# Controller Current vs Battery Current



## major (Apr 4, 2008)

Sunking said:


> OK I know Battery Current does not equal Controller Current. I know Controller Current is higher than Battery Current.
> 
> SO lets say I have an Altrax SPM 72650 a 650 amp controller for a Series wound 48 volt DC motor. How do I convert that to maximum battery current?


With the EV motor controllers, the stated maximum current is the current limit for the motor. The battery current will be less than the motor current until full throttle is applied at or above base speed. Then battery current will equal motor current. At base speed and full throttle, the PWM will reach 100% and it will just be coming out of current limit, so that will be the point of maximum battery current, which is equal to maximum motor current. This is also the maximum power input point. 



Sunking said:


> In my minds eye these are PWM controllers and if you switch to 100% duty cycle or full throttle Input Current should = Output Current, unless we have a buck/boost converter in which case would be a power converter from one voltage to another.


The motor controller is a buck converter which happens to use the load (motor) itself for the inductor.


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## dougingraham (Jul 26, 2011)

Sunking said:


> SO lets say I have an Altrax SPM 72650 a 650 amp controller for a Series wound 48 volt DC motor. How do I convert that to maximum battery current?


Since this is a buck converter the only time the motor current will be equal to the battery current would be when the duty cycle goes to 100%. At that time the motor voltage and battery voltage will be the same as well and is the only time the FWD does nothing. 



Sunking said:


> In my minds eye these are PWM controllers and if you switch to 100% duty cycle or full throttle Input Current should = Output Current, unless we have a buck/boost converter in which case would be a power converter from one voltage to another.


All the controllers I know of are simple buck converters. Battery current x battery voltage = motor current x motor voltage - controller losses. Most controllers primarily control based on motor current but sometimes have additional considerations for RPM, motor voltage, battery current and battery voltage. In a control intended for an EV what you are commanding is torque which is essentially equivalent to motor current. Everything is done by lowering the voltage sent to the motor.



Sunking said:


> I know my vehicle running at full throttle up to speed draws about 90 amps which matches the horsepower rating of the motor. Is it peak power I am missing?


You lost me on this. If you have a 650 amp controller is there any place where the current reaches 650 amps? If you floor it and the current goes to 650 amps does the vehicle speed up to a particular RPM and then the current drops off as the motor RPM continues to rise? If that is the case your peak power will occur just before the current starts to drop off. It is at this point that the back emf of the motor is approaching the battery voltage and the current the motor demands drops off as well. To get more speed out of the vehicle you either have to increase the voltage which means adding more cells or change the gearing to increase the load at speed on the motor. Increasing the battery voltage (controller limits apply) will widen the torque band. You can look at dyno plots of your motor to understand what is happening.


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## Sunking (Aug 10, 2009)

dougingraham said:


> You lost me on this.


Don't feel bad, that is where I got lost.



dougingraham said:


> If you have a 650 amp controller is there any place where the current reaches 650 amps? If you floor it and the current goes to 650 amps does the vehicle speed up to a particular RPM and then the current drops off as the motor RPM continues to rise?


First I have two different carts. One is DC, and the other is AC, the DC cart has a 450 amps controller, and the AC has the 650 amp controller to a HPEV AC9 motor which is the one O am trying to wrap my mind around. 

Only metering I have on the AC cart is a Fluke Clamp On and I don't think it is fast enough to catch the peaks. When I floor it from a standing stop I do see current surge and can read up around 400 amps briefly, then taper down to around 90 amps when I hit the 6000 RPM limit I have the controller set at. That 80 to 90 amps is on flat level paved surface. If I go down hill current slacks off to maintain 6000 RPM, down a steep enough hill will reverse direction for braking. If I go up hill, current does go up to maintain 6000 RPM. At full throttle there is no more to have. Cart limits to 6000 RPM on a 7000 RPM motor.

