So now that each element of the board is verified, the time has come to evaluate the power-source switching performance of the board. First with diode-protected method for which the board was designed, and second with a cheaper, capacitor-smoothed method I came up with after ordering the board.
The diode-protected power switching method is best illustrated with this example. There are two power sources connected to different load switches, 5V and 9V. At first, 5V is connected to output, 9V is disconnected, and everything is honky-dory.
But then user decides he wants to switch to 9V power. 9V is connected to power output also in logical-OR configuration with 5V. Protective diode keeps 9V PSU from driving 5V rail high (and thus applying too high of a voltage to other devices connected to 5V rail)
Then finally 5V switch is turned off, and we're back to normal operation at 9V.
The disadvantage of this method is that diodes with sufficiently high maximum reverse bias are a little pricey, at least when you have to buy 100 of them ($25 total), and they have quite large footprints. Additionally, one diode is needed per power rail. Not a problem now with three rails, but I am going to go back up to 8 or so at some point. And there is an efficiency loss due to the voltage drop across the protective diodes.
Whether you're going from low to high or high to low voltage, this method should be safe. In the low-to-high case, the target output voltage is already achieved in the intermediate step. In the high-to-low case, the target output voltage isn't achieved until the final step. But each step is safe at steady state and therefore the timing of toggling the load switches is not important.
The capacitor-smoothed method has the advantage of requiring only a single small capacitor on the power output. Capacitors go on each power rail too, but not per slave board, rather per master board, and so their size and cost is not as important.
I'll explain it in the same manner as the Diode-Protected Method. You still have 5V and 9V power inputs. Initially, the circuit is supplying 5V and it looks like this:
Then, when user wants to switch to 9V, both switches are turned off. We can't switch both on like before since we don't have those protective diodes to keep the higher rail from applying an overvoltage to the lower one. The output capacitor supplies current to the output.
Then, very quickly the 9V switch is turned on and begins supplying the higher voltage to output.
Timing is a lot more important in this scheme. First, no matter what, there will be a little voltage drop on the output during the intermediate step. This is no problem if the transition is high to low, but could be a problem if we're going low-high, as we will briefly supply too low voltage to output. Time in intermediate step tINT, load current iLOAD, and output capacitance COUT together determine the voltage drop. I know that eventually the module should support 10A. If I accept a 0.1V voltage drop and assume tINT = 1ms, then:
COUT = iLOAD * tINT / VDROP = 10A * 1ms / 0.1V = 100 mF
Yikes! I'd better hope tINT is more like 1us. On the STM32G0B0KE you can write to multiple GPIOs in one instruction so long as they're on the same port, so it seems prima facie possible.
But the only way to know is to run some experiments!
Please send comments to blogger-jack@pearson.onl.