Calculating DC Output Current From Unregulated AC Transformer

It's not up to you to determine the rating. The manufacturer *correctly* rated it for its intended purpose.

It's not my argument; it's the manufacturer's.

But there are still lower temperature grades available. We're talking about

*this* transformer, which may not be using the most common insulation system.

It would still depend of the insulation system, as I said, wouldn't it?

"The Phantom" "Phil Allison"

( snip even worse DRIVEL )

** How completely ASININE !!

A transformer with unusually good regulation = a transformer that is being under utilised !!!

No need for exists for transformer makers to make special high regulation models - the customer simply picks a model with more VA capacity than strictly needed !!!

Piss off - you PITA fool.

...... Phil

"The Phantom Congenital LIAR "

** No.

** No *we* are not - asshole.

Using an isolated example to prove a case is false logic.

Using an isolated example that only you know about is a classic debating cheat.

Combining the two, as you have done, is the act of a desperate LIAR.

You have no case to put - f*ck off.

...... Phil

If we assume Irms/Idc 2 in some circumstances, the user should know what those circumstances might be.

Interestingly, with respect to your reply to whit3rd and the behavior he described, the analysis I came up with seems to have a limit to the ratio Irms/Idc. I allowed the transformer internal resistance to become orders of magnitude lower than would ever be possible in the real world, without superconducting wire, and the calculated Irms/Idc never got over 4, no matter how peaky the waveform. I wonder what simulation would show. It might be numerically difficult for a simulator to correctly calculate the RMS value of a spike of current whose duration is one billionth of the period!

--
Be fair. :-)

Explain that a transformer that might be seen as being under-utilized
while feeding a resistive load  might have to be as big as it is in
order to drive reactive loads which pump current back and forth
instead of just forth.

"John Fields"

** What is the heading on this thread ???

So what is the context ?

** What is the heading on this thread ???

So what is the context ?

Is being a context shifter related to being a shirt lifter ?

...... Phil

Sounds like you have a quibble with the manufacturer, not with me. I only report the transformer's measured parameters and labelling.

"The Phantom"

Piss off - you PITA BLOODY TROLL.

..... Phil

Could well be that the manufacturer de-rated the transformer slightly to reduce their liability in case of transformer failure. For example, if the transformer could really take up to 100 VA, the manufacturer said the transformer is good for up to 80 VA

Michael

What will limit the narrow high current spikes you are talking about will ultimately be the inductance of the other circuit components and their internal resistance, and the diode characteristics. Also, the current spikes are a function of the rate of rise of the waveform at the time of onset of conduction (which also is not immediate), and the size of the capacitor. For low values of capacitance, the conduction occurs on a faster rising portion of the waveform, but the peak current is limited because the capacitance is smaller. As you increase the capacitance, the spikes are greater, and RMS current can be high enough while charging the capacitor to exceed the transformer's specifications and overheat or even instantaneously blow the windings, but that would be an extreme case.

For real life, reasonable circuits, after a large capacitor has charged to its final value, the diodes will conduct only when the waveform is greater than the diode forward drop and the minimum output voltage excursion (Max DC - P-P ripple), and this will occur very close to the nearly flat top of the sine wave, so the slew rate is very slow. Eventually the lower slew rate is balanced by the high capacitance so the peak current will be limited, and the Irms/Idc reaches a maximum value, such as the figure of 4 you mentioned.

Ordinary waveform distortion (and normal transformer saturation) will usually cause a flattening of the peaks and an even slower slew rate. If the waveform somehow has sharp peaks, like a triangle wave, this indicates high frequency harmonics, and peak currents can become extreme. But that is a very rare situation, and it may cause transformer overheating even with an AC load, due to excessive saturation. We must assume reasonable power quality.

Paul

"Paul E. Schoen"

** Where in time are the current peaks from Imag ( or Isat) ???

Engage brain, before putting mouth in gear.

.... Phil

The transformer will have current peaks that coincide with the peaks of applied sinusoidal voltage. The output of the transformer will have the voltage peaks flattened where the current peaks occur.

Paul

"Paul E. Schoen" "Phil Allison"

** No way - there is a 90 degree phase lag ( ie inductive load ..... ) that means currents peaks are at or near the supply voltage zero crossings. ** See above.

Hard to believe that someone who spends so much time * wrangling * with some very serious transformers was unaware.

..... Phil

Yes, you do. Your question reveals that you do not understand the introductory material.

Sorry, but you seem unable/unwilling to understand the answers you've been given.

In your case, you can try a different path. Rather than posting and fighting, build the supply with your transformer. Construct a variable load. Run the thing, set the current to wherever you want, monitor the voltage to see what you can get. Watch the temperature rise and keep it within safe limits - or test to destruction if you have deep pockets and time to keep burning out transformers. Arguing with the replies has gotten you nowhere, so far, so the testing approach might be more productive for you.

Ed

I thought I ought to consider the effect of the transformer's leakage inductance in the analysis. I measured the 192 VA transformer I've been using and it has 80 uH leakage inductance as seen from the secondary with primary shorted. So, in your simulation, you should make the transformer's equivalent impedance .125 ohms in series with 80 uH. Also, I used 80 amp Shottky diodes in the bridge, so the peak voltage drop across the diodes is about .4 volts at the peak current of about 20 amps.

Going further to determine the effect of the flattening of the grid waveform, I got a high power audio amplifier out of the basement and drove it with a low distortion 60 Hz sine wave. I used the bridged power amp output to drive the transformer. This way I could see the effect of flattening of the grid waveform. I was able to get a clipped sine by slightly overdriving the power amp. I took 3 sets of measurements with varying degrees of clipping. I've posted scope photos of the current and voltage waveforms over on ABSE.

You can see the effect of the leakage inductance in the fact that the peak current is slightly delayed in time with respect to the peak of the applied voltage.

OK, you are right on this point, and the flattened voltage waveforms and peaked current waveforms I have seen may be due to the fact that the circuit breaker loads may be highly inductive, or contain current transformers with non-linear diode and capacitor loads that draw more current at the peaks.

For breaker testing, I'm usually looking at the current, and I often see a peaked waveform, while the voltage appears to be flattened. But it could be a reflection of distortion on the input waveform. When drawing several hundred amps from a 480 VAC line, to pump 30,000 amps or more into a breaker, strange things can happen.

The following waveforms show flattening as I have seen, but maybe it was a clipped waveform to begin with:

This paper shows saturation occurring near the zero crossing, when the volt-seconds of the primary is reached:

and this also shows current peaks just before each zero crossing, and no apparent flattening.

Here is some good information (but mostly a copy of the above):

I seem to recall ferroresonant regulating transformers having flattened waveforms.

I found this interesting:

Paul

I've posted a derivation of a formula over on alt.binaries.schematics .electronic