Over-Volting a DC Motor

Sorry, but heat is always amps times volts and so increased voltage matters just as much as current. Simple as that. You can look at the causes of a particular current flow, but it doesn't change the formula. You can look at efficiency, which will put more energy into the shaft and less into motor heat, but increasing the top end RPM will almost always reduce it (ie, put more heat into the windings, even if you held current AND voltage constant).

 

jonr,
Volts (over the motor) * amps (into the motor) will give you the power (into the motor). Hopefully, the motor is doing something useful so most of that power should go out through the shaft.

And yes, volts * amps also creates heat in the windings, but it's not the same volts as goes into the motor. If you want to figure out the losses by doing P = U*I then you first have to figure out which voltage to use in the calculation. The voltage that drives the current through the windings is not the same as the voltage across the motor inputs. As Yellowjacket pointed out, the rotor will generate a voltage while rotating in the magnetic field created by the current. This is the back-EMF. The voltage driving current through the windings is the input motor voltage minus the back-EMF.

If you stall the motor, the back-EMF will be zero and all of the applied voltage will be used to generate heat. There is no mechanical power output. At the other end of the scale you have the no-load speed where the back-EMF almost equals the applied voltage. The input current will be low, and the voltage needed to drive that current is also very low, as are the losses.

Erik

 

Well put Erik... Jon doesn't understand that volts times amps is the power into the motor, the winding resistance times the amps squared are the resistive losses, and the heat generated by the motor is caused by the losses, not the total power put in.

 

Your arguments involve efficiency, which as I said, is likely to be declining (ie, MORE winding heat) as you apply a normal load and increase RPM beyond the normal limit (this is what the topic is about). Erik's not so pertinent examples are a stalled motor and a no-load case which are even lower efficiency.

> You want the current to be the same so that the motor doesn't overheat.

This is the original wrong comment and yet "the same current", increasing voltage, more power output and decreasing (or steady) efficiency will certainly create more winding heat.

 

I've tried to explain this in relatively simple terms so that folks who don't really understand the subject can try to grasp the fundamentals, but basically if you keep the current the same you will not burn up the windings even if you are running a slightly higher voltage and have increased the speed of the motor. I'm truly sorry that you just don't get it, but if the current through the motor is the same, the resistive losses will be the same (remember it's I squared R again) and the heat load will be the same. And yes there will be an increase in power made by the motor, and there well may be a decrease in efficiency, but the result of that decrease in efficiency will just be a bit less torque produced by the motor.

Erik is trying to convey the extremes of what back EMF is in his examples so that you can understand how back EMF effects the total current in the motor. You are failing to understand that voltage across the windings is a function of speed. The current through the motor varies with load and resulting speed. That is, if you unload the motor, the speed rises and the current decreases (because back EMF got higher and there is less voltage available to drive current through the windings).

For each speed, load and voltage there is a corresponding current. That current is based on the back EMF of the motor (and varies with speed) plus the winding resistance. You are failing to understand this relationship and are thinking that the voltage into the motor relates directly to the current at all times. It does not. You have to consider the effects of back EMF on the current flow and then determine the losses.

Fortunately it can be a lot easier to understand losses if you just look at the current. Since the losses in the windings are directly related to the current if the current is kept the same, the amount of heat generated in the windings is the same. The efficiency of the machine isn't the issue, as I said the shaft output will be the output and yes you won't get as much power out (the efficiency of the motor may not get better) since there are magnetic losses, but if the current is maintained then the amount of heat generated is the same. That is why my statement is correct.

 

Hp curve

I atached a typical DC motor curve for hp vs torque and current. This is from electric motor I used in the electric conversion of my push mower.

The dark dotted line is the power output of the motor. The light grey dotted line is the efficiency of the motor. The loss of the motor (heat) is the input power minus the losses of the motor. Or you could also say it is the input power times (1-efficiency) of the motor. Work out the shaft is the input power (volts time amps) times the efficiency. Loss in the motor is not simply volts time amps unless you have a locked rotor or no load. In both cases, your efficiency is zero and all of the input power goes into heat (and windage, noise, vibration... at no load).

All PMDC motors have the same shape power vs torque curve when held at constant voltage. The curve is the same shape but higher and wider with increasing voltage. Increasing the voltage AND changing where you are on the curve can actually get you more power output and less losses in the motor especially if you are operating past peak efficiency at the lower voltage!

It's safe to say that some people will never get this and the concept of back emf in dc motors and that's ok. Not everyone is an Engineer.

I think that I will set up a bench dyno with my 36v 109 MG motor this weekend and get some data. The benchmark is about 36v and 36amps for this motor as the sticker says the rating is 1.3 kW (input). I might even try to mount a tach inside the lower unit to help with some in-water testing later.



