# Electric motor output wrt voltage



## major (Apr 4, 2008)

roflwaffle said:


> Does anyone know how motors like ADC's 9"er change the power curve, specifically the power curve before peak power, when the voltage increases? Based on what I've seen, peak power moves to a lower speed, but does power before that speed (and peak power) also increase proportionally?


Hi rofl,

This is a series motor. When you have the RPM and power plotted against torque (on the x-axis, horizontal), then both the RPM and HP are proportional to motor voltage change. In other words, a 10% increase in motor voltage at a given torque value increases RPM and HP by 10%. The current (amps) does not change with voltage at that torque value. This is an approximation, but very accurate in the middle of the performance plot. Becomes less accurate at very low and very high loads (torque).

Regards,

major


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## roflwaffle (Sep 9, 2008)

It also seems that an increase in voltage decreases the rpm at which peak power is reached, which is what I'm really interested in. I'd like to stick an 8-9" motor in series behind the transmission in an old diesel, so that I can get a good ~20-50+hp, maybe more if the diff can take it, in every gear. The lower I can make peak power with a motor, the better, since for instance 1500/2500rpm motor speed would be ~30/50mph, so if I could get an additional 40-70hp from a 10kWh pack at 5C rates during that time I would be one happy camper.


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## Bowser330 (Jun 15, 2008)

From the sounds of things you are looking to get into something like this...

http://www.go-ev.com/EMIS.html

It may have worked for some but the EMIS didn't really work for this customer...

http://www.hrivnak.com/



Now my question is why cant we sandwich the e-motor between the engine and the transmission...


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## roflwaffle (Sep 9, 2008)

Sticking the motor between the trans and the engine would be a total pain in the arse and would probably require a ton of unibody fabrication. Otoh if I stick it after the trans all that's required is a shorten driveshaft, the appropriate linkage on the motor, and another trans cross-member that's modded so the motor can bolt to it. 

It'd be better in terms of power output, but I'm not going for tire squeeling performance, just a ~50hp boost, since the car only came w/ ~50hp from the factory, and the ability to go electric only when I want to during low load situations.

EMIS on conventional automatics is kind of a catch 22 situation, since adding electric power and throttling back the engine tends to reduce engine efficiency, and being in an automatic means most drivers can go electric only at low loads and shut off the gas engine, which is where most gains would be seen. Using LAs only makes the situation worse, since they don't pack a lot of energy, especially at higher discharge rate, and weight a ton. I was thinking about a 9" motor, w/ 40 100ah lfps, so range at 55mph should be about 40 miles, and in the lower load situations where I would want to use it, about 1.5-2 times that, so maybe 60-80 miles in the city or on a slow moving freeway. I should also be able to get about 60kW from the pack, so about 52-53kW/70hp peak at the motor, on top of the ~50hp peak from the engine, so in 2nd and 3rd I should have a good 100+hp, which would make for "modern" acceleration as opposed to the 0-60 in ~25-30s in the car's original fit. Granted, there will be some downsides, like dropping carrying capacity form ~1150lbs to ~750lbs, but it should do pretty good considering the motor power/gearing/vehicle specs (curb weight, CdA) are nearly identical to the RAV4 EVs, except there will be an extra 50hp coming from a nearly 50 year old diesel up front, so 0-60 will probably be around 14-15s instead of the stock 18s and while all electric range would only be about half what it is for a RAV EV, I can just flip on the diesel engine and drive for another 600 miles or so.


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## Bowser330 (Jun 15, 2008)

you are right, it would require some expanding of the tunnel, extra mounts/braces, dual adapter plates and a custom driveshaft...

But the benefits are not only electric boost, since the electric power kicks in so low, it would essentially be powering the car through the city and you would not be required to use as much gas to get up to speed. Additionally with the regen activation, which could be setup to be triggered by the brakelight swith, could also help save on brakes...

I dont know I know it seems like a lot of fab work, But i see the benefits as very attractive...maybe Im wrong...


