Continue to Site

Welcome to EDAboard.com

Welcome to our site! EDAboard.com is an international Electronics Discussion Forum focused on EDA software, circuits, schematics, books, theory, papers, asic, pld, 8051, DSP, Network, RF, Analog Design, PCB, Service Manuals... and a whole lot more! To participate you need to register. Registration is free. Click here to register now.

Current output regulated Buckboost converter feeds BLDC Motor and its inverter drive

Status
Not open for further replies.
T

treez

Guest
Hello,
I am doing the above as part of my hobby.
What will the combination of BLDC Motor (its for a pump) and its inverter drive look like electrically to the buckboost converter which sources current to it?
I assume it will not look like a resistive load.
Total Pout = 650W
Vin to buckboost = 18 to32V, 28V Nominal
Vout to motor inverter drive is 5 to 34v
 

The output capacitor is close to zero ohms, and it is the most immediate load on the converter.

The capacitor absorbs a large portion (perhaps the majority) of the amperage coming out of the coil.
 

thanks, so you think there should be no output cap on the output of the buckboost?..do you agree that the load of an inverter and a bldc, will look like an RL network, and that every 250us or so, the load will be open circuit as the motor inverter has dead time?
How do you deal with the regular spiking up of buckboost output voltage due to the dead time of the inverter?
 

The output capacitor is needed to absorb inductor current bursts. This is essential at moments when the load is not drawing current. If the coil sees high impedance, then output voltage can spike to the kV level.
Being very low impedance the output capacitor prevents this.

It's easy to guess an inverter will draw varying current, depending on the load.

As for a BLDC motor (3 or more winding), I think it has smoother commutation than a brushed type, hence it pulls a relatively smooth current waveform. I get the impression a BLDC appears almost as a constant resistive load (looking at waveforms in the link below). I am not a motor expert, so I could be wrong.

https://dcacmotors.blogspot.com/2010/03/waveform-of-brushless-dc-motor.html
 
  • Like
Reactions: treez

    T

    Points: 2
    Helpful Answer Positive Rating
How can a BLDC motor, or any motor, look like a resistive load?
 

How can a BLDC motor, or any motor, look like a resistive load?

A brushed DC motor has a number of inductors (windings)...
each of which is switched (commutated) into and out of the power loop, one after the next.
The effect is sort of like a mechanically-switched buck converter, when you think about it. (However sparks may occur, there being no diodes.)

Slow speed, greater current drain (more time for a winding to be exposed to power supply.)

The L/R time constant governs on a magnified scale, however there isn't just one inductor, but many inductors. Thus the waveform is sawtooth-like, uniform. On a long-term scale, it averages out to a more or less constant current.

A brushless type can have less abrupt commutation. Current draw can be less sawtooth-shaped.
 
  • Like
Reactions: treez

    T

    Points: 2
    Helpful Answer Positive Rating
I once tried putting two 12V BLDC fan motors in series to run of 24V. It did not work!, even putting a 1000MF across each motor it still did not work. I had to get a proper 12V PSU for them to work. The fans I used must have had a an extremely spiky wave form, so that if one motor was not taking current, there was not enough voltage available to keep the other going.
Frank
 
  • Like
Reactions: treez

    T

    Points: 2
    Helpful Answer Positive Rating
I once tried putting two 12V BLDC fan motors in series to run of 24V. It did not work!, even putting a 1000MF across each motor it still did not work. I had to get a proper 12V PSU for them to work. The fans I used must have had a an extremely spiky wave form, so that if one motor was not taking current, there was not enough voltage available to keep the other going.
This might happen due additional requirements of the control circuit (e.g. a minimal operation voltage) and must not necessarily be related to BLDC principle. But it should be taken as a warning that unconstrained current feed may involve problems.

In contrast to post #4, I'm sure that a reasonable understanding of "current regulated" refers to averaged quantities and thus implies a bus capacitor between boost output and BLDC bridge. A simple reason is that neither boost output nor BLDC bridge are designed for current fed operation.

Current fed bus is nevertheless an inverter design concept.
 
