Yes.
The 120V top side mosfets see 60A 1/6th of the time, the 60V topside mosfet see 60A 1/3rd of the time. In contrast, the bottom side mosfets will see 60A 5/6th of the time vs 2/3rd of the time. That 60A is ALWAYS going through a mosfet. Especially on HW 4.x it doesn't matter much if it's a high side mosfet or a low side one: they dissipate heat into the same piece of the PCB.
Current limit of motor

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 Joined: 19 Jan 2016, 10:54
 Location: Australia
Re: Current limit of motor
rew wrote:Yes.
The 120V top side mosfets see 60A 1/6th of the time, the 60V topside mosfet see 60A 1/3rd of the time. In contrast, the bottom side mosfets will see 60A 5/6th of the time vs 2/3rd of the time. That 60A is ALWAYS going through a mosfet. Especially on HW 4.x it doesn't matter much if it's a high side mosfet or a low side one: they dissipate heat into the same piece of the PCB.
How much of an influence do DC bus capacitors have on low speed performance? At low speeds with high motor currents is performance limited by how many amps the capacitor can discharge, most electrolytic caps don't have terribly high ripple current ratings.
Re: Current limit of motor
FYI, I don't know how to calculate the ripple current in the presence of multiple different capacitors. Besides the "big" electrolytic capacitors, there are also a few ceramics on the VESC. In theory, you'd assume something like: battery is doing 10A continuous, so 5/6th of the time the electrolytic capacitor is charging at 10A, 1/6h of the tim it discharges at 50A. Well... Our capacitors cannot handle that at all. Not even close. I'm using 4700uF 3A rated capacitors. Most people use 3 2A rated capacitors. That's 6A total, but with 50A motor current at 50% duty cycle, the capacitors are constantly doing about 25A!

 Posts: 171
 Joined: 19 Jan 2016, 10:54
 Location: Australia
Re: Current limit of motor
rew wrote:FYI, I don't know how to calculate the ripple current in the presence of multiple different capacitors. Besides the "big" electrolytic capacitors, there are also a few ceramics on the VESC. In theory, you'd assume something like: battery is doing 10A continuous, so 5/6th of the time the electrolytic capacitor is charging at 10A, 1/6h of the tim it discharges at 50A. Well... Our capacitors cannot handle that at all. Not even close. I'm using 4700uF 3A rated capacitors. Most people use 3 2A rated capacitors. That's 6A total, but with 50A motor current at 50% duty cycle, the capacitors are constantly doing about 25A!
I have been reading about this a bit, a few things I read suggested electrolytic capacitors are poorly suited to suppressing ripple on motor controllers. They mainly became common on low voltage motor controllers because they usually go into fairly cheap products. Metalized film capacitors are common for higher end motor controllers, a 15uf one can deliver up to about 20A, however they are bulky and targeted at 600V+. Ceramic caps have been improving in capacity at higher voltages. The datasheets don't usually list ripple current ratings, supposedly they are only limited by heat, I did find one datasheet with a rough estimation of around 60A.
One interesting this is using film or ceramic caps the needed capacity is drastically reduced.
For example my rough calculation, if a 60V 40A pulse is needed, 60V 40A = 2400 watts or 2400 joules for 1 second. Reducing the time period to 1/40,000th of a second leaves you with 60 millijoules, while the capacity of 45uf 60V ceramic capacitors is 60millijoules. Assuming 50% duty cycle the capacitor would have to supply about 30 millijoules to convert 60V 40A into 30V 80A?
In comparison large electrolytic capacitors rated for 60V 6A could deliver 360 joules if it could run for one second, divide that by 40,000, leaves you with 9 millijoules per pwm, when the requirements are halved by 50% duty cycle you still need to draw 3x it's rated current. I'm not sure as to what happens exactly in this situation, do the electrolytic caps still deliver the required energy but incur large losses due to resistance?
Atleast on the motor controller side the large electrolytic capacitors are'theoretically using very little of their total capacity. I don't know how it works when it comes to inductance in the power cables, is this also better suited to smaller caps with very low ESR and high ripple currents? Or is the larger capacitance needed.
I'm not sure if my math is remotely correct, I don't fully understand capacitor ARMS ratings, looking at some electric car inverters I have noticed they almost exclusively use the metalized film caps which are closer to 100uf or so than 4000uf on large electrolytics.
Re: Current limit of motor
You need to use "the capacitor formula".
I = C dU/dt
the worstcase is when PWM is at 50%. With a 40kHz PWM frequency, that would be 12microseconds. The dU is the drop in voltage at the capacitor. and that is what we want to know.
