AC run capacitance reactance voltage divider

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Mosaic

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Hi all:
I am going to have to run a 110VAC 0.21A Fan (brush less 120 mm), measured at 0.23A @120VAC.
Resistance is 158 Ω and XL = 440mH @ 120hz.

https://www.amazon.com/110Vac-Muffin-Cooling-Fan-120mm/dp/B00MNTEX0E

on a 220VAC supply.

Reactance calcs show a 5uF run cap to be 530 Ω reactance @ 60Hz which *should* be close to the fan's impedance (120/0.23 = 522Ω) but...when adding it in series the FAN speeds UP and draws 0.31A on 120VAC. The RMS voltage across the fan jumps to 142VAC, across the cap is 153VAC. So the phase shift/resonance is causing a lot more voltage to manifest across the load, which will probably wear it out before its time..

Placing two 5uF caps in series (2.5uF) now passes 0.15A into the fan at 120VAC. This drops about 76 VAC across each 5uF cap and 85VAC on the Fan
3 x 5uFs in series ( 1.66 uF) drops the 120VAC to 63VAC across the FAN and 48VAC across each cap.....which looks about right for use at 220/230 VAC.

I guess I can try for a 1.5uF/450VAC run cap to do the job, as it sees 144VAC manifest across it with 120VAC supply, I est. about 265VAC with a 220VAC supply.
Any thoughts on this?
 
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There is a well known experiment applied to students in their first courses of electrical engineering:

A)
-apply a DC voltage to three different value resistors in series.
-measure with a DMM the voltage drop in each resistor
-demonstrate that Kirchkoff's mesh law applies, by summing all three voltage drops and comparing that it is equal to the source voltage.

B)
-apply an AC voltage to a resistor, inductor and capacitor in series.
-measure with a DMM the voltage drop in each component.
-puzzle yourself to death, because the sum of the three voltages is completely different to the source voltage.

The course instructor will then show you that the DMM is only indicating the scalar part of the voltage. Without knowing the phase angle of each voltage, decomposing it into its real and imaginary portions, sum them, and re-composing them into a vector, one cannot perform an accurate mesh analysis.

Perhaps this is what is happening in your circuit.
 

What are you trying do do here? Are you trying to add impedance to the circuit to drop the voltage across the fan? A clear definitive statement addressing the previous questions would help enormously. Putting a cap in series will cancel out the inductive reactance until it resonates. What you need is inductance or resistance in series with the fan.

Ratch
 
The Xc doesn't just cancel XL..it stands on its own as a voltage drop. The 1.66uF drops enough voltage to allow the fan to see only half the scalar voltage and thus achieve the goal on running at nearly 2x the OEM voltage rating.

Inductance to the tune of 450mH is going to generate heat due to winding losses , thus the Capacitive means of limiting the voltage across the fan seems to be the optimum approach given the far lower ESR ( vs inductor ESR) of the capacitor.
 
Mosaic said:
The Xc doesn't just cancel XL..it stands on its own as a voltage drop. The 1.66uF drops enough voltage to allow the fan to see only half the scalar voltage and thus achieve the goal on running at nearly 2x the OEM voltage rating.
Click to expand...
Mosaic said:
The Xc doesn't just cancel XL..it stands on its own as a voltage drop. The 1.66uF drops enough voltage to allow the fan to see only half the scalar voltage and thus achieve the goal on running at nearly 2x the OEM voltage rating.
Click to expand...
The voltage is not scalar. It has a magnitude and a phase. The phase of the voltage across the capacitor is 180° different with respect to the voltage across the inductance. Therefore, the capacitor voltage will cancel some of the inductance voltage. When it cancels all the voltage, you have resonance. You will get less current in the circuit if you leave out the capacitor. Insert inductance or resistance instead.

Ratch
 
Just tested the 1.66 uF cap solution across a 236VAC line.
Current draw @ 0.19A and 109VAC across the fan, 286VAC across the caps.

Looks like a 1.7uF might be the better value than 1.5uF. But that's not really an easily available run cap value.
 
Mosaic said:
Just tested the 1.66 uF cap solution across a 236VAC line.
Current draw @ 0.19A and 109VAC across the fan, 286VAC across the caps.

Looks like a 1.7uF might be the better value than 1.5uF. But that's not really an easily available run cap value.
Click to expand...
Cap value can be significantly different than what is marked on the body. Try a different cap and see if it is closer to the value you want.

Ratch
 
It's for a production device so I'll probably have to augment with a 0.22 uF X1 class 400VAC PET cap.
 
My nearby market is mostly 110/120VAC with 2 phase 240VAC. So no cap is usually required at all. I'll have to track if this becomes a trend and, if so, think about a solution.
 
On another point. The A.C. fan motor runs faster at higher higher currents and more slowly at lower currents. Would this be because more torque is generated at higher currents and thus less load 'slip' is seen?

In that regard, I'd need to run the motor at a bit higher current to offset using it at 50 Hz rather than 60 Hz?
 


No, you have things backwards. In an induction motor, the more slip you have the more torque is produced (up to a point) and the more current is drawn. At zero slip, there is no torque production. Torque is a consequence of the slip, therefore more torque doesn't mean less slip. If anything, it means more. That's not to say that the slip can't technically decrease since it's measured relative to the no-load speed which changes as the voltage/frequency changes (I don't know enough to say whether that sometimes happens or not, but it's definitely not the general case). In any case, even if it sometimes happens, to say that the slip decreases because there is more torque is misleading.

The current isn't the cause of the motor running faster. The cause is the increased voltage/frequency. Increased voltage/frequency means the motor runs faster and since it's a fan, due to air resistance, the required torque increased by the squared of the speed. Therefore, more voltage/frequency = more speed = higher load = more slip = more torque production = more current. So the current is actually a consequence of all this, not the cause.
 
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To add to dknguyen's excellent post, just remember that on an induction motor the rotor speed N is:

N = 120*f * s/poles
 
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