# PWM Controller Current ratings?



## Sunking (Aug 10, 2009)

I must be missing something with respect to series or shunt wound DC motors, and controller current.

For example let’s say we have a 48 volt motor rated at 157 FLA or 7500 watts (10 HP). Control is via PWM controller. What stumps me is I am told to use a minimum 400 amp controller or larger. Why would I need any more than say 200 amps? LRA maybe to get low RPM torque?


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## dimitri (May 16, 2008)

Because motor voltage at 0 RPM is 0 Volts and will rise as RPM rise. So you need more amps at lower volts while you are accelerating to keep max power.


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## Sunking (Aug 10, 2009)

dimitri said:


> Because motor voltage at 0 RPM is 0 Volts and will rise as RPM rise. So you need more amps at lower volts while you are accelerating to keep max power.


Ok not sure I buy that answer; but how would one determine how much current is needed out of a controller?? The motor has a DC resistance, and any current will develop a voltage. Something is not computing.


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## major (Apr 4, 2008)

Hi Sunking,



Sunking said:


> I must be missing something with respect to series or shunt wound DC motors, and controller current.


First off, forget about shunt wound motors for EV traction. 



> For example let’s say we have a 48 volt motor rated at 157 FLA or 7500 watts (10 HP). Control is via PWM controller. What stumps me is I am told to use a minimum 400 amp controller or larger.


By FLA you mean Full Load Amps, right? And by rated, you mean the one hour or maybe continuous nameplate rating for the motor, right? And let's stick to series DC motors for your example. So, if you got a controller which could only put out 157 amps, and your vehicle load for your desire speed (corresponding to 10 HP) was 157 amps, it would take you forever to accelerate up to that speed.

The typical series DC motor is likely to be rated at 150-200% of the FLA for 15 minutes and maybe 250% for 5 minutes, with a peak capability (10-20 seconds) of like 500% of the FLA (or 1 hour rating). So to take advantage of the motor's overload capability and get acceptable acceleration and hill climbing, the motor controller should carry similar overload ratings and current limits on the order of 3 to 10 times the motor's FLA.



> Why would I need any more than say 200 amps?


Back to your 10 HP example, if you need 10 HP for steady state driving power which is 157 amps and only had 200 amps max, you would not enjoy driving the vehicle. Acceleration times would be ridiculous and even small bumps in the road or slight inclines will stall the vehicle. Some might say you can overcome this problem with gear selection. But it is a lot easier to let the motor do what it is good at, overload torque. 



> LRA maybe to get low RPM torque?


What is LRA? The series DC motor is great at low speed torque. But this takes current.

Regards,

major


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## major (Apr 4, 2008)

Sunking said:


> but how would one determine how much current is needed out of a controller??


Easy. Get the motor performance curve and find the motor torque you need for your application and read the current for that torque. Again, stick with series DC motors. It is a 1 to 1 relationship of current to torque, regardless of voltage or RPM. So the more motor torque you need, the more current. And that (thru the gears) relates directly to acceleration and gradeability.

major


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## Sunking (Aug 10, 2009)

Starting to get it, but here is where I am stuck.

Using a PWM controller at full power is the same as connecting 48 volts to the motor right? Current will be limited by the battery cable wiring and the motors winding resistance. So I am stuck on Ohm's Law. How can you force more current than the battery voltage and resistance dictate?


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## Sunking (Aug 10, 2009)

major said:


> By FLA you mean Full Load Amps, right? And by rated, you mean the one hour or maybe continuous nameplate rating for the motor, right?


Yes and Yes


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## major (Apr 4, 2008)

Sunking said:


> Starting to get it, but here is where I am stuck.
> 
> Using a PWM controller at full power is the same as connecting 48 volts to the motor right? Current will be limited by the battery cable wiring and the motors winding resistance. So I am stuck on Ohm's Law. How can you force more current than the battery voltage and resistance dictate?


The motor is not just resistive. There is the back EMF (BEMF) or as I prefer to call it generated voltage (Eg). This is a potential (voltage) which opposes the applied voltage (battery, or Vb). Motor resistance (Rm) is very small, as is battery resistance (Rb) and lead resistance is so small it is usually ignored.

So the voltage equation for the motor circuit is:

Vb = Eg + Ia * (Rm + Rb)

Where Ia is the motor current, and in this case (100% PWM) = battery current. Solve for Ia:

Ia = (Vb - Eg) / (Rm + Rb)

For the batteries and motors likely to be used in EVs at 48 volts, Rm can be about 0.02 ohms and Rb about the same.

Vb stays pretty constant, 48 volts. Eg is proportional to speed (and flux). Eg will be zero at stall (zero RPM). So the maximum current for this motor and battery would be:

Ia(stall) = 48 / (.02 + .02) = 1200 amps.

As the RPM increase, Eg increases and Ia decreases.

But like I said, the torque is proportional to current (Ia), so that stall torque (at 1200 amps) is quite a bit, usually too much for comfort and mechanical stress. So that is where the controller current limit comes into play. Thru PWM, it will reduce the motor voltage and prevent motor current (and torque) from hitting these high values when launching from standstill.

Hope that explains it.

major


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## Sunking (Aug 10, 2009)

Yep that nailed it. Thanks


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## ClintK (Apr 27, 2008)

major said:


> The motor is not just resistive. There is the back EMF (BEMF) or as I prefer to call it generated voltage (Eg). This is a potential (voltage) which opposes the applied voltage (battery, or Vb). Motor resistance (Rm) is very small, as is battery resistance (Rb) and lead resistance is so small it is usually ignored.
> 
> So the voltage equation for the motor circuit is:
> 
> ...


My physics lesson for the day, thanks for the great explanation!


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## major (Apr 4, 2008)

ClintK said:


> My physics lesson for the day, thanks for the great explanation!


You're welcome, ClintK. I appreciate knowing others benefit from my lessons


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