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## roflwaffle (Sep 9, 2008)

Like I mentioned in my edited post, even w/ the motor behind the trans, I should have nearly identical gearing, curb weight, CdA, and peak power output to a RAV4 EV, so performance should be o.k. Regen should be doable no matter where the motor is, and I'll probably go w/ a modded alt off to the side if there's enough room. I think putting the motor in front of the trans would be a fun project on an old Subaru awd car, but IMO the extra work just isn't worth the extra power, especially since most of it wouldn't be usable (the only big power gain I would have is in first and second since third and fourth are at 1.29:1 and 1:1, which would be very limted since first goes to ~15mph and second goes to ~27mph IIRC) except to burn tires, which would certainly be fun, but not worth the extra cost. I could probably go buy a beater V8 mustang for the cost of enlarging the tunnel, moving the trans back and making new mounting points, putting in different clutch setup, and on and on, if I really wanna waste rubber that much.

edited for errors


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## major (Apr 4, 2008)

roflwaffle said:


> It also seems that an increase in voltage decreases the rpm at which peak power is reached,


Hi rofl,

Either I don't follow what you're talking about, or you're mistaken.

Undoubtedly, you will be current limited by the controller. For example, 500 amps. So peak power will occur at 500 amps when the motor controller reaches 100 % duty cycle, or passes full battery voltage to the motor. At RPM higher than that, the torque and HP decrease, as well as current.

So, peak power for you is at 500 amps and battery voltage. If you increase battery voltage, you increase motor RPM at 500 amps and also increase peak power.

Regards,

major


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## roflwaffle (Sep 9, 2008)

It's probably that I don't know what I'm talking about. 

In all the power/torque/amp/eff curves I see, the info is only from peak power and onward. Is performance symmetrical (I've read the electric motor power curves are hyperbolic) about the rpm at which peak power is made. For instance, if an adc 9" motor at 144V makes 90hp at 3k rpm, dropping to 30hp at 6k rpm, would it also climb from ~20-30hp at some very low rpm, say between 0-1000rpm, to 90hp at 3k rpm in the same manner, w/ a nice beefy power curve from ~2000-3000rpm going from ~50-90hp, and going from ~20-50hp from ~1000-2000rpm?


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## major (Apr 4, 2008)

roflwaffle said:


> It's probably that I don't know what I'm talking about.
> 
> In all the power/torque/amp/eff curves I see, the info is only from peak power and onward. Is performance symmetrical (I've read the electric motor power curves are hyperbolic) about the rpm at which peak power is made. For instance, if an adc 9" motor at 144V makes 90hp at 3k rpm, dropping to 30hp at 6k rpm, would it also climb from ~20-30hp at some very low rpm, say between 0-1000rpm, to 90hp at 3k rpm in the same manner, w/ a nice beefy power curve from ~2000-3000rpm going from ~50-90hp, and going from ~20-50hp from ~1000-2000rpm?


Hi rofl,

Plotted against torque, the power curve resembles a parabola with a constant motor voltage. The peak of the power curve will be in the neighborhood of half the stall torque (zero RPM) and half the stall current (voltage divided by motor resistance). So, in your 9 inch motor at 144V example, uncontrolled stall current would be like 5000 amps. So, uncontrolled peak power would at like 2500 amps. Obviously conditions any available controller will not tolerate. So operation of the motor on the high current side of peak power is a non-issue. Even if it was not, you would not want to do so because motor efficiency would be 40 % or lower.

You have to consider the motor and controller as a package deal when it comes to calculating your shaft power. When doing this, peak power will be at current limit (like 500 amps for a Curtis) and when the controller comes up to full voltage to the motor (144V). At that point, the motor RPM is set. At higher RPM, the motor torque decreases as well as power and current. At lower RPM, providing you have it floored and are still in current limit, motor current will be 500 amps, so you get the same torque, but the controller reduces the motor voltage and you get proportionally less power, and battery amps decrease.

So, the motor performance curves you see don't carry the plot out beyond peak power because the motor maker does not want you to operate the motor in that region, for good reason.

Hope you can follow me on this,

major


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## roflwaffle (Sep 9, 2008)

I think I follow most of it, except for the last sentence about not plotting beyond peak power. In all the graphs I've seen they start from peak power and plot that as it decreases and speed increases.


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## Tesseract (Sep 27, 2008)

roflwaffle said:


> I think I follow most of it, except for the last sentence about not plotting beyond peak power. In all the graphs I've seen they start from peak power and plot that as it decreases and speed increases.



Eh? You might be reading the graphs wrong, then. Here's the plot for a WarP 9 from NetGain:

http://www.go-ev.com/images/003_15_WarP_9_Graph.jpg

Torque/current/hp increases as you move from left to right on the graph; rpm decreases. Voltage is constant.


