# How many volts to draw high amps? (long)



## Coley (Jul 26, 2007)

Quote:
"The question:
With a lightweight direct drive car as the load, how many volts are required to get the motors to draw high amps all the way through high rpms?"

The question isn't how many volts, as the amp draw will be larger with the 
acceleration rate, no matter how many volts, at full acceleration.
Higher voltage will reduce the amp draw as in 144 volt over 72 volt.

For those of you with amp gauges in your car and 400+amp controllers, how many rpms does your motor achieve in a 4th gear full throttle run before the amps fall below max? How quickly do they fall off? What voltage are you running as well?"

Mine pulls 200 amps @ full throttle, until 3000 rpm is hit. Then 140 amps to
cruise on flat road. I run 72 volts. 500 amp controller.


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## Batterypoweredtoad (Feb 5, 2008)

Coley said:


> Quote:
> "The question:
> With a lightweight direct drive car as the load, how many volts are required to get the motors to draw high amps all the way through high rpms?"
> 
> ...


Assuming you want the maximum acceleration and are pushing full pack voltage, and that you have more amperage available than you would ever need, wouldnt the only thing that increases acceleration be pushing higher volts (increased pack voltage)?



Coley said:


> Higher voltage will reduce the amp draw as in 144 volt over 72 volt.


To achieve the same watts at the same rpms with twice the voltage and half the amps wouldnt you require a motor designed with 1/2 the rpm constant (rpms/V)?



Coley said:


> For those of you with amp gauges in your car and 400+amp controllers, how many rpms does your motor achieve in a 4th gear full throttle run before the amps fall below max? How quickly do they fall off? What voltage are you running as well?"
> 
> Mine pulls 200 amps @ full throttle, until 3000 rpm is hit. Then 140 amps to
> cruise on flat road. I run 72 volts. 500 amp controller.


Thanks for the info. What speed is your car capable of at 72V? (Im considering using 2 72V controllers for around town use to be a little nicer to the battery pack and wallet)


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## Coley (Jul 26, 2007)

It will hit 50 mph on flat road...


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## MrCrabs (Mar 7, 2008)

Batterypoweredtoad said:


> With a lightweight direct drive car as the load, how many volts are required to get the motors to draw high amps all the way through high rpms?
> 
> 
> Thanks for wading through a long post!


Answer: As many volts as your controller can handle.

If you pull 1000 amps or more from your batteries, the voltage will sag quite a bit, and having a higher voltage pack will let you keep the amps high.


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## Batterypoweredtoad (Feb 5, 2008)

If I can chose the voltage of my controller to be anything I want, how many volts do you think it would take to draw 1000A all the way through motor redline? (110mph at 7k rpm)

Is it possible without going to a 300volt system?


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## MrCrabs (Mar 7, 2008)

You can't get a motor to pull 1000A all the way up to redline. As the motor approaches max rpm, the torque produced will drop off, and it will require less current.

I would guess-timate ~200 volt system could handle it...

Here is an old EVDL thread on voltage sag.
http://www.crest.org/discussion/ev/199906/threads.html#00840
http://www.crest.org/discussion/ev/199906/msg00827.html
You could also try and estimate it based on the internal resistance of the batteries and battery connections, and the current draw.


If your looking to build a racer, check out these 2 Mazda RX-7s, and the White Zombie.

http://www.evalbum.com/120
http://www.evalbum.com/047.html
http://www.plasmaboyracing.com/whitezombie.php
http://www.nedra.com/record_holders.html


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## major (Apr 4, 2008)

Batterypoweredtoad said:


> If I can chose the voltage of my controller to be anything I want, how many volts do you think it would take to draw 1000A all the way through motor redline? (110mph at 7k rpm)
> 
> Is it possible without going to a 300volt system?


Hi toad,

About the only motor curve around going to 1000 amps is one for FB1-4001, 9.1 inch ADC. It is at 96 volts - .03I. At a 1000 amps, about 750 RPM. So to get 7000 RPM at 1000 amps, you'd need 616 volts at the motor, or about 1000 volt battery.

But not a good place to run the motor, even if it could survive. At that point, you'd see 326 hp or 243 kW output for 616 kW input to the motor. That is 39.4 percent efficient.

Good luck,

major


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## Da_Fish (Apr 28, 2008)

Yeah but think of the burnouts!!!! The car cant move because the battery weight and it still has enough power to burn the rubber off the wheels hehehe


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## Batterypoweredtoad (Feb 5, 2008)

major said:


> Hi toad,
> 
> About the only motor curve around going to 1000 amps is one for FB1-4001, 9.1 inch ADC. It is at 96 volts - .03I. At a 1000 amps, about 750 RPM. So to get 7000 RPM at 1000 amps, you'd need 616 volts at the motor, or about 1000 volt battery.
> 
> ...


Thats cool information. What is the ".03I"? Id be trying to push around 500amps each to 2-7inch motors instead of through one bigger one. Do you have any information or links to the torque production of smaller motors and how you calculated those figures? I chose 500amps each because while interpreting one of the graphs I have it looked like a 6.7" motor reached peak torque between 500-700amps.