Here is what is throwing me off. I am an EE, don't work with motors a lot other than those in HVAC, pumps, elevators, door openers, etc you would find in a commercial environment. But all I need to know is LRA and FLA. In my minds eye FLA is continuous power and LRA peak power at a given voltage. LRA in this case is lower than the controller rating. So how does a controller pump more current in than what Ohm's Law works out to be without boosting the voltage? 

If you look at the HPEV AC9 power graphs you can operate it with 48/72/96 volts with a 350/550/650 amp controller. I understand the controller can limit current in both AC and DC motors. But how can it increase current without increasing voltage into a fixed impedance or resistance? That is where I am stuck. 

FWIW I know the AC controller is a VFD and speed is controlled via frequency, and DC is voltage. I got that part. It is just the whole Ohm's Law thing that has me dizzy.


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## dougingraham (Jul 26, 2011)

Sunking said:


> and the AC has the 650 amp controller to a HPEV AC9 motor which is the one O am trying to wrap my mind around.
> 
> Only metering I have on the AC cart is a Fluke Clamp On and I don't think it is fast enough to catch the peaks. When I floor it from a standing stop I do see current surge and can read up around 400 amps briefly, then taper down to around 90 amps when I hit the 6000 RPM limit I have the controller set at. That 80 to 90 amps is on flat level paved surface. If I go down hill current slacks off to maintain 6000 RPM, down a steep enough hill will reverse direction for braking. If I go up hill, current does go up to maintain 6000 RPM. At full throttle there is no more to have. Cart limits to 6000 RPM on a 7000 RPM motor.
> 
> ...


Ok, Lets look at this HPEVS Graph.

AC-9 48V 650 amps

Note that the HP peak occurs at 2200 rpm when it is pulling 698 battery amps. This gives 27.75 HP at the motor shaft. Battery voltage is about 48 and current is 698 giving 33.5 kw (45 HP). This is about 62% efficiency. At this point the battery voltage and motor voltage are equal because the only thing the controller is doing is commutation. As the motor RPM increases the motor generates its own voltage which counters the voltage coming from the controller and the motor current diminishes which equates to a decrease in torque. At 6000 RPM the battery current is down to 250 amps and power output is down to 12 HP. This would give an efficiency of about 76% at this operating point.

If you double the voltage you move the HP peak up to around 4400 RPM.

I think you are confused because there are too many variables. Pick the 2200 rpm point where we see the max HP. Battery voltage is 48, current is 700 and watts is 33600. If the motor were a pure resistance it would be 0.0686 ohms at this point. But it isn't a resistance. And unless you lock the shaft it isn't an inductor either. If you take the special case where you lock the shaft in place and turn on the controller what you see on the battery side is the battery current goes up just a little and if you look at the current in the windings you will see the controller current limiting on the motor side. The motor voltage will be very low but the current the motor is commanding is essentially infinity. The motor controller will be at just a few percent duty cycle because it is limiting the motor voltage in order to keep the motor current under control. The free wheeling diodes are working overtime in this situation and if the controller cooling is not good enough it will eventually cut back the voltage even more to limit self heating.

From what you are saying you imply you want a higher top speed. You can get this by increasing the battery voltage which will push your torque band farther to the right and increase the effective voltage at 6000 rpm which will in turn make the motor want to turn faster which in turn increases the current it wants to draw which is torque. You can also do this by changing the final drive ratio so the wheels want to turn a little faster. This increases the load on the motor at 6000 RPM so the current will go up and more power will be delivered. Of course going from something like a 5:1 reduction to a 4:1 reduction also lowers the torque at the face of the tires so acceleration will be slower.

If you knew the PWM duty cycle you could approximate the motor voltage and more of this would make sense. With 48 volts and a 10% duty cycle the motor would see about 4.8 volts. If it was pulling 10 amps at 48 volts on the battery side and the PWM is 10% so the motor is seeing 4.8 volts the motor current would be 100 amps (excluding losses).

I don't know if any of that helps. I don't feel like I explained it too well.

If you look at it from the motor side it might help to pretend it is an ideal motor. In an ideal motor the voltage controls the RPM.