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## roflwaffle (Sep 9, 2008)

I don't think so. Starting from from peak power (not the left side of the graph, but peak power at the right side), power decreases and rpm increases, right? I'm still confused about the not plotting beyond peak power part, since the graph only covers peak power and beyond from ~2k rpm to 5k rpm, not the power curve before that, presumably because it isn't constant voltage.


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## major (Apr 4, 2008)

roflwaffle said:


> I don't think so. Starting from from peak power (not the left side of the graph, but peak power at the right side), power decreases and rpm increases, right? I'm still confused about the not plotting beyond peak power part, since the graph only covers peak power and beyond from ~2k rpm to 5k rpm, not the power curve before that, presumably because it isn't constant voltage.


Thanks for the help, Tesseract.

rofl,

The common way to read these motor plots is to use the horizontal (x-axis) as the load (torque), starting at zero on the left and increasing moving to the right. As the load (torque) increases (moving to the right), the RPM decrease, amps increase and power (HP) increases. The plot actually extends much further to the right than what is shown. As the torque continues to increase off the page (to the right), RPM continues to decrease until eventually is reaches zero (called stall torque). About half way between zero torque and stall torque, the HP will peak. This is the true peak power of the motor with the specified applied voltage.

Now, I think I see what you're getting at and think you may have a grasp on it, although your terminology is confusing. The motor maker does not show on these graphs what you are looking for because it is not a function of the motor alone. It is a function of the motor and controller combination. The motor maker and graph drawer does not know what controller you are using. So, he draws the motor performance plot as if it was connected to the battery directly and only shows the region he expects it to be used in.

Regards,

major


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## roflwaffle (Sep 9, 2008)

How do motor/controller combos tend to behave from stall torque to peak power in terms of efficiency/power? For instance, if I had a ~144V pack and a 500A connected to a 9" motor that made ~60+kW at 2500rpm, would (depending on efficiency) making ~20-30+kW at 1250rpm at 50% pulse width result in roughly the same efficiency (and power) drop as detailed here? Are there any graphs/data points for motor/controller efficiency at different loads/speeds below the peak power?

Edit - Would this also imply that something with lower voltage and peak power would have better low speed efficiency because the duty cycle would be greater and the peak current lower given the same output? In other words, given the same power output/speed, would having half the voltage and double the duty cycle improve efficiency?


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## major (Apr 4, 2008)

roflwaffle said:


> How do motor/controller combos tend to behave from stall torque to peak power in terms of efficiency/power?


Hey rofl,

Let's use the Warp9 graph Tesser posted up. 

http://www.go-ev.com/images/003_15_WarP_9_Graph.jpg 

It only goes out to 75 lb.ft. of torque. See that at the far right? But the motor will actually operate at greater torque. It is just not shown on that particular graph. But let's use this for the example.

Let's say your controller has a current limit of 300 amps. Find the line labeled "Amps". It crosses the 300 amp scaled line at torque of 58 lb.ft., on the x-axis. Got that?

So, the controller gives you a max of 300 amps, you can get a max of 58 lb.ft. from the motor, regardless of RPM, or power (HP).

If you're in your EV sitting there at 0 mph and 0 RPM and floor it, the controller goes into current limit and puts out 300 amps to the motor. The motor, still at 0 RPM, puts out 58 lb.ft. of torque. Which is 0 HP. This torque starts the motion of the vehicle (acceleration). As the motor accelerates, the controller increases the motor voltage to increase the RPM while maintaining 300 amps. So the torque remains at 58 lb.ft. The motor is increasing RPM at 58 lb.ft., so is increasing power. This continues until the controller is "full-on, or 100% duty cycle", at which time, the motor voltage is 72 volts (per the curve). 

At the torque value of 58 lb.ft., draw a vertical line. That line will intersect the RPM curve at 2250 RPM. See that? The vertical line will also intersect the HP curve at 27 HP. See that? This is the maximum power point for the motor/controller combination with a 72 volt battery.

You can go faster, higher RPM, but the torque will decrease and so will the power. If you take that vertical line (at 58 lb.ft.) and the RPM curve to the left of it, and shade in the area below the RPM curve and to the left of the 58 lb.ft. line, it will give you the operational region for this motor and controller with a 72 volt battery. By using your throttle input, you can run at any torque and RPM in the shaded area. But you never exceed that 27 HP.

