# Basic Question Regarding Electric Motors



## rangerdev (Oct 25, 2015)

Hey all,

I have what I can imagine is a basic question about electric motors and was hoping you could help me out. 

Just say for this example I have a 90v motor and a bunch of 12v batteries. I obviously can't make 90v with 12v batteries. I can go over or under. 

What happens when you use less or more voltage than what is listed on the motors nameplate? Is it going to destroy the motor? Does it even matter? Speed difference, torque difference....? If you could spare some time for an explanation of this I would appreciate it. Something I am trying to wrap my head around. 

In a more extreme case what would happen if I ran 72v worth of batteries (6)? I ask because I have a really nice 90v that was given to me for free, which I would like to use to get my feet wet on EV building but I can't fit 7 or 8 12v batteries on my application (dirt bike). 

My intention is to order a golden motor 48v 5kw motor for this project when my tax return comes in, but I wouldn't mind playing around with what I have for now. Thanks!


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## Yabert (Feb 7, 2010)

All depend of the motor controller... and the motor controller will depend of your motor type.
Picture of the motor? Spec?


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## miscrms (Sep 25, 2013)

Assuming we're talking about basic brushed dc motor:

At a very basic level (about where my understanding of motors is) voltage = rpms, and current = torque. Voltage * Current = Electrical power, and rpm * torque = mechanical power. 

Running less voltage is generally fine, that's what a motor controller essentially does. Reduce the voltage to the motor at low rpms to keep the motor from drawing too much current. At 0 rpm the motor will essentially appear as a short across the battery and try to draw infinite current. Running more voltage will let you maintain torque to higher rpms, but exceeding the rated voltage will eventually cause problems. Arcing at the brushes, overheating the windings, breakdown of insulation, etc. You also don't want to exceed the max rpm rating of the motor or it may shake itself apart. 

It all get more complicated of course when you start talking about real motors, where the strengths of magnets (or field windings), finite resistances and impedances, etc come in to play to impose limits on how much can be gotten out a given motor.


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## rangerdev (Oct 25, 2015)

Yabert said:


> All depend of the motor controller... and the motor controller will depend of your motor type.
> Picture of the motor? Spec?


Motor is a LEESON 098000 C42D17FK1C 

1/2 hp 18 inch pounds of torque 1750 RPM

See http://www.leeson.com/leeson/search...ProductDetails&motorNo=098000.00&productType=

With such a small amount of torque and hp I could never see it pushing a bike, but the guy who gave it to me said it would without hesitation... Am I missing something? Maybe you could explain a bit more about controllers for me so I have a better understanding. Thanks for the reply, appreciate it.


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## rangerdev (Oct 25, 2015)

miscrms said:


> Assuming we're talking about basic brushed dc motor:
> 
> At a very basic level (about where my understanding of motors is) voltage = rpms, and current = torque. Voltage * Current = Electrical power, and rpm * torque = mechanical power.
> 
> ...


That's interesting... So really without a controller the motor would get pretty hot under large loading? Huge amount of amps passing through it? And the controller just offsets that by keeping voltage flowing to bring the amps down? So does this mean that the power put out by an electric motor is constant, amps and voltage just vary depending on rpm? I think I'm slowly piecing this together....


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## miscrms (Sep 25, 2013)

Yeah, this is what you sometimes see referred to as "locked rotor amps" on a motor spec. If a bearing seizes or something else prevents the motor from turning it will try to draw a huge amount of current (ie putting out as much torque as it can to try and spin) and quickly cook itself, particularly without a speed controller to limit current but often even with one as the turning of the motor is often part of the cooling design.

