I want to increase the range of the current that my program can measure.

[LABVIEW TECHNICIAN]

I have NI-DAQ 6024E and CB-68LP connector.I'm using pins 68 & 67 for input current that is to be measured.It somehow reads 20 mA when it's 200 and 30 mA when it's 300!! Also please tell me how can I increase the current range that can be read by this program and hardware as it currently reads upto 500 mA and I need to read upto 1 A.

[markwysong]

All it's really doing is measuring a voltage across some known resistor value, and calculating a current based on that.

i.e. If you put a 1 ohm 1 watt resistor in series with your circuit, then you'd configure your DAQ analog input channel such that 1 volt = 1 amp. Therefore, the voltage it sees is actually the current.

You don't have to use a 1 ohm resistor, you can use anything you want. Keep in mind, though, that any resistor you use to monitor current has a voltage drop, so other parts of your circuit may react to a lower voltage. Therefore, CVRs (current viewing resistors) are usually low resistance values (1 ohm or less, if possible).

Keep your DAQ analog inputs configured as differential for such a measurement, because typically when
measuring current, you want both wires floating. Grounding one side may cause problems.

That means that you're positive lead should go to ACH0, and your negative lead should go to ACH8, not AIGND.

Mark

[LABVIEW TECHNICIAN]

Thanks Mark for your nice explanation.
I have my input differential only so that should not be a problem.Now after thinking on your suggestion,I think it would work.But I have got bit confused.I have resistors of 6 ohm,9, 15, 22, 30 & 100 ohms through which I'll be passing 6 V to 24 V(resistor is actually a valve with a coil of above resistances).So now when I try reading the current on ACH0 & 8, I get exactly the same voltage value that I use in the circuit.Here could you please elaborate your explanation how I should connect external resistor(if at all required) and what value?

[markwysong]

Well, that depends. You can use the valve. First, measure its exact resistance. Then, connect it's terminals to ACH0 and ACH8 as well as in the circuit. Take a reading on analog input channel 0. Since V=IR, the voltage you read is equal to the current through the valve times the resistance of the valve. Since the voltage and resistance are known, then I=V/R, so take the voltage reading you get and divide it by the resistance of your valve. THIS is the current through the valve.

Mark

[LABVIEW TECHNICIAN]

Hi Mark, basically if I connect the terminals of my valve to ACH0 and ACH8,then it'll be a parallel connection.Here surely I'll get the voltage to read and i can divide by my resistance to calculate the current, but I dont want to do that.That is possible even with a regular calculator.Here I have a program that reads current directly but somehow I'm not able to read proper current.And my valve is a current controlled device so the more current I pass through it, more does it open.Now with the attached program,I want to know how to exactly wire my valve with the CB-68LP connector so that I can read the current.Please also suggest me how exactly to put the resistance in the circuit.

[markwysong]

If you take a look at the Constant Current Measurement VI that you attached, it is performing the exact operation that I just described.

It first reads a voltage from the channel, then divides it by the resistance value to get the current. That's how you take a current measurment!

So, using your Constant Current Measurement VI, you'll attach the (+) and (-) in parallel with your valve (where your valve IS the resistor in series with the rest of the circuit) on channels ACH0 and ACH8, and enter the resistance of the valve into the control on your VI. Then your measurements will be correct (in mA).

Mark

[LABVIEW TECHNICIAN]

Hey Mark,
That was a real quick reply.Thanks a lot.