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## Batterypoweredtoad (Feb 5, 2008)

MrCrabs said:


> You can't get a motor to pull 1000A all the way up to redline. As the motor approaches max rpm, the torque produced will drop off, and it will require less current.
> 
> I would guess-timate ~200 volt system could handle it...
> 
> ...


Thanks for the links, I will read through them this evening. I know theres no way I can match the performance of any of those, but they are inspiring.


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## major (Apr 4, 2008)

Batterypoweredtoad said:


> Thats cool information. What is the ".03I"? Id be trying to push around 500amps each to 2-7inch motors instead of through one bigger one. Do you have any information or links to the torque production of smaller motors and how you calculated those figures? I chose 500amps each because while interpreting one of the graphs I have it looked like a 6.7" motor reached peak torque between 500-700amps.


Hey toad,

The "-.03I" is the battery droop. So the graph for the motor shows a voltage at the motor terminals which compensates for battery droop. In this case, V=96-.03I, or 96 volts open circuit voltage at the battery minus a voltage drop due to internal resistance of 0.03 Ohms times the current (I). So at 1000 amps, 96-.03I = 66 volts at the motor.

You have to get motor performance curve for the particular motor you have. Or find such on the web. I'll help you if you can get that. I don't have it. And, peak torque (for series motors) is at peak amps. Usually controller limited.

Regards,

major


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## Batterypoweredtoad (Feb 5, 2008)

Lol-at myself! Looks like im going to ask 15 questions in one thread.

If a motors peak torque is at peak amps, what determines a motors peak amps? I know you can limit them with the controller, but at what amps do you saturate an average 7" motor? One of the graphs I found for torque curves shows one model 6.7" rising up to about 150n/m at 500A then going flat over the next 100A, while the curve for another 6.7" rises to over 200n/m at about 700A where the plot stops. There is no information on what voltage was used for the tests. The 2-6.7" motors plotted were the ADC 6.7 X91 and L91. Is that flattening of the curve indicating saturation of the X91 meaning no more amps will help?


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## major (Apr 4, 2008)

Batterypoweredtoad said:


> Lol-at myself! Looks like im going to ask 15 questions in one thread.
> 
> If a motors peak torque is at peak amps, what determines a motors peak amps?


Hi toad,

Resistance and voltage determine peak amps. And torque is determined by the product of current and flux. So, even if saturated (meaning no more flux to be had), increasing current will increase torque.

Got to go, I'll check back tomorrow,

major


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## Dennis (Feb 25, 2008)

> Ok, im just getting my mind wrapped around how an electric motor works. My basic understanding is that Volts determine how many rpms a motor wants to turn. My understanding of AMPs is that the motor pulls amps in response to a load resisting the motors ability to achieve the volt driven rpms.
> 
> With that said:
> -Im building a stripped miata (2000lbs final weight) for short racing. (Drag racing, maybe autocross)
> ...


The motor torque curve for a series wound motor on manufacturers websites is the condition when the motor is hooked up to fixed voltage supply with no current limit controller and the load applied to the motor's shaft is increased in increments to near stall.

In a vehicle though with the series-wound motor hooked up to a controller with current limit, and with the pot at the full throttle position with the vehicle at a dead stop, the torque curve will look nothing like what the manufacturers show for the motor, but instead the torque curve will be horizontal if graphed on a RPM versus torque graph from zero RPM to the cutoff RPM point in where the torque curve falls off with a similar shape to what the series-wound motor torque would be when not hooked up to current limited controller. Coley's is a fine example of the current limit of his controller keeping the current at 200 amps from 0 RPM to 3,000 RPM then it starts to fall off when he pegs the accelerator pedal to the floor. If he ran higher voltage like 96 volts then he could extended his current value of 200 amps into higher RPMS before it falls off. Since torque in a DC motor is proportional to current then his torque curve is also constant in those RPM ranges.

Here is a visual of what I am talking about:










So you want to have high voltage so that the torque curve can be extended as far as it can go in the RPM range before it starts to fall off. Once torque falls off your acceleration will start to decline which you want to occur as late as possible in the 1/4 mile drag race. However, if you set the current limit of the motor controller too high then you will have a very short horizontal LARGE value torque curve but it will fall off too quickly resulting in you losing the 1/4 mile race. So you have to balance your controller current limit and the how many volts you plan to use. I would also use dual motors wired in series to each other so you can gain lots of torque with small amounts of current increase versus using just one motor.


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## major (Apr 4, 2008)

Dennis said:


> Here is a visual of what I am talking about:
> 
> 
> 
> ...


Hi Dennis,

How do figure an increase of 33% in voltage will get an 83% increase in RPM?

Regards,

major


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## Dennis (Feb 25, 2008)

I cannot find any generated voltage per 1000 RPM data of series wound motors when spinning with a battery connected to it, nor can did I find the total resistance of a series wound motor from the terminal's measurement point on the spec sheets.

