160 Va Toroidal with 15V Secondaries

crutschow said:

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.

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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.

I've gone back and looked at RMS secondary current. The RMS current does depend a bit on the series resistance. If there is very little series resistance the capacitor charges very fast, increasing the RMS current. With an equivalent resistance of 0.152 Ω, there is 13.3 W of heating at 5 A DC, which represents 74% of the maximum heating. The RMS secondary current is 9.35 A, which gives 77% of maximum heating.

I think that either RMS current or transformer heating is a valid approach, but estimating the current waveform is not all that easy, and it makes a big difference to the peak current.

Diver300 said:

......................................
I ran some figures for this transformer, and I've attached a graph of the results.
...........................................

Click to expand...

What is the output current for those graphs?

crutschow said:

What is the output current for those graphs?

Click to expand...

I should have said. They are all at 5 A d.c.

Hi Guys

What can I say to here that have helped except thank you
And a little Rep from me.

Anyway, the big battery is not here yet. Will be arriving probably Thursday or Friday.

And then I am going to work with it and the Toroidal and see how all pans out.

As always, you Guys are the best out there. No exceptions.

You all know your stuff. And are prepared to show me but not shove it down my throat.
Now, that is true for all Pro's.

Final decision is mine. God, I love this place and all it's Members

It just makes me so happy to talk Technical stuff here (I do it very rarely) and have the likes of you all trying to help me without guessing. I love this kind of attitude.

Thank you all.

Your Friend,
tvtech

Diver300 said:

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.

Click to expand...

At full load, the temperature of the primary and secondary will be considerably higher than 25 degrees, and hence so will the resistance be.

Diver300 said:

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

Click to expand...

The no load Vsec should be used to calculate the turns ratio, so the expression should be:

The primary current is 5.333 * 16.63 * 2 /230 = 0.7712 A

Diver300 said:

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.

Click to expand...

Taking these additional two factors into account gives copper loss much closer to 18 W.

Two good points there.

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

tvtech said:

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

Click to expand...

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 Diver

Public Holiday here today so I am working on my project's battery charging reality. You have been a great help.

Kudos to you man

Regards,
tvtech

Hello there,

I see you got a lot of help already so i'll just add a few notes.

First, since this kind of circuit comes up so much on forums (full wave rectifier and cap filtering) i decided one day to look into it and write a program that would calculate everything. What i found out was troubling because there are a lot of variables, and we must know them all if we wanted to calculate anything with any reasonable accuracy. What is unfortunate is that we dont usually know all these variables or we dont want to take the time to measure them all. That means we cant get the exact result without a real life controlled practical test.

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

Why all this? That's because the peak current (as we all know) of the diodes and the secondary is much greater than the average, and that means we need the spice model of the diodes and also the inductances and winding resistances, but also the line resistance because even that small amount could make a difference when the cap is charging.

Adding an inductor in series helps the charging characteristic, so you could also look into that, but it requires a decent size inductor.

So what to do? Probably the only way to do this practically is to measure the transformer temperature with different loads on the rectifier/filter. After several hours run time and a given ambient temperature if the transformer temperature rise stays within spec then it should work for a long time. If it goes too high, then you know you've got to lower the load current and let the temperature stabilize again before making another temperature measurement. Also note that if you move the transformer to a different location it may change the temperature rise if the line impedance is significantly different. To get a handle on this you might also include a measurement of the line voltage drop with full load, and compare that to any future installation.

Ive seen very large transformers melt down...it's not pretty.

One of the simplest circuits that we all probably learned when we first started out, but not the easiest to analyze if we want really good results.

Thanks MrAl

I appreciate your input here.

You see the Toroidal I am going to use for it's mammoth task ahead is well known to me. I know years ago with the Cyclon thing it could bring the hungry 8Ah up to speed within an Hour. So, I could take all it could give (4x LM317T in parallel) giving 8.8 Amps into what the 317's saw as basically a short for the initial load.

Quickly things equalized and the lot (battery and 317's) kind of balanced each other. 317's quickly came to their 1.5A limit each after the initial rush of Current. And all was happy.

Different animal this new project with a 65Ah battery. I needed to know what this Toroidal can do continuously working a big battery and keeping it happy.

No quick charge and cooling stuff anymore. Continuous duty to get the battery charged...and then Float mode and all is happy

So, going from what I have been guided with by Diver and others above, I am going to ask only 5A DC output from this Toroidal on a 100% Duty Cycle.

I reckon that's the safe, conservative way to do stuff. If the 65Ah is flat, the Toroidal is going to work it's butt off getting it charged. Could be Hours.

Don't want fires here

All the best,
tvtech

Hi,

The ratio of RMS of a limited time rectangular wave to RMS of a square wave of equal area is:
r=sqrt(1/t)

where t is the fractional percent time high.

Looking at this for a few values of t:
t=1, r=1
t=0.5, r=1.4
t=0.4, r=1.6
t=0.3, r=1.8
t=0.2, r=2.2
t=0.1, r=3.2

So to get an idea what is happening we can see that if the diode conducts for 50 percent of the time we have an RMS current that is about 1.4 times higher than when it conducts all the time, and when it only conducts for 1/10 (one tenth) of the time the RMS current would be about 3.2 times higher than the normal RMS current. What this means for 5 amps with an 8 amp rated winding is that the ratio is about 1.6, so we are banking on the diode conducting for about 40 percent of the time.

