Timeline for Which voltage uses the "Kv" parameter in a BLDC (Brushless DC) motor?
Current License: CC BY-SA 3.0
15 events
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| Jul 23, 2011 at 6:23 | comment | added | Diego | True, but there are too different things: \$K_v\$ which is a commercial constant for [rpm/V] and, on the other hand, a more technical constant given by books like Electric Machinery Fundamentals by Champan, which is refered to the voltage phase in the following equation: \$E_A=K\phi\\omega\\$ | |
| Jul 22, 2011 at 17:33 | comment | added | user924 | Its Vpeak rectified. The book is wiki about Lorentz force and Maxwell equations. So its fundamental and deep, unless it is too boring for you | |
| Jul 22, 2011 at 17:15 | comment | added | Diego | haha all right... What about the question? Is it Vpeak, Vpeak-peak or Vrms? Any book or something like that as a source? | |
| Jul 22, 2011 at 16:48 | comment | added | user924 | I know that high quality (>$100..$200) Mitsubishi BLDC motors can get nearly completely demagnetized if say you get to much high frequency AC and stall it. So yes. | |
| Jul 22, 2011 at 16:41 | comment | added | Diego | That much?? It was new when I made the test. Is there any chart of life of the motor vs the Kv? | |
| Jul 22, 2011 at 16:38 | comment | added | user924 | The most well known common reason, why kV is way below spec is that motor got demagnetized, or had flux path degraded. Say bad glue, epoxy. Aged, thermally shocked magnets, etc. | |
| Jul 22, 2011 at 16:36 | comment | added | Diego | lol ok So for you is exactly 1858.84 rpm/V. You're aprox is enough but I prefer using the measures of the machine. | |
| Jul 22, 2011 at 16:35 | comment | added | user924 | No. no. I also made another mistake, by not rectifying AC. The peak voltage on your scope is 1V exactly. so its 2200 kV | |
| Jul 22, 2011 at 16:34 | comment | added | Diego | So, for you \$K_v=929.42\$, a value quite far than the one the company ensures and with the same deffinition as Wikipedia has. I need more people and sources to ensure that. | |
| Jul 22, 2011 at 16:19 | comment | added | user924 | Agree actually. Right the poles are in pairs. So it will be divide by 7 pairs of poles. Then for 2 Vpp you have 2200 rpm. Makes it 2200 kV when you Abs(voltage) | |
| Jul 22, 2011 at 16:15 | comment | added | Diego | You forgot that \$\theta_e=\frac{P}{2}\theta_m\$. So the equation is \$n_m=\frac{120*f_e}{P}=255.9*120/14=2193.4286\$ rpm. | |
| Jul 22, 2011 at 15:53 | comment | added | user924 | I multiply Hertz by 60 sec/min and divide by 14 poles. It makes rpm =1096. | |
| Jul 22, 2011 at 15:48 | comment | added | Diego | Actually, the Kv value is 2600. And why do you say that the velocity is 1096 rpm? Are you using \$\f_e = frac{n_m}{120}\$? I'm not sure yet about the topology, I guess I'll make sure of it measuring the resistance between lines and with lines two shortcuted. I'm guessing the "\$K_v\$" value has to do with the voltage between lines, no matter the topology of the motor. But that doesn't solves my question about which voltage should be (peak or RMS). | |
| Jul 22, 2011 at 15:08 | history | edited | user924 | CC BY-SA 3.0 |
added 12 characters in body
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| Jul 22, 2011 at 15:03 | history | answered | user924 | CC BY-SA 3.0 |