# How to modify a small power supply to not require a 20W load to turn ON? [closed]

I have a couple of obscure german “AT60” power supplies I scavenged from some designer lamps. They’re small, and don’t get hot even under ~30W load — but for my purposes, I need them to be always on, or at least switch ON at a load of 0.25W. A microcontroller needs to be always on and do some things, which may bring the load up to 10-15W, by my estimates.

I have no diagrams, only the images enclosed... There are two components related to (presumably) SMPS, these are named U1P220 (BULK 38D CRW). The large coil leads directly to the 12V terminals.

To clarify: I need this powersupply to either turn on at lower load, or be always on. It needs to power a microcontroller at ~0.2W, and this micro will in turn switch “things” on, up to a max load of ~15W.

## closed as unclear what you're asking by pipe, Elliot Alderson, Chupacabras, RoyC, Warren HillJan 13 at 10:39

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• If it requires a minimum load, then it isn't a simple transformer. It is a power supply. – JRE Jan 12 at 15:11
• Allow me to hazard a guess, it’s a “halogen transformer”? – winny Jan 12 at 15:30
• @JRE Shine your light on me, please. How can I reduce the required load? – user2497 Jan 12 at 15:32
• If that large coil indeed directly connects to the bulbs from the lamps, then what is output there can't be DC, so you'll have to at least have a bridge rectifier, followed by stabilizing caps, and most likely followed by yet another regulator to get the microcontroller voltage. Are you sure this pays, and it wouldn't be much easier to just either add an old phone charger supply (got a stash of 5V 250 mA supplies from the chocolate-bar-sized-phones days), or to make your microcontroller run off e.g. a NiCd cell that you only charge when you enable the large load anyway? – Marcus Müller Jan 12 at 18:32
• why would you remove the casing? If you keep the supply intact, you might not have to re-certify your product as grid voltage product! – Marcus Müller Jan 12 at 19:12

If they need a 0.25 W load to keep them awake then you can just add a permanent resistor load. From $$\ P = \frac {V^2}{R} \$$ we can calculate the required resistor value, $$\ R = \frac {V^2}{P} = \frac {12^2}{0.25} = 576 \ \Omega; \$$. 560 Ω would be the nearest standard value below this.