That is my take on how to use a motor curve.

major


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## roflwaffle (Sep 9, 2008)

We can assume torque and current draw are proportional, but whether or not torque is proportional to current draw and power increases linearly with rpm/speed depends on what low speed motor and controller efficiency are like, which is what I'm interested. I've seen motor power curves that are parabolas, so it seems that either the power draw from the pack is being regulated for more efficient operation, or efficiency is fairly low duing low speed operation. For instance the link in my last post mentions a drop in efficiency w/ PWM duty cycle, but I can't find much specific about it or low speed motor efficiency in general.


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## Tesseract (Sep 27, 2008)

roflwaffle said:


> ... For instance the link in my last post mentions a drop in efficiency w/ PWM duty cycle, but I can't find much specific about it or low speed motor efficiency in general.


I looked at that link - the initial author (Kunz) would be correct about the peak to average ratio getting worse as duty cycle goes down IF the switching frequency and/or the motor's inductance were so low that current drops all the way to zero during the time the switch is off. This is highly unlikely with most modern PWM motor controllers and dc series motors. In general, there is very little ripple in the current waveform across the motor even when our prototype controller is operating at 8khz and 1% duty cycle. That is to say, the current waveform (measured with a Hall Effect current probe) is nearly DC.

Now, the wild card in the controller group is the Curtis which drops down to 1.5khz to emulate low duty cycles. 1.5khz was the switching frequency that most industrial VFDs used until very recently but AC motors typically operate at higher voltages and lower currents, so correspondingly have higher winding inductances than equivalent continuous power rated DC series motors. By my calculation there will be significant current ripple at 1.5khz with a Warp motor, but still not enough to cause more than a 5-10% increase in motor heating over and above what the losses would be if the motor were supplied with DC.


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## major (Apr 4, 2008)

roflwaffle said:


> I've seen motor power curves that are parabolas,


Hi rofl,

An electric motor power curve which resembles a parabola is for the motor alone on a constant voltage or battery source without a controller. Once you put the controller into the picture, the power curve is truncated at the current limit point.



> efficiency is fairly low duing low speed operation. <<snip>> but I can't find much specific about it or low speed motor efficiency in general


Efficiency is power out divided by power in. For the motor, power out is mechanical and equal to torque times speed. So, at zero speed, the electric motor can still have torque, but the power is zero. It still has power in, volts times amps = watts. So the efficiency is zero, zero power out divided by the watts input.

So, at low motor RPM, just as it starts to rotate, the efficiency will be just a percent or two. As the motor RPM increases, the efficiency increases up to the point shown on the motor curve which is a maximum RPM for that given voltage at the particular load (torque) in question. This function of efficiency vs RPM for reduced voltage at load is not linear and is difficult to depict. But you can bet that if you have a significant load and very low RPM, efficiency will suck. And, not much you can do about it. Except shift to a lower gear and run the motor faster.

So, I guess I don't know why you're hung up on this.* Every time you start the motor at load from stand still, you start at zero percent efficient. That is where your transmission comes in handy. Get the motor up to speed sooner and shift to keep the motor speed high (up to a point) and the motor amps down.

Look at the plot on page 3 of this 

http://www.acpropulsion.com/tzero/AC150_Gen2_specs.pdf 

This is one of the few power/RPM/efficiency plots for electric motors I have seen. Also, a very efficient system. Not something you're likely to find for the motor you can afford. But you can see from that curve how all the efficiency lines get bunched up down at low RPM. They don't even bother to go to zero RPM, starts at 1000.

Regards,

major

* Looking back, I see you want to try to use a motor in a post transmission hybrid. Sorry for the wise crack. I just get used to EV thinking here, not HEV. Anyway, the same motor physics apply. Hope I helped.


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## roflwaffle (Sep 9, 2008)

Tesseract said:


> I looked at that link - the initial author (Kunz) would be correct about the peak to average ratio getting worse as duty cycle goes down IF the switching frequency and/or the motor's inductance were so low that current drops all the way to zero during the time the switch is off. This is highly unlikely with most modern PWM motor controllers and dc series motors. In general, there is very little ripple in the current waveform across the motor even when our prototype controller is operating at 8khz and 1% duty cycle. That is to say, the current waveform (measured with a Hall Effect current probe) is nearly DC.
> 
> Now, the wild card in the controller group is the Curtis which drops down to 1.5khz to emulate low duty cycles. 1.5khz was the switching frequency that most industrial VFDs used until very recently but AC motors typically operate at higher voltages and lower currents, so correspondingly have higher winding inductances than equivalent continuous power rated DC series motors. By my calculation there will be significant current ripple at 1.5khz with a Warp motor, but still not enough to cause more than a 5-10% increase in motor heating over and above what the losses would be if the motor were supplied with DC.


