DC motor and generator modes

suppose we have a PM DC motor
we have A and B terminals

if we conect (+) to A and (-) to B the motor turns to one side
if we disconect power supply and turn the motor to the same side like above
( i.e generator mode )

will the A (+) and B (-) ?
or A (-) and B (+) ?

A will be +. just breaking the supply to the motor will result in the motor instantly becoming a generator until it slows down to a stop. The motor is actually a generator all the time, it only takes power when the external voltage is higher then the voltage it is generating. For instance, 12 V motor with a resistance of 2 ohms, would at first sight take 12/2 = 6 A, but as it spins it will generate say 10V, so there is only 12-10 =2 V to push the current of 2/2 = 1A through it.
Frank

hello thanks for the answer
I was thinking the same answer too
why I asked this question ? becouse I am planning to build a BLDC servo driver
and how I will gonna brake the motor confuses me much

you say A will be (+) while braking in the above example
ok - but how the DC link voltage is increasing while braking then ?
the motor back EMF and DC link voltage is opposite - they doesnot add - they subtract

I read about profesional BLDC drivers and it says in the many documents
while braking the DC link voltage increase and
a braking MOSFET "PWM" this voltage through a braking resistor

can you explain the mechanism of braking in BLDC motor driver ?

If the motor is generating power while it is running down, where does this power go? If it connected to a battery, it charges the battery (regenerative braking) if it's connected to a power supply, its voltage will continue to increase until it can pump current back into the source. This is where your braking resistor comes into effect. If you want to brake a motor fast, put a short circuit across it, there will be a massive amount of current flowing so the motors stored energy is dissipated rapidly.
Frank

I draw a picture to show if I understand you correct
in the above picture the motor is "motoring", the current flows through IGBT -> motor phase -> IGBT -> back to DC link

now the below picture
if I understand you correctly , the motor current , while changing from motoring to generating
cannot reverse due to its inductive nature
and continues to flow in the same direction
but there is only one way for the current to flow through the freewheeling diodes into the DC link

now my question
if I understand it correct I dont need to PWM the 3 phase IGBT 's while braking ( keep them OFF )
I only need to PWM the braking IGBT until the DC link voltage is down to 310 V again ? ( ~220*sqrt(2) )
or do I need to measure also the current through the braking resistor ? ( if it is down to zero )
to finish braking mode

please correct me if I am wrong

while changing from motoring to generating cannot reverse due to its inductive nature and continues to flow in the same direction

Click to expand...

Unfortunately, the assumption is wrong. The current direction is changing from motor to generator operation. The shown circuit isn't able to "brake" for this reason. You would need a full bridge, that handles both current polarities.

You're of course complicating the discussion by introducing a BLDC, which is basically an AC motor and must be driven by an inverter instead of a DC chopper. The circuit confusingly shows an AC motor connected to a DC chopper, which doesn't make sense.

P.S.: Regarding your question, because the motor voltage will be usually below DC bus voltage, the inverter has to enter boost operation to "pull" power from the motor to the DC bus during brake operation.

A regular VFD inverter is synchronously switching, no matter of the power flow direction. So decreasing the inverter output voltage below the motor voltage will simply start braking.

hello - thanks for the answer

I drew only one phase of BLDC motor + inverter operating in one direction
for the simplicity I eliminated the other IGBTs and diodes of 3 phase bridge

if I decrease the inverter output voltage below motor voltage which way the current flows ?
through the IGBTs or through the freewheeling diodes ?
can you edit my second graphic ?

what is the mechanism which increases the DC link voltage during braking ?

Observing a motor supplied by a sine voltage, e.g. connected to the mains supply, we can simply assume, that in motoric operation, motor voltage and current have the same direction, resulting in a positive instantaneous power P(t) = V(t)*I(t) > 0, while the signs are opposite in generatoric mode, so P(t) < 0. The motor voltage is mainly representing the rotor emf. More specific, the magnitude represents the rotational speed, and the phase the instantaneous electrical rotor angle. If you change to generator respectively braking, the voltage won't change much, but the current immediately reverses it's sign.