So I just guessed that since Coley is only using 200 amps that the differential voltage (Vsupply - Vgen) does not have to be very high to maintain 200 amps through the motor. He did manage to hold 200 amps up to 3000 RPM with 72 volts so it seems possible with 96 volts that his torque curve could extend to that RPM point of 5500 RPM or maybe less. I mean it is a series wound motor which is basically a dead short ( a few milliohms of resistance!) which cannot be easily measured with a standard ohm meter, so getting large amounts of current to flow is not that hard. Now the back emf voltage may increase non linearly which be the case, but I am not for certain.


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## mattW (Sep 14, 2007)

Your peak current is determined by the heat your motor can withstand, As you increase the current the torque increases linearly until you start melting wires... For example the motor I'm using for my motorbike produces 0.095 Ft-LBS per amp with a 400A limit giving 38Ft-lbs peak. Your peak voltage is determined by the brushes which commutate the motor which will arc like a welder at too high a voltage. Also since high voltage means higher rpm you have to be careful not to rip your motor apart. Peak Rpm is also linearly determined by voltage, my motor gives 72 RPMs per volt which will give me 5184rpm (theoretical) peak.


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## Dennis (Feb 25, 2008)

> Your peak current is determined by the heat your motor can withstand, *As you increase the current the torque increases linearly until you start melting wires.*.. For example the motor I'm using for my motorbike produces 0.095 Ft-LBS per amp with a 400A limit giving 38Ft-lbs peak. Your peak voltage is determined by the brushes which commutate the motor which will arc like a welder at too high a voltage. *Also since high voltage means higher rpm you have to be careful not to rip your motor apart. Peak Rpm is also linearly determined by voltage, my motor gives 72 RPMs per volt which will give me 5184rpm (theoretical) peak.*


Series wound motor torque is not linearly proportional to a linear increase in current as you have stated, but instead the torque is approximately equal to the square of the current. The actual formula will include other terms that account for losses and magnetic field saturation so the actual torque curve will not follow the ideal one. 

The usefulness of the series-wound motor torque curve for drag racing applications is the user can set the current limit on their controller to the value that corresponds to ONE POINT on the torque curve of the series wound motor so that if the user pegs the throttle, then the torque will be limited to that point and therefore will start to become horizontal because of the feedback loop of the current sensing device formed with a comparator circuit, controlling the voltage chopper circuit of the motor controller.

Voltage increase does result in the RPMs increasing in a series wound motor, but it is not linear either. An unloaded series wound motor will speed up till the armature windings come apart under no load. Likewise, a linear increase of load placed upon the motor's shaft will not result in a linear decrease in motor RPM, but instead the RPMs will decay exponentially.


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## mattW (Sep 14, 2007)

Sorry I forgot my motor was PM, I understand how the relationship is the square of current since both sets of coils would increase their EM force not just the one as in a PM motor... Cheers


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## Mastiff (Jan 11, 2008)

If your looking for some serious Torque you should check out the AC drive systems from Azure Dynamics:
http://azuredynamics.com/productsComponents.htm


Azure Dynamic's AC90 motor + DMOC645 controller drive system has a peak Torque of 665 Newton Meters or 490 Foot Pounds at about 1600 RPM:
http://azuredynamics.com/pdf/AC90_DMOC645 Product Sheet.pdf

They also build the AC55 motor and DMOC445 controller:
http://azuredynamics.com/pdf/AC55_DMOC445 Product Sheet.pdf
This has a Maximum Torque of 280 Newton Meters or 206 Foot Pounds at 2000 RPM.

Both of these Torque Charts show a totally Flat Torque Curve until the drop-off RPM limit.

The Azure Dynamics AC55 system is what I'm considering for a performance conversion of my own. (assuming I somehow get the money)


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## Wirecutter (Jul 26, 2007)

Mastiff said:


> If your looking for some serious Torque you should check out the AC drive systems from Azure Dynamics:
> 
> [snip]
> 
> The Azure Dynamics AC55 system is what I'm considering for a performance conversion of my own. (assuming I somehow get the money)


 Ok, I'll bite. Have you gotten a price for any of the Azure systems?

-Mark


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## Wirecutter (Jul 26, 2007)

Here's a way to think about volts and amps that might make it easier to get your head around:

Voltage is a measure of "electric pressure" just like psi can be a measure of water pressure.

Amps are a measure of the speed of the electricity down the wire (electron drift velocity), and for water it's how fast the water is moving down the pipe.

For a given wire/pipe size, the higher the pressure, the faster you're going to get the water to move down the wire/pipe. That's why power transmission lines use very high voltages. I think I heard that most (or a lot of) commercial power is generated at 4000-some volts, but then it's stepped up to hundreds of kilovolts for transmission. Then of course it's stepped down for end use.

Watts are volts x amps, or for water, gallons per minute.
Total energy (gallons) would be kilowatt-hours.

In electric motors, the voltage generally determines the speed, and the amps determine the torque. The motor needs a load - something to push against - to produce torque. Of course, if the motor doesn't have the load, it just spins faster and torque doesn't rise appreciably.

