So I was thinking that an AVR-driven lab power supply from trash (except for the microcontroller) would be a good idea. You need a lab power supply for a lot of electronics tinkering.
Lots of discarded electronics have power supplies in them. They usually have a few problems we would need to fix for them to be decent lab power supplies:
Problem #1 is really tough to solve, so I think I’ll ignore it. I think I can solve the others pretty reasonably, though, with a generic microcontroller and a relatively simple circuit plugged into the output of some generic salvaged power supply, like an ATX power supply, say.
The basic idea is that you drive a buck-boost converter (a diode, an inductor, a capacitor, and a MOSFET) using a microcontroller’s digital output. To its output you hook, in series, an inductor, a sense resistor, and the load, with a capacitor and megohm resistor across the load; then you add some diode protection. You hook each end of the sense resistor to a separate high-impedance voltage divider to scale down the voltage to a range adequate to the microcontroller’s analog inputs. Calibration trimpots form part of these voltage dividers.
Two additional potentiometers are attached to other analog inputs of the microcontroller for use as user input, and an old Nokia serial LCD is attached to five digital I/O pins of the microcontroller.
Actually, I think maybe the inductor and capacitor on the load end aren’t necessary. The AVR can respond to changes in load in about 50 μs. I was thinking that the inductor would keep the short-circuit current from rising too quickly for the Arduino to respond, but in fact the buck-boost converter can’t respond instantly either — if its output impedance suddenly approaches zero, it won’t dump an unlimited amount of energy into it because it doesn’t contain an unlimited amount of energy in the first place! It isn’t until the next cycle that it has the opportunity to charge up, but of course the microcontroller then has the opportunity to not charge it up.
ATX power supplies have ±12V lines, but they are very limited in power. The 5V lines are where the real power is at, typically hundreds of watts of it. A pretty minimal bench power supply should be able to produce up to 30 volts and at least an amp or so, say 1.5 amps, which works out to 45 watts. (You can run four to eight of these off a single ATX power supply.)
Consider first the simple inverting topology, even though that isn’t the right thing for a bench power supply.
If we’re running the buck-boost converter at 31.25 kHz (the highest frequency an Arduino running an AVR at 16MHz supports with its native PWM with full 8-bit resolution) then each cycle is 32 microseconds and involves converting up to 1.44 millijoules from the input voltage of 5 V up to the output voltage of 30 V.
At this case, the duty cycle is 17%: the inductor (on the 5V side of the freewheel diode) is getting its current ramped up with 5V for 83% of the time, then getting its current ramped down with 30V for the other 17% of the time when the transistor is turned off. During that time it dumps 83% of its 1.44 millijoules into the capacitor and 17% of it into the load, I guess, which means that the capacitor actually needs to be able to hold several times those 1.44 millijoules in order to keep ripple reasonable.
In particular, say our output ripple is 3%, one volt. That means that the difference between the capacitor energy at, say, 29V, and 30V, is that 1.44 millijoules. C(30²-29²)V²/2 = 1.44mJ, so C = 2·1440μF/(30²-29²) = 49 μF. This means that the total energy in the capacitor is CE²/2 = 22 mJ. If you were to short it out with a ¼Ω short circuit (about the internal resistance of a D cell), you would briefly get 120 A and 3.6 kW — but with a time constant of 12 microseconds.
Suppose we use a 10Ω sense resistor to measure the current, though. Now at 1.5 A, we drop 15 V across the resistor, and at 30 V, it’s impossible to draw more than 3 A, no matter how we short the output. This, however, requires that the capacitor charge up to 45 V, in order to get 30 V to such a hefty load. So we could draw as much as 4.5 A actually, and a rather alarming 45 W on the sense resistor very briefly, coming down to 15 W after the control circuit kicks in. Gonna need a hefty heatsink on that sucker.
So now the duty cycle is 11%: 5 V to spin up the inductor for 89% of the time, then 45 V as it charges up the capacitor the other 11% of the time. And instead of 1.44 millijoules, because of the extra power in the sense resistor pushing us up to 60 W, it’s 1.92 millijoules, so our capacitor goes up to 65 μF, and the total stored energy is 29 mJ. And if you short out the output, the time constant is 290 milliseconds.
This is sort of alarming because it means you can’t get a reasonable slew rate on the output voltage, whether in response to changing load or in response to user input. I think the solution is a combination of a higher switching rate and a somewhat smaller sense resistor, basically in order to keep the energy stored in the capacitor down.
(No, wait, all of the above is wrong. The time constant was 650 microseconds.)
It’s useful to have some kind of scale for how much energy we're talking about here. 2.3 kJ is enough to boil a gram of water, a cubic centimeter. 2.3 J is enough to boil a milligram of water, a cubic millimeter. 2.3 mJ is eough to boil a microgram of water, a cube 100 microns in diameter. This would still be enough to vaporize the bond wire of a chip if it could somehow be released there, but the bond wire will never be more than 0.1Ω, so even with a sense resistor of 1Ω we can ensure that the majority of the short-circuit energy is dissipated in the sense resistor and not the bond wire. So tens of millijoules should be fine, but not more.
Let’s suppose we can manage a higher PWM frequency — say, 125 kHz, maybe by using 6-bit resolution, since we’re only shooting for 3% ripple anyway. And a 4.7Ω sense resistor, so at 1.5 amps we only dissipate 10.6 watts in its 7 V, for a total of 55.5 W, which means 0.4 millijoules per cycle, and 15 microfarads is enough to keep the ripple acceptable, and our time constant when shorted is 70 μs. Our duty cycle then is 5V/37V = 13.5%.
You might also want some kind of minimum load there, in parallel with the real load, to keep the time constant reasonable when you're trying to adjust the voltage under open-circuit conditions. 50 ohms or something.
Okay, enough about the capacitor and sense resistor. What about the inductor? It needs to spin up to some current Imax at 5 volts in 6.9 microseconds, then spin back down at 37 volts in 1.1 microseconds, delivering 0.4 millijoules. 0.4 millijoules in 1.1 microseconds at 37 volts is an average of 9.8 amps, so the max should be 19.6 amps. Let’s say 20 amps. That means the inductor needs to be 1.7 microhenries.
Such inductors exist; Digi-Key lists a 12.8 × 12.8 × 7 mm item that costs US$1.20.
What happens if the inductor is some other size? Like, will it work better if the inductance is bigger or if it’s smaller?