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What is the output of Regulated power supplies? whether Peak DC or Average DC V?
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Brumby:
The OP seems to be trying to make sense of the input to the regulation circuitry - and does, indeed, appear to have a gap in basic understanding.

To the OP: Talking only about voltage will make no sense here.  The conversation needs to include the current involved as well.
xavier60:
The ".9" from techguru's post came from the Hammond guide for FULL WAVE BRIDGE Capacitor Input Load.
The "V (Peak) D.C. = 1.41 X Sec. V A.C." makes sense for the unloaded voltage.
 I assume that "V (Avg) D.C. = 0.90 X Sec. V A.C." means the fully loaded DC output voltage. I would expect it to be a bit higher than this for a  transformer sized by their guide. "I D.C. = 0.62 X Sec. I A.C."
Has anyone taken measurements lately?
I guess that the 0.9 comes from 0.9 x RMS = Average, ignoring diode drop which should't be ignored really.
The loaded DC output does not have to strictly equal the Average.
AngraMelo:
I think what he doesn't see is that there are PS where you can set a specific voltage and some that you cant
If you have a linear PS, using a regular transformer you will have around 1.4*voltage of the transformer's secondary on the Bridge rectifier. If you then decide to use a regulator you will now be able to choose from 2 types: fixed regulators or variable regulators. The fixed regs will do all it can to maintain a specific voltage say for example 12V, it will keep constant 12V on the output. There are also the variable regulators, those let you "dial" a specific voltage. Like those variable voltage power supplies. So the regulated voltage is what the PS is designed to produce on its output.

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xavier60:
In post  #6, the op says "DC regulated power supply: 0-24V @ 3A". 
The information in the Hammond guide is confusing.
For the FULLWAVE BRIDGE Capacitor Input Load, the statement "V (Avg) D.C. = 0.90 X Sec. V A.C." is irrelevant.
fsr:
Any signal has a peak, average, RMS voltage, etc. Here you will find a very good explanation of the concepts. All the education section in that site is very good, i recommend that you read it from the beginning DC section, it's very good: https://www.allaboutcircuits.com/textbook/alternating-current/chpt-1/measurements-ac-magnitude/

Now, if you take the output of a transformer and pass it thru a full-wave rectifier, at the output you will have a wave similar to the sinewave, but the negative parts now are positive too:



Also, as the current has to travel across two diodes, you lose about 1.4v of voltage at the output of the bridge.

Now, if you add a big capacitor to the output of the diode bridge, the capacitor will charge to the peak value of the output voltage from the bridge, and almacenate chage, so that it will power your load during the "gaps" of the output wave from the bridge. And so, the output voltage will tend to be more like a DC value. But at any time that the capacitor is supplying current to the load, it's discharging, so your output voltage will get more and more ripple, and less like DC, at higher currents. Bigger capacitors discharge less under the same current.

That kind of PSU is an UNregulated one. A regulated PSU could be one built by adding a linear regulator like the LM7805 at the output of the UNregulated supply. The LM7805 will will output a 5v voltage if you supply it with a higher voltage (higher than the dropout of the LM7805). The capacitor will dischage the same than before, but as long as it doesn't discharges below the minimum voltage requirements of the 7805, the IC will continue to provide a stabilized 5v output.

As you can see, the output of the regulated PSU doesn't matches the peak or average values of the input voltage.
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