DC motor closed loop control through labview using usb 6008

Hello
i am doing a project in which i want to control the speed of a DC motor (PID control) through labVIEW 2010. I am having a DC motor, a drive unit that regulates the voltage to the motor by getting analog voltage from 0 to 3.3V ( it can be used for both directions) and i also have a NI usb 6008 board. The problem is that i can't feed the usb 6008 with the digital signal from my hall effect speed sensor. Is it posible with this setup to control the motor?? I can also try to use the pulse as analog input and trasform the frequency into rpm's. Then i am thinking of generating an analog signal to feed the motor drive. If you have any further suggestion of a probably better hardware setup would be more than helpfull to me.
Sincerely
Jason Chaloulos

Hello Michael
Thanks for the reply. I came across those topics before and all of them are trying to generate a PWM signal as output i want to use just an analog signal output so timing on the output is not that important i guess. i am struggling on getting the frequency from the digital input signal that my hall effect sensor generates. Since the maximum speed of the motor is 3000 rpm and with my tooth wheel the maximum output frequency of the sensor will be 300Hz which i see its way less than the limitation of my ni board. Is there any tutorial documentation that might help me with this one ? Thank you in advance.
Kind regards
Jason

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Now for example I got five points: (0,0) (10,20) (30 40) ( 31 60) (50 65). The request time interval I set as 10ms. This mean that in the first 10ms, the machine should move from (0 0) to (10 20). But for some reasons the machine can not move to (10, 20), it moves to (8,16) and the first 10ms passed. In the second 10ms, how it move? Move from (8 16) to (30 40), If like this, the following error will accumulated. In practise, it is not accumulated. Is there anyone can explain something about the closed loop control of stepp motor? Thanks a lot!

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My question is like this. For example I got five points: (0,0) (10,20) (30 40) ( 31 60) (50 65). The request time interval I set as 10ms. This mean that in the first 10ms, the machine should move from (0 0) to (10 20). If it is impossible for the machine to move from (0 0) to (10 20), there are two ways to solve this problem. First is the time interval is 10ms and assume it move to (8 16) within 10ms and take (8 16) as the start point and (30 40) as the end point of the second move. Second is extend the time interval, assume it takes 14ms to move from (0 0) to (10 20). After 14ms, the (10 20) is the new start point for the second move. I think it works as the second ways.If not, the following error will accumulated. Is there anyone can explain something about the closed loop control of stepp motor in detail? Thanks a lot!

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Hi
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Hi Premenanda
The freq data receiverd from the daq assistant has the units of Hz, or in terms of motor speed, rps (rotations per second). I can see that you are trying to convert that to rpm (rotation per minute). which can be done by dividing the rps speed by 60. Instead, you are multiplying it by 60 and also comparing with 2000. the resultant multiplied value is displayed in your program only if it is smaller than 2000.which is causing the problem here.( as seen in file 'original code.png)
I have attached a file named 'modified code.png' that shows how the conversion from rps to rpm is done. Please do the required changes in your code and let me know if it works.
Attachments:
modified code.png ‏18 KB
Original code.png ‏15 KB

Closed loop control of step motors possible with 7344?

Hi All
I have been looking around here for a while for an answer to this question.
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If the load on a system changes, there will often be a need for compensation in order to keep the position right. Then you need a closed loop control system.
I have made a system using DC motors with a 7342 controller. We are in the process og getting a new similar system, where the manufacturer chosed stepmotors in the believe that it can be used in closed loop.
Is it really true that the NI version of closed loop is only performing a what you call a "pull-in" at the end of the move in order to correct for lost steps or flexibility of the connection between motor and object/encoder?
I would call this "backlash compensation" or something like that.
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/b

You are right. NI motion control boards don't apply a control algorithm
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If you really want to do PID control with stepper motors you still
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The most important issue is the fact that you could easily exceed
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Best regards,
Jochen Klier
National Instruments Germany

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Solved!
Go to Solution.

Hi,
the attached example I've made should demonstrate what you need to do.
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Regards
Florian
Attachments:
Example01.vi ‏24 KB

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Message Edited by Conseils on 05-08-2007 08:51 PM
Message Edited by Conseils on 05-08-2007 08:52 PM
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46185785AD594_fbs.gif ‏17 KB

Using USB-6008 software timimg Generate waveform

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DephinTW,
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http://img177.imageshack.us/my.php?image=howtocreatethesetpointnx2.jpg
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Hi Cyrus
Have you had the opportunity to see our
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Thank You
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Solved!
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Attachments:
WINUTIL.LLB ‏609 KB
DisableCloseButton.vi ‏44 KB

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at

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Attachments:
pneumatic diagram.jpg ‏33 KB

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