Do you have the info, or links to the info, you used to calculate heating losses due to ripple? Thanks btw!


major said:


> So, at low motor RPMs just as it starts to rotate, the efficiency will be just a percent or two. As the motor RPM increases, the efficiency increases up to the point shown on the motor curve which is a maximum RPM for that given voltage at the particular load (torque) in question. This function of efficiency vs RPM for reduced voltage at load is not linear and is difficult to depict. But you can bet that if you have a significant load and very low RPM, efficiency will suck. And, not much you can do about it. Except shift to a lower gear and run the motor faster.
> 
> So, I guess I don't know why you're hung up on this.* Every time you start the motor at load from stand still, you start at zero percent efficient. That is where your transmission comes in handy. Get the motor up to speed sooner and shift to keep the motor speed high (up to a point) and the motor amps down.


I'm very curious about how much efficiency will suck because that determines what kind of operating parameters I'm looking at with my project, or whether or not I go ahead w/ it in the first place. For instance at 35mph/2k rpm, ~87% efficiency is great, but if I'm at ~17mph/1k rpm, where does efficiency go? I've seen the AC150 map before, but I'm wary of applying it to DC motors since I'm not sure if they share the same characteristics, eg worse efficiency at high power compared to low power at the same speed as well as decent low speed efficiency. If I could see ~45-65% efficiency from .5k-1krpm (~7-17mph), ~65-80% efficiency from 1k-2k rpm (~17-36mph), and 85+% efficiency from 2k-4k rpm (~36-72mph), over the power output/speed/eff range, I'd be a happy camper, but if efficiency was in the ~20-40% region at low speed (~.5k-1.5k rpm) that'd change my design approach.


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## major (Apr 4, 2008)

roflwaffle said:


> I'm very curious about how much efficiency will suck


Hey rofl,

If you select the motor and controller and get the resistance figures, brush voltage drop figures and controller voltage drop figures, you can calculate such things for the particular load of interest. At low RPM, you can ignore rotational loss in the motor and just work with electrical loss, because that will be the bulk of it.

And you're right; you cannot apply the AC150 eff map to a DC motor. I have never seen a similar plot for a DC motor. It was just an example.

I have seen simulation programs where you can model the motor and controller and run (pretend) thru a load cycle. I've never used such a thing and feel such simulations are of dubious value unless you can verify them against some real world tests. But see if you can search out such a simulation package. This may have been done by some university project.

Regards,

major


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## roflwaffle (Sep 9, 2008)

Sweet! What equations would I use to calculate the electrical losses at different loads/low speeds? All the current losses would be (I^2)R sub whatever (weight average of all the winding resistance), are brush contact losses and/or stray losses a significant portion of low load efficiency?


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## major (Apr 4, 2008)

roflwaffle said:


> Sweet! What equations would I use to calculate the electrical losses at different loads/low speeds? All the current losses would be (I^2)R sub whatever (weight average of all the winding resistance), are brush contact losses and/or stray losses a significant portion of low load efficiency?


Hi rofl,

Going back to my original response to you, post #2: 




> This is a series motor. When you have the RPM and power plotted against torque (on the x-axis, horizontal), then both the RPM and HP are proportional to _*motor voltage*_ change. In other words, a 10% increase in motor voltage at a given torque value increases RPM and HP by 10%. The current (amps) does not change with voltage at that torque value. This is an approximation, but very accurate in the middle of the performance plot.


This is an approximation because the RPM (and power) are actually proportional to the generated voltage (Eg) in the armature, not the motor voltage (Vm). The approximation works well for most of the curve, but falls apart at low speeds because Eg and Vm diverge.

Take the worst case, zero RPM. Looking at the motor performance curve, the torque/current relationship will still hold true. Let's continue to use the curve for the Warp9. http://www.go-ev.com/images/003_15_WarP_9_Graph.jpg 

So, even at zero RPM, you can get 58 lb.ft. at 300 amps. At zero RPM, Eg = zero. So all the applied motor voltage is dropped in the motor resistance (copper and brushes). Let's guess at the Warp9 resistance. Let's say 0.01 ohm in the copper and 0.01 ohm in the brushes. Together, 0.02 ohms in the motor. Now use Ohm's Law (V=IR) to figure the needed motor voltage (Vm). Vm = 300A * 0.02 ohm = 6V. So, to get 58 lb.ft. you need to apply 6 volts to the motor and it will draw 300 amps and not rotate.