If you think of the case of a dyno / acceleration curve with a given motor / controller it would generally look something like this. At 0 rpm, torque and current will be large, but voltage out of the controller will be near zero. Power is essentially 0 as big torque * 0 rpm = 0, and big amps * 0 volts =0. As the motor starts to turn it generally runs in a constant torque / current condition as rpms increase, and the controller increases voltage to maintain current. So power is increasing with increasing rpm/voltage times constant current / torque. Above some critical rpm (a fundamental parameter of the motor / controller combination) torque / current will start to decrease as voltage / rpms continue to increase, resulting in roughly constant or decreasing power output above that rpm. Increasing volts / rpms * decreasing current / torque = roughly constant or even decreasing power. The max speed will either by when you hit the rpm limit of the motor, or if the "back EMF" generated by the motor becomes large compared to the battery voltage. In this case there is a critical speed at which output power (volts - back emf) * current is in equilibrium with the mechanical power required to maintain that speed.

That's still a pretty generic / simplified picture, but a useful one in my mind. And about as far as my motor knowledge goes 

Rob


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## Yabert (Feb 7, 2010)

rangerdev said:


> Motor is a LEESON 098000 C42D17FK1C
> 1/2 hp 18 inch pounds of torque 1750 RPM


With this kind of motor, you can simply connect a 12v battery to the terminals (one to the + and the other to the -) and it will spin.
More volt (battery in serie) and it will spin faster.
Add a controller and you will control his speed.


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## ken will (Dec 19, 2009)

rangerdev said:


> With such a small amount of torque and hp I could never see it pushing a bike, but the guy who gave it to me said it would without hesitation... Am I missing something?


In most of U.S. bikes are limited to 750 watts.
In most of Europe bikes are limited to only 250 watts.

If you mean motorcycle,.. well that is different!


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## rangerdev (Oct 25, 2015)

miscrms said:


> Yeah, this is what you sometimes see referred to as "locked rotor amps" on a motor spec. If a bearing seizes or something else prevents the motor from turning it will try to draw a huge amount of current (ie putting out as much torque as it can to try and spin) and quickly cook itself, particularly without a speed controller to limit current but often even with one as the turning of the motor is often part of the cooling design.
> 
> If you think of the case of a dyno / acceleration curve with a given motor / controller it would generally look something like this. At 0 rpm, torque and current will be large, but voltage out of the controller will be near zero. Power is essentially 0 as big torque * 0 rpm = 0, and big amps * 0 volts =0. As the motor starts to turn it generally runs in a constant torque / current condition as rpms increase, and the controller increases voltage to maintain current. So power is increasing with increasing rpm/voltage times constant current / torque. Above some critical rpm (a fundamental parameter of the motor / controller combination) torque / current will start to decrease as voltage / rpms continue to increase, resulting in roughly constant or decreasing power output above that rpm. Increasing volts / rpms * decreasing current / torque = roughly constant or even decreasing power. The max speed will either by when you hit the rpm limit of the motor, or if the "back EMF" generated by the motor becomes large compared to the battery voltage. In this case there is a critical speed at which output power (volts - back emf) * current is in equilibrium with the mechanical power required to maintain that speed.
> 
> ...



Thanks for an amazing response. This stuff is madly interesting. I have been a petrol head my whole life, and spent a long time drag racing big petrol guzzling motors so I really love to know the small nuances that make these things tick. In the case of electric motors, it seems they pack a lot more power for their size than a similar size gasoline motor, at least as far as torque goes. 

I've done a bit more research, and from what I gather;

Power (in watts) = Motor Efficency * Voltage * Current

This is just a simple Ohms law statement I think. 

So from this I can find that

Current = Power / Voltage * Motor Efficiency

This gets me the amps. Now, a few more things I need to clear up to have a full understanding of this concept. 

Say we are using a 48V pack with a 90V motor. 

I know that I am supplying 12V (or close to it). 

Do I use the power value in Hp stamped on the plate, convert it to watts and then use it in the equation? Is this Hp rating only at no load and max motor rpm? 

Say theoretically that I load a motor so much that it can't spin. I gather that the amps are going to be extremely high, and the voltage will be 0. Is this because no electricity is able to travel across the motor? Its all just pushing wanting to try and turn the motor? I try and think of it like water, but I would think that there would still be pressure (voltage) and no flow (amps). Not sure... I would really appreciate an explanation on why this happens and how it works if possible. 

