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Interpreting MOSFET driver datasheet

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Rusttree

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I'm looking at the L6384E MOSFET driver and I'm having difficulty parsing the datasheet.

Specifically, I want to know if I can run the chip from a digital output pin of a microcontroller. The voltage on the IN pin is not really very clear. Can I use the 0-5V from my uC to control it? The closest thing I can find in the datasheet is the Vil and Vih logic inputs on page 7. If I'm interpreting those correctly, then any voltage below 1.5V on the IN pin will turn off the driver and anything above 3.6V will turn on the driver. In which case I can use my uC directly to control the driver. Is that correct?

Also, what is that resistor across the top of the circuit on page 1?

Thanks,
Dan
 
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Yes, the inputs are logic compatible with a microprocessor digital output.

The resistor is apparently part of a RC filter the provides voltage to the bootstrap circuit (which provides voltage higher than the supply voltage to fully turn on the high-side (HVG) N-MOSFET output transistor).
 
A follow up question that just occurred to me. Ultimately, I'd like to use this driver to drive a 3-phase brushless DC motor. Conveniently, STMicroelectronics provides an application note that contains exactly the circuit I need. See page 12 in the PDF.

The L6384 provides only one pin to drive both the high-side and low-side MOSFET in the half-bridge. How does that circuit work if I can't independently control either MOSFET to allow current to flow through the correct sequence of coils in the motor?
 
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I think I figured it out. The driver inverts the signal between HVG and LVG. Therefore, if I send a PWM signal to one driver and the inverse of that PWM to another driver, current will flow through the high-side MOSFET of the 1st driver, through the motor coils, and then out the low side MOSFET of the 2nd driver.

Am I on the right track?
 
Your are correct.
 
Thanks for your help so far, Carl.

I'm stuck again.

Looking at the 3-phase bridge on page 12 of the application note again. Simplifying it to just DC, let's say I apply a logic high to IN1 and a logic low to IN2. This would drive the high-side MOSFET on the top driver and the low-side MOSFET on the middle driver. But what about the bottom driver? If I put a logic high on it, then current is going into two different coils. If I put a low on it, current is coming out of two different coils. I can't see how to get current to only flow into exactly one coil and out of exactly one coil.

I thought about deactivating the bottom driver by setting pin SD/DT pin low for just that driver, but that's not feasible in the provided circuit. And with the dual-functionality of that pin, it's not trivial to do that anyway.

Is there something obvious I'm missing?
 
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You have a good point. The schematic would appear to be in error. The only obvious way to do what you want is to deactivate the unused driver using the SD/DT pin as you suggested. If you also want to use that pin to determine the dead time, then you will need to switch the pin between 0V and the desired voltage for the dead time.
 
Ok, that makes me feel better that I do understand the fundamental concept of the circuit.

I put everything together and found some success using the SD/DT pin to turn off one of the three drivers in each phase. Instead of using a single resistor and connecting all the SD/DT pins together, like in the circuit, I put an individual resistor on each driver to ground. Then I connected digital I/O pins from the uC to each SD/DT. When I want the driver on, I change the digital I/O pin to input mode, which puts it in a high impedance state. That effectively makes the digital I/O line non-existent to the SD/DT pin. When the I want the driver off, I change the the digital I/O pin to output and drive it low. That overcomes the resistor and drives the SD/DT pin low, which turns off the chip.
 
hi,
I don't think the schematic is wrong.
The micro has to be coded to cause each pair of coils to be switched in the correct sequence and polarity to create the rotating magnetic field.
The direction is determined by the switching sequence and the speed is determined by the sensor lag angle.

pilko
 
Pilko has a good point. If you look at a normal digital 3-phase signal the phases sequentially alternate between high and low (see attached). Thus you do not necessarily need to disable one phase to generate a 3-phase signal and a rotating field for the motor.

3-Phase.jpg
 
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Looking at Carl's attached image above, at about the 1ms mark, the top two coils will both have current flowing into them and out of the bottom coil. Similarly, at the 6ms mark, current will into the 3rd coil and out of the top two.

So it is expected to get overlap when current is flowing into or out of two coils simultaneously?
 
It's a Six Step system, and the waveform looks like this:-

pilko
 

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  • pilko.jpg
    pilko.jpg
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Mosfet switching sequence:-

Phase A Hi \ Hi \ Off \ Lo \ Lo \ Off

Phase B Lo \ Off \ Hi \ Hi \ Off \ Lo

Phase C Off \ Lo \ Lo \ Off \ Hi \ Hi

pilko
 
Right, I understand the phasing sequence. My question relates to how you turn "off" one of the MOSFETs using an L6384 driver. If I could individually control each MOSFET, that would be trivial. But the L6384 half-bridge drivers do not allow for individual control. Using the circuit in the app note I linked to above, there does not seem to be a way to do so.

However, the circuit could achieve the phasing sequence that Carl attached in post #11. But then all three coils would always have current running through them in all 6 steps of the sequence. Either into 2 coils and out of 1, or into 1 and out of 2. If that's desirable, then we're good to go. But it's contrary to the theory that I've learned about 3-phase motor control.
 
And just to further clarify, because I think this is the source of confusion: The high and low-side drivers on the L6384 are always inverse of each other. So if I, for example, desire to turn off the low-side MOSFET of the half-bridge, the driver will automatically turn on the high-side MOSFET. And vica versa. The only way to turn off both MOSFETs on the half-bridge is deactivate the entire chip via the SD/DT pin. But the circuit in the application note does not allow for individual control the SD/DT pins. They're either all on or all off together.
 
Current towards star point is positive
Current away from star point is negative

Step 1--- Mos 1 and 4 on
Step 2--- Mos 1 and 6 on
Step 3--- Mos 3 and 6 on
Step 4--- Mos 3 and 2 on
Step 5--- Mos 5 and 2 on
Step 6--- Mos 5 and 4 on

pilko
 
pilko, please reread my posts #15 and #16 above. I am not confused in any way about the sequence of turning MOSFETs on and off. As I already stated, I understand the concept.

The problem I'm talking about is how to achieve the sequence with three L6384's using the 3-phase motor circuit provided in the application note.

Step 1--- Mos 1 and 4 on
What are the states of MOSFETs 5 and 6 in Step 1 using an L6384?
 
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I've used my digital sequence to drive a 3-phase AC motor. The main problem is that the induced current harmonics from the square-wave do not contribute to the motor power but create an additional I²R power loss in the motor windings. These losses do not occur when driving it with a sine wave.
 
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I figured that by appropriate switching between Shutdown and Deadtime modes, the six-step timing sequence could be achieved. I think now I was asking too much of the chip. I hope I havn't wasted too much of your time.

pilko
 
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