PWM operation is superimposing switch frequent voltages and currents to the above sketched basic motor waveforms. It's average values are however corresponding to the said low frequency waveforms. For a basic PWM model, you don't need to distinguish between transistors and diodes, you can simply assume ideal switches for the equivalent circuit. Furthermore, synchronous switching should be assumed, in other words, in each half bridge, either the high side or low side switch is closed. The duty cycle is representing an average voltage, e.g. 50 % means half of DC bus voltage. This simple model is representing a voltage source, the voltage is set by the PWM generator. By making it smaller or larger than the motor emf, you set the energy flow direction and amount. Generatoric operation generally implies charging the DC bus from the motor. So far a behavioral PWM inverter model, that doesn't care if IGBTs or diodes are carrying the current and how the PWM current waveforms looks like.

To go back to the original problem of IGBT versus transistor current, the current will communitate between both permanently, both in motoric and generatoric operation, but with different duty cycle. In motor operation, the IGBTs will carry the current for more than 50%, so the average DC bus current goes to the motor. In generator operation it's the opposite case.

hello , thanks for answers

I have 2 more questions

1 ) do I have to measure back EMF of motor or can I calculate it from the speed of motor ?

2 ) I want measure the motor current with a shunt resistor
during motoring the motor current flows in one direction
during braking , does it flow in the opposit direction ?
do I have to make an absolute value amplifier for measuring motor current
both in motoring and generating ? ( becouse microcontroller has unipolar ADC )

1. You can't measure it directly, only terminal voltage can be measured. And of yourse you can calculated it.
2. One would prefer a bipolar measurement that also gives the current sign. The ADC range can be shifted.

hello - please your advice

I_soll = (Speed_soll - Speed_ist) x S_gain
Vpwm = ( I_soll - I_ist ) x I_gain + EMF

is this aproach correct ?

is this aproach correct

Click to expand...

I understand, that you are suggesting a control algorithm. I guess, it can work. If it's good depends on the operation conditions of the motor, particularly the disturbing influences. If constant speed is an objective, most people would start with a PI speed controller. Because EMF is directly linked with speed, the purpose of using an separate estimated EMF in the control algorithm isn't obvious to me.

just like what you said , I am planning to derive EMF from speed
i.e. aproximately EMF = ( Speed_ist x 310 V ) / nominal speed of BLDC motor
< 310 = sqrt(2) x 220 V >

what I want ask is , if my second equation ( Vpwm) in the previous mesage is ok ?
this equation takes into account the reversal of current sollwert I_soll when generating

If you want to estimate the required Vpwm setting for a given operation point,
Vpwm = ( I_soll - I_ist ) x I_gain + EMF
can't be correct. The difference I_soll - I_ist (I_targetvalue - I_processvalue) doesn't actually exist in the motor state. It may be reasonable as error input to a controller, then I_gain would considered as a proportional gain. But as I already mentioned, I don't see a clear purpose for it. Won't a simple speed control loop achieve the same?

hello

I have almost completed the PCB for my brushless DC motor driver
what I want ask is

is the motor coils UVW conected as star or delta ?

if it is conected delta
when I energize for example UV then the current not only flows
through UV coil but also flows thrugh the coils UW and WV as well

but isnot this an error ?

is the motor coils UVW conected as star or delta ?

Click to expand...

Both circuits can be found.

but isnot this an error ?

Click to expand...

No.

hello dear forum ,

at last I could finish my servo driver
as you can see photo in the attachment

the servo driver works good if I control it with 0-10 V analog speed command
( +/ -10 V from servo itself )

however I try to conect it to the PLC board with RS485 comunication
becouse the power circuitry of the driver not isolated from microcontroller circuitry no success in the RS485 comm.

if I conect the both grounds of driver and PLC it gets worse
If I disconect grounds of driver and PLC there is comunication
but if I conect an operator panel to the PLC the comunication is again disabled

now I understand why the profesional drivers have isolation between control and power circuitry in the driver :-(

what can I do now ? I cant connect the ground of the driver to neutral line
becouse I use 220 V network to power the driver - and between neutral and ground I measure about 130 VAC - I am afraid I can blow up the all circuit if I conect them