What you say makes sense for voltage but not for current.

No.
It applies to both voltage and current.
Why makes you think there's a difference?

2A rms AC has the same heating effect as 2A rms DC.
Hence the power from 2A AC is the same power as 2A DC.

That is the definition of rms.

That's true as far as it goes.
But the current out of the transformer is not 2Arms in your example, it is 4.5Arms.
As you pointedly noted in post #12 "RMS is the square root of the mean squared" which applies to current as well as voltage.

So if you look at the high peak current drawn by the rectifier-capacitor in post #9, you can see that this causes the transformer RMS current to be ≈1.9 times the DC output current (in the small Waveform window).
The is because of the high I²R loss in the transformer windings from the high peak currents as compared to the average current.
If you don't believe the simulation calculated RMS value, you can do your RMS calculation of that waveform to see its high RMS value.

Why do you think Hammond derates the transformer by 60% if that weren't true?

 

Taking a big step backwards- let's also consider what you are expecting in terms of performance.
The circuit shown in the first post will not regulate voltage very well, it will sag under load.

There are much better regulator circuit designs out there.

 

Ac voltage and current are specified in rms values..after rectification and filtration they tend to go higher in magnitude and direction cause the waveform tends to attain maximum amplitude..so to my own opinion if a transformer is rated 2.5A rms after rectification it should be able to deliver more than that..althouugh I Understand perfectly what u meant by the transformer burning out since there is no limiter there..

RMS voltage of a sine wave is the amplitude times 0.7.
Another way of putting it, rms x 1.414 = amplitude.

We know that when you charge a capacitor via a rectifier diode, the voltage on the capacitor will seek the maximum voltage, i.e. the amplitude of the sine wave.

However, the reservoir capacitor cannot maintain that peak amplitude because the diode conducts for only a small fraction of the cycle. Hence if the transformer is rated at 2.5A rms, don't expect it to be able to deliver 3.5A DC. That is why there is ripple in the DC supply voltage.

If you convert both voltage rms and current rms to peak voltage and current, then you are asking the transformer to deliver twice the amount of power that it was intended. The transformer is limited by the amount of power it can deliver.

How much power the transformer can actually deliver depends on the construction of the transformer and not what the manufacturer puts on the label and specifications. The output voltage of the transformer will droop as the output current is increased.

Hammond transformers are known to be very conservatively rated.

 

No.
It applies to both voltage and current.
Why makes you think there's a difference?
That's true as far as it goes.
But the current out of the transformer is not 2Arms in your example, it is 4.5Arms.
As you pointedly noted in post #12 "RMS is the square root of the mean squared" which applies to current as well as voltage.

So if you look at the high peak current drawn by the rectifier-capacitor in post #9, you can see that this causes the transformer RMS current to be ≈1.9 times the DC output current (in the small Waveform window).
The is because of the high I²R loss in the transformer windings from the high peak currents as compared to the average current.
If you don't believe the simulation calculated RMS value, you can do your RMS calculation of that waveform to see its high RMS value.

Why do you think Hammond derates the transformer by 60% if that weren't true?

Your simulation on post #9 shows a signal generator consuming 12V x 1.9243A = 23W
while delivering 5V x 1A = 5W to a resistor load. Hence 18W is wasted somewhere in the components given in the simulation. (If the 12V indicated is really 6V amplitude or 6V rms it still doesn't solve the problem.)

I have provided a real world example of a transformer rated for 31.5VA delivering 35.7W with no evidence of overheating.

I will believe a real demonstration over a simulation any day.

 

Your simulation on post #9 shows a signal generator consuming 12V x 1.9243A = 23W

Again, not true.
Multiplying RMS current time RMS voltage only works when both waveforms are sinusoidal.

The simulation shows the power from the supply is 10.5W.

I have provided a real world example of a transformer rated for 31.5VA delivering 35.7W with no evidence of overheating.

How long did you run it?

If if were true that you don't need to derate the transformer then I'll again ask, why does Hammond derate the transformer by 60% for that circuit?

 

Again, not true.
Multiplying RMS current time RMS voltage only works when both waveforms are sinusoidal.

The simulation shows the power from the supply is 10.5W.
View attachment 138007
How long did you run it?

If if were true that you don't need to derate the transformer then I'll again ask, why does Hammond derate the transformer by 60% for that circuit?

