Voltage amp to amp conversion

[vulcom1]

I thought I would ask on here and it probably is a dumb question for most but for me it does not sound right the answer.
I have a stepper that is rated 40 vdc at 5.1 voltage amps(VA) and would like to know the conversion to Amps for the motor for calculating the setup.
I got 40vdc divided by 5.1 VA = 7.84 amps. This seems high so if somebody can set me straight it would be appreciatted.
John

[Mariss Freimanis]

You are mixing apples and oranges (volts and amps actually). Your motor has a rated 'Amps per phase' on its nameplate. That is what you use to set the current output on any drive you plan to use. Power supply voltage only determines how fast you want the motor to go. The higher the voltage the faster and hotter the motor. My guess is you have a PacSci motor; they are one of the very few motor manufacturers that list a maximum drive supply voltage for their motors.

Mariss

[vulcom1]

Mariss,
The motor is a Astrosyn Miniangle Type 23LY. It is marked 40 VDC, 5.1VA
and 0.9 deg/step. I know I won't be anywhwere near the 40 volts as my power supply is 24 DC. But I would like to know what to set the Vref before I let the smoke out of the board.
Thanks, John

[Mariss Freimanis]

Well, the problem is '5.1VA' makes absolutely no sense when used with a step motor.

Here's how you can figure out what current to set a 'mystery motor' to. A single-stack round NEMA-23 motor is 5W. Maybe that's where the 5.1VA comes from. Measure the motor's winding resistance, use the resistance reading in the following:

I = SQRT (W / R)

Example: Say you measure 2.5 Ohms resistance. Plug the numbers in:

I = SQRT (5W / 2.5 Ohms) = SQRT ( 2 )= 1.41A

You would set the drive to 1.4 Amps and be pretty assured nothing would burn down or blow up.

Mariss

[vulcom1]

Mariss,
Thanks for the info and equation.
John

[segar]

hello sir

[Smartbomb]

Originally Posted by Mariss Freimanis Well, the problem is '5.1VA' makes absolutely no sense when used with a step motor. Here's how you can figure out what current to set a 'mystery motor' to. A single-stack round NEMA-23 motor is 5W. Maybe that's where the 5.1VA comes from. Mariss

"Volt Amps" is really a synonym for Watts, i.e. power (watts) is the product of voltage and amperage. Just like torque in Ft-Lbs is the product of distance and force.

Put another way, Watts = Volts x Amps. So 1 Watt is 1 Volt-Amp.

So for example, if you put 5 volts across a motor and it's drawing 1 amp of current, it's consuming 5 watts. Put 40 volts across it, it should only draw 1/8 of an amp (125mA) for 5 watts.

Hope that helps.

Dan

[Mariss Freimanis]

That's not the way it works.

1) Step motors are current driven at low speeds. It is helpful to know what that current is. 5.1VA yields any result for amps (5.1 million amps at 1uV to 1uA at 5.1MV) so it's not particularly useful in solving that question.

2) Step motor coils model as an inductance in series with a resistor. Typical values for a 1A motor are 8mH and 5 Ohms. Putting 40VDC across such a load would result in an initial current slope of 5,000 Amps/second, a 1.6ms L/R time constant and 8A of steady-state current after 4.8ms (3 time constants). The results wouldn't be pretty for the motor or its owner (320W dissipated in a 5W coil).

Mariss

[Smartbomb]

Originally Posted by Mariss Freimanis That's not the way it works. 1) Step motors are current driven at low speeds. It is helpful to know what that current is. 5.1VA yields any result for amps (5.1 million amps at 1uV to 1uA at 5.1MV) so it's not particularly useful in solving that question. 2) Step motor coils model as an inductance in series with a resistor. Typical values for a 1A motor are 8mH and 5 Ohms. Putting 40VDC across such a load would result in an initial current slope of 5,000 Amps/second, a 1.6ms L/R time constant and 8A of steady-state current after 4.8ms (3 time constants). The results wouldn't be pretty for the motor or its owner (320W dissipated in a 5W coil). Mariss

Sorry my response wasn't clear... I wasn't so much addressing the amount of current the OP would channel through the motor (your previous answer was in my opinion satifactory regarding that), I was simply addressing the VA nomenclature.

I wasn't advising he put 5V, 10V, 40V or 100V across the motor, I was only indicating that the product of the current and the voltage across (any) load is the power. With a motor coil, the series resistance is what dissipates the power (and limits the current), and the inductance governs the rate of change of current.

Maybe I'm sensitive to the issue of unit labeling because one of my EE professors from way back would often use units of VA on a test after we'd all done many problems thinking watts. Sometimes it helps to re-inforce the fundamentals.

Just didn't want to get into reactive components & time constants, expanding/collapsing magnetic fields, transients, back EMF, etc... however I think your explanations were great.

[Mariss Freimanis]

Smartbob,

I agree with you. Unambiguous communication is a fundamental requirement for EEs and other technical professionals. My understanding is VA and Watts are not synonymous except for resistive loads. VA is normally used for reactive loads where there is a non-zero phase relationship between current and voltage.

For your interest, I have attached a graphic showing data for a 5W / coil motor over a 0V to 100V supply range. Losses should be independent of supply voltage if resistance is the only dissipative mechanism.

The graph shows it takes 14W to turn the motor at 3,000RPM (8VDC supply and 1.75A supply current). Motor dissipation rises to 55W at 100VDC and the curve that connects those two points is parabolic. The difference, 41W, is dissipated as eddy current losses in the motor laminations and accounts for 75% of losses in the motor at 100VDC.

Mariss

[gar]

070711-0843 EST USA

Smartbomb:

Integrating the instantaneous product of voltage times current over some time and dividing by the time period gives you the average power over that period.

Some voltage reading times current from arbitrary meters does not necessarily equate to power even in a steady state condition and a resistive load. Even with sine waveforms for voltage and current in a circuit that is not purely resistive V * I is not power, but something greater than power. Vrms * Irms is only equal to power in a purely resistive cicruit.

Consider a simpson 260 meter which measures the full wave rectified average voltage but is calibrated in RMS of a sine wave. A 10% duty cycle pulse measured with this meter will not read the RMS value. The RMS value of this pulse is not 10 % of Vpeak which is the average value, but is the sq-root of ( Vpeak squared * 0.1 ) or 0.316 * Vpeak. Or the Simpson reads about 1/3 of the actual RMS value of the pulse.

On another point: The unit of measure for torque is #-ft and not ft-#. ft-# is a measure of work. See my discussion and references at
http://www.cnczone.com/forums/showthread.php?t=38724[url]
you will need to wade thru the various pages to view the discussion on #-ft. My comments on this start at post #26 on page 3.

.

[Mariss Freimanis]

1) Torque is a force applied perpendicularly to the end of a moment arm. Its unit of measure is the product of that force and the moment arm length (T=F*L). The commutative property of multiplication makes no distinction between F*L or L*F. Convention has torque units of measure as Nm or oz-in even though the clumsy-looking mN or in-oz is just as valid.

2) Work is a force applied over a distance. Its unit of measure is also the product of force and distance or a distinct unit of measure like Joule. By definition, 1 Joule = 1 Nm.

The point is both torque and work are measured in Nm. The meaning of Nm (ft-# for that matter) depends entirely on its context.

Mariss