This is based on two square waves however, where one is 100 percent duty cycle and the other has very limited duty cycle but maintains the same area. I think to be more accurate we'd have to look at sine waves and partial sine waves instead. I just wanted to show a quick idea about what is going on by looking at how the diode duty cycle affects the RMS current.

It would also be interesting to simply do a simulation and look at varies duty cycles and calculate the RMS current for each case.
Anyone feel up to the task?

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.

Hi Guys

I am very happy that I now know my 160Va Toroidal can do 5A DC continuously with it's 15VAC Secondaries wired in Parallel.

That's all I was worried about. Equations and stuff are not my forte. I am a simple Guy. That likes results

I am not much of a Maths person. I am much more into actual stuff. Like walk the talk.

You see how much I have learned here though. And graciously I have been helped and guided as to what is the limit for this Toroidal...6ADC is cutting it close and 5ADC is fine. Always my choice what to do though.

So, I have been given guidelines. And very good advise. Now that is worth Gold to me. Not do this or do that. And it gets force fed to me. And the feeder keeps shoving stuff down my throat..

And I am Gobsmacked. Seriously.

I never expected any help here. I mean I have Friends here that regard me as a Joker. I clown around. A lot sometimes.

Here is the thing that I have said from Day One when I joined here in 2010.....and my first post was about Transformerless Power Supplies..

One of you here (maybe Ron or Jim or another person) never wrote me off. Rather they listened.

And then I realized I am amongst good people. Not braggarts..just well, good people.

That is when I dug my heels in and decided ETO for life. That's it.

And I have never regretted a moment of it.

Here is the thing. ETO stands for accurate advise. No guesswork. People here don't talk much...but they know stuff

And I have made this a long post so that you can all feel proud of yourselves.

If I don't do it..who will?

Love every one of you,
tvtech

Diver300 said:

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.

Click to expand...

Hi,

When you want to convey information as quickly and simply as possible it's always best to use the least known file type so no one else can read it, hint, hint

Series inductance is used on high current power supplies, designed to reduce ripple and strain on the diodes and caps which is worse with higher current. It sounds like you are most familiar with low current power supplies. But even a little smoothing effect can change the peak current drastically.



Also, i worked out an example just for reference. It's a bridge rectifier made up of common 1N400x diodes.
The series resistance is 0.2 ohms, the input sine 100 volts peak, the output load 100 ohms, the output filter cap is 1000uf to 100000uf, and the line frequency is 1 Hz to 100 Hz.
At 1 Hz, 100000uf, the RMS current is 2.5 amps.
At 10 Hz, 10000uf, the RMS current is 2.4 amps.
At 100 Hz 1000uf, the RMS current is 2.4 amps.

Since the load current is always close to 0.95 amps, the ratio is close to 2.4 to 1.
Doubling the series resistance to 0.4 ohms brings the ratio down to about 2.2 to 1.
Doubling the series resistance once more to 0.8 ohms brings the ratio down to about 2 to 1.
Just for reference, adding a 1mH inductor in series brings the ratio down to 1.8 to 1., and a 10mH inductor brings it down to 1.1 to 1 which is almost equal to the RMS load current now. (at 60Hz 1.3 to 1 which isnt that bad either).

This makes it seem like an 8 amp rms transformer would be used at 4 amps DC output. There is also series inductance to consider however which smooths out the cap current. Even a little smoothing changes everything, which would allow a higher RMS current.

BTW, once an inductor is added it's easier to calculate.

MrAl said:

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

Click to expand...

What about the capacitance and ESR of the filter capacitor?

MrAl said:

So what to do?.

Click to expand...

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.

I would assume that the ratings are for the transformer in free air. Putting it in an enclosure would cause it to get hotter and a measurement of the actual transformer temperature in that case would be a good precaution to stave off eventual transformer failure.

A small muffin blowing directly on the transformer would greatly reduce the operating temperature.

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

Regardless of whatever other considerations there may be, a small fan blowing directly on the transformer helps a lot!

^^^^^I know

Shush now. Till at least I get the Battery

And then we can go on.

Regards,
tvtech

tvtech said:

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

Click to expand...

Hi,

Well that's just too darn bad for you, you are going to take in all the extra information and like it ! (ha ha)

On the more serious side, what i was doing was investigating how well we can determine what the 'linear' (as you call it) technique could be and if there was too much of an error to be relied upon.

Electrician, yeah sure, the ESR of the cap, the capacitance itself, all play a role.
The RMS meter isnt too bad of an idea, as long as it takes enough samples over one or one half cycle because those diode pulses sure can be short being a fraction of the total half cycle period. The rest of the time it's zero.
There's a chance i can try this myself but then again the clamp on meters arent that good down low.
Maybe later this month or next month when i'll then have another piece of test equipment that will help with this test.

Another interesting idea: water cooling.