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I'm trying to build my own BLDC motor controller. I'm wondering if I really have to implement dead time between switching the high- and low-side FETs. My switching frequency is 30 kHz, my motor is 5 kW, 48 V and the max. RPM of the motor is ~3000 RPM.

Attached is the switching scheme for my motor in forward direction. It also shows the state of the Hall sensors.

enter image description here

For example, the state of phase U never suddenly changes from HIGH to LOW. As far as I understood dead time insertion, I don't need this for my switching scheme, am I right?

ocrdu
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BornToPizza
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    Dead times are required in PWM signal on the high and low sides of your half-bridge inverter (if you have one) and not on the signals you are showing here. And yes, these are absolutely required to prevent short-circuiting. – Eugene Sh. Apr 14 '21 at 21:17
  • As Eugene said, dead-time is on the gate signals for each individual transistor in the motor driver, not the phases of the motor (the effects will manifest there, but it is not where they are meant to have an effect and not where you examine them) Your motor driver definitely has situtations where one transistor in the half-bridge turns on immediately after the other one turns off in order to produce the phase state waveforms you posted. – DKNguyen Apr 14 '21 at 21:20
  • Alright, thank you guys. How should I implement this dead thime then? For example if I switch from Step 01 (U Low, V 0, W HIGH) to Step 02 (U 0, V LOW, W HIGH), the first thing my program does is swith off all Transistors. Then I switch on the Low Side Transistor of Phase V and after a certain dead time the HIGH Side Transistor of Phase W. Or is this wrong? – BornToPizza Apr 14 '21 at 21:29
  • Are you using PWM to control the phases? Or you just fully energize them? – Eugene Sh. Apr 14 '21 at 21:32
  • If you are using complimentary PWM peripherals in an MCU, these have a deadtime setting. – DKNguyen Apr 14 '21 at 21:33
  • yes im using PWM. The PWM Signals are generated by a STM32 Microcontroller. (Blue Pill Board) – BornToPizza Apr 14 '21 at 21:35
  • I think, as @DKNguyen said, the PWM controller on STM32 might have dead time setting. – Eugene Sh. Apr 14 '21 at 21:36
  • Only the complimentary timers have this I think. You need (and should) be using those. If you don't have those then it's a massive PITA to do in software I think and you might just be better off using diodes and resistors at the gate to make the turn of time faster than the turn on time. – DKNguyen Apr 14 '21 at 21:38
  • Ok thank you. Though I want to understand what the MCU does by using the dead time setting. So are my thoughts from above correct? – BornToPizza Apr 14 '21 at 21:38
  • The deadime setting in the complimentary PWM module just makes it so that neither output of the signal pair can both be HI within a certain time period of each other. So if it's 100ns, it means that after one signal in the pair goes from HI to LO, the other signal cannot go HI until 100ns has passed. – DKNguyen Apr 14 '21 at 21:40
  • Ok im using Diodes and resistors at the gate to source. Do you think I can leave out the Dead Time in Software? – BornToPizza Apr 14 '21 at 21:41
  • Yes if you are doing it in hardware it shouldn't be needed in software. – DKNguyen Apr 14 '21 at 21:44
  • Alright, thank you very much. You really helped me a lot. :-) – BornToPizza Apr 14 '21 at 21:45

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I don't need this for my switching scheme, am I right?

Correct. You only need dead time if you are switching the high and low sides of a phase terminal alternately, eg. when doing synchronous PWM or 'active freewheeling'.

Bruce Abbott
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  • Could you elaborate your answer. I might be interpreted to mean that it is OK to have have both high and low sides of a phase active simultaneously. ("You only need dead time if you are switching the high and low sides of a phase terminal alternately" does not rule that out). However, if interpreted correctly, then isn't it the case that what you are suggesting is an example of dead time? The poster's scheme goes "off" between each high and low. – Math Keeps Me Busy Apr 15 '21 at 00:27
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    Obviously it isn't OK to have both high and low sides of a phase on simultaneously, but that won't happen with properly implemented 6 step commutation. The 'off' times in that scheme are not 'dead' time as it is usually defined, they are part of the commutation sequence. – Bruce Abbott Apr 15 '21 at 12:54
  • @MathKeepsMeBusy Things are only problematic when you switch directly from "low closed" to "high closed" or vice-versa. In 6-step commutation such state transition doesn't occur: you only switch between "low/high closed" and "open" or "braking". There's never a chance for both high and low sides to be closed simulataneously, since such states are never adjacent in time. – Kuba hasn't forgotten Monica Jul 25 '23 at 19:54