03-30-2011 10:39 PM
Is there a way to measure the current from a motor with the NI USB-6009? I understand that I would have to use I = V/R. But it seems that with this concept, I would have losses to the motor, if I used a small resistor and measured the voltage accross it. I am trying to get a reading of the spike in current with a voltage step input to a motor. Thanks for the help.
-Nick
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03-31-2011 01:58 PM
Hi Nick,
The only way would be to use a small shunt resistor. If you use something in the 1ohm of 0.5ohm range, it shouldn't effect the motor too much.
Eric L
04-01-2011 02:28 AM - edited 04-01-2011 02:34 AM
You can use current sensors like from LEM or Allegro . They will also provide a galvanic isolation, but will need some Vcc to run. Insertion losses are nearly zero since they measure the magnetic field due to the current.
They are not as cheap as a shunt resistor (well, depending on the performance you need 😉 ) but the output can be measured by the 6009 directly without additional signal conditioning.
It would have been helpfull to tell us the current range, AC or DC? and the needed uncertaincy.
04-01-2011 09:39 AM - edited 04-01-2011 09:46 AM
Hi Nick,
You may also want to measure the voltage at the motor when doing the spike current measurement. Depending on the motor and connections, you may have power supply/line losses that will cause the voltage at the motor to fluctuate/drop when it is switch on.
A shunt resistor is not the only way, but its the only simple and cheap way of making a current measurement with the 6009. The use of the external current sensors may be best for your application, but that can't be said without knowing you application. Depending on the configuration, the losses in a shunt resistor may be negligible. As Henrik has stated before, more details on your application will allow the people on the forum to give you suggestions suited to you application.
Eric
04-01-2011 10:32 AM
Thank you for the replies. The application is for a 6V dc motor. I know there is saturtation occuring due to the power supply, but that is one thing I would like to confirm. My theoretical model says that the startup current spike is about 3A. I have done stall torque tests with the motor and power supply to check what the current spikes to, and it is under the theoretical model at around 1.36A. But I am wanting to check current spike vs. load so I can verify my theoretical model to help me validate it.
Thanks,
Nicholas
04-01-2011 11:44 AM
Nick,
I looked at some of the data sheets and did a couple of quick calculations. If you set the voltage range on the 6009 to -1to1v you can get a theoretical resolution of (1.22*10^4 / Shunt Resistor) amps. If you use a 0.1ohm shunt, that would be about 0.0012A resolution. If you can accurately measure the resistance of the lead wire, you can use that like the shunt.
You can double check the calculations to be sure.
Also, you can add some capacitors in line with your circuit to help with the power supply saturation.
Eric
04-01-2011 12:14 PM
And I would place the shunt resistor in series with the current path to the motor?
Thanks
04-01-2011 12:48 PM
The shunt would just go inline with the with the motor then measure the voltage drop across the resistor.
Eric
01-28-2012 11:53 AM
Hello, I have a similar application, I need to measure a sinusoidal current of 11 µAPP from this gauge,
http://www.lna.br/~det/Projetos/TCSPD/doc/HeidenheinEXE602.pdf
I need to transform the sinusoidal current into a sinusoidal tension close to 5V, since I cannot define an analogic current input in AI0.
Is it correct if I use a shunt resistor like in the attached JPG? is calculation correct?
Actually there are two signals with 90 phase shift, they will be sent to CHA and CHB on a XY graph to display a circle.
In that case should I use 2 shunt resistors, both connected to GND and to AI0 and AI1 respectively?
Thank you
Luke
01-30-2012 07:47 PM
Hi,
I looked at your calculations and that should provide approximately 5 V so that would be correct. However, be careful that the input impedance on your DAQ is greater than the resistance of your shunt resistors or else you will lose some of your current into your DAQ reducing your voltage across your shunt resistors, distorting your reading.
Regards,
Marcus