The amps that make the torque do so by generating magnetic fields. Think of the magnetic field as the breeze created by all those electrons rushing by. Get those puppies moving fast enough, and the breeze can become a pretty stiff gale. Anyway, the magnetic field is where all that electric energy goes mechanical. Ok, heat, too. The breeze/gale/torrent can also create heat. (Heck, the SR-71 got pretty hot traveling at mach 3...) Heat is a motor killer. So is the centrifugal force of running at a zillion RPM. 

Beyond that, ditto what Dennis, MattW, toad, major, and MrCrabs said.

Oh, and I have verified Coley's assertion that 72v "...will hit 50 mph on flat road..." I hit 55 everytime I take out my gokart. I drive on some hilly roads, so I need the torque, but if I stuck to flat ground, I could gear it taller for higher top speed.

The funny thing is, I've often considered tweaking up the kart for more speed/power. It seems like everytime I do that, I soon find myself in a situation where the kart is actually overpowered. Last time I was out, a light mist began to fall. With slick tires and even the smallest amount of moisture, the driving got pretty sporty.  I put it away as soon as I could, but that was only because parts of the kart aren't waterproof. Gawd it was fun!

-Mark


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## Batterypoweredtoad (Feb 5, 2008)

What nobody mentions thats something really hard to wrap your head around when youre first learning is that you cant "Push Amps" to a motor. You have to get the motor to "Suck Amps" (now I know how that car got that name). The problem is that a lower voltage controller wont try to spin the motor fast enough to make it draw enough amperage to accelerate rapidly beyond its lower rpm range.

Next question in my neverending series of questions on one thread:
-If you assume a 72V 500A controller on a full throttle run will hold 500A through 3000 rpm before falling off, will a 144V 500A controller be any faster from 0-3000 rpm on the same motor?

Related to the same question:
-Will the 144V 500A controller hold peak torque to 6000rpm (or somewhere close)?


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## Dennis (Feb 25, 2008)

> Next question in my neverending series of questions on one thread:
> -If you assume a 72V 500A controller on a full throttle run will hold 500A through 3000 rpm before falling off, will a 144V 500A controller be any faster from 0-3000 rpm on the same motor?


In the real world I do not see 500 amps being held to 3000 RPM with just 72 volts, but you did say lets assume. So in that case if 500 amps could be held to 3000 RPM with just 72 volts, then 144 volts will extend the current held to a higher RPM range and since current is what gives the motor torque, it would also be constant up to that new point. The motor would be faster with the higher voltage as well. 

If you graph the horespower of a constant torque curve you will notice that it increases linearly. So not only did you extend your constant torque-speed range, but you also increased horespower.



> Related to the same question:
> -Will the 144V 500A controller hold peak torque to 6000rpm (or somewhere close)?


I could be very wrong about what I am about to say since someone on here probably has held 500 amps to 6,000 RPM under full "throttle" conditions with just 144 volts, but I believe at 500 amps that such a scenario would not happen. The series-wound motor would have to be some type that is terrible at developing decent back EMF. I suppose you could remove a few turns of the series field windings which reduces the magnetic field strength of the field poles, but that would lower the torque value for the current of 500 amps versus leaving it stock and the motor will run hotter.


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## Batterypoweredtoad (Feb 5, 2008)

Dennis said:


> In the real world I do not see 500 amps being held to 3000 RPM with just 72 volts, but you did say lets assume. So in that case if 500 amps could be held to 3000 RPM with just 72 volts, then 144 volts will extend the current held to a higher RPM range and since current is what gives the motor torque, it would also be constant up to that new point. The motor would be faster with the higher voltage as well.


 Ahah! Now its getting close to the answer for the question I started the thread with! Do you or does anyone else have a guesstimate of what volts will maintain high amps to high rpm's?(lets say 500a for this question)



Dennis said:


> If you graph the horespower of a constant torque curve you will notice that it increases linearly. So not only did you extend your constant torque-speed range, but you also increased horespower.


The extension of constant torque into the higher rpms to achieve higher hp is exactly my goal! Im fairly convinced my motors will handle 500A for brief bouts which will give me great torque. The problem is that my horsepower will suck if I dont actually increase the rpm range that torque is carried to. Im just looking for some guidance on how many volts are needed to actually do it. So far its looking like the more volts the better up to flashover.


Dennis said:


> I could be very wrong about what I am about to say since someone on here probably has held 500 amps to 6,000 RPM under full "throttle" conditions with just 144 volts, but I believe at 500 amps that such a scenario would not happen. The series-wound motor would have to be some type that is terrible at developing decent back EMF. I suppose you could remove a few turns of the series field windings which reduces the magnetic field strength of the field poles, but that would lower the torque value for the current of 500 amps versus leaving it stock and the motor will run hotter.


Do you have a guess as to what rpms the torque would start to fall off with 144 volts? (full throttle 4th gear run, 120lbs of 7" motors, 2k lb car, battery amps not the limiting factor)


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## major (Apr 4, 2008)

Dennis said:


> So I just guessed that since Coley is only using 200 amps that the differential voltage (Vsupply - Vgen) does not have to be very high to maintain 200 amps through the motor. He did manage to hold 200 amps up to 3000 RPM with 72 volts so it seems possible with 96 volts that his torque curve could extend to that RPM point of 5500 RPM or maybe less.