I don't know. Why don't you ask Hammond?
I used a Hammond transformer in my demonstration and clearly I did not have to derate it.

 

Again, not true.
Multiplying RMS current time RMS voltage only works when both waveforms are sinusoidal.

Geez, where do you get your crazy ideas from?

RMS has nothing to do with the shape of the waveform.

 

I don't know. Why don't you ask Hammond?
I used a Hammond transformer in my demonstration and clearly I did not have to derate it.

My, we're getting kind of snarky here.
My data is in line with Hammonds derating value, so I have no reason to ask them.
It's your information that doesn't agree, so it would seem you should be the one concerned about the discrepancy.

A short test of a transformer doesn't "clearly" indicate that you don't have to derate it for long term operation.

Geez, where do you get your crazy ideas from?

RMS has nothing to do with the shape of the waveform.

RMS values are determined by the shape of the waveform, and it does effect how you determine the power from those values.
Suppose that peak current occurred at the 45° phase of the sinewave instead of the 90° polnt.
Obviously you couldn't still multiple the RMS voltage by the RMS current to get the power since power is only being generated where the current and voltage are in phase.

My "crazy ideas" agree with the simulation results and Hammonds derating value so that's sufficient for me.
If you don't believe it, that's your problem.

I think we're through here.

 

My, we're getting kind of snarky here.
My data is in line with Hammonds derating value, so I have no reason to ask them.
It's your information that doesn't agree, so it would seem you should be the one concerned about the discrepancy.

A short test of a transformer doesn't "clearly" indicate that you don't have to derate it for long term operation.
RMS values are determined by the shape of the waveform, and it does effect how you determine the power from those values.
Suppose that peak current occurred at the 45° phase of the sinewave instead of the 90° polnt.
Obviously you couldn't still multiple the RMS voltage by the RMS current to get the power since power is only being generated where the current and voltage are in phase.

My "crazy ideas" agree with the simulation results and Hammonds derating value so that's sufficient for me.
If you don't believe it, that's your problem.

I think we're through here.

Nice of you to switch arguments.
You make a statement that voltage and current must be sinusoidal, which is incorrect.

Now you introduce phase difference, trying to obfuscate your statement.

I will provide you with more real data on the power delivered by the Hammond transformer. Any transformer company can derate their own specs on current carrying capability and power with impunity in a real circuit application. That is the conservative thing to do. In my experience, Hammond transformers perform very well according
to their labeling and do not require to be derated.

 

RMS values are determined by the shape of the waveform, and it does effect how you determine the power from those values.

since power is only being generated where the current and voltage are in phase.

Any person not familiar with RMS and power factor should disregard these two statements.

 

Solutions of the problem type 0B-15V and current 3A or 1.2V-15V and current 2A usually do not take into account a very large dissipated power by a regulating element (transistor or stabilizer microchip). Requires an ocher large cooling radiator and possibly a fan. It is more correct to use a pre-dc/dc stabilizer, which gives a voltage of 2 V more than is necessary at the output.

 

See

 

I was going to stop this little tete-a-tete due to your condescending remarks, but I could not let your false (and possibly dangerous) statements stand.

Nice of you to switch arguments.
You make a statement that voltage and current must be sinusoidal, which is incorrect.

Now you introduce phase difference, trying to obfuscate your statement.

You are the one that is obfuscating by using only part of my statement.
I said they had to be sinusoidal to multiply them together (to get the power) as you did in post #24.
Phase shift always is included when calculating power in an AC circuit.
I don't understand why you imply otherwise.

Any person not familiar with RMS and power factor should disregard these two statements.

Since when does the waveform shape not affect the RMS value of the wave?

I should have been more clear about the power.
I meant that the power is determined by the parts of the waveform that are in-phase (volts times amps at any point in time).
That is where phase-shift comes into the equation.

Hammond transformers perform very well according
to their labeling and do not require to be derated.

But you are not using them according to their labeling.
You are using them well above their RMS rating.
Hammond transformers may be sufficiently robust that they don't have to be derated for rectifier-capacitor outputs (even though their data says they do, which is what I follow) but that is not the norm, as most transformers are not built with that much margin.

You state that you understand RMS and yet you seem to argue that the RMS values caused by the high peak currents of a rectifier-capacitor filter don't have to be used to derate a transformer.
That is a false view, and can lead to overheating of a transformer.
You should not be encouraging people (especially as a moderator) to design a circuit that could lead to damage (or worse) to an electronic device.

