Jonathan S.

There's a bit of a problem with your requirements: You get maximum power at the output of a generator (generic power source) when the source and sink impedances are matched. This means that, at most, you can get (360V)² / 16 Ohms = 8100W into the robot, no matter what you do, unless you use a buck converter. At the same time, 8100W would be lost in the wire (that's bad, even in water). Your 30A requirement equates to 9600W. A buck convertrer is the only viable option. You need to reduce the current flowing in those wires.

No problem. Good luck with your motor driver!

I also learned programming first before I started tinkering with electronics! :)

You should be able to just solder the BTN8982 onto the PCB instead of the BTS7970 without any changes.

Yeah, it's amazing how powerful those tiny FETs have become!

For the low side, you only need a single FET as it will only see short spikes of current, which aren't enough to heat it up significantly.

Two of them in parallel should be able to drive your motor just fine without any cooling at all - 1W per FET is fine.

When you parallel two FETs, their resistance halves, the overall power dissipation halves, the power dissipation of each FET goes down to 1/4 (half the power through twice the FETs), and the current handling capability doubles.

Yeah, I saw that. If you're going to make a PCB, put two or three of these FETs in parallel for the high-side FET. (You can just connect their drain, source and gate terminals all to each other and treat them as a single big FET.)

If you want to do it cheap, just solder the drain tab to a bit of copper plate.

All good! :)

If you really push your motor to the stall rating, the high-side FET might dissipate up to 5 Watts, which will require some cooling. Without cooling, the FET will overheat and die.

Also, calculate the power dissipation of your FETs. It's not going to be insignificant at 20A or more.

I don't, I'm just a random engineer helping people out here on the site :P

Anything above a microsecond is probably fine.

The main goal is to move the braking energy from being dissipated in the low-side FET to being dissipated in the resistor, which is much more robust.

The resistor value isn't actually that critical. You could just see how it works with 1 Ohm. Use a high-power resistor, 10W at least.

If your motor pulls at most 5A, for example, 2 Ohms would work.

The resistor value depends on your motor. It has to be able to pass the maximum expected motor current at just below 12V - let's say 10V. It also has to be able to handle the braking energy without overheating.

The 2101 can drive those FETs just fine without one.

In the end, the circuit's goal is to create a supply voltage that's relative to the FET's source, not relative to ground, since the gate voltage also has to be relative to source.

All good! :P

The high-side MOSFET sees all voltages relative to its source terminal.

Let's assume the MOSFET's source is at 12V (motor on). That means OUT- is at 12V since it's connected. As a result, OUT+ is at 24V, and so is VB.

By connecting OUT- of the DC/DC converter to the MOSFET's source, it generates a voltage at OUT+ that's always 12V higher than source, regardless of where the source potential is "floating". As a result, that voltage can always be used to turn the MOSFET on cleanly.

If you want to turn the MOSFET on, gate has to be at least 5V higher than source. If source is already at 12V, gate must be at least at 17V.

The MOSFET only cares about the voltage difference between its gate and source terminals.

When the high-side MOSFET is on, the driver connects HO to VB. When the high-side MOSFET is off, it connects HO to VS.

VB is the supply voltage that the chip uses to generate the "HO" signal. If VB is at 12V with respect to ground, the driver can only output 12V at HO. However, since the source of the high-side MOSFET is also at 12V (when the motor is on), the difference between its gate and source voltage (which are both at 12V) is zero. As a result, the gate driver can't actually drive the high-side MOSFET.

Don't omit the capacitor between VS and VB! A 1µF ceramic cap would be ideal there. Don't use an electrolytic. If you don't have 1µF on hand, 100nF will do too.

Yeah, all "12(V)" nodes in that schematic are connected.

You connect it just like its datasheet suggests, but replace the bootstrap diode with the DC/DC converter.

Please excuse the crappy hand-drawn schematic. That's how you can use an isolated DC/DC converter (such as the 2nd one I linked) to power the high-side section of the IR2101S.

Well, maybe this one would be better since it can tolerate a wider input voltage range: digikey.de/en/products/detail/flex-power-modules/PUC0512S1B/…

In fact, all you need to add to your existing IR2101S circuit is one of these: digikey.com/en/products/detail/cui-inc/PDSE1-S12-S12-S/14673231

Normal MOSFETs need 10V to turn on, your logic-level ones need 5V, but it doesn't hurt them to drive either of them with more gate voltage (up to 20V at most).

You didn't waste the MOSFETs. You can still use them even with a driver that can also drive non-logic-level FETs.

That's why you can't use a regular bootstrap circuit. Instead, you'll have to somehow supply the high-side driver in a way that doesn't rely on the H-bridge switching frequently enough to top up some capacitor. An isolated DC/DC converter can do this - it can continuously supply power - and the LT1336's boost converter can also do this.

The problem arises when you drive the high side at 100% - that is, you have the motor on for a long period of time ("long" being "0.1 seconds"). Diode-capacitor bootstrap circuits work like this: Whenever the low-side FET is on, the capacitor recharges through the diode. When the high-side FET is on, the charge on the capacitor is used to supply the high-side driver. If the low-side MOSFET isn't turned on for a while, the charge on the capacitor runs out and the high-side driver malfunctions.

It's all good! It takes a while to develop intuition for this kind of stuff. :)

In that case, you need something to "translate" the ground-referenced voltage from the MCU to a source-referenced voltage for the MOSFET's gate. High-side drivers do this. In fact, it's their only purpose.

The video you linked only uses N-MOSFETs as low-side switches, in which case the MOSFET's source terminal is tied to ground and there's no problem driving it directly from the MCU. The MCU's reference voltage is ground, and the MOSFET's reference voltage is also ground, so it all works. The problems arise when a MOSFET's source terminal isn't tied to ground - like with your high-side MOSFET.

You can use MOSFETs that require up to 10V on their gate, yes. Resistors marked "Rsense" are for measuring the load current - if you don't need to measure it, you don't need those resistors.

It's even available in a DIP package. All you need is a few diodes, capacitors, and resistors and off you go.

Actually, Analog Devices has a chip that integrates a boost converter for the high-side (bootstrap) supply voltage, the LT1336. It is specifically designed for driving motors via a half-bridge. Here's its product page, complete with a schematic that shows how to use it to drive a motor from 12V: analog.com/en/products/lt1336.html

Isolation is not at all needed. You can directly implement the circuit topologies presented in that article.

You already have the correct MOSFET driver for this. The IR2101 is intended to do exactly that: Drive a half-bridge made from two N-MOSFETs. All you have to do is wire it up according to its datasheet. See: analog.com/en/technical-articles/… (You need a dedicated isolated supply, such as a 1W, 12V-to-12V isolated DC/DC brick.)

That's correct. The problem is that voltages are always relative - you always need a reference point. The microcontroller outputs 5V with respect to ground, while the MOSFET expects 5V on its gate with respect to its source terminal. If you connect source to ground, that's totally fine, both have the same reference. In your case, though, source is connected to the motor, which is at 12V (to ground) when switched on. Those 5V with respect to ground are actually -7V with respect to the MOSFET's source in that case.