Hi Dennis,

I don't know what Coley is doing, but I know motors pretty well. So, let's say you have the series motor at 72 volts, 200 amps and 3000 RPM. You guess that if you upped it to 96 volts at 200 amps, you'd see 5500 RPM. Let's see how that plays out.

Assume the motor is 80% efficient at 200 amps, 72 volts. Then at 72V and 200 amps, power in is 14.4 kW. At 80%, power out is 11.52 kW, or 15.46 hp. At 3000 RPM, it is 27.1 lb.ft.

When you increase the voltage to 96, the torque stays the same at 200 amps, 27.1 lb.ft. So at 5500 RPM, you now have 28.35 hp, or 21.1 kW output power. 96 volts times 200 amps is 19.2 kW input power. So efficiency would be Po/Pi, or 110%. 

It is fairly accurate to proportion the speed by the change in voltage on series motors unless you're at very high currents or low speeds. So in this case it would be safe to say that a 33% increase in voltage would cause a 33% increase in RPM. The error in the estimated RPM will probably be on the order of a few %.

As you had mentioned, the series resistance for the motor is very low, milliOhms, so at 200 amps, Eg will be within a few volts of the applied voltage, 72 or 96. The RPM is proportional to Eg. Using the applied voltage is pretty close in these types of examples.

Regards,

major


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## major (Apr 4, 2008)

Batterypoweredtoad said:


> Ahah! Now its getting close to the answer for the question I started the thread with! Do you or does anyone else have a guesstimate of what volts will maintain high amps to high rpm's?(lets say 500a for this question)


Hi toad,

If you can follow my posts on this thread, you will see how to do that for yourself. First you need the motor performance curve for the motor you want to use. Find the torque for the current you intend to use as the limit. Draw a vertical line. See what the RPM is and what the applied motor voltage (including sag) is for that torque. Use that RPM and your desired RPM to form a ratio and multiply that by the actual voltage to find your new voltage.

Simple as that.

major


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## Batterypoweredtoad (Feb 5, 2008)

major said:


> Hi toad,
> 
> If you can follow my posts on this thread, you will see how to do that for yourself. First you need the motor performance curve for the motor you want to use. Find the torque for the current you intend to use as the limit. Draw a vertical line. See what the RPM is and what the applied motor voltage (including sag) is for that torque. Use that RPM and your desired RPM to form a ratio and multiply that by the actual voltage to find your new voltage.
> 
> ...


Ah but until your post two up, no one had definitively said that it is a fairly linear relationship between volts and rpms and the torque. Now you have and I can use that information accordingly. I sincerely thank you very much for the info, im off to dig up more information on the motors I have or at least some similar ones.


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## Dennis (Feb 25, 2008)

> Assume the motor is 80% efficient at 200 amps, 72 volts. Then at 72V and 200 amps, power in is 14.4 kW. At 80%, power out is 11.52 kW, or 15.46 hp. At 3000 RPM, it is 27.1 lb.ft.
> 
> When you increase the voltage to 96, the torque stays the same at 200 amps, 27.1 lb.ft. So at 5500 RPM, you now have 28.35 hp, or 21.1 kW output power. 96 volts times 200 amps is 19.2 kW input power. So efficiency would be Po/Pi, or 110%.
> 
> ...


Wow, so the motor is 80% efficient which you made up to justify your argument against *MY GUESS* of 5500 RPM at 96 volts. Way to go! I suppose it could not be 50%, 57%, 61%, etc.. efficient? Lets not forget that for different horespower figures the efficiency is not the same.



> It is fairly accurate to proportion the speed by the change in voltage on series motors unless you're at very high currents or low speeds. So in this case it would be safe to say that a 33% increase in voltage would cause a 33% increase in RPM. The error in the estimated RPM will probably be on the order of a few %.


If you apply a load on a series wound motor or have it unloaded where it speeds up to destruction, then the accuracy is terrible. I hope you realize a series wound motor has extremely poor speed regulation and therefore this is not the best way to to come up with a hard data figure that is meaningful.


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## major (Apr 4, 2008)

Dennis said:


> Wow, so the motor is 80% efficient which you made up to justify your argument against *MY GUESS* of 5500 RPM at 96 volts. Way to go! I suppose it could not be 50%, 57%, 61%, etc.. efficient? Lets not forget that for different horespower figures the efficiency is not the same.
> 
> If you apply a load on a series wound motor or have it unloaded where it speeds up to destruction, then the accuracy is terrible. I hope you realize a series wound motor has extremely poor speed regulation and therefore this is not the best way to to come up with a hard data figure that is meaningful.


Hi Dennis,

We are talking about the example you gave in your post #15. A 72 volt series motor at 3000 RPM, and 200 amps (stated later). 80% efficient is a pretty good estimate for the type of motors used in EVs. However, one could do a similar calculation using different efficiency.

And we are talking about your plot from post #15 which shows the same load for both 72 and 96 volts (3000 and 5500 RPM) which you said was 200 amps. So the context of using a voltage ratio to estimate RPM is at the same load.