Because this is a series motor, the flux changes with load. It is therefore impossible to use a universal RPM constant like with PM motors. However, you can calculate a RPM constant for each load point because the current (and therefore flux) is fixed.

Let's continue working at the 300A load point. The graph shows 2250 RPM when Vm = 72V. At that point, Eg = Vm - IR = 72 - 6 = 66V. Now, divide 66V by 2250 RPM. You get 0.0293 V/RPM, where it is the Eg voltage, not Vm.

If you want 100 RPM at 58 lb.ft. (300A), then you need Eg = 100RPM * 0.0293 V/RPM = 2.93V. Vm = Eg + IR = 2.93 + 6 = 8.93V. At a load of 58 lb.ft., it requires 8.93 volts to be applied to the motor to get 100 RPM.

Losses would be IsquaredR = 300^2*0.02 = 1800 watts. Motor output power (Po) = 58 lb.ft. * 100 RPM / 5250 = 1.10 HP = 823 watts. Motor efficency = Po / Pin = 823W / (8.93V*300A) = 0.3072 = 30.7%. Or you can use eff = Po / (Po + losses) = 823 / (823 + 1800) = 31.3%. Or because all the losses are resistive, you can use eff = Eg /Vm = 2.93V / 8.93V = 32.8%. 

You notice some differences in the percentage, due to rounding errors and such. Also, this method ignores rotational losses in the motor. But at low speeds, like 100 RPM, such losses will be low. Rotational losses will increase with RPM and become a factor about equal to resistive losses at high speed and lower currents.

The above is just an example. And when I double check it against the Warp9 curve at 2250 RPM, I get 91.6% and the graph shows 87%. This could be attributed to an incorrect guess at motor copper and brush resistance, as well as ignoring rotational losses.


Regards,

major


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## roflwaffle (Sep 9, 2008)

Neat! So if I reduce current and torque I can also increase low speed efficiency, correct?


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## major (Apr 4, 2008)

roflwaffle said:


> Neat! So if I reduce current and torque I can also increase low speed efficiency, correct?


Generally speaking, yes. To a point. And you are also reducing power, unless you raise voltage and RPM.

major


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## roflwaffle (Sep 9, 2008)

Sweet! One last question, thanks for all the help btw, as a general rule, does a higher voltage pack tend to improve efficiency all things (resistance, load, and speed) being equal?


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## DIYguy (Sep 18, 2008)

major said:


> Hey rofl,
> 
> Let's use the Warp9 graph Tesser posted up.
> 
> ...


Very nice explanation Major! Thank you for that.


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## major (Apr 4, 2008)

roflwaffle said:


> does a higher voltage pack tend to improve efficiency all things (resistance, load, and speed) being equal?


Hi rofl,

When you say "all things (resistance, load, and speed) being equal", then the motor voltage must be the same. So a higher voltage pack will mean a lower duty cycle on the controller. Meaning the motor volts and amps are the same as before, but now the battery voltage is higher and battery current lower. Any efficiency improvement would be only in the battery circuit. Depending on how the higher voltage pack was achieved, there may be efficiency improvement, although it is not a given.

Regards,

major


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## roflwaffle (Sep 9, 2008)

Cool. What if motor voltage was increased? Since Cu losses go according to current, would I be able to see better efficiency given the same power output?


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## major (Apr 4, 2008)

roflwaffle said:


> Cool. What if motor voltage was increased? Since Cu losses go according to current, would I be able to see better efficiency given the same power output?


rofl,

There is no easy answer here. Sorry. Generally speaking, within the operating parameters common to the EV DC series motors being used, it won't make much of a difference until you push the limit somewhere. Like way too high RPM, or way too low RPM. But if you're talking about 10 HP at 3000 RPM vs 10 HP at 4000 RPM, there is no big difference. Now if you're talking about 10 HP at 3000 RPM for a 48 volt motor vs a 96 volt motor, then you're talking about two different motor designs. And if those two motors had the same amount of steel and copper, they would be about the same efficiency.

Regards,

major


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## roflwaffle (Sep 9, 2008)

I suppose the only thing to do some time in the indefinite future is testing to see how efficiency compares to the published and extrapolated figures. Thanks a bunch major, you've been very helpful!


