I'm trying to understand a paragraph in Horowitz and Hill, "The Art of Electronics", under the sub-section "Transformers":
"Since a rectifier circuit draws current only over a small part of the cycle (during the time the capacitor is actually charging), the RMS current, and therefore the I^2R heating, is likely to exceed specifications for a load current approaching the rated RMS current of the transformer. The situation gets worse as you increase capacitor size to reduce pre-regulator ripple' this simply requires a transformer of larger rating."
What I do understand is that given a resistive load, for example, where ripple is small, current will only flow through the rectifiers for a small proportion of the cycle, therefore the current flowing in that small proportion will be a lot higher than that through the load at any time.
What I don't understand is why this would cause I2R problems with heating. Perhaps my understanding of RMS current is flawed. My contrary reasoning is that If for example 1 amp is drawn 90% of the time, the I2R heating would be the same as if 9 amps are drawn 10% of the time, since the heat dissipation is relatively slow compared to the 50Hz cycles.
I hope I've given enough information for someone to spot what I'm failing to understand.
Thanks
Best Answer
If 1A were drawn 100% of the time, this is an RMS current of obviously 1A. If 10A were drawn for 10% of the time and zero otherwise, what is the RMS current?
I'm not using 9A at 10% because I'm doing the math in my head.
First convert to a "power" into an imaginery 1 ohm resistor - 10 amp is squared and power is 100W for ten percent of the time which means average power is 10W. Take the square root and the RMS current is 3.16 amps
What is the power for 1A 100% of the time? Answer 1W or 1A RMS into the same imaginary 1 ohm resistor.
Both currents average at 1A of course.