Transformer capacity with constant current load - Page 1
Forum section: electronics / projects

author: Circlotron
author_rank: Super Contributor
timestamp: October 04, 2021, 11:38:34 pm
content: Say we have a transformer rated at 100VA. It will safely supply 100 watts into a resistive load. Now we change the load to something that draws a constant current for the whole of each AC half cycle. The secondary current is effectively a square wave. Would it be possible to now draw slightly more than 100 watts from the transformer with the same temperature rise as before? My gut feeling says no.

author: bob91343
author_rank: Super Contributor
timestamp: October 04, 2021, 11:58:53 pm
content: Gut feelings notwithstanding, you need to get analytic. If the current is constant, you can see what the copper loss is. The core loss will not be much of a factor at such a high current. In spite of that, you need to discover what that loss is. Then you need to evaluate the copper losses, primary and secondary. Finally you must decide what hot spot temperature you will allow. You can run a transformer above its ratings if you can tolerate shorter life and higher internal temperature. You can also cool the transformer with a heat sink and some forced air to get more out of it.

author: Circlotron
author_rank: Super Contributor
timestamp: October 05, 2021, 12:39:11 am
content: Having more of a think about it, I can see I2R losses in the windings would more or less track I2R dissipation in the resistive load, so I probably wouldn't be able to get more out of the tranny after all.

author: Wolfram
author_rank: Frequent Contributor
timestamp: October 05, 2021, 10:21:35 am
content: Let's say 100 VA represents 10 V and 10 A RMS. If you have a resistive load, you get 100 W out. The secondary current rating concerns the RMS current, as the losses are mainly resistance loss in the wire. A square wave current with an amplitude of 10 A RMS will only provide 10 V * 10 A * 2sqrt(2)/pi = 90 W, so it presents a power factor of 0.9 to the transformer. Any current shapes other than sinusoidal will present a lower than unity power factor, and consequently less power for the same amount of resistive loss in the secondary.The easiest way to show this is to average the first half cycle of the power waveform, subsequent half-cycles are identical, so just calculating the first one will do. Current is a constant 10 A and voltage is 10 V * sqrt(2) * sin(x). Power is therefore 10 A * 10 V * sqrt(2) * sin(x). Averaged from 0 to 180 degrees by integrating and dividing by the integration interval, this gives 90 W.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: October 05, 2021, 12:08:06 pm
content: The power factor is about 0.81. Which is the case for a rectifier with choke-input filter (well above critical current), or CCS load. Better than a cap-input rectifier (typically 0.4-0.6) though still far from perfect.Tim

author: Wolfram
author_rank: Frequent Contributor
timestamp: October 05, 2021, 12:30:13 pm
content: Quote from: T3sl4co1l on October 05, 2021, 12:08:06 pmI double checked my calculation with an LTSpice simulation, and it also suggests a PF of 0.9 for a constant current load after a rectifier. Where do you disagree with my analysis?

author: T3sl4co1l
author_rank: Super Contributor
timestamp: October 05, 2021, 12:37:08 pm
content: See the table here,https://www.vintage-radio.net/forum/showthread.php?t=66951&page=2The PF is 0.9 and the voltage is 0.9 so you get 0.81 for the same DC voltage, or transformer rating or whatever.At least, that's what it looks like it's saying...Tim

author: Wolfram
author_rank: Frequent Contributor
timestamp: October 05, 2021, 02:06:18 pm
content: With a sinusoidal voltage source, an in-phase square wave current will have pi/(2*sqrt(2)) times the RMS value of a sinusoidal current, for the same power. The voltage is the same in either case.If you look at the case with a rectifier followed by a choke-input filter, the output voltage will be the rectified average (ignoring diode drops), or 2 * sqrt(2)/pi = 0.9 times the RMS input voltage, while the DC load current will be identical to the RMS secondary current. In both cases the power factor is 0.9 rather than 0.9^2.I had to do a few simulations and calculations to gain confidence in this conclusion, and I'm still open to being wrong of course.