RPM = Kv * Volts

Kv is the motor constant which gives RPM per volt. Lets say Kv is 50 and you apply 100 volts. An ideal motor will turn at 5000 rpm. It doesn't matter what load is placed on the motor because it is ideal. It will draw any amount of current to get the shaft to that speed.

Torque = Kt * Amps

The motor will draw current to maintain the commanded RPM. Kt is another magic constant and its value will depend on the units used for Torque.

It is important to remember that the only thing we control is the voltage the motor sees. The current is what the motor commands. The motor controller makes this fuzzy because it changes the voltage to keep a whole bunch of things it is watching under control. RPM, Motor current, Battery Current, Battery voltage, Controller temp, Output power and commanded torque are some of the things a controller watches.


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## Sunking (Aug 10, 2009)

dougingraham said:


> Ok, Lets look at this HPEVS Graph.
> If you double the voltage you move the HP peak up to around 4400 RPM.


FWIW you are pretty darn close. *HPEC AC15* is the exact same motor as the AC9 Only difference is AC9 graphs are for 48 volts, and the AC15 graphs are for 72/96 volts



dougingraham said:


> I think you are confused because there are too many variables. Pick the 2200 rpm point where we see the max HP. Battery voltage is 48, current is 700 and watts is 33600. If the motor were a pure resistance it would be 0.0686 ohms at this point. But it isn't a resistance.


 Should be impedance Resistance + Reactance? With the understanding Reactance is going to increase with RPM/Frequency. 




dougingraham said:


> From what you are saying you imply you want a higher top speed. You can get this by increasing the battery voltage which will push your torque band farther to the right and increase the effective voltage at 6000 rpm which will in turn make the motor want to turn faster which in turn increases the current it wants to draw which is torque. You can also do this by changing the final drive ratio so the wheels want to turn a little faster. This increases the load on the motor at 6000 RPM so the current will go up and more power will be delivered. Of course going from something like a 5:1 reduction to a 4:1 reduction also lowers the torque at the face of the tires so acceleration will be slower.


I am understanding this part, and you are correct speed is the issue. Fortunately I kind of lucked out on half arse educated guesses. I have a golf golf cart I converted to AC and wanted a 40 mph top speed with decent acceleration and I lucked out and got both. As you may know golf carts are direct drive with only a Differential Gear Ratio. 6000 RPM, 10:1 ratio, 11 inch radius tires gives me 40 mph. I get really great acceleration up to about 30 mph, then tapers off. On flat level paved surface pulls roughly 90 amps. Once i go up hill current increases to about 120 amps and holds speed. Increase the grade a bit more, current creeps up but RPM stats to slip off. Right now running 48 volt 100 AH LFP and considering adding 8 more cells to get 72 volts which should push the torque band up in RPM right? 



dougingraham said:


> I don't know if any of that helps. I don't feel like I explained it too well.


Actually you have helped and I thank you for that. I think where I went off track is I have a grasp of DC Series Motors where Torque is directly proportional to current vs RPM, and Voltage is directly to RPM. Switching to AC motors breaks those relationships I had figured out.


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## MPaulHolmes (Feb 23, 2008)

For a series wound DC motor, it's 

batteryCurrent = pwm_duty * motorCurrent

For a 3 phase motor, it's:

batteryCurrent = duty1*phase1Current + duty2*phase2Current + duty3*phase3Current.


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## major (Apr 4, 2008)

MPaulHolmes said:


> For a 3 phase motor, it's:
> 
> batteryCurrent = duty1*phase1Current + duty2*phase2Current + duty3*phase3Current.


Hi MPaul,

I've never seen this expression before. Can you give me your source or is it something you developed? 

And I don't see any usefulness to a guy like Sunking. Motor phase currents are typically expressed as RMS values and considered equal to each other. And with the Curtis or any other AC controller of which I am aware, PWM duty on the phase is not a parameter which is available.

major


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## major (Apr 4, 2008)

Sunking said:


> But all I need to know is LRA and FLA. In my minds eye FLA is continuous power and LRA peak power at a given voltage. LRA in this case is lower than the controller rating. So how does a controller pump more current in than what Ohm's Law works out to be without boosting the voltage?