Another questions, because it seems that amps and voltage just trade off to produce power ( P = V * I ) If I have a 0.5hp motor (372.85 watts) and I load it so it can't move (stalling) Then P Watts = 0 Volts * X Amps, therefore P = 0 . Is this true? 

Is there an equation that relates Volts and Amps to engine speed? SO MANY QUESTIONS! 

I really appreciate all the help. This is really informative and is stuff I have always wanted to learn and implement. Thanks again you guys.


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## galderdi (Nov 17, 2015)

rangerdev said:


> This is just a simple Ohms law statement I think.


Yeah on the surface its all fairly simple. But I get asked by friends and family all the time, What HP or Torque does the motor produce? I usually refain from answering because its totally misleading as they are trying to compare to a petrol or Diesel motor. Comparisons don't work well as the electric motor delivers the power and torque in a totally different, and more useful way. For my application (Electric Autocross car) this is even more important. In an ICE vehicle to produce power and torque its necessary to rev the motor, when you take off you rev and drop the clutch which usually results in unwanted wheel spin. But with the electric equivelent you can harness all the power and torque without over reving and spinning the wheels. So the initial launch is awesome, I am counting on that as my advantage in the autocross events. From your questions it seemed like you were trying to do a comparison back to a petrol motor. I caution against that.


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## miscrms (Sep 25, 2013)

For the water flow example, I think its more like this. When the motor is prevented from spinning, you really just have a short circuit from one side of the battery to the other. Resistance is nearly zero. In other words the diameter of your nozzle becomes large relative to the feed pipe, so even if you have a lot of water flowing through it, you can't really build up any pressure and the water just kind of dribbles out the end of the pipe. Think of a garden hose with no nozzle on it. Flow is fairly high but there's really not any pressure behind it, so when you put your hand in front of it you don't really feel any "push" (power). Now if you put a nozzle on to increase resistance, you have flow and pressure and can feel the power of it pressing against your hand. Electrical power is Volts times Amps, just as the power of the water spray is pressure times flow rate.

The problem with putting a short across a battery is a charged battery never puts out 0V. If its a 12V battery its always putting out ~12V as long as its charged by nature of its chemical reaction. So if current is big, but volts at the load (motor) are 0, this is just another way of saying that the resistance of the load is much smaller than the resistance of the wiring between the cells and the load. Another basic electrical law, Kirchhoff's Voltage Law, states that the sum of all the voltage drops in a circuit (V=I*R) must sum up to the voltage of the source. So if we have Vbat = 12V, then I*R for each resistance in the circuit must sum up to 12V. If the voltage drop at the load / motor is ~0V, then the voltage drop in the wiring (internal and external to the battery) must = ~12V. And even though power delivered to the load (motor) is 0, there is lots of power being dissipated as heat in the wiring. This can quickly cause something to burn out. In a way this is the same as the garden hose case, where its the ratio of the nozzle size to the hose size that allows you to build up pressure at the nozzle rather than just wasting it all along the length of the hose.

You can also think of the controller as a pressure / flow regulator. It allows the pressure and flow rate on one side to be different than the pressure and flow rate on the other. The power out must still be less than or equal to the power in, but you can now decouple the two. So battery volts * battery amps is still >= motor volts * motor amps, but motor volts can be < battery volts, and motor amps can be > battery amps. 

This simplified case really only looks at the true DC case of a locked rotor though so far. A motor, even a simple dc brushed motor is not really a DC device though. As it starts to spin, the commutator starts switching creating a pulsed or AC waveform with a frequency component proportional to rpms. The idea is the same, but now instead of just the DC resistance of the motor winding wire, now you have a complex impedance Z related to the DC resistance R + w (omega=2*Pi*freq) * L, where L is inductance related to the winding and magnetic field. So now V=IZ, where Z becomes larger with increasing rpms causing the current draw from a given voltage to be much lower than the V=IR locked rotor case. As mentioned above, V also becomes more complicated as the voltage applied to the motor is reduced by the back EMF generated by the turning motor.

