I am confused about how the Kv rating of a bldc motor affects the 3 phase ac it generates because, the Kv rating is based on the dc voltages that are applied to the motor by its esc. So for example, if i had a 1000 Kv motor and spun it at 1000 rpm would the motor generate 1 volt rms or something else and what would the peak voltage be?
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1\$\begingroup\$ "KV" is meaningless. "kV" is kilovolts and "Kv" is probably what you are trying to express. \$\endgroup\$– Andy akaCommented Jul 22, 2018 at 19:50
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1\$\begingroup\$ Don't be so harsh, Andy. Some people do capitalize the 'V'. as long as they don't say "kV," it is all good. \$\endgroup\$– user57037Commented Jul 22, 2018 at 21:09
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For a BLDC, Kv is based on half of the peak-to-peak voltage. In other words, it is based on the amplitude of the sine-wave voltage. The units of Kv are rpm/Volt.
So to measure it, spin the motor at a known RPM, and measure the amplitude of the phase-to-phase voltage. So let's say you measure the voltage and it is 10V peak-to-peak, or 5V in amplitude. Let's say the motor is spinning at 1000 RPM. Then the Kv is 1000 RPM / 5V = 200.
This measurement is done when the current in the windings is zero. If you are using the motor as a generator, then the output voltage will be lower than the Kv predicts due to resistive losses in the copper windings. If you are driving the motor, the phase-to-phase voltage will be higher than the Kv predicts also due to resistive losses in the motor windings.
The wikipedia page on motor constants can be used as a reference, however I have experience measuring motor Kv to compare with manufacturer specifications.
https://en.wikipedia.org/wiki/Motor_constants#Motor_velocity_constant,_back_EMF_constant
In your example, you say you are spinning at 1000rpm and the Kv is 1000rpm/V. So you take 1000 rpm and divide by 1000 rpm/V and the result is 1V.
Note: that is not the RMS voltage. It is the peak voltage of the sine wave, which is also half of the peak-to-peak voltage.
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\$\begingroup\$ Kudos for the correction to term for \$K_v\$ But is it really "Konstant of voltage " as in DE language origin or just K as in the EE std. control system variable letter? even though it is commonly published as KV \$\endgroup\$– D.A.S.Commented Jul 22, 2018 at 22:08
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\$\begingroup\$ If a BLDC motor is ideal and lossless and has \$K_v=1000 RPM/5V=200\$ not 20 , so if the motor spins at 1000 RPM with 5V peak sinusoid 3 phase drive, WILL IT RUN FASTER with Full DC 3 phase drive? or the same RPM? So is it average BEMF, Peak EMF or RMS EMF? If peak, any references? \$\endgroup\$– D.A.S.Commented Jul 22, 2018 at 23:06
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1\$\begingroup\$ @TonyEErocketscientist, I corrected my math error. Thank you! Kv is a property of the motor. It tells you how to estimate the magnitude of the back EMF. That is it. However, I believe motors will generally spin a bit faster using 6-step commutation vs sine wave drive, assuming the DC-link voltage is the same in both cases. However this can be compensated to some extent using field weakening techniques. So a motor can spin faster than Kv predicts by deliberately applying the drive voltage earlier than the EMF. \$\endgroup\$ Commented Jul 23, 2018 at 1:01
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\$\begingroup\$ @TonyEErocketscientist, I don't know what the 'K' really stands for. I think of it as the motor voltage constant, and I know that 'K' is often used as a constant in science notation. \$\endgroup\$ Commented Jul 23, 2018 at 1:04
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\$\begingroup\$ Yes I think so. But the real question is this constant of RPM/V based on sine (pk,rms or average back EMF or DC as a motor) Those who use DC as a motor get tend to get the wrong result from the thesis topics I read. \$\endgroup\$– D.A.S.Commented Jul 23, 2018 at 2:46