Simple probe PCDU

Question

I am currantly working on an aerospace engineering project at my university which will fly on a sounding-rocket and I really suck at electical engineering, so I basically need a complete explanation of how this works:

I have a sort of small "space-probe", which has a built-in 3.7V LiPo-cell (500mAh) and I need 2 power-rails, which provide 3.3V at 50mA and 5V at (max) 750mA to the probes dual RP2040 flight computer assembly, the sensors which are connected to the computer via an I2C bus and 2 servos, which will not run simultaneously.

The only thing I came up with yet, is a resistive divider for the 3.7 to 3.3 part, which is pretty wasteful though and will drain the cell way too fast. Although there are a couple of explanations of how to build step-up and step-down converters, I've not found one yet, which is simple enough for me to understand and I don't just want to copy something I see and not get how it works. It does not need to be adjustable or anything. Just 3.7 to 3.3 and 5.0

Answer 1

So you need a step-up converter to generate 5V from a Li-ion battery. I will assume the battery has built-in protection, otherwise you will need to add such protection to avoid over-discharging and damaging the battery.

750mA from an 85% efficient switching DC-DC converter will draw about 1.25A from the battery. You can buy a chip, inductor, and some passive components and build this, you can buy a module or a PCB-mount DC-DC converter. The latter may be a good choice for you- switching converters can be a bit finicky about layout, especially those that operate at MHz frequencies and are thus relatively small/light. I'm not going to suggest any particular product, but if you peruse a distributor catalog you can find many options. PCB modules (as opposed to PCB-mount devices) from dubious sources (Amazon, eBay, Ali etc.) aimed at hobbyist applications are extremely cheap but not always the best quality or performance (perhaps not meeting claims, even), so a few extra dollars may be well-spent and you'll get proper specs and curves to work with.

For 50mA maximum at 3.3V, a linear regulator (sometimes called an LDO, rightly or wrongly) can be used. Nominal power dissipation at 50mA is only 85mW so even a tiny SOT23 regulator can be used. That will add about 80 mA maximum to your 750mA so you need 830mA at 5V at the input current might be 1.4A. You could operate the 3.3V regulator directly from the battery and save a bit of power/battery capacity but the regulation would fail when the battery voltage dropped below the dropout voltage of the regulator (at best a bit more than 3.3V). Again, you can find many such products in distributor catalogs. They're fairly easy to use but you need to pay close attention to the recommendations for capacitors, in particular the output capacitor characteristics. I suggest choosing one that specifically says you can use ceramic capacitors.