That refers to the input.
VFD Long Lead Lengths - Correctly Calculating Voltage Drop
- Thread starter wonderdave
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I have no idea.
clueless
gotcha
if it's not R it must be X
and at fund freq X is negligible
so the logical conclusion is harmonics whose freq increase X exponentially
In my example (without knowing drive specifics)
X with only fund 0.11 Ohm or same magnitude as R 0.14
total drop 7%
my guess 12 pulse drive
X increase to 0.45 due to harmonics
R 0.14
drop 18% or 86 vac
If that what floats your boat..........
12-pulse refers to the input...........and you call me clueless???
Any possibility OP VFD malfunctioning so that its output carrier frequency above design value to increase cable impedance and voltage drop?
Increasing the carrier frequency would decrease the current harmonics.
But fundamental current component will remain same (if load is same) and that would increase voltage drop due to increased cable impedance.
That makes no sense.
Do you think fundamental current component will not remain same with change in carrier frequency even when load is same?
No, I don't.
Disagree.
winnie
Senior Member
- Location
- Springfield, MA, USA
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- Electric motor research
If the carrier frequency increases, then the voltage drop _at the carrier frequency_ will increase. The voltage drop at the _fundamental_ will remain essentially unchanged. The sinusoid should get _cleaner_.
If the carrier frequency increases, then you might see an increase in 'zero cross distortion', which will add harmonics, however in a properly designed drive this should be compensated for.
If the parameters are set correctly, then the VFD will compensate for voltage drop...until you reach the maximum voltage capability of the VFD. So with a VFD excess voltage drop doesn't result in the motor seeing low V/Hz, but limits the maximum speed at which the motor will see correct V/Hz.
-Jon
If the carrier frequency increases, then you might see an increase in 'zero cross distortion', which will add harmonics, however in a properly designed drive this should be compensated for.
If the parameters are set correctly, then the VFD will compensate for voltage drop...until you reach the maximum voltage capability of the VFD. So with a VFD excess voltage drop doesn't result in the motor seeing low V/Hz, but limits the maximum speed at which the motor will see correct V/Hz.
-Jon
So increased voltage drop in OP case may be due to change in carrier frequency. How to check it?
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If the PWM frequency increases, the ripple current content will reduce. Not that it's normally much anyway with switching frequencies usually in the the kHz range. We made some special drives up to about 100kW for high speed motors. We switched at around 10kHz. Not a lot of current gets through at that frequency.
How about simply measuring it?............:roll:
We've been here before.
Which post?
- Location
- Massachusetts
The OP has not been online since 7/21, I think it is safe to say they are no longer participating here and all I see left is bickering.
We are done here.
We are done here.
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