Hope that clarifies it.

major


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## major (Apr 4, 2008)

major said:


> And we are talking about your plot from post #15 which shows the same load for both 72 and 96 volts (3000 and 5500 RPM) which you said was 200 amps. So the context of using a voltage ratio to estimate RPM is at the same load.


Hi Dennis,

Check out this curve for ADC 8 inch motor at 72 and 96 volts. 

http://www.evparts.com/img/mt2116torque.PDF 

200 amps is 24 lb.ft. Draw a vertical line at 24 lb.ft. It intersects the 72 volt curve at 3600 RPM and the 96 volt curve at 4650 RPM. That is a ratio of 1.29.

Calculate the voltage including the sag for each. V=75v-.03(200A)= 69 volts and V=96-.03(200A)=90 volts. 90/69 = 1.30. That is within a few percent of 1.29.

So using the voltage ratio to estimate the RPM difference for a given load is valid in most cases. Like I said before, at the extremes of current or speed, accuracy declines.

Regards,

major


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## ngrimm (Oct 19, 2007)

So what is the relationship between amps, volts and torque in a nearly stalled condition in a series wound motor? For instance if the amps are limited by the controller and you double the voltage, wouldn't the torque nearly double? I realize that emf limits torque at higher rpms. The title of this topic being volts to draw high amps doesn't make sense to me.


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## 3dplane (Feb 27, 2008)

Hi Norm!
The original posters question basically is how many volts would be needed to ensure that the torque(and amps) does not "fall off" during a drag race on a direct drive setup where the rpms go high. Barna


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## major (Apr 4, 2008)

ngrimm said:


> So what is the relationship between amps, volts and torque in a nearly stalled condition in a series wound motor? For instance if the amps are limited by the controller and you double the voltage, wouldn't the torque nearly double? I realize that emf limits torque at higher rpms. The title of this topic being volts to draw high amps doesn't make sense to me.


Hi ngrimm,

If the controller has a motor current limit, that sets the maximum torque from the motor, regardless of the voltage. Torque is the product of flux and current. Voltage does not enter the torque equation. The generated voltage (Eg) is proportional to RPM, so is higher at high RPM. The current is the [applied voltage (Vm) minus Eg] divided by the equivalent resistance. So at higher speeds, Eg is higher, current is lower and torque is lower. Torque (and current) is highest at stall which is zero RPM and zero Eg.

Hope that helps.

major


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## Dennis (Feb 25, 2008)

> Hi Dennis,
> 
> Check out this curve for ADC 8 inch motor at 72 and 96 volts.
> 
> ...


What you did here was apply the fact that the shape of the speed torque curve will be the same, but at higher voltage the curve will shift up some X vertical units with some RPM value at a known torque and dividing it by the desired RPM value gives you the multiplier that is used to give you the X amount of units the new curve will shift up vertical for the new voltage when the multiplier is multiplied by the original voltage....

That is true. What is not true is you saying that the speed torque curve of a series wound motor will follow a linear relationship that is in the form Y=mx+b which is NOT true. At least that is what I gathered from what you were saying or maybe it was a miscommunication.




> Hi ngrimm,
> 
> If the controller has a motor current limit, that sets the maximum torque from the motor, regardless of the voltage. Torque is the product of flux and current. Voltage does not enter the torque equation. The generated voltage (Eg) is proportional to RPM, so is higher at high RPM. The current is the [applied voltage (Vm) minus Eg] divided by the equivalent resistance. *So at higher speeds, Eg is higher, current is lower and torque is lower. Torque (and current) is highest at stall which is zero RPM and zero Eg.*
> 
> ...


Just so I am not misunderstanding you here. Are a telling him this under *non-controller conditions*? In that case it is true. If you mean under *full throttle controller, current limited conditions* then what you are saying is false. The torque curve will become constant from zero RPM to the cutoff point where the torque will start to decline.


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## major (Apr 4, 2008)

Dennis said:


> What is not true is you saying that the speed torque curve of a series wound motor will follow a linear relationship that is in the form Y=mx+b which is NOT true. At least that is what I gathered from what you were saying or maybe it was a miscommunication.
> .


Hi Dennis,

Yes, I never said that. We all know the shape of the speed torque curve for series motors at a fixed voltage. The whole discussion was to the change with regard to changing voltage.

Regards,

major


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## major (Apr 4, 2008)

Dennis said:


> Just so I am not misunderstanding you here. Are a telling him this under *non-controller conditions*? In that case it is true. If you mean under *full throttle controller, current limited conditions* then what you are saying is false. The torque curve will become constant from zero RPM to the cutoff point where the torque will start to decline.


Hi Dennis,

I think it is true in both conditions including full throttle with a controller. If the controller maintains that current limit (and maximum torque) up to the cutoff point (which I gather you mean 100% duty cycle), it was still a max at stall, wasn't it? Torque never gets any larger.

Regards,

major


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## Dennis (Feb 25, 2008)

> Hi Dennis,
> 
> I think it is true in both conditions including full throttle with a controller.* If the controller maintains that current limit (and maximum torque) up to the cutoff point (which I gather you mean 100% duty cycle), it was still a max at stall, wasn't it? Torque never gets any larger.*
> 
> ...


