transformer questions

[autir]

1) What measurements can we make in an AC/AC transformer with nothing but a digital multimeter (and diodes, capacitors, resistances etc.)?
2) Suppose we have an AC/AC transformer rated 12V/800mA. Do the values refer to rms, peak or p-p?
3) After rectifying the AC with a bridge and connecting the output to an electrolytic capacitor, we can see that the voltage of the capacitor is greater that the V(peak) of the AC transformer. Why does it happen? Shouldn't it be V(peak)-V(d)? (Where V(d) is the voltage drop in one of the four diodes of the bridge).

[audioguru]

Hi Autil,
A transformer is designed to output its rated RMS voltage with its rated load current. Since its windings have resistance they cause a voltage drop when current flows through them. The designers have adjusted the output voltage so that is is what it should be when the voltage drop caused by the current is included.
With less load current the voltage drop is also less. Therefore the output voltage is higher. Usually, a large powerful transformer will have less voltage change by its load current. But a little 200mA transformer will nearly double its output voltage with a light load.

A rectifier bridge (4 diodes) has 2 rectifier voltage drops in series on any half cycle. Their total voltage drop is from 1.2V to about 2.5V depending on the peak charging current of the filter cap.
The peak voltage (RMS times 1.414) is applied to the bridge that drops it by 1.2V to 2.5V, and any ripple also drops the DC output.

[autir]

I suppose that the current rating is Irms as well?

[audioguru]

Yes, the transformer's current rating is RMS.

[Kevin Weddle]

Somewhere I read that the rms value is the DC voltage that produces the same amount of power. To begin, the rms affects the equation of the signal inherently. The average does not. For rms, the equation is squared and the area found. Divide that by the length and you get the average of the equation squared. To make a long story short, if the average of the equation is .636 then the average of the equation squared is .5 because you must take the square root and arrive at .707. The average of both equations makes sense if you look at the graph of the sinewave compared to the graph of the sinewave squared. The average is less but the square root makes it greater. So what does this mean for our DC power dissipation? Well I think it has to do with squaring the voltage. Like in the power equation. But I can't prove this. It is interesting that if you have an amplitude of one, the difference between the rms and the average is only .071. Bump that to five and the difference is .355. So the farther you increase the amplitude the greater the difference between the rms and the average. Hum, amplitude dependent difference between the average and the rms. I hope I am able to make sense to somebody when I say the average is looking like the RMS all the time.