# transformer resistances

Discussion in 'Electronic Basics' started by davidt, Feb 25, 2008.

1. ### davidtGuest

I have an transformer from a old power supply.
It has 2 red wires and 2 yellow wires.
I think the red wires are the high voltage (mains) in and the yellow
the low voltage (12v ?) out,
To check this I touched a 6V battery across the yellow wires and got a
high voltage across the red wires (enought to spark across a 3mm gap)
But when I measure the resistance across the red wires I get 31.1ohms
and across the yellow 10.4ohms.
This is not what I expected.
I thought the high voltage winding would be thicker wire and less
turns than the low voltage winding and so would have a lower
resistance.
Could anyone explain where I am going wrong?

2. ### amdxGuest

You have it backwards. The high voltage side will have more turns
of smaller wire. (it carries less current) The low voltage side will have
less turns if thicker wire. (it carries more current)

3. ### davidtGuest

Yes that makes sense now when think about the current.

4. ### Jon SlaughterGuest

Only smaller wire because the current requirements are smaller. There is no
restriction on the wire size for a transformer except that it be able to
handle the current. (so it definitely could have been larger)

5. ### John FieldsGuest

---
Sure, but since what a transformer does is transfer _power_ from one
winding to another, the product of current and voltage have to be
the same for both windings.

So, if the voltage in one winding is higher than in another, then
the current in the high-voltage winding _must_ be less than in the
low-voltage winding and it would be pointless to use wire with a
diameter more than that required to carry the current.

More precisely, for a lossless transformer:

Np Ep Is
---- = ---- = ----
Ns Es Ip

Where: Np is the number of turns on the primary,
Ns is the number of turns on the secondary,
Ep is the voltage impressed across the primary,
Es is the voltage induced across the secondary,
Ip is the current forced through the primary, and
Is is the current into the load

6. ### JamieGuest

The resistant you're measuring has nothing to do with the operation
of the transformer other than lowering its efficiency and Q. It's simply
a by product of the gauge wire used along with the length of wire used
for each winding.

The number of turns ratio between the Red windings and yellow windings
tells the whole story along with the gap spacing between them..

At this point, maybe you should reference some data from the net on
transformer theories.

8. ### JamieGuest

my point was short and simple. difference in current mode xformers and
potential types along with turn ratio's. also the construct of the
cores. With all this put together leads to PF (power factors), current
ratio's, form factor, wet/dry, efficiency etc..

Keeping it simple since the poster obviously needs to research some
transformer basics.

9. ### Phil AllisonGuest

"davidt"

** If it is from a 240 volt mains power supply, then the red wires are for
240 volts AC and the yellow ones are for 140 volts AC. The unit is rated

If from a 120 volts supply, the yellow wires are for 70 volts AC and the
unit is rated for about 25 VA.

........ Phil

10. ### Jon SlaughterGuest

And why not Np/Ns = D_Rs/D_Rp where D_Rx is the diameter? or even Np/Ns =
Rs/Rp?

The fact of the matter is that it doesn't matter as long as the as the wire
can carry the current(which essentially means heat dissipation).

You can use any wire size in any circumstance and its not *pointless*. It
depends on the application. True that you pretty never need a larger
diameter on the secondary than the primary in a step up... but potentially
there is still the possibility in certain applications. (such as if the
secondary happens to be in a much much hotter environment)

The idea transformer equations say nothing about the diameter's of the wire
used and for good reason. (not to mention that most wire sizes are
standardized)

So going off the restance or the diameter of the wire is by no means correct
and technically has nothing to do with a abstract ideal transformer. (I'm
not saying you can't use that relationship to find the optimal diameter for
a given number of turns but it is not necessary)

11. ### John FieldsGuest

---
Because the turns ratio has nothing to do with the resistance ratio.
---
---
Well, no. What _really_ matters is the resistance of the winding,
since that'll determine the voltage dropped across it, and that
voltage won't be able to be used by the load. While the I²R losses
increase the winding's resistance, it's not usually enough to
matter much. For a clue, take a look at the temperature coefficient
of resistance of copper and translate that to the change in
resistance of a winding due to the change in temp. In any case, the
temp rise because of I²R losses should be taken care of during the
design of the transformer and the proper wire size chosen to give
the desired regulation at that maximum temperature.
---
---
You _cannot_ use any wire size in any application and expect the
application to work. For example, it would be pointless to wind a
secondary expected to supply 12 volts at 10 amperes with a single
strand of #40 AWG.
---
---
Precisely. And, for the transformer the OP was talking about,
(which he was using as a stepup transformer to get the results he
did) it would have been pointless to have wound the high-voltage
winding with larger diameter wire than the low-voltage winding.
---
---
Like where??? You're grasping at straws.
---

12. ### Jon SlaughterGuest

No! Why you think he is winding anything? They are already wound and it was
wound for a specific application. In general you are correct in that for a
step up the wire diameter will be smaller because there is less current. The
point I'm trying to make is that it is not a necessary or sufficient
requirement for step up transformers. maybe in 99.999% of transfomers it
will hold but even then there is no "formula" to calculate the diameter of
the secondary from the primary or vice versa.

I could easily make a transformer that would violate your "principle" and it
would work fine! Sure it might be cost or space inefficient but the
tranformer itself doesn't care.

Now you say "Well if its too small then it will over heat"!! This has
nothing to do with basic ideal transformer theory which says nothing about
wire's resistances or anything. Hell, if I wanted I could then just use
super conducting material and use infinitely thin wire on both sides!!

I think thats what you are doing. You are trying to make the OP think that
the diameter matters and gives a good approximation to the turns ratio and
it doesn't. Go take a transformer and measure the diameters then take the
ratio and I bet you won't get a good approximation to the turns ratio! I
DARE YOU!

Since I know you are chicken I did it for you. I have a tranformer on my
desk. Its a step down from 120:12(approximately) so its about 10:1.

The diameter of the primary is 100mil, the diameter of the secondary is 40
mil.... hmmm, is this 10:1 or 1:10? not even close!

Of course now you will claim that the person who made the transformer is
nutz and has no clue about transformer design.
WTF? I asked why that isn't shown in the books... I'm ASKING you why that
isn't true because you are essentially claiming it is. Thats why I have
question marks there.

Forget it... I can see this is a lost cause.
Damn, I really thought I put you on my ignore list... along with the other
morons. Well, you won't escape this time.

14. ### amdxGuest

So your transformer has #10 wire on the primary and #18 wire on the
secondary? I think your secondary will have about 60% or less of the
resistance of the primary.
Please post the resistances of the primary and secondary windings, that way
I can see how I misled the questioner.
What is the use for this transformer? It's not like anything I'm used to
seeing.
Thanks, Mike