Calculating the specific torque / power relationship for a particular motor unfortunately gets very complicated very quickly, and is unfortunately mostly beyond me 

The basic idea for a DC motor is that at a given voltage it will free run at a particular rpm with fairly minimal current draw. Power consumption is fairly low, but there is no load on the motor so no work is being done and efficiency is 0. As you apply a load to the motor it will reduce the rpms, and increase current drawn to supply the required torque for the load. Basically the motor is "fighting back" and trying to get back to equilibrium at its free running rpm. The relationship between rpms, torque and current are fairly complex and specific to the motor. This is basically what is plotted in the torque curves for a given motor. The rating you mention with a given HP, RPM and voltage is basically one point on this curve but unfortunately doesn't tell you about the shape of the curve. 

What you probably can say pretty safely is that at 1/2 the rated voltage, the torque curve will be similar but shifted to roughly 1/2 the rpm and consequently 1/2 the power with about the same current draw and torque. This doesn't always apply when going to higher voltage though, as the motor is likely to be more non-linear or saturate beyond its design spec.

Maybe not the answer you were hoping for, but hopfully it helps a bit.

Rob


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## Duncan (Dec 8, 2008)

Hi Ranger
The easy way to understand what is going on with a brushed DC motor

Torque is proportional to current (amps)

Your motor can be thought of as two things 
A simple resistor - low ohms
A generator that develops a voltage dependent on speed and current

With your motor stationary (zero speed) the current is controlled by the internal resistance - which is very low maybe 1/50th of an ohm

So when you put 12v across it there is a surge of current - 600amps (if your battery can supply that much)
This causes a lot of torque and the motor accelerates
If you have jammed the motor so it can't turn then it will continue to take that current until something melts

If the motor is free it will accelerate - once it is spinning it will produce a "Back Voltage" 
The back voltage will effectively reduce the voltage available to drive current through the motor
So your motor may end up spinning at 2000rpm and drawing 10 amps
This would be
12v supply minus 11.5v back voltage = 0.5v which is pushing 10 amps (remember 50 ohms) through the motor
At that stage the 10 amps would be producing enough torque to keep the motor spinning at that rpm (driving against friction and it's built in fan)

Stalled I would be absorbing 12v x 600 amps = 7200watts 
But that power would all be going to heat
Mechanical Power = Torque x Rpm - so at zero rpm zero power is being developed

You controller works by controlling the voltage in order for you to achieve the demanded current

So my 500amp controller would try and deliver 500amps at full throttle
At half throttle it would try and deliver 250 amps


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## rangerdev (Oct 25, 2015)

miscrms said:


> For the water flow example, I think its more like this. When the motor is prevented from spinning, you really just have a short circuit from one side of the battery to the other. Resistance is nearly zero. In other words the diameter of your nozzle becomes large relative to the feed pipe, so even if you have a lot of water flowing through it, you can't really build up any pressure and the water just kind of dribbles out the end of the pipe. Think of a garden hose with no nozzle on it. Flow is fairly high but there's really not any pressure behind it, so when you put your hand in front of it you don't really feel any "push" (power). Now if you put a nozzle on to increase resistance, you have flow and pressure and can feel the power of it pressing against your hand. Electrical power is Volts times Amps, just as the power of the water spray is pressure times flow rate.
> 
> The problem with putting a short across a battery is a charged battery never puts out 0V. If its a 12V battery its always putting out ~12V as long as its charged by nature of its chemical reaction. So if current is big, but volts at the load (motor) are 0, this is just another way of saying that the resistance of the load is much smaller than the resistance of the wiring between the cells and the load. Another basic electrical law, Kirchhoff's Voltage Law, states that the sum of all the voltage drops in a circuit (V=I*R) must sum up to the voltage of the source. So if we have Vbat = 12V, then I*R for each resistance in the circuit must sum up to 12V. If the voltage drop at the load / motor is ~0V, then the voltage drop in the wiring (internal and external to the battery) must = ~12V. And even though power delivered to the load (motor) is 0, there is lots of power being dissipated as heat in the wiring. This can quickly cause something to burn out. In a way this is the same as the garden hose case, where its the ratio of the nozzle size to the hose size that allows you to build up pressure at the nozzle rather than just wasting it all along the length of the hose.
> 
> ...


