What is a usual value for a toroid transformer resistance?
Primary coil and Secondary coil?
In my SPICE you can set values for primary and secondary.
By default both is set to 10 mOhm.
Is there anyone here that has measured on a real typical toroid transformer?
Primary coil and Secondary coil?
In my SPICE you can set values for primary and secondary.
By default both is set to 10 mOhm.
Is there anyone here that has measured on a real typical toroid transformer?
Depends on the current and voltage rating of the winding. Copper is expensive, so transformers are optimized
for minimum copper by weight. Check the mfr specs for the particular transformer.
for minimum copper by weight. Check the mfr specs for the particular transformer.
There is hardly a usual value. It depends on how thick and how long a winding wire is. For example, i just measured one of mine i have on hand. 9Vac 0.6A, gives me 0.6R.
I guess bigger trafos have less than 0.6R.
What values are we talking about?
And is there a differance between primary and secondary?
What values are we talking about?
And is there a differance between primary and secondary?
I wrote the above 0.6A 9Vac. If it were for example 3A and 9Vac, thicker wire would be used, while number of windings would be the same, and you'd have less R value.
Of course, they won't be the same value. Secondary and primary aren't connected.
They are indeed connected - by the turns ratio squared. Suppose the turns ratio is 5. So there are 5x as many turns on the primary as compared to the primary.
So to get the effective primary resistance you (in this case) you add 1/25 of the primary resistance to the secondary resistance.
Again just as a for instance with these numbers, lets say that the secondary resistance is 5 ohms. The primary resistance is 100 ohms and the turns ratio is 5. So the effective secondary resistance is 5+ 100/25 = 9 ohms.
Now that rather extreme example would be a poorly designed transformer.
But picking up a transformer I've just bought, the primary resistance is 31 ohms at 240V, and the 250V secondary winding (this is a tube/valve transformer) is 126 ohms. Now the turns ratio is about 1, so I simply add the primary resistance to the secondary resistance and get 126 + 31 = 157 ohms. And that is the effective resistance you have to use when designing a power supply.
So to get the effective primary resistance you (in this case) you add 1/25 of the primary resistance to the secondary resistance.
Again just as a for instance with these numbers, lets say that the secondary resistance is 5 ohms. The primary resistance is 100 ohms and the turns ratio is 5. So the effective secondary resistance is 5+ 100/25 = 9 ohms.
Now that rather extreme example would be a poorly designed transformer.
But picking up a transformer I've just bought, the primary resistance is 31 ohms at 240V, and the 250V secondary winding (this is a tube/valve transformer) is 126 ohms. Now the turns ratio is about 1, so I simply add the primary resistance to the secondary resistance and get 126 + 31 = 157 ohms. And that is the effective resistance you have to use when designing a power supply.
There is no physical conductor contact/connection. To write it in obvious manner, which can be deducted from "not the same value".
There is no connection by a piece of wire, but they are connected by magnetic coupling. That is how a transformer works.
Read a book about it if you don't understand.
Read a book about it if you don't understand.
I understand correctly, and i'll repeat myself, no physical connection. Without going further into how transformer works. It was you who jumped the boat with bad interpretation, now just being rude.
Edit: Since i have to state it in a more obvious manner, regarding bad interpretation, connected =/ coupled. Not my main language, but should be obvious to natives.
Edit: Since i have to state it in a more obvious manner, regarding bad interpretation, connected =/ coupled. Not my main language, but should be obvious to natives.
Last edited:
the winding resistance is correlated to the copper losses. Suppose you have a 200VA transformer having 10% total losses. Ideallly, copper and iron losses should be balanced, thus 10W for the copper, balanced between primary and secondary, meaning 5W for the primary.
The current for 200VA and 230V is 200/230 is 0.87A. To dissipate 5W at 0.87A the resistance needs to be 6.6ohm.
The copper losses depend heavily on the size of the transformer: a miniature 1VA type might have 40%, but a 1kVA might have <4%.
Some additional details here:
The current for 200VA and 230V is 200/230 is 0.87A. To dissipate 5W at 0.87A the resistance needs to be 6.6ohm.
The copper losses depend heavily on the size of the transformer: a miniature 1VA type might have 40%, but a 1kVA might have <4%.
Some additional details here:
I think I have not got an answer.
What is the typical resistance for say a 2x15VAC 100VA toroid? Primary and secondary.
Or like that.
6.6 Ohm like Elvee mention seems very high ...
What is the typical resistance for say a 2x15VAC 100VA toroid? Primary and secondary.
Or like that.
6.6 Ohm like Elvee mention seems very high ...
No it's not very high, but the variation can be high, depending on the materials, construction and configuration.
A top-notch toroid using the best magnetic tape and optimal winding will outperform this example, whilst an older EI using traditional techniques will incur twiice that loss level.
Not all transformer are made equal, but using some techniques, you can guesstimate what the end result will be
A top-notch toroid using the best magnetic tape and optimal winding will outperform this example, whilst an older EI using traditional techniques will incur twiice that loss level.
Not all transformer are made equal, but using some techniques, you can guesstimate what the end result will be
I tested a big potted multi secondary transformer. It was worth it as a number of them were being offered FOC at the time.
I measured the DC resistance and the AC output for each secondary winding.
By ohms law I calculated the copper loss at 5% and assumed the iron losses to be another 5%.
Time flies. It was more than 10 years ago.
https://golbornevintageradio.co.uk/forum/showthread.php?tid=4809
I measured the DC resistance and the AC output for each secondary winding.
By ohms law I calculated the copper loss at 5% and assumed the iron losses to be another 5%.
Time flies. It was more than 10 years ago.
https://golbornevintageradio.co.uk/forum/showthread.php?tid=4809
You have not got an answer because you are asking all wrong questions.
Are you interested in dc resistance? Measured by digital multimeter? Because that is very easy to measure, is totally different for primary and secondary winding, is specific for every type of transformer, ...and is totally useless except to check for continuity as for burn out wire.
AC transformers are not meant to see dc, therefore any resistance measured by dmm using dc is useless.
You need to measure impedance using sine wave at 50/60 Hz. Use dats or any other method.
If I add 6.6 Ohm as secondary resistance in my SPICE
the 12VAC transformer puts out only a few Volts at 2A load.
2A x 6.6 Ohm = 13.2 Volt
the 12VAC transformer puts out only a few Volts at 2A load.
2A x 6.6 Ohm = 13.2 Volt
If you draw 2 A the voltage probably gets "negative". A 12 V 6.6 Ohm Rdc transformer cannot provide 2 Amps, even if shorted.
Why not give PSU2D a go? https://groups.io/g/duncanampspsud/files/psud2_windows designed for power supply simulation.
For info , I have an 1 x 9.5 v 100 va that have 1.75 ohm at the secondaries and 32 ohm at the primary ( 230 v ) , and an 2 x 24 v 100 va that have 2 ohm ohm at the secondaries and 30 ohm at the primary ( 230 v ) , all toroids .
.
Last edited:
I have given a method to derive the main parameters based on the power and construction, but IIRC there is a small mistake somewhere. I have posted the correction somewhere, but the search tool of the forum is useless, and I cannot locate it.