saturation happening..

Also, the core material may be non linear and so the transformation would not be according to the ideal transformer equations.

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4. ## Re: Transformer: current loss

Originally Posted by KlausST
Hi
same current does not mean same power...
Klaus, I didn't say that I expected same power with same current. I said the voltage is stepped down by a factor of 10, but the current stays the same i.e. there is a ten fold power loss which includes the phase shift.

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Originally Posted by treez
so they are mains transformers (ie 50Hz)?
No they are both home rolled, "sucky-I" is an air-core, "sucky-II" is ferrite-core should be good up to 40KHz.

Originally Posted by treez
What is the coil resistance of each.
The only one I can measure without making a wheatstone-thingy is the primary of "sucky-I" with about 240 turns it comes to 1.1ohm, secondary must be ten fold less. "sucky-II" I can't measure the resistance is too small.

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Anyway I'm giving up - or at least take a very very long vacation, I'm tired, I have been at it for a long time now.

And when I cut this connection

The secondary current drops by 40% (see scope traces in OP), which could mean two things, either I don't know how to measure, and/or there are some physics going on that I do not understand.

Either case, it seems I'm royally scrćwed. So I think I will mark this thread as "solved" as there is no solution time being within my reach.

I take this opportunity to thank everyone for their help.

Maybe I will revert later with some problems for which there exist known solutions, always wanted to make a VFD and a servo motor controller...

5. ## Re: Transformer: current loss

The transformer are behaving as expected, for 1:1 construction.

p.s. you are lucky the insulation ( turn - turn ) has not broken down ...

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6. ## Re: Transformer: current loss

Sorry 10:1 constrction. At very high frequencies the coupling is very important, high coupling = low leakage inductance - for the Tx on the right there is very poor coupling and the high input current will saturate the ferrite quite early on - making it a very low coupled air cored Tx

the one on the left - I assume it is air cored? the magnetising inductance is very low ( have you measured it..? ) so there is a great deal of current in the pri that will never flow as load current in the sec - also depending on the separation of the windings the coupling will be poor for such high frequencies.

This job is a sitter for a matrix style ( or distributed ) transformer - where you need good coupling at very high frequencies - but also good voltage withstand ....

7. ## Re: Transformer: current loss

Originally Posted by Easy peasy
p.s. you are lucky the insulation ( turn - turn ) has not broken down ...
Not luck, but fingerspitzengefühl and rubber tape

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Originally Posted by Easy peasy
the one on the left - I assume it is air cored? the magnetising inductance is very low ( have you measured it..? ) so there is a great deal of current in the pri that will never flow as load current in the sec - also depending on the separation of the windings the coupling will be poor for such high frequencies.
The left one aka "sucky-I" is the air-cored one. I measured the "inductance" of the primary (with the secondary left open) which was 970uH but I'm not sure that is the same thing as "magnetising inductance" or is it?

8. ## Re: Transformer: current loss

How many turns did you need to get 970uH? on an air core - a fair few I'm guessing. Yes Lmag is that inductance measured with other windings open circuit, whether air cored or otherwise.

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B = u.H = Uo.Ur I. N^2 / Lmag. For air Ur = 1. Uo = 4.pi.E-7 or 1.245.E-6 Lmag = magnetic path length around winding (m).

once B reaches 400mT in ferrite it begins to saturate, and you have an air core ... and almost no coupling for the UU core construction shown ...

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