Skip to main content

Kollmorgen Support Network

To better serve our Kollmorgen users with questions too complex to be addressed in this space, we made the painful decision to close this Community forum.

Please submit your question using the contact us form to reach our application engineering team.

For the immediate future, you can still access and read past posts.

Thank you for understanding and for participating in the community over the years.

AKD2G and power data confusion | 25 Mar 2020 | |

AKD2G and power data confusion


I look the data about AKD2G, and I am a little bit confused with Power data.

Let's take the AKD2G-6V06S in 240 Vac.

We could read in the commercial datasheet a continuous current of 6 Amp and a Typical Shaft Power of 2 kW


Well, I am rather happy if I look the DC bus power (for 240 Vac)   :   6*240*Sqrt(2) = 2036 Watt

But when I look on the manual, we could read : (chapter an output power for 3*240 Vac at 1.25 kW


Which value is OK ????

It seems that 1.25 kW is the power in case where the AKD2G is power supply directly in 340 VDC (with 4 Amp in this case)

But when the drive is power supply in 3*240 Vac, we have for the power input : P = Sqrt(3) U I = Sqrt(3) * 240 * 5.3 = 2203 VA



Could you please clear this point ?

Thanks a lot.

Best regards


Comments & Answers

toddevans01 said ...

toddevans01 | Fri, 04/03/2020 - 12:19

Your questions are under review by engineering.
We will post an answer upon receipt of their response.

hurley.gill said ...

hurley.gill | Wed, 04/22/2020 - 16:00

Clarification on continuous AKD2G output capability: AKD2G-6V06S in 240 Vac.


Wherein the newest AKD2G installation manual presents: AKD2G-6V06S in 240 Vac, capability per the following:

Input power of 2.2kVA can be confirmed by Input calculation:

Input power = 3 x 240/√3 x 5.3 x cosϴ = 2203, assuming ϴ = 0°.
Based on the results : 2203, we are actually calculating Watts (W) with cosϴ = 1.
However, this calculation can also be viewed as the maximum possible continuous Apparent-Power (VA) capability regardless of the ϴ angle between: voltage & current, drawn by the drive; thus our presenting it in terms : kVA, versus watts(W).

Remember Watts loss will produce heat, and VA produces the same heat in terms of losses, if the product : VA, is the same value.
This is also confirmed by the fact that a well-designed full-wave rectified 3-phase circuit will have a Power Factor approaching 100% (i.e. 100 x watts/VA), especially between: 0 & Ic(drive).

Side note:
Power Factor during initial bus power application, can be considerably less than 100%; while the soft-start circuit is implemented to protect DC-bus capacitors from an excessive surge current.

The Drive's published output data is in the terms : 'Continuous motor shaft output power at rated current'.

Since this continuous capability is presented at the motor's shaft output [not the drive' output], it is in the equivalent term of mechanical power; and, this physical power output must be estimated for the same Watts = 746 x [Ta(Nm) x Na(RPM)]/7121, based on worst-case assumptions and Ic(drive) published capability.
Engineering for the published condition: Ic(drive), has determined this worst-case condition to be cos(60°), wherein the 60° would be the worst-case delta between: voltage & current, effected by the interaction of the drive (e.g. d-q currents, controlled field weakening), motor (e.g. specific motor's design/characteristics), and the specific load under evaluation.

Output power (W) = 3 x 240/√3 x 6 x cos(60°) = 1247 Watts, at motor shaft.

If one wanted to know the best possible continuous Apparent-Power output (VA) of the dive, when not published; it can be estimated by the following:
Output power {maximum possible continuous Apparent-Power (VA) capability} = Input power(W) - Drive internal losses(W) = 2200 - (80-12) = 2132 VA.


toddevans01 | Fri, 04/03/2020 - 12:18

Your questions are under review by engineering.
We will post an answer upon receipt of their response.

About this Question