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I am using a ZVS induction heater like this:

ZVS induction heater

Image Source

The schematic looks akin to this (schematic has not been tested)

Schematic

Schematic Source

I modified it with IRF3205 N-Channel MOSFETs for better heat dissipation and more efficiency (lower RDS(on)), and wound the 56 μH inductor on the same iron dust core with multistrand PTFE wire to reduce the impact of the skin effect.

Photo of modification

The circuit now works better, with less heating. The MOSFETs warm up just a little bit.

On the input side, I'm using 0.75 mm copper wire. For the output, I'm using 2.5 mm multi-strand copper wire connected to the 10-12 AWG induction copper coil. The induction coil uses solid copper wire, which might be subject to the skin effect, reducing its efficiency.

I've noticed something odd: the input 0.75 mm wire doesn't warm up, but the 2.5 mm output wire, even though it's thicker, gets very hot. This is not due to the heat of the coil itself, but rather the output wire gets really hot for some reason.

I've measured the input and output current and voltage with my scope and a current probe:

Input side:

  • Voltage: 14.1 V DC
  • Current: 3 A (no load), 8-9 A (with load inside the induction coil)

The input looks like this (Yellow line on Channel 1 showing voltage waveform, and blue line on Channel 2 showing current waveform):

Input Current / Voltage

Output Coil side:

  • Voltage: 42 V (sine wave, 165 kHz)
  • Current: 22 A (sine wave, 165 kHz) with or without load. Without load, the voltage and current stretch a bit.

The output looks like this:

Output Current / Voltage

The power input is 14 V × 3.1 A = 43.4 W

The apparent power on the AC 165 kHz output seems to be:

Given Vpp = 82 V, Ipeak = 25 A (approx)

  • Vrms = Vpp / (2 × √2) ≈ 29 V
  • Irms = Ipeak / √2 ≈ 17.7 A
  • Apparent Power = 29 V × 17.7 A ≈ 513 VA

I've measured the product of CH1 and CH2 with the oscilloscope, and it looks like this:

enter image description here

Unfortunately, I can't directly add math to the measurement on this scope (unlike what Keysight scopes and other high-end scopes offer). My scope's Math function operates on the voltage of channels even if a channel is set to measure current. During peak power consumption, CH1 showed 22.4V, and CH2 indicated a current draw of 22.4A (224mV with a 10mV/A current probe). So, the value slightly over 5VV is actually 22.4V * 0.224A = 5.017VV = 500W. This confirms a peak instantaneous power draw of > 500W at 165kHz.

There is a significant discrepancy between the input power (43.4 W) and the coil apparent power calculations (~513 VA). How can the input power be smaller than the output power?

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  • \$\begingroup\$ I'm looking for my own answers, but this seems like a good place to start, in regard to the originally posted question. How do we get from Input: 14.1V 3A to Output: 42V, 22A. I'm looking for the formulae needed to achieve this through the ZVS circuit. What factors are considered and what terminology should I be researching? \$\endgroup\$ Commented Jul 2 at 21:15

2 Answers 2

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Indeed, you have measured the apparent power S, which is composed of real and "imaginary" (reactive) elements S = P + jQ. (Note, this is not the 'Q' of "quality factor"; there just aren't enough letters to label everything uniquely I'm afraid. It perhaps doesn't help that quality factor is related to these as their ratio, 'Q' = Q/P.)

For V and I being sine waves of arbitrary phase, the peak or RMS measurement discards phase -- it's only sensitive to amplitude. When you multiply their RMS values, you get a product with units of power, but you do not get power exactly. We use the special label VA for units of power at an unknown phase. It is correct to say your unit is outputting around 500VA. (Or for the remaining component (reactive power), we use the units VAR, to emphasize they're a VI product with phase for which no power is dissipated.)

For real power, you have to prove that V and I are also in phase.

In this case, as you can see from the waveforms, V and I are nearly 90° apart -- V is peak when I is near 0, and vice versa -- and their (instantaneous) product goes up and down, alternately drawing power and returning it a half-cycle later; it averages nearly zero.