  • Like
Reactions: treez

    T

    Points: 2
    Helpful Answer Positive Rating
Please advise if we can feed the ML4425 BLDC Inverter driver with an upstream regulated current source, instead of using the ML4425 to regulate motor speed and current? (as in the block diagram attached)

I suspect our method is bogus, and that we are all totally wasting our time here in our Dutch engineering company

ML4425 datasheet:
http://pdf.datasheetcatalog.com/datasheet/fairchild/ML4425.pdf

We are going to use the ML4425 BLDC Inverter driver IC to drive a BLDC in a water pump.
The inverter obviously needs a big supply capacitor at its input, and we do not have room for this, therefore, we will instead put a 150KHz buckboost converter upstream of the inverter, and ‘feed’ the inverter from that. We will in fact use the buckboost converter as an output current regulated power source, and so it controls the current that gets fed to the ML4425 based inverter. We will feed whatever current to the Motor/Inverter that makes it spin at 8000rpm.
The ML4425 is capable of regulating the motor current and speed itself, using its current sense pin (pin 1), etc.. however, we will simply ground the ML4425’s current sense pin, so that the ML4425’s internal current sense and current limitation is bypassed. (since we are limiting the current with our upstream buckboost converter)

We have a closed loop on the motor speed, whereby the motor speed is regulated to 8000rpm, however, if in so doing, the motor current goes over 20 Amps, then our current limiter kicks in and keeps the current regulated to no more than 20 Amps. The current limiter function is done by the buckboost converter and its feedback loop.

The Vin to the buckboost is 18-32V
The motor will be ~34V when at 8000rpm, and will draw approx 20 Amps at this speed of 8000rpm. The inverter switches at 500Hz with 200us dead time.

Do you agree that we are wise to act in this way? We will experience high voltage spikes at the output of our buckboost converter whenever we are in the inverter commutation dead time intervals, but should that worry us too much? We do have an overvoltage clamp there (converter shuts down till Vout falls below 35V).

Or would we be better off making the buckboost SMPS an output voltage regulated SMPS, with output current limitation?

Alternatively, is there a Current Source Inverter version of the ML4425? (ie something like ML4425 which drives an inverter with switch “overlap time” instead of switch “dead time”)?
 

Attachments

  • Motor speed regulator.pdf
    206.9 KB · Views: 176
Last edited by a moderator:

I understand that the BLDC operation is self-commutated, similar to a brushed DC. I would expect a resistive behaviour with parallel bus capacitor. Load inertia torque will act like an additional capacitance.
 
  • Like
Reactions: treez

    T

    Points: 2
    Helpful Answer Positive Rating
The following
**broken link removed**
..just above the waveoform graph, says that BLDC inverters run cooler (less switching losses) if driven at 100% duty cycle and with the DC link voltage regulated to give the required speed, why is this?

according to Dave Wilson, technical lead for Texas Instruments Motor Solutions Group. "Instead, use 100-percent duty cycle all the time and only switch the inverter transistors at the commutation boundaries. To control the motor’s speed, simply change the DC voltage driving the inverter transistors. This approach reduces the switching losses to almost zero on all phases, and it mitigates high frequency losses inside the motor."
 

If you had asked me about the problem, I had given the same answer without knowing the article. It's just plausible.

Most important, you get lower Irms for some torque generating average current. Less important with fast transistors, switching losses can be reduced if switching occurs nearer to voltage zero crossing.
 
  • Like
Reactions: treez

    T

    Points: 2
    Helpful Answer Positive Rating
-and you mean that when it (the inverter) is always on 100% duty cycle, then you do tend to get the zero voltage switching in the inverter's transistors?

..point taken about the Irms being less...that is a good point which I did not think of.
 

May I ask another question about our water pump drive?
The PDF below again shows the block diagram of our setup.
....it goes like this....VIN...Variable output buckboost SMPS....Inverter.....BLDC.

The point is that our DC link capacitor is only 300uF, and we have no room for more capacitance than this.
The problem is(?), that the resonant period of the DC link capacitor, and the motor coil inductance is just 814us....and this is less than than the inverter commutation period of 1.05ms.........
therefore, do you believe that we will suffer overly high resonant currents in our BLDC motor coils?

Please remember that usually, the inverter would switch the motor coil current at high frequency, such that the motor coil current is in control at all times, and the resonant LC situation doesn't emerge....but we are using the buckboost to effectively control the motor current, and the problem is that the inverter switchs a motor coil in for a time period that is longer than the LC resonant period of the dc link capacitor and the motor coil inductance.

