How to improve efficiency of a "dead simple" motor driver - are more MOSFETs better or not?

Question

This is a follow-up of that question.

The only reason I've got 8 transistors on that schematic is efficiency. I'm now trying to figure out whether switching losses are making it pointless to have so many transistors.

Theoretically only 1 of those transistors will be enough to run the motor, but power lost in the DS path is current squared times resistance so at 30A that's:

$$ 30^2*0.025 = 22.5 W $$

That's not a small loss and will probably require a huge heat sink. Contrary, if I put 8 transistors in parallel as in the schematic, each resistor will carry 30/8 = 3.75A, which will result in a total loss of:

$$ 3.75^2*0.025*8 = 2.81 W $$

However switching losses will increase 8 times. The answer to the other question has helped me determine that if I want to switch at 100ns the current needed to energize the gate will be 350mA, so times 8 (transistors) that's 2.8A or 14W, which is getting close to the loss with just 1 transistor. But thinking about it, 350mA is not consumed constantly rather only during gate energizing, which lasts 100ns. At a frequency of 22khz that's obviously 22k switches per second, times 100ns equalling 2.2ms of total time draining current per second.

Thus, the real power "wasted" in switching gates will be:

$$ 14 * 0.0022 = 0.031W $$

Which is substantially better than 14W, so having more transistors is clearly the better choice. Unfortunately I'm nowhere near an electrical engineer and I'm completely uncertain about my theory above, that's why I'm posting this question.

EDIT:

For future readers, since the comments on the answer are quite long: It turned out switching losses are not at all what I thought they were, what I was referring to as switching losses is apparently gate driver losses. The real power wasted is the sum of resistive losses and real switching losses, which in my case turned out to be optimal at 3 transistors, in theory.

Answer 1

Ideally, you want conduction losses to equal switching losses. But conduction and switching losses change with load (ala duty cycle) so the crossover point where switching losses = conduction losses depends on the operating point you choose.

Don't forget that MOSFET resistance rises with temperature. You should be using that value obtained from a graph in the MOSFET datasheet You need enough gate drive.

You already know how to calculate conduction losses. Calculating switching losses is more difficult and depends on your gate drive. That means if you do not have enough gate drive then more parallel MOSFETs will hurt more than it helps.

There's a lot to be said about how to calculate switching losses and I don't understand a lot of it (modelling MOSFET and all that junk), and frankly I've never had calculations match up with real world results so I am probably not doing something correctly, but I use a simplified calculation anyways since it's something to go by.

The calculation I use is the one that assumes that when the MOSFET turns on and off the Vds and Ids ramp in linearly opposite directions at equal rates between 0% and 100% voltage and current. So a more symmetrical version of this. An actual switching waveform looks a lot messier with lots going on, but whatever.

https://www.digikey.com/en/articles/a-review-of-zero-voltage-switching-and-its-importance-to-voltage-regulation

The rise or fall time is assumed to be the same and is approximated using Q = IT, where Q is the MOSFET total gate charge and I is the gate driver current. So \$t_{rise/ fall} = \frac{Qgs}{I_{gate.driver}}\$

The switching loss during one transition is the Vds multiplied by the Ids during that ramping. So on a graph it's a triangular shape which works out to be power.

Since we are assuming the voltage and current rise or fall between 0% and 100% and equal rates then one way to calculate the power lost in one transition is to multiply the average voltage by the average current during this transition period (which is half the value of each): \$P_{one.switch}=\frac{1}{4}V_{ds}I_{load}\$

Then to get the total switching power lost you do:

\$P_{switching} = P_{one.switch}(t_{rise}+t_{fall})f_{PWM}\$

Or you can calculate the energy lost during one transition geometrically from the area under the power graph over the switching region obtained by multiplying the voltage and current waveforms multiplied together. Then divide that by the PWM period to get the power.

ADDENDUM: The 0.031W calculated by the OP is the power used to charge and discharge the MOSFET gates. We don't normally care about this when calculating switching loss because the value is so miniscule compared to the power being burned off by the MOSFET as in transition between on and off states. But for interest sake, the energy in one charge or discharge would be calculated as \$\frac{1}{2}CgsVgs^2\$, then averaged over the PWM period to get the power. Neglecting little annoyances such as the gate resistances, losses in the gate driver components and the way gate capacitance varies with Vgs.

• Full on = low Vds x high Ids = low loss.
• Full off = high Vds x low Ids = low loss
• But when switching you are in between the same way a linear regulator is = lots of heat.