True. The figures that I used were for the combined primary and secondary windings. I used the regulation quoted in the data sheet to workout the equivalent secondary resistance, and used that alone. The total copper loss is quoted as 18 W for primaries and secondaries.But note that the secondary copper loss for the rated RMS current is 8.36W total (both windings) so if you go above that value you may cause overheating and deterioration of the winding insulation.
What is the output current for those graphs?......................................
I ran some figures for this transformer, and I've attached a graph of the results.
...........................................
I should have said. They are all at 5 A d.c.What is the output current for those graphs?
True. The figures that I used were for the combined primary and secondary windings. I used the regulation quoted in the data sheet to workout the equivalent secondary resistance, and used that alone. The total copper loss is quoted as 18 W for primaries and secondaries.
At full load of 5.333 A, the secondary copper loss is 5.333^2 * 0.1469 = 4.178 W per winding.
The primary current is 5.333 * 15 * 2 /230 = 0.696 A, so the primary copper loss is 0.696^2 * 5 * 2 = 4.84 W
It doesn't quite add up, because those figures add up to around 13 W, while the data sheet quotes 18 W total copper loss.
Although I did calculations at 5 A load, they showed that the heating wasn't on the limit. You could go to 6 A and the RMS current is around the 10.666 A that is the limit for the transformer.Thanks Guys
From what I understand above this Toroidal is only good for 5Amps DC output when used in a Battery charging role ie: 100% load all the time it is working?
Regards,
tvtech
The time the diode conducts for is not easy to work out, nor is it a good measure of the ratio of RMS to average current. The current waveform isn't an easily categorised one. The attached file attempts to work out the waveform, assuming some series resistance. The RMS current is calculated.
To get a really accurate model you would have to include the diode models and inductance from various sources. However, I think that a simple resistive model is likely to be good enough, unless the other components are very marginal.
Series inductance, for instance, can be useful for reducing EMC, but will have virtually no effect at 50/60 Hz. Similarly, if the diode characteristics make any difference, the diodes are too small.
What i found (not surprising now) is that we need at least the following:
1. The spice model of the diodes
2. The primary and secondary resistances
3. The equivalent series inductance
4. Any additional inductance introduced on purpose
5. Line impedance
6. Of course the load resistance
I would just use a clamp-on current probe or a shunt and measure the (paralleled in this case) secondary current with a true RMS meter as Schmitt Trigger said in post #3.. If it's less than 10.6 amps (rated secondary RMS current) it should be ok.So what to do?.
OK ..I know you Guys mean well..
This is a thing though that was sorted many posts ag0
I do not need specifics of Diode Bridge drops et al..
This is a simple Linear thing. Like a nice question that I could not answer myself. Because I forgot.
Diver gave me the simple answer I was looking for. I really don't need anything more
Unless I plan on making this complicated...
Love you all,
tvtech
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