FrightRisk
commented
1 year ago You are using load current in your calculation instead of ripple current. You are also off by a decimal point. So for 30% ripple on your theoretical 1.2A load, it is 1.2 x .3, which is .36 ripple. And correcting 410Hz to 410kHz, Your formula would be 7.2(12-7.2)/12.36410000). That is 1.95e-5 or 19.5e-6 which is 19.5uH.
Using our values calculated from 12V to 18V, the most efficient current would be 400mA at 12V and 600mA for 18V (give or take). Higher inductance gives us more efficiency at the cost of transient response, so pulling more current to 1.2A is not an issue. Since we are connecting to boards that all have plenty of capacitance at Vin and all have their own regulators, we really don't need to care about any transients.
I haven't done the heat transfer calculations for trying to pull more than 1A out of that regulator while both H-Bridges were near maximum, but we don't think most of our users would need more than 500mA. Powering accessories are best done on their own bus anyway, so all we have to do is power the microcontroller the shield is connected to. Of course there are situations where you would isolate the power between boards and use dedicated power supplies if you need to, and that is what the jumper pads are for.
murarduino
I looked at the data sheet for the MP9942GJ-Z and it supports up to 2A output with FSW of 410Khz. Usually VIN(9942 VOUT) to MCU Shield power supply, the general power supply needs in 100mA~500mA, in order to leave enough redundancy, I think 1.2A is enough. According to the inductance value formula: VIN 12V VOUT 7.2V: 7.2(12-7.2)/(12 1.2 410)=5.8uH VIN 20V VOUT7.2V: 7.2(20-7.2)/(20 1.2 410)=9.4uH VIN 12V VOUT 9V: 9(12-9)/(12 1.2 410)=4.6uH VIN 20V VOUT 9V: 9(20-9)/(20 1.2 410)=10uH
Based on the above calculations, the required inductance range is between 4.6uH and 10uH.So, what was the reason for choosing a 56uH inductor?