If i want a full bridge rectifier to output a rectified signal @4A, then must the AC input be 4A? Or 8A?
Assume sine input.

 

hi jr,
4amp will be OK.
E

 

This is a handy DC power supply design guide to have.

https://www.hammfg.com/pdf/5c007.pdf

 

There is only one way for the current to leave, and that includes the same loop as the current in. So with 4 amps out you will have 4 amps in.
This IS an interesting question, though. Different rectification schemes CAN lead to different peak currents, but the total energy is the same.
I did read an article that presented the consideration of the heating values of the different peak currents and the heating effect that they would cause. THAT aspect, heating more because of the greater peak currents, is a much more complex issue. The transformer heating is due to the POWER delivered, Transformer heating is also proportional to the power delivered.
That article was published in the "QST" magazine quite a few years ago.

 

If i want a full bridge rectifier to output a rectified signal @4A, then must the AC input be 4A?

For a rectified-filtered DC output the transformer RMS current rating must be higher than the DC output current due to the high peak currents drawing by the rectifier-filter.
For example, as shown in MrChips referenced guide, a transformer derating of about 60% is generally used for a bridge-capacitor circuit.

 

An example estimating 1A output used in sizing a transformer for standard linear.
In years past some would double the 1A to 2A but ≥3A RMS in a surge.
As rule of thumb for AC sine wave. An average is 0.62 current and 1.2 is used in the voltage below
The Hammond shorthand came from thermal testing into design practice that dates back to RCA, Philips, and military handbook.

Vpeak = √2× VAC(rms) ≈1.414×VAC(rms)

Idc≈0.62×Iac(rms)
we can rearrange and solve for Iac (or AC current)
The transformer winding should be rated for at least 1.6 Amperes RMS on the secondary and
the diodes will drop the voltage(RMS) Vac(rms) ≈ Vdc /1.2
rearrange to solve for Vdc and allowing sufficient capability with regulator overhead.

This works in linear full bridge rectifier and was fairly standard it takes into account that the
diodes and filter capacitor do add some thermal dissipation. This thermal stability under heavy
load conditions can be helpful in determining if the power supply meets pre-regulator spec.
Some power run cool because of the application's specified requirements.
We can heed the admonition to be safer rather than sorry.

In simulation it is possible to address the multiple concerns that power supply can have.
It can take several different simulations in optimizing a supply for an end use.
This is where keeping notes can help to open the right folder and choose the right simulation.

 

Average current is the integral over time of the instantaneous current.

You've specified a sine wave for voltage. Assuming a mostly resistive load R and that the diode drop is small compared to the peak, the average current is just the average voltage over R, which is the integral of that sine wave over R (I = V/R), evaluated from zero to pi over 2 (90°), times four. This works out to the result you've already been given that Vdc = 0.62Vac and thus Idc = 0.62Iac

 

For a rectified-filtered DC output the transformer RMS current rating must be higher than the DC output current due to the high peak currents drawing by the rectifier-filter.

If i understand the answers, everyone is saying there's reduction in current due to resistance or voltage drop, correct?
I = V/R

So, assuming no smoothing filter, no resistance, and ideal diodes, the full-rectified sine can deliver the same current to the load as the original AC can deliver?

 

So, assuming no smoothing filter, no resistance, and ideal diodes, the full-rectified sine can deliver the same current to the load as the original AC can deliver?

Yes.