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Does will over-volting destroy a motor? If so, why?

#1
Will over-volting destroy a motor? If so, why?

Here's an interesting question:

So let's say I take a motor designed for 3s and I give it 6s. When I do this, I prop way down so the motor only draws the same amount of current (or less) as it drew on the 3s setup with a bigger prop.
As I understand it, most of the heat a motor generates is due to high current in the thin wires that make up the windings.
If I'm not wrong, increasing voltage without increasing current shouldn't generate any additional heat, so would it damage the motor?

I ask this because I'd be interested in trying a 6s setup on my Viggen, but the motor I have is rated for 3s. It's an 1100kv and calls for a 11x5 prop, so on 6s i would put a 6x5 or 7x4. This should draw about the same current.
 
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joshuabardwell

Senior Member
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#2
You may exceed the max rpm of the bearings. All else being equal, you will double the rpm. If you increase the voltage and decrease the prop size, you will more than double the rpm.
 

joshuabardwell

Senior Member
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#3
Also, don't forget that if your prop tip exceeds the speed of sound, bad things happen. 7" prop = 22" circumference. Speed of sound is nominally 803,834 inches per minute. Divided by 22" is 36,538 rpm. Realistically, you'd need to approximately triple the rpm of a typical RC motor, but then again, you're going to double the rpm based just on the voltage, and you're going to a smaller prop, so...
 

joshuabardwell

Senior Member
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#4
The final thing that I can't evaluate is that if you double the voltage and keep the amperage the same, you will have doubled the wattage. When doing heat dissipation calculations, you calculate the wattage dissipated by the component. But when sizing wires based on heat build-up, you use amperage alone, regardless of the voltage (which allows getting more power through smaller wires by going to higher voltages). I don't know how to resolve this, though.
 

Craftydan

Hostage Taker of Quads
Staff member
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#5
The final thing that I can't evaluate is that if you double the voltage and keep the amperage the same, you will have doubled the wattage. When doing heat dissipation calculations, you calculate the wattage dissipated by the component. But when sizing wires based on heat build-up, you use amperage alone, regardless of the voltage (which allows getting more power through smaller wires by going to higher voltages). I don't know how to resolve this, though.
In typical power disapation calculations for sizing, either the voltage is fixed for the operation of the device (usually for something non-linear) or the impedence of the device is fixed (which in turn usually fixes the voltage/current tradeoff). Simply put, if I were to do this for a resistor (really any linear circuit), if I increase the current, the voltage must simultanously increase to force that current through. Current, voltage and power are inseperably linked.

This is not the case for an electric motor.

An electric motor will have a back EMF altering the total impedence depending on the load. *Assuming* the load were reactive to input voltage (the builder changes props between changes in voltage), such that current was kept constant, and *assuming* the voltage was kept well under the insulation's breakdown voltage, while the power transfered from the input source to beaten air would increase, the current would not . . . by definition.

An electric motor can have a variable current/voltage relationship, depending on the load.

As you've seen, it is the current flow that generates the waste heat, not the voltage -- a broken wire generates no electrical heat regardles of the voltage difference (up to the air's break-down voltage), while a near-0-volt hard short generates sufficent heat to melt and weld steel (and not only do the welding points where the current is concentrated heat up, so does the whole welding aparatus even though it's a very low resistance path).

While waste power can be used in calculating power transfer efficiencies -- it's convienient to express things in terms of power -- however do not expect the waste power to increase with voltage increases. there are emperical connections between current and heat, but voltage does not play in this.

Therefore, if you fix your current by varying your load with voltage, you'll find increasing the voltage will actually increase your power transfer efficiency because the waste component remains the same . . . because we've fixed the current.




To go back to the OP, and summarize (from Josh's posts) . . . so long as the prop size is changed to keep it efficient and limit current, the next failure mode is from overspeeding the bearings. If the bearings can take it, and the prop is sized right, kicking up the voltage will allow you to get proprtionally more thrust-power out of the same motor, at a greater overall efficiency.

I would not assume, however, that doubling the voltage will cut the required prop size in half. prop disk area, maybe-kinda-sorta, but size, no. Most of the Rules of Thumb I have at hand for comparing props are for a set voltage/RPM, and power draw for a given prop is very nonlinear with voltage . . . so I'd caution, test carefully with a power meter/current meter before you lock into that prop and fly extended periods WOT.
 

willsonman

Builder Extraordinare
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#6
Good write up Dan. And yes... props exceeding the speed of sound = very bad stuff. Although it would make a good challenge for the FT guys... or Asbjorn :)
 

joshuabardwell

Senior Member
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#9
What happens exactly when the prop tip exceeds the speed of sound?
You need an expert in aeronautics to give you the specifics, but, basically, you create a sonic boom. Air pressure at the front of the shock wave increases sharply and abruptly.

Air acts much differently at supersonic speeds than it does at subsonic speeds.

When an aircraft approaches the speed of sound, the airflow over the wing reaches supersonic speed before the airplane itself does, and a shock wave forms on the wing. The airflow behind the shock wave breaks up into a turbulent wake, increasing drag.

When the airplane exceeds the speed of sound, a shock wave forms just ahead of the wing's leading edge. The shock wave that formed on the wing is now at the trailing edge.

When the wing is tilted upward, a shock wave forms below its leading edge, and an expansion wave forms above its leading edge. The higher pressure behind the shock wave and lower pressure behind the expansion wave result in a single force that pushes the wing up and back.