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whitbuzben (Electrical) (OP)
19 Apr 12 10:08
Dear Sirs

I am working with a Control Techniques Permanent Magnet AC Servo Motor. I am looking to feed the motor through a full bridge rectifier (Six Diodes) and would like to workout what the rectified DC voltage will be. On the data sheet the the Ke(voltage constant) is given as Ke = 98Volts RMS/Krpm.The rated speed of the motor is 3000RPM. I have no idea if this is volts line to line or line to neutral. I am a mechnical guy and am quite new to this any advice would be useful.

Many Thanks

iop995 (Electrical)
19 Apr 12 11:25
Seem to be line to line voltage. There are not data about nominal voltage? Nominal current and winding resistance may help to calculate DC voltage needed. Nominal EMF is 294Vrms (417Vpeak), so DC voltage must be above this value, maybe 430-440V.
sibeen (Electrical)
19 Apr 12 18:17
General rule of thumb is that the unfiltered (no capacitor)DC voltage out of a 6 pulse rectifier is 1.35 times the line to line voltage, so in this case about 397 volts.
ScottyUK (Electrical)
29 Apr 12 5:54
You are going to feed an AC servomotor with DC? Hmmm...

This link might be of interest.
  

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If we learn from our mistakes I'm getting a great education!
 

whitbuzben (Electrical) (OP)
30 Apr 12 8:06
Hi ScottyUk.

I am not feeding the servo  with DC. We have a motor/generator rig setup. Wereaby one of the motors is used as a generator that is driven by the other motor. It is the generator side that is being fed through a rectifier.  I could give a full description of why we are doing this but I didn't think i was relevant. My question was simply about calculating the rectified DC value.

Thanks   
ScottyUK (Electrical)
30 Apr 12 15:07
No problem, just sounded a bit odd. smile

If you want to get in to the maths, this paper from CERN gives some good notes on rectifiers from single diode half-wave types right up to multiphase designs, then on to thyristor-controlled types.
  

----------------------------------
  
If we learn from our mistakes I'm getting a great education!
 

cswilson (Electrical)
1 May 12 19:13
I believe the OP's question is whether the listed Ke of 98Vrms/krpm is 98Vrms(phase-to-phase) or 98Vrms (phase-to-neutral). I went to the on-line specification document for these motors at:

http://www.emersonindustrial.com/en-en/documentcenter/ControlTechniques/Brochures/unimotor_technical_data.pdf

and found on page 10 a class of motors with a listed Ke of 98Vrms/krpm and Kt of 1.6 N-m/A.

The Ke and Kt for a motor are really the same thing (or conservation of energy would not apply). In consistent units, they will have the same numerical value. The consistent SI "analytic" units are N-m/A(rms) for Kt and V(rms) / [rad/sec] for Ke. The document gives the Kt in these units, but we will have to convert Ke.

98 (Vrms / [krev/min]) * 1/1000 (krev/rev) * 60 (sec/min) * 1/2Pi (rev/rad) = 0.93 (Vrms / [rad/sec])

This does not match the value of the 1.6 N-m/Arms spec. But if you multiply it by 1.732 (sqrt(3)), which is the factor for converting phase/neutral voltage to phase/phase voltage in a 3-phase Y-circuit, it does match. (In a Y-circuit, line current equals phase current.)

So it looks to me that the Ke spec is for phase/neutral voltage per unit speed. Anybody agree/disagree?
whitbuzben (Electrical) (OP)
2 May 12 5:19
Hi Guys

Firstly thank you for to ScottyUK for the link to the paper.

cswilson i worked through the same calculation that you have done and came to the same conclusion. So i think you are correct. Although possibly someone could verify it for us.

98V(rms)/Krpm
98/1000 = 0.098V(rms)/rpm

1 rpm = 0.104719755rad/sec
1/0.104719755 = 9.54939669 rad/rpm
therefore 9.54939669 * 0.098=0.936 Volts(rms)/rad/sec
0.936 * sqrt3 = 1.62
  

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