Control of DC Motor Driver - Help 1

[yanco]

I am using BA6219 reversible motor driver from ROHM. The Resistor connected to the BA6129 is 10 ohm 5 Watt.

The 10 ohm is connected to input of the 18 V unregulated adaptor which is supply to Pin 8 of the motor driver. The 7812 output is supplying 12 V to the rest of the circuit.

7812 is not hot. It is cold enough to touch.


18 Vin o-------------7812 --------
| |
| |
| R= 10ohm 5 watt
| |
| |
Pin7 Pin 8 of BA6219 Rohm Reversible Motor Driver


The Resistor 10 ohm is too hot until it melt the solder joint. I have try placing it vertical far from the PCB and using 50 ohm 5 Watt, it is also very hot. I can only touch the solder joint for 3 second.

If I change the resistor to higher value, like 100 ohm. The current to the BA6219 is too low. There is not enough energy to drive the motor. So, where is my mistake. Pls advice.

Thanks.
James

 

[jbartos]

Suggestion: It appears that the motor driver is BA6219 not BA6129. Also, clarify if the 7812 output is supplying 12V or 18V.
Visit

http://www.donberg.ie/descript/b/ba6229.htm

etc. for more info

 

[jbartos]

Suggestion: Visit

http://www.rohm.com/

http://www.rohm.com/products/databook/motor/pdf/ba6219b.pdf

(see if the connection is according to schematic. The current drive is BA6219B.)
etc. for more info

 

[RadarRay]

In a very very very quick read of the data sheet, it appears Rc is used to limit current for the unit if the output is shorted. Therefore you may have 18 volts across 10 ohms, which is 32 watts. No wonder it is hot. Also from the data sheet, the 6219 can supply 2.2 amps. So one would wonder what the resistance of your load is.

If you put a 100 ohm resistor in, your dropping all (most) of your supply across the 100 ohms, which is why that doesn't work. I would double check your wiring of the circuit and measure the ohms of the motor you are trying to drive. Again, that resistor is there to prevent a short circuit, so is it a short?

 

[yanco]

jbartos,

7812 is supplying 12 V to the rest of the circuit.

radarray,
I don't understand why the datasheet ask to ensure that the 10 ohm must be there to prevent a short circuit, the datasheet never mention the wattage require for the resistor. I have try to eliminate the 10 ohm by connecting a wire but it burn out the 6219B IC.

How to measure the ohm of the motor I am driving ?

 

[RadarRay]

I'd feel better about giving advice if I worked this all out but unfortunately I don't have the time right now.
Disconnect the motor and measure the motor resistance with and ohmmeter. Then take the supply volts and divide that by the total ohms. Example 18volts/(10 ohms resistor pls 5 ohms motor) = 1.2 amps. I doubt your motor has a resistance than high but this should give you an idea of what the motor would see at zero speed. Keep the zero speed amps under the max chip specs. Now if you can find what the motor CEMF will be at the speed you want the motor to run try to size the resistor small enough to still allow max motor current.
As a general rule, size the wattage of the resistor to twice your calculated watts.
For testing purposes, get a 10 ohm, 100watt resistor and mount if away from your circuit board until you get a feel for what is going on. You should also make sure the motor is unloaded during this learning experience.
Your motor probably has low ohms, which makes it draw a lot of current at zero speed, thus exceeding the 2.2 amp limit of the chip.

 

[yanco]

The motor resistance is 12 ohm when I measured. So total amp is 18V/22 ohm = 0.8 Amp. For the resistor wattage calculation, P= 0.8 x 0.8 x 10 = 10 watt. That means I need a 20 watt resistor. This is crazy, because 20 watt resistor will be damn big.

Is this means that the ROHM driver is contracdicting by itself, because datasheet max current is 1.6 A. Even use for 0.8 A, I need a very big wattage resistor ??

Am I Right ?

 

[RadarRay]

Ok, you've got my interest. I'll read the complete data sheet and see what if I can figure it out. Just one thing, make sure it is actually wired like you think it is wired. Would not be the first time a misconnection prevented a circuit from working.

 

[jOmega]

Yanco,

I think before much more time is spent head scratching trying to understand what you have and what you are doing, you need to tell us something about the motor.

V =
Amps =
RPM =

Oh, and one other piece of vital information, .....

The data sheets for the BA6219 show a ZENER diode connected at pin 4..... However, in the app. notes, it states that if an output voltage High limit is not used, then pin 4 (VR) can be left open or short circuted to Vcc1... So, for our understanding, tell us what you have connected to pin 4... a zener diode of what value, nothing, or it is jumpered to pin 7...

And what is connected to pins 5 & 6... the input control... and how is this being manipulated....?

This will help to fill in the picture of your application.

Thanks.

 

[schwarz]

The o/p current max given in datasheet for BA6219B is for 500microsec pulse at 1% duty cycle.You will probably need high watt RES without chopped drive.

I made a stepper motor ( Perm magnet ) drive long time back.It needed bulky current limiting resistors .The solution was to chop the drive at high frequency and voila very low wattage resistors did the job. Average watts got reduced.

