Average Current

sheet.

Almost certainly. But I never use a component at its absolute maximum rating, voltage, current, or power.

Thanks John

I probably knew that, just wanted to get assured of it.

Ok now it will be wise to ask, suppose i want to get 3amp dc out of it using bridge rectifier, what should be the transformer amps rating.

By calculation it should be 3.33amps,

Iav=2.82*Irms/pi

so, Irms=3.33amps

It that correct? And will it depend upon full/half wave rectifier?

The diodes have an average current rating because they drop a nearly constant voltage over a wide range of current. So the heat produced is fairly closely proportional to the average current, as long as you don't get into extreme cases (of very large pulses widely spaced). Transformers are heated by the square of current, like resistors, because their windings are resistive. So they have an RMS current rating. RMS stands for square root of the mean (average over a representative time period) of the square of the instantaneous current. The proportion of the RMS current from the transformer to the average DC rectifier output current depends a lot on what follows the rectifier. If the load is a resistor, then the RMS load current is also the RMS transformer current. In this case, the RMS load current is 1.11 times the rectifier average current.

If the load is primarily capacitive, all the rectifier current is confined to the brief periods when the transformer voltage is higher than what is stored in the capacitor, so the RMS current can get quite a bit higher than the average current. This affects the current rating required in the transformer.

It is correct for a resistive load and a full wave rectifier (load current proportional to absolute value of instantaneous transformer voltage). Half wave rectifiers pose additional problems for transformers that involve core saturation that don't occur with full wave rectification. And the ratio of RMS to average current is also different (higher).

What load will your rectifier be driving?

Just two more questions, John

1. Reactance offered by a capacitive load is always higher that a resistive load, then why should it extract more current?

1. Why do capacitor charges to peak ac voltage if placed across a bridge rectifier? Without a cap i measured 11v, and it increased to

Thanks for your help

Reactance is a linear concept that applies to sine waves. To get your mind around a capacitive input rectifier filter, you have to think in the more general differential description of capacitance. I=C*(dv/dt) current equals the capacitance times the time rate of change of voltage across the capacitor.

As long as the capacitor voltage is equal to or greater than the transformer voltage, the rectifier isolates the two. But the moment the transformer wave rises above the capacitor voltage, the rectifier is essentially a short circuit, and the voltage on the capacitor must rise as fast as the transformer wave is rising, regardless of how much current that takes. So the current into a capacitive filter is narrow sort of half sine wave pulses that occur on the part of the transformer voltage wave just before the peak voltage. Since the transformer windings are heated by the RMS current, a pulse waveform like this has a much higher RMS value than the average of the current in those pulses. This is what the 'squared' part of the RMS does. It is not unusual to have to double (or more) the transformer RMS current rating relative to the DC output average current when using a capacitor input rectifier filter to take care of this higher winding RMS current. The exact ratio depends on how much leakage reactance there is between primary and secondary windings that tends to spread out the charging pulses, by sagging the waveform a bit while the cap is charging, lowering the slope a bit. If you look at the transformer waveform with a scope, you can see the flattened spot on the wave where the cap is charging just before peak voltage.

The diodes act as check valves, pumping the cap up all the way to the transformer waveform peak voltage and then turning off, leaving that voltage trapped in the capacitor. A resistor load on the rectifier keeps it on the whole waveform so the resistor voltage is the same as the transformer waveform (except for the inversion of one half). your meter reads the average of the rectified transformer waveform instead of the peak value.

Thanks a lot.