Here are some more measurements I took on the setup described in the previous post.

I adjusted the variac for exactly 12.0 VAC at the secondary. I then captured scope traces showing the voltage on the caps with various loads. The same 46,000 μF is in place, giving about 100 mV P-P of ripple when loaded. This amount of ripple can barely be seen at the scale of these images.

The first image shows the secondary current in purple and the cap voltage in orange, with no load. The cap voltage is 17 VDC.

The second image shows the voltage with no additional resistance in series with the secondary and with a 1.0 amp DC load. The cap voltage is about 15.1 VDC, with what may be just barely visible ripple.

The third image shows the voltage with an additional resistance of 2 ohms in series with the secondary winding. The cap voltage is about 10.9 VDC, with no visible ripple.

The additional 2 ohms in series with the secondary reduced the loaded DC voltage on the caps from 15.1 to 10.9 volts, a decrease of about 4.2 volts. It's more than just the drop the DC load of 1 amp would induce in 2 ohms because the current when the rectifiers are conducting isn't just 1 amp; it's more like 2+ amps since the rectifier current occurs in pulses rather than a steady DC current.

This transformer is rated for much more output than it's actually loaded with. This means that its regulation will be very good in this situation.

Transformer regulation is usually worse the smaller the transformer. A typical transformer rated for this DC load of 1 amp at 12 volts probably wouldn't have such good regulation. Adding the additional 2 ohms in series with the secondary simulates a transformer with inferior regulation. It can be seen that for accurate prediction of rectifier performance, the transformer parasitics should be taken into account.

Usually, the winding resistance is all that needs to be considered if the transformer is concentrically wound. But, transformers with split bobbin construction can have leakage inductance of about the same impedance as the resistance of the windings, and it should be taken into account.

 

Without going into a lot of math and testing, this chart that AudioGuru posted pretty well covers uF vs ripple voltage vs current:

 

The formula shown at the top of the chart is the well known formula: ripple = I/(2fC). This formula is nicely described on Wikipedia
http://en.wikipedia.org/wiki/Ripple_(electrical) , and is referenced in Ryder's classic textbook.

I wonder why the graph doesn't follow the formula. For example, on the graph, .1 amp and 2000 μF indicate a ripple of .3 volts P-P, but the formula gives a result of .4167 volts P-P.

As explained on the Wikipedia page, there are some assumptions underlying the formula such as the diode conduction time being short compared to the capacitor discharge time, the discharge being linear, etc.

The grid voltage these days is rather flat-topped, which lengthens the diode conduction interval. This also reduces the peak diode current, but makes the ripple formula have greater error.

For example, in my earlier post, I had 1 amp of DC load current and 46,000 μF of capacitance, and a measured ripple of 100 mV P-P, but the formula suggests the ripple should be 181 mV P-P.

With only 2000 μF of capacitance and 1 amp of load current, the formula suggests the ripple should be 4.17 volts P-P, but the experimental value is 3 volts P-P with .127 ohms transformer winding resistance. If an additional 2 ohms is added in series with the secondary winding, the ripple is reduced to 2 volts P-P.

The formula can be substantially in error for large ripple.

I've attached two more images, showing the case with 1 amp DC load current and 2000 μF capacitance.

The first image is with only the inherent .127 ohms winding resistance and the second is with an additional 2 ohms in series with the secondary.

PS the forum software has problems with the brackets in the links, so I changed that for you.Bertus

 

I tried to do some simulations in order to see if it could be used in education. Particularly to illustrate how current is drawn from the transformer then a standard setup with a filtercap is used. But my problem was that my transformer model was far from real world. So it was hard to simulate different load conditions. And my result could therefor only be used to illustrate normal load conditions. I still think the topic how to dimension the power transformer for a given DC voltage and current specification is important. But then it comes to the current specification I think the thumb rules will do fine.