Indeed, we can use the supply voltage and current to infer approximately the power flowing -- or at least an upper limit, less efficiency (but, considering the transistors aren't getting very warm, it's a good estimate in this case). The calculation P = VI always works at DC, because DC can only be in-phase. ("Out of phase DC" would be trivially zero, because, having by definition no frequency, its phase is always 0°, and \$\sin \omega t = \sin 0 = 0\$. Since we measure a nonzero quantity, we conclude phase cannot be perfectly "out".)

It's not uncommon for inverters to measure their power output this way: the DC input is so much easier to measure, the efficiency can be assumed high in suitable designs, and computing the real-time power product of a high-frequency signal is especially nontrivial.

Just to explore that further: an accurate real-time measurement of power requires sensing output voltage and current at least as accurately, including phase and amplitude tolerance, as the P and Q we wish to measure. If a load has say P = 1kW and Q = 100kVAR, a 1% phase error (~0.6°) might completely corrupt our measurement: instead of dissipating 1kW, it might read double -- or zero, or even negative! The inverter output is also rich in harmonics, so we need to preserve phase matching across all relevant frequencies -- and, while harmonic power drops off quickly, that's still several MHz of pristine bandwidth we might need this tight phase matching over. Basically we'd end up making a fixed-range oscilloscope just to measure a little oscillator!

Which, as you already have that -- an oscilloscope, I mean -- you may be able to use that to your advantage. If it is equipped with a MUL option, set MATH = Ch1 * Ch2, then measure the average of the MATH trace: this is real power. But beware phase shift between what probes you are using; you may wish to calibrate them by reading the voltage across, and current into, a noninductive resistor and verifying they read zero phase shift.

For example, consider this point:

enter image description here

28V * 40A = 1120W. This is close to 45° phase after zero crossing so should be near power peak. The corresponding point in the MATH waveform is only a div, so it must be around 1000 W/div.

Anyway, whatever the phase shift between voltage and current, the current in and of itself is very real indeed! This, combined with skin effect, and conduction from the coil, explains why the larger wire still gets much hotter on the output.

Which is, in part, why the power company frowns upon reactive loads on the AC mains, for example: it increases current draw, that their wiring has to supply, that they might not otherwise be able to bill for; it increases system losses and reduces real power capacity.

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  • \$\begingroup\$ Thank you for such a deep insight! I've added CH1 * CH2 waveform shown in the scope to the question... \$\endgroup\$
    – 15 Volts
    Commented Jun 17 at 8:11
  • \$\begingroup\$ Looks like I miscalculated 5VA to 500VA during multiplying CH1 with CH2. The scope shows a 5W peak instantaneous max power, multiplied values as the voltage and current cancels each other out! \$\endgroup\$
    – 15 Volts
    Commented Jun 17 at 10:01
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    \$\begingroup\$ No, the peak power should be around 500 (more, actually) -- I'm not sure how to resolve units in the scope's case, though if you can dump the raw (Ch1/2) waveforms and do it in a spreadsheet, that might be more illuminating. \$\endgroup\$ Commented Jun 17 at 13:33
  • \$\begingroup\$ My oscilloscope displays measurements in VV (possibly, Vertical V/div but not documented in the manual). For example, if you multiply CH1 and CH2, and both are set to 10V/div, it will show 100VV. In my setup, the multiplication result displayed 5VV. It seems like my scope is multiplying CH1's 40V with CH2's 100mV (10mV/A), so it's not directly multiplying voltage and current as in 50V/div * 20A/div. At the peak, it's actually multiplying 22.40V * 224mV or 22.4A = 5018mV or 5VV or 500W peak instantaneous power at that point. I'll update this post to incorporate the correct values. \$\endgroup\$
    – 15 Volts
    Commented Jun 17 at 15:58
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    \$\begingroup\$ See edit -- yes, that checks out. \$\endgroup\$ Commented Jun 17 at 17:45
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How can the input power be smaller than the output power?

Because you have not measured output power accurately. The calculation you applied (Pav = Vrms x Irms) is only valid if the voltage and current waveforms are in phase at all times (ie: for resistive loads); you can clearly see from the scope image that they are not in phase.

Does your scope do math, in particular, multiplication? If "yes", then I suggest show a trace that shows instantaneous voltage x instantaneous current in real time; you will see that it will be a sine wave (or cosine, depending on your point of view) with a DC offset; and frequency twice that of the input waveforms. The DC offset (ie: the average value) is the average power.

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