So any person not familiar with RMS should ignore any statements made that imply you don't need to significantly derate most transformers if you have a rectifier-capacitor at the output.
You do need to derate, as shown in the Hammond chart that was referenced in post #8.

.

 

RMS is the square root of the mean value of the voltage squared.
This is independent on the shape of the curve.

If the voltage and current are not in phase that is a different matter. Power is still delivered even if the current and voltage waveforms are not in phase.

There are two important criteria for selecting the proper power rating of a power transformer (besides other things such as construction and form factor).

1) Is the output voltage sufficient when supplying the required power output?
2) Is the temperature rise acceptable over the long term when supplying the required power output?

The purpose of derating the transformer is so that both criteria above are met.

I don't think I can make it any clearer than that.

 

Let's get back to the original argument about rms of voltage and current.

10VAC rms of a sine wave means that the peak amplitude is 14V. We know that in capacitor-filter circuit the voltage on the capacitor will seek the peak voltage.

You cannot use the same analysis with current.
1A rms does not immediately translate to 1.4A peak.
The power delivered by 1A rms is the same power delivered by 1A DC. That is the definition of RMS.

Am I clear on this?

Both you and I are both experienced and knowledgeable enough to know all of this.
What the argument has been diverted to is how much should one derate the power rating of the transformer.
As proper electronic design engineers, we always derate our electrical and electronic components for safety and reliability.

I know we both can agree on that.

 

The power delivered by 1A rms is the same power delivered by 1A DC. That is the definition of RMS.

Am I clear on this?

Perfectly clear.
I never had a problem with the basic definition of RMS.

But you can't equate 1Adc from a rectifier-capacitor bridge supply as being equal to 1A RMS coming from the transformer (even though it is 1A average).
The RMS current is typically 170% higher.

Am I clear on this?

As proper electronic design engineers, we always derate our electrical and electronic components for safety and reliability.

I know we both can agree on that.

Yes we can agree on that.

It's the magnitude of the derating (which is not arbitrary, but determined by the transformer RMS rating and actual RMS current as generated by the rectifier-capacitor DC current) that seems to be the source of our disagreement.
Thus if the RMS transformer current is 170% of the DC output current, then a typical transformer should be derated by a minimum of 60% of the maximum DC current to avoid overheating.

 

I do not understand what you mean by:

"The RMS current is typically 170% higher."

higher than what?

To me 1A rms is 1A rms is equivalent to 1A DC.

 

I do not understand what you mean by:

"The RMS current is typically 170% higher."

Higher than the DC output from the rectifier-capacitor supply.

To me 1A rms is 1A rms is equivalent to 1A DC.

I'm not arguing that, but you seem to be confusing the definition of RMS with what is happening in a transformer with a rectifier-capacitor output.

The rectifier-capacitor conversion process means that the transformer RMS current is higher than its DC average output current.

Do you not understand that?
It's a straightforward RMS calculation of the peak current out of the transformer that charges the capacitor.

 

My understanding is if we were to assume 100% efficiency for the moment (which we know is not the case),
10V rms x 1 A rms = 10W.

On the DC side, we would get 14VDC x 0.7A = 10W (ignoring rounding errors) given a 20Ω load resistor.

Depending on the size of the capacitor, we know that the diode conducts for a fraction of the cycle.
Hence the peak diode current can be much higher that the DC current. The shape of the diode current is not sinusoidal.
The peak diode current could be 2 to 5A depending on the size of the capacitor and total resistance in the charging chain, which includes the transformer secondary winding, the diode resistance and interconnections.

That is how I understand it.

In the real example that I presented, the average DC current was 2.73A while the peak diode current was 7A.

 

My understanding is if we were to assume 100% efficiency for the moment (which we know is not the case),
10V rms x 1 A rms = 10W.

On the DC side, we would get 14VDC x 0.7A = 10W (ignoring rounding errors) given a 20Ω load resistor.

Okay.
So that's not far from the 60% derating between the transformer RMS current rating and the DC output that is found by calculating the transformer RMS current from the peak current.
This derating from calculation or simulation includes some of the losses due to transformer resistance and diode forward drop so it gives about 0.6A DC output instead of 0.7A, for a 1A RMS transformer rating.