Timeline for How to improve efficiency of a "dead simple" motor driver - are more MOSFETs better or not?
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| Dec 25, 2022 at 16:45 | comment | added | php_nub_qq | Oh I was dividing by 350uA, my bad, thanks again! | |
| Dec 25, 2022 at 9:22 | comment | added | DKNguyen | @php_nub_qq 8.6nC/350mA = 25NANOseconds. And that would be without things like gate resistances The listed rise and fall times are around there at 84ns and 15ns. It is also stated in the right column that the conditions these timings were measured under was with a 5V gate drive through a 3.5 Ohm gate resistance which is an RC charging/discharge which is not a constant current drive which the Q=IT approach assumes (which I guess I never mentioned and is one of the inaccuracies of the simplified approach. A gate driver is usually not constant current). | |
| Dec 24, 2022 at 20:54 | comment | added | php_nub_qq | I'm sorry to continue posting on here but there's still something I don't get. In your answer you say that trise/fall (switching speed?) is equal to Qgs/Igatedriver, for this transistor Qgs is 8.6nC and according to the answer in the linked question the gate current needed for 100ns switching is 350mA (Qg/100). But if I calculate with your formula I get 8.6/350000 = 25us, which is very far from 100ns. | |
| Dec 24, 2022 at 18:55 | history | edited | DKNguyen | CC BY-SA 4.0 |
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| Dec 24, 2022 at 18:43 | history | edited | DKNguyen | CC BY-SA 4.0 |
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| Dec 24, 2022 at 18:26 | history | edited | DKNguyen | CC BY-SA 4.0 |
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| Dec 24, 2022 at 18:23 | comment | added | DKNguyen | @php_nub_qq It's just kind of something to go by. It's part of the reason I used a simplified switching waveform for the calculation and not something more complex. Also note I made some corrections to the equations. There are other calculation methods and spreadsheets using other methods supplied by vendors. Not sure if they are more accurate or not but they take different (and confusing) parameters. They didn't seem any more or less accurate to me but maybe I was using them incorrectly. | |
| Dec 24, 2022 at 18:20 | comment | added | DKNguyen | @php_nub_qq Yeah it's a design problem so you're picking an option from a multi-dimensional space. You play around with things until you find something you can both achieve and live with. And if you increase your gate driver current, then you can reduce your switching losses which then means you can reduce your conduction losses for a better overall crossover point. But remember what I said...I've never had the calculations match up with real world measurements, though measuring that can be tricky to begin with. | |
| Dec 24, 2022 at 18:19 | comment | added | php_nub_qq | Using the information provided in your answer I quickly spun up a spreadsheet and determined the sum of conduction and switching losses decreases until we reach 3 transistors, after which it begins raising. This was very useful, thank you! | |
| Dec 24, 2022 at 18:18 | history | edited | DKNguyen | CC BY-SA 4.0 |
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| Dec 24, 2022 at 18:13 | vote | accept | php_nub_qq | ||
| Dec 24, 2022 at 18:12 | comment | added | DKNguyen | @php_nub_qq Then you do what I am describing. And yes there is also a switching loss in the DS path in addition to the conduction (resistive) loss and in doing so you usually neglect the energy used to energize the gates since it is miniscule compared to the heating the MOSFET experiences while transitioning. | |
| Dec 24, 2022 at 18:10 | comment | added | php_nub_qq | I was looking at the gate driver as part of the motor driver, I'm aiming at improving efficiency of the motor driver as a whole. If I'm understanding correctly, with a switching time of 100ns, Vds of 36V and Iload of 30A, then Pswitching works out to be 19W which is pretty much the same as using 1 transistor. | |
| Dec 24, 2022 at 18:10 | history | edited | DKNguyen | CC BY-SA 4.0 |
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| Dec 24, 2022 at 18:09 | comment | added | DKNguyen | The energy dissipated by the gate driver should be something like \$\frac{1}{2}C_gsV^2\$ then you multiply that by frequency to get the power. | |
| Dec 24, 2022 at 18:05 | comment | added | DKNguyen | Do you want to increase the efficiency of the gate driver itself? Or of the motor driver as a whole? Most people would try to increase the efficiency of the motor driver as a whole. If you want to do that, then you do what I described. If you want to increase the efficiency of the gate driver that's something else and in that case you would be unconcerned with the conduction losses of the MOSFET since those have nothing to do with the power that the gate driver is dissipating. | |
| Dec 24, 2022 at 18:03 | comment | added | DKNguyen | @php_nub_qq I read over your post more carefully and what you call a switching loss is not what is normally understood to be a switching loss. You're sort of chasing after the wrong thing. You're chasing after the energy of the gate driver, not the losses incurred in the drain source of the MOSFET when switching. The drain-source of a MOSFET experiences both conduction losses and switching losses. | |
| Dec 24, 2022 at 17:59 | history | edited | DKNguyen | CC BY-SA 4.0 |
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| Dec 24, 2022 at 17:59 | comment | added | php_nub_qq | I suppose that's why the accepted answer on the linked question suggested a short as possible switch time. But by you saying the energy needed to charge the gates is negligible I'm assuming that my thoughts were correct, which was what I was aiming to learn with this question. Unfortunately I used poor terminology, apparently. | |
| Dec 24, 2022 at 17:56 | comment | added | DKNguyen | @php_nub_qq It does take energy to charge and discharge the gates but that is negligible in most circuits compared to the massive amounts of power being burned when the transitions between on and off. 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. When people talk about switching loss of MOSFETs, that's what they are talking about, not the tiny amount of energy used to charge and discharge the gate capacitance which is basically just the power consumption of gate driver | |
| Dec 24, 2022 at 17:54 | history | edited | DKNguyen | CC BY-SA 4.0 |
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| Dec 24, 2022 at 17:47 | comment | added | php_nub_qq | I'm picking up that there's something I'm completely unaware of. As switching losses I'm addressing energy used to energize the gates, there's also a switching loss in the DS path? | |
| Dec 24, 2022 at 17:45 | history | edited | DKNguyen | CC BY-SA 4.0 |
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| Dec 24, 2022 at 17:39 | history | answered | DKNguyen | CC BY-SA 4.0 |