What I meant was did you mean the torque curve would look like what is shown on the manufacturers spec sheet will be the same as what it would look like when the motor is connected to a current limited controller? That was how I took what you said with this:



> *So at higher speeds, Eg is higher, current is lower and torque is lower. Torque (and current) is highest at stall which is zero RPM and zero Eg.*


I want to make sure you meant this statement for fixed voltage supply, *non-current limited* starting from a standstill of the vehicle.


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## ngrimm (Oct 19, 2007)

Another dumb question. Let's say you were climbing a hill at let's say 50 volts at maximum rpm it that it would go since your current was limited to say 200 amps and you suddenly doubled the voltage to 100 without increasing the amps. Wouldn't the motor speed up? If so, how could it do so if the voltage doesn't add any torque? Norm


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## ww321q (Mar 28, 2008)

ngrimm I'm thinking a little like you . 400 amps from a start at 12 volt will give you a certain amount of torque . At 120 volts from a start I would think would be less then 400 amps for the same torque . J.W.


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## ngrimm (Oct 19, 2007)

ww321q said:


> ngrimm I'm thinking a little like you . 400 amps from a start at 12 volt will give you a certain amount of torque . At 120 volts from a start I would think would be less then 400 amps for the same torque . J.W.


I might be willing to try and hold the wrench on the shaft at 12 volts but I would prefer to give someone else a turn for the 120 volt test.


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## ww321q (Mar 28, 2008)

ngrimm said:


> I might be willing to try and hold the wrench on the shaft at 12 volts but I would prefer to give someone else a turn for the 120 volt test.


hahahahahahahahahahahahahahahahahaha!!!! thats for sure ! I about got twisted off the ladder more then once with an 18v drill when I was putting my shop up . J.W.


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## Dennis (Feb 25, 2008)

ngrimm said:


> Another dumb question. Let's say you were climbing a hill at let's say 50 volts at maximum rpm it that it would go since your current was limited to say 200 amps and you suddenly doubled the voltage to 100 without increasing the amps. Wouldn't the motor speed up? If so, how could it do so if the voltage doesn't add any torque? Norm


Doubling the voltage will increase the torque range to further RPM's so that means now you can meet the torque demand at higher RPM's. It should be no surprise that horespower is what the major player here is.

Now if you stall the motor from the hill requiring more torque than what is being produced at 200 amps, then doubling the voltage will do nothing to get you out of the stall. The controller puts in so many dead time pulses that increasing the voltage will just make the controller put in even more dead pulses or "misses" to keep the voltage average such that it's just enough to allow 200 amps to pass through the motor.


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## major (Apr 4, 2008)

Dennis said:


> I want to make sure you meant this statement for fixed voltage supply, *non-current limited* starting from a standstill of the vehicle.


 
Hi Dennis,

Sure, when you're in current limit, there is a fixed current to the motor and the motor voltage is adjusted to maintain that current as the RPM and Eg rise. Therefore the non-controller speed torque motor performance curve is modified, chopped off, so to speak, at that current or torque value. But the torque is still maximum at stall, just that it stays at that maximum until the controller goes to 100%.

O.K. So that particular statement I made was for non-controller case. You happy? But the statement had the qualifier that the current would fall as Eg increased, so it should have been obvious that it was not a current limit case.

major


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## ngrimm (Oct 19, 2007)

Thanks Dennis. Your final sentence cleared it up for me. You are only talking about the volts/amps characteristics when a motor is powered by a controller which isn't the same as just connecting to a battery. Norm


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## Dennis (Feb 25, 2008)

> Thanks Dennis. Your final sentence cleared it up for me. You are only talking about the volts/amps characteristics when a motor is powered by a controller which isn't the same as just connecting to a battery. Norm


I do not know of a reason why anyone would want to run a series-wound motor with no current limiting at higher voltages. The dangers in not doing so can lead to catastrophic failure of the motor or battery terminals exploding off the battery, spewing acid everywhere or anything else failing that cannot handle the high current surge. Hopefully before that occurs the fuse will blow. 




> Hi Dennis,
> 
> Sure, when you're in current limit, there is a fixed current to the motor and the motor voltage is adjusted to maintain that current as the RPM and Eg rise. Therefore the non-controller speed torque motor performance curve is modified, chopped off, so to speak, at that current or torque value. But the torque is still maximum at stall, just that it stays at that maximum until the controller goes to 100%.
> 
> ...



What through me off was you used the word peak which to me means one point that is the highest value of all other points which is true at stall for a series wound motor with no current limit. But with a torque value of the same numerical quantity from zero RPM to the cutoff RPM, I do not call it a peak at zero RPM as their are infinite number of points that correspond to the same value from zero RPM to the cutoff point.

Glad you cleared up what you were saying though..


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## Bugzuki (Jan 15, 2008)

Let me see if I can reword what has been said to see if I understand. Please correct me if I am wrong.

In the example a 200 amp current limiting controller is used. So for ease lets say you have a 100 volt battery. 