Hi Sun,

By LRA do you mean Locked Rotor Amps? If so, we typically refer to stall current. LRA, I think, is usually needed for industrial applications where you see across the line motor starting. This isn't done with EV propulsion motors. You always have the motor controller providing "soft start", so to speak. That is, at stall or zero motor RPM, even with WOT (full throttle), the motor controller limits the motor current by reducing the voltage applied to the motor. So that initial stall current on the motor will have a maximum value of the controller current limit but will occur at an much lower value of battery current due to the controller behavior as a buck converter. There is no voltage boost. Ohm's Law is satisfied, as always  

major


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## Sunking (Aug 10, 2009)

major said:


> Hi Sun,
> 
> By LRA do you mean Locked Rotor Amps? If so, we typically refer to stall current. LRA, I think, is usually needed for industrial applications where you see across the line motor starting. This isn't done with EV propulsion motors. You always have the motor controller providing "soft start", so to speak. That is, at stall or zero motor RPM, even with WOT (full throttle), the motor controller limits the motor current by reducing the voltage applied to the motor. So that initial stall current on the motor will have a maximum value of the controller current limit but will occur at an much lower value of battery current due to the controller behavior as a buck converter. There is no voltage boost. Ohm's Law is satisfied, as always


Yes Major LRA is Locked Rotor Current, and FLA is Full Load Current. In the NEC world you need to know those two things for sizing conductors and over current protection devices. Doug really helped me out and I am starting to get my brain wrapped around it. 

Right now I am tinkering with a fuel gauge for the cart which should help me a bit as it will be a Coulomb Counter with current transducers so I can see the peak currents vs my Fluke Clamp-On which I do not think is fast enough to see the quick peaks, not to mention 650 amps is over-range.


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## major (Apr 4, 2008)

dougingraham said:


> RPM = Kv * Volts
> 
> Kv is the motor constant which gives RPM per volt.
> 
> ...


I think it should be mentioned that these so called constants are only constant for motors operating with constant flux. That does not include series wound motors or motors using field weakening where the flux will vary depending on conditions.


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## dcb (Dec 5, 2009)

MPaulHolmes said:


> For a 3 phase motor, it's:
> 
> batteryCurrent = duty1*phase1Current + duty2*phase2Current + duty3*phase3Current.


Interesting. I confirmed with some spot-checks in ltspice, seemed to work pretty well for average current, though not RMS (rms is basically flat, until you set mutual inductance close to %100). Also it starts to fall apart if there is any overlap in the pwm pulses (i.e. two bottoms on at the same time with the third high).


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## MPaulHolmes (Feb 23, 2008)

major said:


> Hi MPaul,
> 
> I've never seen this expression before. Can you give me your source or is it something you developed?
> 
> ...


It's like the power in = power out for the DC motor case.

Instantaneous Power in = Vbatteries * Ibatteries

Instantaneous Power out = Vbatteries * duty1 * current1 + Vbatteries * duty2 * current2 + Vbatteries*duty3*current3. 

Divide both sides by Vbatteries and you get the formula. I recently made a couple 150kw PMAC/ACIM controllers for some beta testers, but I didn't include voltage monitoring on the control board, so I had wanted to limit battery amps both ways (regen and not). So, I was asking around on the EV Tech list, and Edward Cheeseman mentioned it. 

I tried it out. I measured battery amps with an external shunt, and I had the duties and phase instantaneous currents, and gosh dang it, they are the same!

I actually output phase currents and duties, Id, Iq, Vd, Vq, etc..., and a bunch of other stuff in a serial stream, but that doesn't help anything in this context.

