I usually work with 12VDC or less, but I needed to get a quick ~350mA@5VDC supply out of a 24VAC transformer available to me. "No problem," I think: "I'll bridge rectify, filter a bit, and feed it to this Recom 7805-like switching regulator I have in my parts box."

So I do that, and test it by feeding the rectifier both ways from my 12V testing battery, everything looks good. Hooking it up to the 24VAC transformer, though, releases the magic smoke.

Turns out, you gotta figure you'll see ~80Vp-p when you're working with a 24VAC transformer (24 x 1.414 = 34, but unloaded you might see 28VAC or 30VAC). Rectify and you might still be looking at 40V. Recom is only good to 28V, so ... lesson learned.

Next thought: I can use a trusty 33063. I have some of those, and I know how to work them at lower voltages. But Vmax is 40V, and I know that's close to what's lurking in the rectified transformer.

So my question: what do people who work with AC in the range from ~25-75 do here?

It seems like I could use a 5-10V linear regulator to drop 40V down to a comfortable range for the 33063. If I use an external transistor to feed 40V to the inductor, I should be OK, and the linear regulator only has to provide ~5mA to run the buck controller.

Is there a smarter approach, or is this something an actual competent designer might do?

  • \$\begingroup\$ Plenty of options! Simple switchers are available up to 100 V. Viper12AS is overkill in terms of voltage but never fails me. \$\endgroup\$
    – winny
    Dec 31, 2016 at 20:27

1 Answer 1


Output power: 5V x 0.35A = 1.75W. Estimated efficiency is %85, so the required input power is 1.75 / 0.85 = ~2W. At 38VDC, input current is 2W / 38V = ~55mA. After rectification and filtering, if you get 40VDC then you can place a resistor with a value of RS = (40-38)/0.055 = 36R (Place 39R). Power dissipation of this serial resistor is about PR = 39 x 0.055² = 0.11W. So you can put a 39R/0.25W resistor. You can also make a simple RC low-pass filter by putting a 220uF elco after this resistor to filter-out some ripple.

  • \$\begingroup\$ Ah, much simpler than a linear regulator! I figured that approach would require a big ceramic resistor or something, but when you run the numbers, you can get away with a regular TH device or even a 1206 SMD. Thanks! \$\endgroup\$
    – Dave M.
    Dec 31, 2016 at 20:34
  • \$\begingroup\$ You're welcome. But please note that if you draw lower currents (say, 50mA at 5VDC) then the voltage drop on the series resistor will be smaller. This can lead getting input voltage of the IC closer to 40VDC. If this is the case then you need to recalculate the series resistor. \$\endgroup\$ Dec 31, 2016 at 21:18
  • \$\begingroup\$ I've just been thinking about that now...wondering if there's a simple overvoltage lockout circuit I can add. The 33063 max draw is 4-5mA, but I suppose it might draw less. I need a circuit that draws enough to keep the supply below 40V (easy but wasteful) or one that disconnects the supply if the minimum draw is not met. \$\endgroup\$
    – Dave M.
    Dec 31, 2016 at 21:39
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    \$\begingroup\$ @RohatKılıç - Vin on light load will be an issue. You want eg a zener after the resistor tyat limits Vmax that the IC sees but which does not conduct during normal operation. \$\endgroup\$
    – Russell McMahon
    Dec 31, 2016 at 23:29

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