Rob,

This was great. That really helps understand with the hose example. Thanking of the controller as a pressure flow regulator is also a great thing to keep in mind. 

What I am gathering is that the motor will always try to fight back up to its own, and its batteries limitations. Essentially if the batteries can provide the amps, the motor will take them. So "Technically" a little tiny motor, provided with enough amps could put out the torque to turn whatever it wants. The issue is that it would melt before it could. So (correct me if I'm wrong) when I ask if a motor can provide the torque I need, I need to be more concerned about how many amps my batteries can put out, and if the motor can handle that many amps. 

If this is true, than I think I finally understand. In this case it is similar to a gasoline motor in the respect that if you pressurize air (turbo/supercharger) and pump more fuel into it, you can get the motor to put out as much torque as you want, but eventually something is going to break under the pressure. You would have to either build the motor to take the power, or choose a stronger motor, just as with an electric motor you would have to choose one that could handle the higher amps. 

This is really cool what you guys have here. I'm learning a ton and I really appreciate all the input. 

Looking forward to showing you guys the progress of my project! 

Thanks again.


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## rangerdev (Oct 25, 2015)

Duncan said:


> Hi Ranger
> The easy way to understand what is going on with a brushed DC motor
> 
> Torque is proportional to current (amps)
> ...



Duncan, 

Please see my reply to Rob. Your reply helped me come to these conclusions. 

So basically, if my batteries can put out 600 amps (car batteries) and I am using a 24 volt motor then;

24v * 600 amps = 14400 watts. 

14400 watts = 19.31 hp 

19.31 hp is about the same hp that the 2 stroke motor produces at peak rpm. 

The controller will limit the amps, so really I can expect less than that but its obvious that the motor can accelerate the bike. At this point I would just need to look for a motor that can handle the amps the controller is going to allow through. 

Then as the bike accelerates, the load will decrease which will drop the amp draw and increase voltage production in the motor and everything will kind of balance out. 

Is this all correct thinking? Thanks for the great reply. Much appreciated.


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## Jayls5 (Apr 1, 2012)

You're pretty much correct. There are different types of motors, each with different advantages and disadvantages.

Incredibly dumbed down version of most of the stuff you'll see around here:

In practical application, higher voltage allows higher RPM and the ability to produce more torque under the RPM curve. Amps fed to the motor produce the torque. Both combined is the power.

On brushed motors, over-volting can lead to arcing - which rapidly destroys your commutator and renders your motor dead. People get around this somewhat by rotating the brushes position on the commutator. This move affects motor performance as well. Some motors have built in structures (interpoles) to also allow higher voltages without arcing.

Series wound motors have no magnets. They produce excellent torque with poor speed regulation from varying load (hills). Shunt motors produce less torque but have better speed regulation. Compound wound motors meet in the middle. Most people opt for series wound if they are going brushed. It uses simple "chopper" controllers that are basically rapidly turning on and off the connection from the battery to the motor to regulate current flowing into it.

These controllers are often called "Buck" converters, DC controllers, Series controllers, brushed controllers, etc. It essentially means the same thing.

Advantages: Cheap to control, lots of torque.
Disadvantages: Maintenance of brushes, removal of load (neutral slip and max throttle) leads to motor speeds that rapidly destroy the motor, so some kind of RPM sensing controller cutoff is desirable. Motor RPM capability tends to be on the low side (though this isn't exclusive).

Brushed permanent magnet motors have good torque and can run off the same cheap controllers, but there really aren't any of appropriate size for a full sized car. They're great for compact power on a bike though. They still run into arcing issues with too high voltage, and people still "advance the brushes" on these types to partially get around it. They are naturally speed limiting, unlike series motors.