DCB: Thanks for checking that! I actually used an average of 128 different readings (at 10KHz current measurement frequency) of of the instantaneous output power as the battery amps. Also, my PWM duties are from SVPWM:

```
if (Vc <= Va && Vc <= Vb) {  // Vc is smallest
		// Q1, Q2
		pdc1 = Va-Vc;
		pdc2 = Vb-Vc;
		pdc3 = 0;
	}
	else if (Va <= Vb && Va <= Vc) { // Va is smallest 
		// Q3, Q4
		pdc1 = 0;
		pdc2 = Vb - Va;
		pdc3 = Vc - Va;
	}
	else { // Vb is smallest.
		// Q5, Q6
		pdc1 = Va-Vb;
		pdc2 = 0;
		pdc3 = Vc-Vb;
	}
```


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## major (Apr 4, 2008)

MPaulHolmes said:


> It's like the power in = power out for the DC motor case.
> 
> Instantaneous Power in = Vbatteries * Ibatteries
> 
> Instantaneous Power out = Vbatteries * duty1 * current1 + Vbatteries * duty2 * current2 + Vbatteries*duty3*current3.


Thanks for the info. But I don't accept the validity. It doesn't account for the reactive component. In other words, it assumes a constant P.F. with respect to load. Duty1, 2, & 3 will essentially be fixed at the same point on the sine wave for a particular phase voltage and frequency. Now at the that particular V & f (and point on the sine wave), compare the battery current vs phase current for no load and full load. It's not possible for your current equation to satisfy both conditions.

Here is an attempt I made a while back if you're interested  
http://www.diyelectriccar.com/forums/showpost.php?p=376274&postcount=9


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## MPaulHolmes (Feb 23, 2008)

See this application note, page 5:

http://www.mouser.com/pdfdocs/55w_28943_0_hr_letter.pdf

It's just kirchoff's law. The power factor could be zero. You are still going to have 3 currents and a battery pack. Forget about sine waves for a minute. What other voltage and current is there that is not being accounted for?


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## major (Apr 4, 2008)

MPaulHolmes said:


> See this application note, page 5:
> 
> http://www.mouser.com/pdfdocs/55w_28943_0_hr_letter.pdf
> 
> It's just kirchoff's law. The power factor could be zero. You are still going to have 3 currents and a battery pack. Forget about sine waves for a minute. What other voltage and current is there that is not being accounted for?


The inverter bridge cannot be considered a single node between the battery and 3 phase motor. There are other paths and loops and active elements.

The reference you supplied shows an interesting item. Blondel's Theorem shows that:

Instantaneous Power out = Vbatteries * duty1 * current1 + Vbatteries * duty2 * current2.

Not:

Instantaneous Power out = Vbatteries * duty1 * current1 + Vbatteries * duty2 * current2 + Vbatteries*duty3*current3. 

as you used 

I don't know for sure. I've never seen a duty cycle percentage associated with 3 phase AC synthesis before. It's a new one on me. But on the surface, it looks like scalar math to explain vector sums and has me confused. 

Like I said before, it is kinda academic because I doubt anyone but you could pull out the parameters to make the calculations.


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## MPaulHolmes (Feb 23, 2008)

With space vector modulation, one of the 3 duties is always zero.



```
if (Vc <= Va && Vc <= Vb) {  // Vc is smallest
		// Q1, Q2
		pdc1 = Va-Vc;
		pdc2 = Vb-Vc;
		pdc3 = 0;
	}
	else if (Va <= Vb && Va <= Vc) { // Va is smallest 
		// Q3, Q4
		pdc1 = 0;
		pdc2 = Vb - Va;
		pdc3 = Vc - Va;
	}
	else { // Vb is smallest.
		// Q5, Q6
		pdc1 = Va-Vb;
		pdc2 = 0;
		pdc3 = Vc-Vb;
	}
```


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## MPaulHolmes (Feb 23, 2008)

You can just do this also:
Ibus = (Vd*Id/2 + Vq*Iq/2)/Vbus

Where Id, Iq, Vd, Vq are the direct and quadrature components. I wonder if curtis makes those available? I mean, Vd/Vbus and Vq/Vbus are available as part of the FOC process. Id and Iq should be knowable if it is set up for throttle = Iq = Id?


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