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I am working with an RF transformer. I would like to understand the voltage rating of this transformer. How can I do that. The only voltage related information ios the inter winding isolation of 300V. What does that signify please ? The part number is AE458RFW01B1SZ and the screen-shot of the datasheet is as shown below -

Transformer datasheet

The other thing could be the derivation of voltage from the wattage.Wattage is 0.254W rating and a current rating of 250mAmps. This makes the max voltage as 1V, which is ridiculous.

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  • \$\begingroup\$ What do you mean by the voltage rating? \$\endgroup\$ – Leon Heller Sep 24 '15 at 15:29
  • \$\begingroup\$ Maximum operating voltage. The max voltage you can possibly apply on the transformer. \$\endgroup\$ – Board-Man Sep 24 '15 at 15:30
  • \$\begingroup\$ Do you mean RF voltage? \$\endgroup\$ – Leon Heller Sep 24 '15 at 15:31
  • \$\begingroup\$ Yes, the RF voltage applied. \$\endgroup\$ – Board-Man Sep 24 '15 at 15:34
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    \$\begingroup\$ You have the DC resistance of the windings and the max. current. What does that give you? \$\endgroup\$ – Leon Heller Sep 24 '15 at 15:45
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It's solvable but not directly....

Start with the maximum DC imbalance current of 36 mA. This figure is all to do with possible core saturation. A mismatch in ampere turns on the primary compared to the secondary severely bumps up the core flux density.

The 36mA figure is telling you what the magnetization current is that will start to cause problematic core saturation. Magnetization current occurs irrespective of any load but it's easier to calculate things assuming no-load on the secondary. It's worth noting that load currents in the primary and load currents in the secondary produce exactly opposite levels of ampere turns so they totally cancel out leaving the mag current.

At the lower limit of operating frequency (250kHz) the reactance of the primary (or secondary because it is a 1:1 device) is this: -

X\$_L\$ = \$2\pi f L\$ = 14.9 ohms

36mA is the peak current that causes saturation and the sinewave equivalent is \$\sqrt2\$ smaller than this at 25.5 mA RMS.

Voltage = 14.9 ohms x 25.5 mA = 0.38 V RMS or 1.07 Vp-p

This is the voltage that should not be exceeded when operating at 250 kHz or you will risk core saturation issues irrespective of load or no-load.

At (say) 2.5 MHz, the p-p voltage will be 10x higher (because X\$_L\$ = 149 ohms) at 10.7 Vp-p and if you go to 25 MHz, theoretically the voltage could be 107 Vp-p.

Clearly at 250 MHz you could apply a ridiculous level of voltage and this is barred by the isolation voltage limited at 300V - in other words you have to put a cap on how far you can go.

Trying to calculate this from the maximum power figure is problematic because if the load is 50 ohms and the power is 0.25 watts then the voltage would be 3.53 volts RMS (10 Vp-p)and this would cause major saturation at the low frequencies all the way up to ~2.5 MHz with or without a load.

I've looked at it from a standpoint of what would cause the core to saturate when driving an open circuit and assumed that the maximum RMS current in the primary is 25.5 mA RMS. If you factor-in load impedance (hence power) then the power formula will prevail at some point in the frequency spectrum but for the low frequencies it is the saturation current that dominates.

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  • \$\begingroup\$ Edited because I misread the inductance at 10uH instead of 9.5 uH and the operating frequency (low) is 250 kHz not 400 kHz. \$\endgroup\$ – Andy aka Sep 24 '15 at 17:27
  • \$\begingroup\$ You wrote: Start with the maximum DC imbalance current of 36 mA. This figure is all to do with possible core saturation. A mismatch in ampere turns on the primary compared to the secondary severely bumps up the core flux density. You do understand that the imbalance refers to unequal DC currents in the two halves of the secondary, right? \$\endgroup\$ – EM Fields Sep 24 '15 at 19:39
  • \$\begingroup\$ @EMFields Absolutely, and that directly relates to core saturation - examine note 5 on the picture in the question. Core saturation is caused by peak current whether it be steady state or just an instantaneous (almost) glitch. \$\endgroup\$ – Andy aka Sep 24 '15 at 20:11
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The inter-winding isolation voltage is the maximum difference in voltage allowed between the primary and the secondary.

The maximum allowable RF input voltage will be:

$$ E= \sqrt {P \times Z} $$

Where P is the allowable RF power into the transformer and Z is the transformer's impedance seen by whatever's driving the transformer.

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    \$\begingroup\$ You need to accommodate core saturation problems at lower frequencies so you can't just hope this formula will work sub 10MHz because it won't. \$\endgroup\$ – Andy aka Sep 24 '15 at 17:02
  • \$\begingroup\$ It won't? I don't see why not, since if power into the transformer is the criterion limiting its input voltage and its impedance at any frequency determines how much charge can flow through the transformer, then voltage is the dependent variable and will always be governed by \$ E = \sqrt {P \times Z} \$ \$\endgroup\$ – EM Fields Sep 24 '15 at 19:57
  • \$\begingroup\$ Consider the dc case producing 36mA of current - the core is 10% into saturation and it says this in the spec sheet of the inductor although anything that talks about current imbalances in transformers is (always I reckon) hinting at saturation problems. \$\endgroup\$ – Andy aka Sep 24 '15 at 20:14
  • \$\begingroup\$ I agree, but since the DC bias on the core will lower the primary's inductance, and thus the impedance the driver sees when it's driving the primary in the same direction as the bias, wont \$ E= \sqrt{P\times Z} \$ still describe the limit? \$\endgroup\$ – EM Fields Sep 24 '15 at 22:01
  • \$\begingroup\$ The dc bias lowering the inductance is basically the onset of saturation. Power can only be transmitted into a resistor and this (ignoring DC resistances of the winding and eddy current losses) has to mean the load on the secondary hence I Z, in your equation is load resistance (proper ohms). At 250kHz and (say) 50 ohms load, V will be \$\sqrt{0.25\times 50}\$ = 3.53 V RMS. My answer says at low frequencies there is a new upper limit on voltage that is governed by the saturation irrespecive of load resistance and, at about 2.5MHz there is an intersection with the power curve. Does this tally? \$\endgroup\$ – Andy aka Sep 24 '15 at 22:16

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