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

 

[yanco]

Yes. I purposely make it forward and reverse.

 

[RadarRay]

When the motor is running in just the forward direction, how much current does it draw? Is it a steady current?

When you are switching the directions, how fast do you switch them? In other words, how long are you in forward mode and how long in reverse?

 

[yanco]

1 sec for forward and 1 sec for reverse continuouly. If the speed is fast, it could be 0.8 sec each. I don't think it is a steady current as the speed is so fast. When I measured with DVM, the current I can see is range from 0.2 A to 0.6 A.

 

[RadarRay]

Have you noticed that you are never under the 0.2Amps which is the full load current according to your motor data.
I would run it in one direction and not reverse it, then measure the current. It appears you do not have enough torque for whatever it is you are doing. I am not familiar with motors this small but I would think that somewhere when you are running it the current should be below 0.2 Amp.
Your RMS current in the motor, I would think would lead to decreased motor life.
You could increase the ohms on your limiting resistor but I am not sure you would get enough current to get the motor up to speed.

 

[jbartos]

Suggestion: Visit the link with datasheet for BA6219B. Then, under Motor Driver ICs notice:
(8) Back-rush voltage
Depending on the ambient conditions, environment, or
motor characteristics, the back-rush voltage may fluctuate.
Be sure to confirm that the back-rush voltage will not
adversely affect the operation of the IC.

 

[schwarz]

Hi,
Hope that you have resolved all issues, please share the info with us.

REgards.

 

[yanco]

I have not resolve the issues, it is still hot using 10 ohms. If I use 50 ohms, the current is not enough to drive the motors. BTW, 50 ohms also get too hot. I wonder may be this IC is not suitable for continuous drive. Anyone experience any driver IC for continuous drive for 10 hours a day ? Forward and Reverse in every sec.

 

[RadarRay]

If you had a plus and minus supply I would suggest you use a power op amp. However for bi-directional operation, you must have a plus and minus supply. This is one of the unique features of the chip you have chosen.

One thing you should note is, with the 18 volt supply you have chosen, any linear device you use will have to drop the difference between the 18 volts and your load voltage. It will do this at the current you are drawing. This creates watts, which then creates heat. The power op amp would do this internally but perhaps being a "can package" it might be easier for you to dissipate the heat.

Your chip is a linear device. You either drop volts across a limiting resistor or across the chip. You already know the effect and problems with each approach.

I am not sure you understand that in the data above, you are suppling 3 times the rated current to the motor when you change directions. Is the motor capable of handling this indefinitely? Perhaps your motor is undersized. This doesn't change the chip problem, as a larger motor will draw roughly the same amps. You never did state the motor manufacturer. Who is it?

 

[jbartos]

Suggestion to yanco (Electrical) Jul 23, 2003 marked ///\\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
///How can there be 18V current over 20 ohms. Should not that read 18 V voltage across 20 ohms?\\ (10 ohm resistor plus 10 ohm motor resistance), the current always 0.9 A will blown the IC.
///Would you post the motor nameplate data or manufacturer data sheet data. Some motors have very low resistance when measured at disconnected.\\Where is the mistake ?
///They will be found as soon as you post requested information.\\\