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## GTX_SlotCar (Jul 29, 2009)

I'm hoping someone here can explain something to me. It's just about small dc motors, batteries and voltage in general. It's actually r/c model helicopters.
I use a LiPo battery that has 2 cells in series giving it 8.4v fully charged. The battery capacity is 1300mah. At a motor speed of about 16,000 rpm, under load, it'll run for about 20 minutes. The exact time isn't important. 
Now, if I use a 3 cell (again, in series) battery I have 12.6v, or 50% more voltage. Each cell is still 1300mah. At the same motor speed, 16,000rpm under load, it'll run for about 22 minutes. According to Ohm's Law, I should only be drawing 2/3 the amps at 12.6v as I'm drawing at 8.4v, so it seems the run time should be a lot longer.
Can anyone explain why?
I don't know if this makes any difference, but the motor speed is controlled by PWM (pulse width modulation), so full voltage is sent to the motor in ON and OFF pulses and the duration of the pulses controls the speed of the motor.
My contention is that Ohm's Law doesn't apply here for some reason. Maybe the R (resistance) of the motor changes with more voltage, or maybe it's EMF. Or maybe the Torque required under load is governing the current (amps) draw regardless of the extra voltage and 1300mah is 1300mah no matter what.
I sure would appreciate some help.
Thanks,
Gary


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## major (Apr 4, 2008)

GTX_SlotCar said:


> I sure would appreciate some help.


Put an ammeter on it and find out for sure.


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## GTX_SlotCar (Jul 29, 2009)

Major, Thanks, I'm about to do that, but in a way it still won't answer the question fully. I already know the run times so I'll have to assume that the amps will be about the same (I'll find out for sure soon). The question of why Ohm's Law doesn't work here will still be open.


I forgot to mention, the ending voltage is about 7v on the 8.4v battery and about 8.7v on a 12.6v battery. There is no safety cut off, so nothing to stop the 3 cell battery from running down to 7v. I have to assume the tank is dry, so to speak (the 1300mah capacity is used up).

Gary


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## major (Apr 4, 2008)

GTX_SlotCar said:


> The question of why Ohm's Law doesn't work here will still be open.


Hi Gary,

Ohm's Law always applies, but so does Kirchoff's. How do you know the RPM is exactly the same? Running a prop, a small difference there can make a big difference in power. Does the extra mass of the third cell increase the load? Are all 3 cells good and equal? Could be a lot of different things throwing you for a loop. Don't blame Mr. Ohm 

Regards,

major


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## GTX_SlotCar (Jul 29, 2009)

Major, I meant that Ohm's Law might not work well in general on electric motors, that as a device they may be none ohmic. It was just a guess as the resistance of the motor winds changes.
RPM is the same on a test bench using a tach. In flight, the extra weight (about 20 grams) has been added to an 8.4v model so the load should be the same for hovering. The increased weight shortened flight times by about 50 seconds. All batteries are new and several were used.
I'm going to try to measure the current now. If it really is reduced by 1/3 using 12.6v, it will create even more questions as to why flight times don't increase from 20 to 30 minutes, and why the 3 cell won't drain to 7v like the 2 cell does. Obviously the motor is capable of operating at 7v (it's rated at 6v).

Gary


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## GTX_SlotCar (Jul 29, 2009)

I did the tests. The only variable was the batteries (2 cells and 3 cells). I did the test several times and used fresh batteries often. Each time the results were very similar, but I averaged them anyway. The batteries were beside the heli, not installed, so weight would not have been an issue. But, in most of the testing I used a tach for motor speed so weight wouldn't matter anyway.
At 1100 rpm rotor speed:
8.4v = 4.85A
12.6v = 4.30A

If Ohm's Law held true, and 12.6v was 4.3A, then 8.4v should have been 6.45A. Far more than the 4.85A measured. So, it explains the reason why there is so little difference in flight times between 2 cells and 3 cells, but not why Ohm's Law calculations didn't work.
I also assumed that the greatest difference in current would show at the extremes of the rpm range (very low and very high), but they didn't.
At very low rpm 8.4v was 1.50A and 12.6v was 1.38A, and at very high rpm it was about 5.3A vs 5.1A, but I admit I didn't spend much time on these last measurements.

Gary


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## Amberwolf (May 29, 2009)

Are those battery currents or motor currents?

Does the controller get hotter under the higher voltage? 

How is your controller designed? Does it have internal voltage shunting that is drawing current not in your measurements (if those were done on the motor, rather than battery side). 