I WOULD BE CURIOUS TO KNOW THE FINAL SOLUTION.

 

[yanco]

OK, let me tell all the information :-

The spec of the motor said that Norminal Voltage of the Motor is 12 V. RPM (freeload) is 5800, current 0.05 A. With Load, Speed is 5000 rpm, current 0.2 A.

The Pin 4 is connected to a Variable Resistor of 10 K. It is use to control the speed of the motor thru' the BA6219. The VR max voltage is 5V, to turn the motor, 0.5 V is sufficient. When I measure the Pin 4 voltage, it is not control by the VR voltage, even VR voltage is 0.5 V, when I measure the Pin 4, it is 3.5-5 V at running condition.

Pin 5 and Pin 6 is used to change the direction of the motor. There isn't any problem over 5 & 6.

Again I clarified that the BA6219 is not very hot with heat sink even continued use for 24 hours/day, it is the resistor that shall be get very hot and melt my solder joint, the high wattage that make me crazy. Why the design of the BA6219 was in such a way ??

 

[yanco]

Another surprise is the spec said that the BA6219B power dissipation max is P=2200 mW and that means 130mA constant current into IC (at 18V).

And with the resistor is 10 ohm must be there, the current is 18 V over 20 ohms (10 ohm resistor plus 10 ohm motor resistance), the current always 0.9 A will blown the IC.

Where is the mistake ?

yanco

 

[schwarz]

"Motor is 12 V. RPM (freeload) is 5800, current 0.05 A. With Load, Speed is 5000 rpm, current 0.2 A."
This translates into 12/0.05 = 240 Ohms at freeload and 12/0.2 = 60 Ohms Interestingly BA6219B lists R(L)=60 Ohms in test conditions.

 

[RadarRay]

Schwarz, motors don't change their resistance. It is as yanco measured previously, 12 ohms. The reason for the difference is when the motor is loaded it droops in speed. The Cemf goes down as the motor slows down and therefore more volts are available to the 12 0hms of the motor.

To further clarify, 12 * .05a no load is .6 volts for IR drop. When motor is at 5800 rpm, that means cemf is about 12-.6 or 11.4. Therefore Kv is probably close to 11.4/5800 or 0.00196 volt/rpm. Thus under load at 5000rpm Cemf is 5000*0.00196 = 9.82. That means the volts available to motor is 12-9.82 or 2.18, which when applied to 12 ohms is 0.18 amps, which is pretty close to the rated 0.2A

 

[RadarRay]

Take that ohmmeter you used to measure the ohms of the motor and put it on the Ampere scale ( or milliamp scale) and measure the current flowing through the resistor.

If the motor saw the current of the wide open circuit 12/(12+10) = 0.54amps, it would probably overheat. Is it?

Pins 5 and 6 select direction, but pins CD1 and CD2 actually turn on the transistors. They have to be tied to Grnd with capacitors to prevent "both output transistors from being turned on at the same time."

So if you are drawing high current in the 10 ohm resistor, it has to be going to either the motor or through the transistors meant for reverse. If you have one, put a scope CD1 and CD2 and find out what they are doing. If no scope at least a volt meter.

Refer to my post to Schwarz to see how the motor should be behaving.

Make sure you have the other caps mentioned in the Data sheet. They are all there for a reason.

This is a very good problem to take a lot of measurements and then learn how the components involved actually work. Let us know of the outcome.

BTW, right now I am betting you are at least partially turning on the reverse transistors.

 

[yanco]

Pin CD1 and CD2 are tied to ground with capacitors of ceramic 0.1 uF. Measured voltage is range between 0-3-8 V when motor oscillation.

The current thru' the resistor I measured is 0.55-0.6 Amp at Max setting at the VR (5 Pin) to pin 4 of the BA6219B.

Yes. The resistor is overheat.

 

[RadarRay]

Is it possible to measure the current through the motor either while measuring the current through the resistor or under the same conditions?
I don't understand your statement ". Measured voltage is range between 0-3-8 V when motor oscillation."
The motor is in oscillation?? The volts are on both pins or one?

Lastly, when you measured the .55 amp, was the motor turning? If so at high speed?

 

[schwarz]

Thanks radarray, I am somewhat a learner in 'motors' area.
"both output transistors from being turned on at the same time." is a common issue in our dc/dc converter and other bridge drive designs. I understand that when changing logic modes the driver should go through 'open' mode briefly. We call this 'dead-time'.
I would also like to ask radarray if chopped drive would reduce dissipation here as in stepper motor.

 

[yanco]

What I means is the motor oscillating in forward and reverse direction, and I measured the current thru a DVM connecting is serial with the resistor.

With my VR set to min (ie, 0.55 V at pin 4), Measured voltage of CD1 and CD2, is around 3 V. CD1 and CD2 are oscillating when motor forward and reverse. I means one of them will have 3 V, one of them zero, and vice versa.

If I set the VR to around 4V, the CD1 and CD2 will have around 8 V. Oscillating also.

 

[RadarRay]

Do I understand this correctly, you are telling the motor to go only forward but it goes both forward and reverse? Or are you using logic to turn it on forward and reverse?