Formula for DC Output Current of a Drive

It should be in the manual that came with the drive and will depend on the size of the DC motor (and its voltage) that you are using.

Thanks for your help.

Rick

Although not exact, the 3 phase input current will be approx equal to 85% of output single phase current. This factor comes from taking the √3 / 2 It is an approximation, that is most accurate when drive is operating at full output volts.

In your example the output current would be approx 149Amps / 0.85 = 175Amps.

The actual relationship is calculated using watts. Watts(in) = Watts(out) + Watts(loss) I do not remember the exact formula for calculating 3 phase power, but it is readily available on wiki or google

By the way if you are measuring total input current to the drive, be sure to account for the portion that may be branched to the motor field power supply.

3 phase power = 3 * VL * IL * cos (PF)

___________________

Thank you and good night

That is INPUT(AC side) WATTS .

How/where is DC coming IN/OUT from.

Please draw a schematic and ask ?

Thank you for your help, this was very useful.

Rick

I agree.

149 A means the current on each phase. 149Amps X 3 Phases / 1.732 = 258 Amps.

Also, be aware that the current in and current out will only be close together if the voltages are the same. In a variable speed drive, the voltage on the DC side will vary over a wide range, meaning that at low motor speeds the current might be very high and the voltage quite low. The power in and the power out should be fairly close, with the difference dissipated as heat in the drive.

Other than change in harmonic content I don't see any significant change between DC x .85 = AC amps over the output operating voltage. This is why DC drives use CT's for measuring the DC load.

With AC drives with a constant potential bus in between I would agree.

Remember the power factor of a DC drive running at half (maximum) output volts is approx 0.5.

Suppose a DC motor is driving a conveyor at low speed. The DC motor may be rated 230 VDC. It may be driven by a drive rated 460 VAC, that is intended to drive either 230 VDC 500VDC motors. If the motor demand is 20 kW, and it is running at 100 volts to maintain a relatively slow speed, then the current will be 200 amps. The input current at the drive would not then be 170 amps at 460V, which suggests a power consumption of 78.2 kW.

power in (minus losses) = power out

current in does not necessarily = current out

Power factor, motor rated, ac input etc is being confused.

A 230vdc 20 kw motor that is 90% efficient will have a full load amps of about 20,000 / 230 / 0.9 = 97 amps.

The maximum average voltage output of an scr bridge is only dependent on the AC supply volts, never mind the np rating. For a 230vac input it is Root2 x 230 (6 / 2pi) = 310 vdc.

Now your example with the motor running at 100 volts and at full load the PF is 97 / 310 = .3, and the AC amps aprox. 97 x .85 = 82 amps

The AC KVA = 230 x 82 x root3 = 32 kva

The AC power is 32kva x .31 = 10kw. and this agrees with the DC side of 100 volts x 97 amps = 10kw.

It is all really very simple math.

Correction "PF is 97 / 310 = .3"

Should read "PF is 100 / 310 = .3"

I agree that the math is simple -- it's the concept that I am struggling with.

I think I have watched the ammeters and voltmeters on the supplies and outputs of PWM DC drives that make it look as if the power factor is very near 1 even when the motor is running at half speed or lower. In other words, the KVA input is very close to equal to the kW output to the DC motor being driven. And I think I have seen this on a drive that is supplied by 480VAC feeding a motor at low armature voltage and high current. It was as if the system was entirely resistive, with, let's say, 480 vac at 10 amps going in, and 100 vdc @ 48 amps coming out.

My memory could certainly be faulty -- it's been quite a while. This was almost certainly before the days of switch mode power supplies, but PF correction was widely practiced. Would the ammeter have been reading real power rather than apparent?

Ah ha!

PWM DC drives or AC VFD drives use an intermediate DC bus. The front end is essentially a diode bridge, sometimes with a choke, and a capacitor bank. So the input to this stage is essentially unity power factor. The output then (for an AC VFD) will have a lagging power factor for the magnetizing current, but that is reactive and does not consume (significant) power and is "recirculated" within the DC link.

The PWM DC drive versus an SCR phase controlled bridge are again two different animals. The previous discussions were on SCR phase controlled. The PWM DC will have close to unity PF on the input and the output. DC motors are rather "well behaved" in that the DC volts x AMPs is essentially 100% of the power delivered, especially if you include losses. It is all real power.

Another interesting configuration is the use of an "active" front end to the AC drive that can also provide regenerative capabilities (I am not referring to the SCR front ends). This uses a bridge similar to the output stage of an AC drive, so it can actually be programmed to provide leading power factor correction for a plant with its unused va capacity.

The bottom line for many of the "confusing" discussions on CR4 is lack of specific details or poorly stated questions, so many assumptions are made from the experience of those participating.

The bottom line for many of the "confusing" discussions on CR4 is lack of specific details or poorly stated questions, so many assumptions are made from the experience of those participating.

How true. At times, it can be comical. There have been a few discussions where the debate goes back and forth on esoteric points of engineering or science, and the original poster is absent, so there's no way to know what was originally meant. Then the OP shows up again, and, using this thread as an analogy, says "Oh... by 'DC drive' I meant a car ride through the US capital."

And then there are the questions that appear to be someones homework.

There was one on Bode plots. It was somewhat challenging, and I gave it a short answer with the technique for handling the equations and what the final shape would be, without the numerical answers. If the students had 2 cents to rub together they had a short tutorial.

I think the response was too late for the assignment. They never responded.

I dont think it has a definitely answer.

it will depend on the way of load linking, efficiency, way of measure , way of rectifier, half or whole? and filter at output etc. factors.

once you can definite these factors, evaluation can only be done.