It will be interesting to see if the losses are less at 48 volts and 27 amps which is the same input power. If so, it might make sense to trim the plastic prop down to run 27 amps on a lightweight boat while at 48 VDC input. I would like to believe that a smaller diameter prop with the same pitch could actually make a lighter boat go faster with the same power. This was one of the original questions asked by the author of this thread.

 

You probably want less pitch and perhaps a bit less diameter in your prop. If you just trim the prop diameter you will just get more slip and while it might let the rpm's come up you will likely lose efficiency unless the boat has low enough drag you will still overload the electric motor.

 

The concept of back EMF is difficult to grasp without a lot of background and so is the difference between resistance and impedance, which is quite similar.

A very simple approach is this:
The current through a winding takes time to rise, because it takes time to build the electromagnetic field. With increasing rpm each winding gets less time, so the current drops, the field weakens and so does the torque.

Without load an electric motor of the type we are discussing here, would spin infinitely fast without drawing any current if there were no limiting factors like bearing and air friction losses.

 

A series wound(field) DC motor will overspeed without load. An unloaded PMDC motor will stop accelerating when the rpm approaches the back emf. Look at the attachment I posted and you will see there is a constant called 'KE' in volts per RPM or kRPM.

 

> Increasing the voltage AND changing where you are on the curve can actually get you more power output and less losses in the motor

It can, but not in the high and going even higher RPM case where efficiency is dropping.

It's simple - more power in at the same or lower efficiency = more heat in the motor. If someone doesn't agree, then where do you think the energy is going? Say 100 watts in at 80% efficiency vs 120 watts in, at 80% (or lower) efficiency? 24 watts is more than 20 watts.

> You are failing to understand this relationship and are thinking that the voltage into the motor relates directly to the current at all times.

Absolutely not what I said (or thought).

> if the current is maintained then the amount of heat generated is the same.

You are failing to understand that you can't put more power into a motor at a fixed efficiency without getting more heat. Even IF current is the same.

 

Series motor curves

I attached the performance curve for one of my series (forklift pump) motors. Notice on the attachment the rpm rises rapidly wil decreasing load. You can't run this motor at no-load above about 12 volts or it will frag the armature due to overspeed.

Trolling motors are all PMDC type. The highest rpm I have obtained unloaded on a trolling motor was 3000 rpm by overvolting a 24v unit with 36 volts. This gives a KE of 12V/kRPM.

I suspect the 36volt 109Lb MG motor has a KE of around 15V/kRPM. My guess is that this motor has a KT of around 13 oz.in./amp as well. I'll post some results in a few days.

 

No Load curve

Here is the measured no load speed vs voltage of the 36 V trolling motor. I pushed it to 50 volts with no issues. I know the seals are gonning to get trashed but it will get rebuilt when I am done testing.

I expected 15 v/kRPM and it was just a little higher. Even at 50 volts, it is drawing less than I can measure with my 0-50A meter. Perhaps around 1.5 amps. I have a 'power prop' attached and it is acting like a fan so there is a small load.

I am using that ammeter in the set-up for the loaded testing. I'm not sure how I want to load the motor. I might use either a friction plate or another motor with a variable resistive load on a bearing with a force gauge at a fixed torque arm length.

 

I'm sorry, but you are assuming that the efficiency is a constant. It is not and it does not have to be. The efficiency can be more or less, but the heat generated in the windings is primarily a function of the electrical losses, which is the current squared times the resistance. That is all I have said.

There are a number of other losses including windage, magnetic losses, bearing losses. However, the largest losses in a DC motor are related to the winding resistance and these are the a direct function of the current flow. While windage and bearing losses generate heat, they don't tend to burn out the windings, the heat gets carried away by conduction to the cases. Inductive losses do generate some heat, but they are second order when compared to the resistive losses. All I said is that you can, within limits, increase the voltage a small amount and if the speed increases and the current doesn't increase you won't generally burn up the motor. You may well have a less efficient motor, but you may well gain efficiency since the resistance losses are the same and the power increased.

 

Regarding tip speed for the propeller. I searched for information; excessive tip speed give problems with cavitation. A 25 hp outboard I have has a maximum propeller speed of 3100 rpm. The maximum recommended diameter is 8", so the tip speed will be about 32 m/s (106 fps). I've seen the figure 120 fps somewhere for maximum recommended tip speed.

 

Initial load tests

Based on my locked rotor current and torque tests, we are looking at a KT of 18 oz-in/A. This is better than I thought but the bad news is that armature resistance is measuring higher (0.25 Ohms) than expected. Most likely, the high resistance is due to the wiring and my unknown ammeter resistance. I will put the voltmeter closer to the actual motor next time for better numbers.

I plan on doing some dynamic testing. I will try to set up a load this weekend to get some actual operating performance data.