Some controllers, like the old ScootNGo controller I started out with, have a simple zener/resistor voltage regulator for the internal electronics (not used for the motor output voltage, which is sliced directly off the battery line), which if it is used at over it's originally rated voltage will pull quite a bit more current, wasting it as heat. It should only be a few dozen milliamps at most, but it depends on the power rating of the zener and resistor involved. 

I had to alter that controller to use it at 36V, since it was designed to be used at 24V, and that "regulator" circuit got very hot at 36V before alteration. 

Same is true if they used a linear regulator (like an LM317 or 7805, etc); if it's heatsinking is good enough it could waste a pretty good amount of power when the controller is used above it's rated voltage.

Another possible issue is the ESR and total capacitance of the capacitors in the controller. If they already had a significant ripple, they will have a higher one under the higher voltage, and have to dissipate more power as heat, with a higher ESR. 


Any of those power losses could make what might be a significant impact on your total runtime. Or they might make no difference whatsoever, if they're minimal. 


EDIT: just as an aside, if you are trying to run 3 4V LiPo cells down to 7V, that puts them at about 2.3V each (which for your 2-cell pack would only be down to 3.5V per cell). Is that safe to do for your particular cells? It's might not be good for them. 
________
GOLYOXXL CAM


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## GTX_SlotCar (Jul 29, 2009)

Hi. Yes, running the cells below 2.65v would hurt them. I'm not trying to run them that low. The point was that the heli will fly with about 7v from a 2 cell battery, so it _should_ be able to go that low with a 3 cell battery. But it can't. It runs out of power at about 8.7v which is higher voltage than the 2 cell battery supplies when fully charged. These cheaper helis don't include a low voltage safety cut-off circuit.

The controller gets a little hotter with 3 cells, but it's made to operate on 2 or 3 LiPo cells, and even 10 NiMH cells. The same controller is used on different heli models, some include 2 cell packs and some models come with 3 cell packs. [edit: the *motor* does run hotter with the 3 cell LiPo]

I don't believe there is any internal shunting. I measured the current at the battery.


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## major (Apr 4, 2008)

GTX_SlotCar said:


> I did the tests.


Hey Gary,

Looks like Mr. Ohm is right on the money here and your controller is the resistor. That is why you run out of power at 8.7V on the 3 pack. Big IR drop across the controller. My take on it 

Thought you were a little off topic, _diy electric car_ you know. But your toy is an electric vehicle. You going to take what you learn with that and do an electric car?

Regards,

major


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## GTX_SlotCar (Jul 29, 2009)

major said:


> ....Thought you were a little off topic, _diy electric car_ you know. But your toy is an electric vehicle. You going to take what you learn with that and do an electric car?...


Major, thanks for your very useful information. I am off topic with my vehicle. Hopefully the answers you provided will also be useful to some here, and helpful for their projects. Sometimes when one industry needs answers, the best place to get it is from the experts of another industry. 

I have given some thought to making an electric car. I'd have a lot to learn. In my case, my office is about 6 miles from home. Mostly 40mph speed limit, and I take one daily trip to the post office - another 4 miles. 16 miles daily at very inefficient gas mileage. Here in Maine, I could use a simple electric car - even without heat - for 6 or 7 months (the regular motorcycle season). I would think that a fairly simple electric vehicle, not to be driven in snowstorms or when the windshield might freeze, might be something I could build. 25 to 35 miles per charge is all I'd need. You know, it's not that I can't afford the gas, but it bothers me that there's nothing available that's optimized for this kind of driving. And, a lot of people I know drive the same way I do. 
Anyway, thanks for your help.

Gary


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## GerhardRP (Nov 17, 2009)

I played with some performance curves for an xp1227a motor [a souped up ADC 8", I think]. I got a family of curves from kta-ev. The most interesting plot is efficiency vs. power [in watts on the graph]. The parabolas are for various applied voltages, while the falling lines are for various RPM, 1 kRPM at the bottom and 6 kRPM at the top. The lesson is to keep the RPM's HIGH. Note that the data are a little ragged at low torque and high RPM. The second graph is power vs. torque.
I've tried to attach the graphs, I hope it worked.


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## roflwaffle (Sep 9, 2008)

Those graphs are exactly (Well, except for not having data under 1k rpm or efficiency data at really low power levels ) what I was looking for, and they seem to square pretty well with what major mentioned, thanks Gerhard!


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## GerhardRP (Nov 17, 2009)

I'll edit the graphs later for the low RPM data later. As for low power, I would like to see that also, but I've never seen any. My modeling could use a family of Stalled torque numbers, also including low current... no fancy dyno needed.


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