(This is an analogous situation to a coupled sepic converter's capacitor, whereby the switching period in a sepic should be much shorter than the resonant period of the sepic transformer leakage and the sepic capacitor.)

When the Inverter's BLDC is permanently on 100% duty cycle like in our system, then shouldn't the DC Link capacitor be big enough such that the motor coil commutation period is much less than the resonant period of the DC link capacitor and the motor coil inductance?...otherwise overly high resonant currents will occur in the motor coils
 

Attachments

  • Original DC pump drive.pdf
    279.3 KB · Views: 93
Last edited by a moderator:

Presumed it has sufficient high switching frequency, I would control the buck-boost converter in a different way:
Fast voltage feedback and superimposed slow speed feedback.

The buck-boost converter should be able to keep the DC link voltage almost constant. In any case, the small DC link capacictor will cause higher peak currents in the buck-boost converter and input source, thus reducing overall efficiency. In so far, an appropriate DC link capacitor would be definitely the better solution.
 
  • Like
Reactions: treez

    T

    Points: 2
    Helpful Answer Positive Rating
Presumed it has sufficient high switching frequency, I would control the buck-boost converter in a different way:
Fast voltage feedback and superimposed slow speed feedback.

...yes, that is the way we are controlling it....because as you know, this stops the speed feedback loop , and the voltage feedback loop from "fighting" with each other.

Though that was not the question here.

The question here, is "will we see overly high resonant currents in our motor coils because the resonant period of motor coil inductance and dc link capacitor is less than the motor coil commutation on time interval?
 

I'm not so good in mind-reading. An underlying voltage control loop isn't shown in your block diagram. The effects you previously described, e.g. large voltage variations in the D.C. link during motor current gaps won't happen with a voltage control loop.

The answer to the resonance problem is, if the voltage feedback loop is fast enough, it will damp respectively cancel the resonance.
 
  • Like
Reactions: treez

    T

    Points: 2
    Helpful Answer Positive Rating
I say its the buckboost's output voltage that we are controlling.....and it is that effectively, but its actually the output current of the buckboost that we regulate............though as you know, when delivering a certain current, the buckboost has a certain average output voltage...and that's what I was referring to here.

The resonant frequency of the buckboost output capacitance and the motor coil inductance is 1230Hz.
So are you saying that the buckboost output current feedback loop needs to have a bandwidth of more than 1230Hz in order for overly high resonances in the motor coil *not* to occur.?

Ive attached an updated block diagram of the whole motor drive system in order to show how it works, with the current loop depicted.
As you know, whether you regulate buckboost output current or voltage....you are going to end up regulating both of these things, since they depend on each other.

If the buckboost's current feedback loop must be higher in bandwidth than the resonant frequency of the buckboost output capacitor (= the DC link capacitor) and the motor coil inductance, then how much higher must it be?, for example, would you say that resonances in the DC link capacitor and the motor coil will be gotten rid of if our buckboost current feedback loop bandwidth is three times greater than the dc link/motor coil resonant frequency?...or should it be more than ten times in order to avoid resonances?
 

Attachments

  • DC Pump drive showing current and speed regulation of motor.pdf
    275.5 KB · Views: 52
Last edited by a moderator:

I don't attempt to design the complete controller for you, I only tried to sketch my approach to the control problem.

Your latest post suggest that you don't have a voltage control loop, according to my previous assumptions. Instead there seems to be an output current control loop.

Apart from the buck-boost control capability, LT8705 is a typical current mode PWM controller. The "inner" current control loop does a cycle-to-cycle pulse width control, with a bandwitdh in the switching frequency order of magnitude. The "outer" control loop tries to maintain a constant output voltage in normal operation, but can also work as current limiting controller. It's bandwidth is set by a compensation network at Vc according to application requirements. The output of the outer controller is a current setpoint for the inner controller. According to your block diagram, you only implement current feedback but no voltage feedback for the outer control loop.

To find out what's possible with LT8705 under the presently chosen conditions (switching frequency, inductor, output capacitor), I would setup it as standard constant voltage controller and adjust the compensation for best load step resonse.
 
Last edited:
  • Like
Reactions: treez

    T

    Points: 2
    Helpful Answer Positive Rating
Status
Not open for further replies.

Similar threads

Part and Inventory Search

Welcome to EDABoard.com

Sponsor

Back
Top