Good day
I want to evaluate the behaviour of a DC driver a copper coil (2.2 Ohms resistance) under the following requirements

Amp	31,3	62,5	93,8	125,0	156,3	187,5
Volts	68,8	137,5	206,3	275,0	343,8	412,5

However, the driver, that adjust the Volts to the Amps requirement generates 138 Volts instead of 68.8 Volts when requesting 31.3 Amps (as per Ohms law the resistance is 138/31.3 = 4.4 Ohms instead of 2.2)
It seem there is a inductive reactance of a coil connected to a 6 pulse dc drive, but its value is too high for DC volts.
Questions:
1. Depends on the quality of the driver generating DC currente?
2. How can be measured?
3. How can be reduced to reduce the voltage requirement?
4. This effect is only during the energizing process but become nule when steady?
Thanks in advance

 

I don't really understand what you are doing here. But, taking the first column of your table, if you have a coil of copper wire with 2.2 ohms DC resistance and you apply 68.8V across it, the steady state current will be 31.3A. But this is 2.153kW so the coil will get very hot very quickly!
I don't understand your second sentence. And what is a 6 pulse DC drive?
Yes, the coil will have an inductive reactance. This means when the voltage is first applied (at t=0), the current will be zero. The current will quickly rise - at a rate that depends on the inductance of the coil - until the current is limited by the DC resistance of the coil.
Normally, in a real application, the source voltage will be disconnected well before the current reaches this limiting value.
However, due the the reactance, the coil current will continue to flow without external power and if there is no path for this current (a free-wheel diode for example) a very high voltage will be generated across the coil terminals because the current will have to flow somewhere (probably causing a spark or damaging the controlling equipment).
So your questions make no sense to me.

 

can you share part numbers and datasheets of motor and driver?