1
\$\begingroup\$

I got a 350VA transformer (230V~/24V~) which is directly connected to a heating unit at it's 24 V ends. The 230V end of the transformer are connected to 230 V~ mains over a solid state relay (SSR).

The scenario is that a controller triggers the SSR to switch through, the heating unit heats up to the target temperature and then a controller with a temperature sensor sends a 5 Hz pulse width modulation (PWM) signal to the SSR to keep the heating unit at a constant temperature.

The heating unit has a impedance of 1.0 Ω, the transformer 0.3 Ω (idle). The PWM frequency is as low as 5Hz and the solid state relay has a response time of 10ms (using zero-crossing, still 20 times faster than the PWM). I measured the actual voltage on the transformer with a RMS multimeter and measured 20 V (idle)

From what I understand the expected current would be

I = U / R = 20 V / (1.0 + 0.3 Ω) = 15.4 A

and the power user would be

P = U * I = 15.4 A * 20 V = 308 VA

That all worked well for a while, with a perfectly costant temperature on the heating unit, until the transformer started smoking and eventually failed.

Is there any reactive power generated in the transformer due to the PWM that actually causes a much larger current that the expected 15.4 A? What could I do to avoid this? Does it help if I put the SSR between the heating unit and the transformer instead of between mains and the transformer?

I need the PWM control since constant temperature is crucial in my application, the choice for AC power in combination with a SSR is simply because AC transformers are cheaper than DC power supplys in that range (15+ A).

Thank you for any advice!

\$\endgroup\$
1
\$\begingroup\$

Actually, there's a glimmer of truth in Linards' answer, but not for the reason he thinks.

When your PWM on-time is an odd mutliple of half the powerline's AC period, you are creating a DC component in the current through the transformer. In the worst case, if your duty cycle drops below 5%, you will be putting unipolar pulses through your transformer primary. It isn't going to like this at all.

If you're going to do this kind of AC powerline PWM, you must make sure that your PWM on-times are quantized to whole cycles of the AC waveform, so that there's never any DC component.

If you drop your PWM frequency to 1 Hz, then you can adjust the duty cycle form 0% to 100% in steps of 2% (20 ms).

\$\endgroup\$
  • \$\begingroup\$ Thank you very much for this explanation, that makes a lot of sense. I got a fresh transformer quicker than I thought and can try this out. Does it help if I just put the SSR in between the heating unit and the transformer or does it also give me the DC problem? Kind regards! \$\endgroup\$ – Moritz Walter Mar 24 '14 at 17:57
  • \$\begingroup\$ Yes, that would work. Even with odd numbers of half cycles, it would be no worse than driving a simple half-wave rectifier with a transformer. The difference is that the transformer's magnetizing current can now flow continuously. But the SSR will be handling 10x the current (and dissipating 100x the power) relative to using on the primary side, which will cut into your overall efficiency (in case that matters). \$\endgroup\$ – Dave Tweed Mar 24 '14 at 18:16
  • \$\begingroup\$ Turns out exactly as you said: I tried it with the SSR in the secondary circuit, the new transformer is totally unimpressed now and does not warm up a single degree, while the SSR gets pretty hot with ~60 °C in the initial heating pase. Once the target temperature is reached and the PWM starts the SSR cools down again to ~40 °C. It's rated 20 A however so I guess that's fine with a bit of heatsinking. Thanks a lot for making me understand this! \$\endgroup\$ – Moritz Walter Mar 24 '14 at 21:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.