Watching BigClive’s air “freshener” teardown I was surprised to see an even simpler DC power supply for a microcontroller from the powerline than I’d seen previously, but it turns out that it’s only cheaper for microcontrollers dissipating less than about 25 mW.
In the air freshener it consists of two 15K resistors in series, fed from the powerline by a fuse and a diode used as a half-wave rectifier (two diodes in series actually), across a 5V zener and a 100 microfarad 16-volt electrolytic storage cap, bypassed with a ceramic capacitor for high frequencies. There is no galvanic isolation.
I think this is even cheaper and simpler than a capacitor dropper: fuse (in the air-freshener case, shared with the line-voltage heating elements), resistor, diode, zener, and the two caps, six components in all, and you have your 5 volts out. It’s extremely inefficient (maybe 2% efficient), but that’s nearly irrelevant at the power levels it’s cheaper for; it isn’t very important if it’s wasting 0.1-1 W.
You could cut it to five components if you used a fusible resistor, of which Digi-Key has 140 in stock, like the US$0.05 Vishay NFR25H0001001JR500, a half-watt “flameproof” 1-kilohm jobbie. But how much resistance do we need, and how much power does the resistor need to dissipate?
Suppose the microcontroller draws no more than 10 mA (50 mW), and you’re designing for a 240VAC environment. Your rectum-fried DC will peak at 340V, but what’s more important here is probably the RMS voltage, which is actually only 120V with half-wave rectification, and the mean voltage, since we want the mean charging current to be 10 mA. The mean of an ideally half-wave rectified signal is about 0.318 of its peak, about 108 V in this case, so we’d need no more than 10.8 kilohms of series charging resistance; lower resistance would be fine but would produce more waste heat, so 4.7k might be better. With 4.7k you get 23 mA average current, 26 mA RMS current, and thus 3.2 watts of power burned in the resistor. That’s a curtain-burner!
So you need maybe a 5-watt resistor. These are off-the-shelf parts like the Vishay AC05000004701JAC00, but they’re quite a bit more costly; that one costs 71¢, while the 45¢ TE RR03J5K6TB would almost work at 5.6kilohm and a 3-watt power rating. And, perhaps unsurprisingly, none of the resistors Digi-Key lists in that power range claim to be “fusible”.
Evidently this simplification is only economical for microcontrollers using significantly less power than that, because using a capacitor to drop that voltage instead would be cheaper. The air-freshener circuit being dissected used 30 kilohms instead of 4.7 or 10, so evidently its microcontroller needs 3 mA or less. Accordingly the resistor’s power is only half a watt, and that’s distributed over two resistors, which I think are sized for half a watt each, and are located a substantial distance apart on the board, perhaps with the objective of avoiding a concentration of heat.
The voltage across the dropper resistor is effectively the whole half-wave rectified power supply voltage, so the power it dissipates is linearly proportional to the current: 3.2 W at 23 mA (4.7 kilohms), but 0.32 W at 2.3 mA (47 kilohms), and 0.03 W at 0.23 mA (470 kilohms). There’s actually a lot you can do even at 0.23 mA.
At 10 mA the 100 microfarad capacitor would also be too small. Because of the half-wave rectification, there’s a dead time of just over 10 ms when there’s no current charging up the cap, and in that time, it would drop from 4.9 V to 3.9 V. So 100 microfarads is adequate for maybe a 4 mA sustained current draw. But at a lower power level, like 0.23 mA, 10 microfarads is likely enough.
So here’s a parts list for a 5V 0.23 mA power supply:
That gives us a total bill of materials of 10.75¢, almost exactly half being the electrolytic. At such low currents, it might be feasible to replace both the electrolytic and the ceramic with a ceramic like the 0805 Samsung CL21A106MQFNNNE, which is 10 microfarads, rated for 6.3 volts, and costs 1.9¢, which would drop the BOM cost to 7¢.
The fuse is essential for safety in case the rest of the apparatus fails short, but it’s potentially just a piece of wire that’s free to melt without setting anything on fire. The cheapest off-the-shelf fuse is something like the Eaton C310T-SC-4-R-TR1, which costs 14¢, twice the cost of the whole power supply. PowerStream’s fuse wire chart suggest that 40-gauge copper wire (79 microns diameter) should fuse at 1.8 amps, and anything 28-gauge or smaller (320 microns diameter) should fuse below 15 amps, which is low enough to keep a house circuit breaker from blowing.
Using 58 megasiemens per meter as copper’s conductivity, 79-micron diameter wire (0.0049 square mm) gives you 3.5 ohms per meter. At 1.8 amps, that’s 11.4 W/m, or 11.4 mW/mm, and so at 0.0049 mm³/mm we have 2.3 W/mm³. Copper is about 9 mg/mm³ so that’s about 2300 W/mg, which does seem like the kind of power that tends to melt metal.
Because we have about a factor of 8000 between the fusing current and the normal working current, it should be easy to provide enough insulation to permit fusing without causing the fuse wire to go into thermal runaway under normal loads.
A few millimeters of such thin wire (far enough for safe creepage allowances at 240VAC RMS) in an environment that won’t catch on fire if the wire melts would be a perfectly adequate fuse.