Then, there are brushless designs. There are ones that use magnets and ones that don't. Both of these types require more expensive controllers and fancy programming. If you're at the point where you're just learning the motor basics, programming one of these controllers is going to be exceptionally difficult unless pre-programmed with the right motor parameters. In lower power systems, the permanent magnet brushless design is pretty efficient and lightweight. As it's scaled up (hundreds of kW power), losses grow larger with the magnet motors vs the "asynchronous" A/C brushless motors.

Brushless designs in general have perks and drawbacks. Generally, torque off the line will not be as great vs old school brushed series motors. They are more difficult and expensive to control. They generally require no maintenance outside of bearings when they go. The RPM capability is usually higher. Some of these designs implement regenerative braking, allowing you to harness some of the braking energy and put it back in your battery (though most people grossly over-estimate how much it helps).

A big series wound is like an old big block carburated V8. Gobs of torque, and the curve falls off early in RPM's. Brushless A/C is like a 4 cyl direct injected turbo, less instant torque but power builds and extends far higher in the RPM range... allowing more aggressive gearing to multiply torque to still get decent starting power.


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## PStechPaul (May 1, 2012)

A useful "rule of thumb" is that a 1 HP electric motor will provide about the same performance as a 3 HP ICE. The gasoline fueled motor is rated at peak HP, which it will attain only at a specific narrow range of RPM and load conditions, and it also has a limited range of torque, which falls off quickly toward zero RPM. An electric motor can produce maximum torque at zero RPM, and maintain that over a very wide range of speed. Also, electric motors can be overloaded to get 2-3 times rated torque (or even more for series wound) for short bursts, and they can be driven at higher than rated voltage for higher speed. Thus they can easily produce the 3x power multiple over a similarly rated ICE.

Another thing to remember is that all components of an electric car or bike circuit have some resistance, although it can be very low. This is sometimes specified in terms of a multiple of rated current, or a percent. A battery rated at 10 amps can provide 10 to 20 times that into a bolted short, so it has an internal impedance of 10% or 5%. So a 12V 10A battery that puts out 100 amps into a short has an internal impedance, or resistance, of 12/100 = 0.12 ohms. External wiring, connectors, switches, fuses, and other components also have finite resistance which becomes significant at the high currents of an EV.


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## Sunking (Aug 10, 2009)

I am with Paul on this one. Rare event, but it is based on facts not ideology.

IMO Brushed DC Motors are pretty much antiques. AC motors are the future of motive power. Knowing that you are a Grease Monkey, you have to clear your head of HP ratings as HP does not mean a whole heck of a lot.

With AC motors they have 2 ratings of Continuous and Peak. The ratio between the two ranges a bit but 1:2 and 1:4 is pretty common. 

What is really unique with AC motors is they can be set up to be Constant Torque or Constant HP over a wide band of RPM's. One motor I am working with delivers peak 88 ft-lbs of torque from 0 to 5000 RPM and red lines at 10,000 rpm. If that were a gas burner would be rated a 100 HP engine. In electric it is 40 HP @ 6500 RPM. 

Key in on that Torque being Flat from 0 to 5000 RPM of 88 ft-pounds peak. That is a steady rate of acceleration. Depending on weight you may not need any transmission other than that of a Differential. That is exactly what I am doing with a 6:1 ratio on 22-inch tires. There are not many gassers that beat my 0-60 mph time. 

So as Paul says depending on RPM/Torque range of an electric motor you can replace 300 HP gasser with a 100 HP motor. If you have an motor with 10,000 working RPM no transmission needed. Think about a motor geared to do 100 MPH @ 10,000 RPM with Constant Flat torque of 200 ft-lbs to 7000 RPM (70 mph), and HP would be Flat from 7000 to 10,000 RPM of 266 HP peak. Continuous HP of such an electric motor would be in the 80 to 90 HP range, and a gasser of around 260 to 300 HP.


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## pchelka (Aug 12, 2014)

Guys thanks for the responses in this thread there are more noobs lurking around and reading from the shadows than you think 

And this usefull for getting your head around EV motors


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