You are driving up a hill at half throttle 50v, The controller is currently limiting the amperage to 200 amps which could mean that the motor wants to turn faster. In effect you are not getting the full effect of the 50v. Then if you go full throttle 100v the controller would still regulate the voltage the same as it was to keep the current at 200amps.

If you had a pot you could adjust to raise the current limit to say 400amps, then the controller would be able to make use of more of the voltage to gain power.

This is all due to the effect of the reverse current created in the system that limits the current draw according to voltage applied and current motor RPM.


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## ww321q (Mar 28, 2008)

Dennis said:


> I do not know of a reason why anyone would want to run a series-wound motor with no current limiting at higher voltages. The dangers in not doing so can lead to catastrophic failure of the motor or battery terminals exploding off the battery, spewing acid everywhere or anything else failing that cannot handle the high current surge. Hopefully before that occurs the fuse will blow.


I guess if a woman was driving and just "start and stomp" on the throttle . A lot of people invest in these things called "gauges" and they actually look at them . I have a gauge on my motorcycle . It's called a tachometer ! If I don't use it and shift before 13000rpm I would have a "catastrophic failure of the motor" I encounter this hundreds of times a day when I ride . Maybe I should invest in rev limiter . J.W. 


Ps don't get mad I'm just having some fun with you OK?


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## Dennis (Feb 25, 2008)

The catastrophic failure I am referring to is the current will be so high that plasma fire balls will shoot out of the vent holes at the brush rigging end. If you are lucky then only the brushes burst into thousands of pieces and the rigging melted. If not then maybe the actual commutator melted too and the enamel coating on the windings is charcoal black. This is why current limit is so important in large horsepower DC motors.

Here is a nice video of a motor with a bad hair day. They even add the sad music to give that special effect. LOL. I bet the company was crying when they saw the repair bill for this motor.

http://www.youtube.com/watch?v=-ZO__ebsSpE




> Let me see if I can reword what has been said to see if I understand. Please correct me if I am wrong.
> 
> In the example a 200 amp current limiting controller is used. So for ease lets say you have a 100 volt battery.
> 
> You are driving up a hill at half throttle 50v, The controller is currently limiting the amperage to 200 amps which could mean that the motor wants to turn faster. In effect you are not getting the full effect of the 50v. Then if you go full throttle 100v the controller would still regulate the voltage the same as it was to keep the current at 200amps.


Going up a hill tends to slow down the armature and hence the back EMF will be less and since the controller is set to 200 amps then it will cause the on pulses to appear less often (dead time) to lower the voltage average to a point where the current flowing through the motor is 200 amps. Most likely at 100 volts since you are going up a hill in one gear ratio it will not accelerate any faster. To take advantage of 100 volts you would have to go to a lower gear.

Hills change the outcome quite a bit compared to level ground because now vehicle weight starts to come into play and it's hard to gain momentum. 

To calculate the weight you are moving you take the SINE of the angle of the hill and multiply it by the total weight of the vehicle.



> If you had a pot you could adjust to raise the current limit to say 400amps, then the controller would be able to make use of more of the voltage to gain power.


Raising the current limit to 400 amps will give you more torque and therefore you will gain acceleration at 100 volts without a gear change necessary.


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## ww321q (Mar 28, 2008)

Dennis said:


> The catastrophic failure I am referring to is the current will be so high that plasma fire balls will shoot out of the vent holes at the brush rigging end. If you are lucky then only the brushes burst into thousands of pieces and the rigging melted. If not then maybe the actual commutator melted too and the enamel coating on the windings is charcoal black. This is why current limit is so important in large horsepower DC motors.
> 
> Here is a nice video of a motor with a bad hair day. They even add the sad music to give that special effect. LOL. I bet the company was crying when they saw the repair bill for this motor.



http://www.youtube.com/watch?v=-ZO__ebsSpE
I don't think that motor damage was caused from plasma fire ball . It was caused from over revving . A problem that could happen to any series wound motor in an EV that doesn't have a rev limiter . Simple over amping will usually cause a failure in the armature windings insulation . A plasma ball in the motor is caused by any or all the fallowing , sudden over voltage (extreme) , sudden over amping (extreme) , improper brush timing , improper motor maintenance , (brush residue in motor ) Usually caused by the later 2 with high voltage .  J.W.


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## Dennis (Feb 25, 2008)

> http://www.youtube.com/watch?v=-ZO__ebsSpE
> I don't think that motor damage was caused from plasma fire ball . It was caused from over revving . A problem that could happen to any series wound motor in an EV that doesn't have a rev limiter . Simple over amping will usually cause a failure in the armature windings insulation . A plasma ball in the motor is caused by any or all the fallowing , sudden over voltage (extreme) , sudden over amping (extreme) , improper brush timing , improper motor maintenance , (brush residue in motor ) Usually caused by the later 2 with high voltage .  J.W.


Yep that is what happens if you over rev. I just thought just I would post it too since we are talking about some of the damages that occur with motors like you said with over revving. I cannot find a video on over current damage. But over current damage has nothing on this though lol.


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