So, we rework the formula: dU = I dt/C = 25A . 12 microseconds / 45 uF = 6.7V. Thus the battery voltage at the capacitor would be bouncing up and down about 7V.
So IMHO, a 45 microfarad capacitor is too small. We need about 10 times bigger capacitance to be able to reduce the voltage ripple at the input of the VESC to below 1V.
I don't think your math is correct. As far as I can see, I think you are assuming that you can discharge the capacitor to 0V. I think "that would be bad".
In general, the ESR also plays a role. And for maximum current ratings it is very important. Suppose my 450uF capacitor has an ESR of 40 mOhm. So we're alternating between 12 microseconds of charging at 25A and 12 microseconds of discharging at 25A. In each case there would be 1V of voltage drop across the ESR. 2V * 25A = 25W. You would be putting about 25W of power into the capacitor! That's going to be one HUGE capacitor if it still likes you after ten minutes of that treatment.
I = C dU/dt
the worstcase is when PWM is at 50%. With a 40kHz PWM frequency, that would be 12microseconds. The dU is the drop in voltage at the capacitor. and that is what we want to know.
So, we rework the formula: dU = I dt/C = 25A . 12 microseconds / 45 uF = 6.7V. Thus the battery voltage at the capacitor would be bouncing up and down about 7V.
So IMHO, a 45 microfarad capacitor is too small. We need about 10 times bigger capacitance to be able to reduce the voltage ripple at the input of the VESC to below 1V.
I don't think your math is correct. As far as I can see, I think you are assuming that you can discharge the capacitor to 0V. I think "that would be bad".
In general, the ESR also plays a role. And for maximum current ratings it is very important. Suppose my 450uF capacitor has an ESR of 40 mOhm. So we're alternating between 12 microseconds of charging at 25A and 12 microseconds of discharging at 25A. In each case there would be 1V of voltage drop across the ESR. 2V * 25A = 25W. You would be putting about 25W of power into the capacitor! That's going to be one HUGE capacitor if it still likes you after ten minutes of that treatment.
Re: Current limit of motor
I think you need to compare the capacitor ESR vs the battery internal resistance+wiring, to see how the current ripple will be shared between them. With a good battery (10mOhms?) and short wiring, most of the ripple will come from the battery, unless you have very low ESR capacitors (and large enough capacity to deliver current for a few useconds without too much additionnal voltage drop).
Re: Current limit of motor
Yeah, in reality the battery will take some of the ripple current. But when I design things I like to know, or calculate for the "worst case scenario". In this case that would mean that somehow (ESR or wire inductance) the battery is more or less a current source providing the average current.
Re: Current limit of motor
Sorry for digging up this old thread, but I was reading up on this and doing some calculations:
From this document:
Eq 29 (voltage ripple at 50 % duty cycle): ΔV_0.5t = Vbus/(32⋅L⋅C⋅f²),
where L is load inductance, C is bus link capacitor, and f is switching frequency.
Rearrange and solve for 5 % ripple at 60 V, 25 µH and 25 kHz: (Edit: this is peaktopeak ripple, so 3 V is only 2.5 % ripple.)
60V/(32⋅25µH⋅3V⋅(25kHz)^2) = 40 µF
Eq 18 (capacitor ripple current at 50 % duty cycle): ΔI_0.5t = 0.25⋅Vbus/(f⋅L)
0.25⋅60V/(25kHz⋅25µH) = 24 App ≈ 8.5 Arms (if sine wave, ≈ 7 Arms if triangle wave, according to Wikipedia)
The linked document suggests polypropylene film capacitors, but they are huge and expensive, and the selection is very small at <= 100 VDC. I'm having trouble finding ripple current specifications for other types of film capacitors, but I found one: KEMET R60. Maybe something like an R60DR54705050K would work in the VESC? It would be fun to try it sometime! (To test it I guess one would need a testbench where one could run a motor at 50 % duty cycle at a fixed/limited rpm?)
From this document:
Eq 29 (voltage ripple at 50 % duty cycle): ΔV_0.5t = Vbus/(32⋅L⋅C⋅f²),
where L is load inductance, C is bus link capacitor, and f is switching frequency.
Rearrange and solve for 5 % ripple at 60 V, 25 µH and 25 kHz: (Edit: this is peaktopeak ripple, so 3 V is only 2.5 % ripple.)
60V/(32⋅25µH⋅3V⋅(25kHz)^2) = 40 µF
Eq 18 (capacitor ripple current at 50 % duty cycle): ΔI_0.5t = 0.25⋅Vbus/(f⋅L)
0.25⋅60V/(25kHz⋅25µH) = 24 App ≈ 8.5 Arms (if sine wave, ≈ 7 Arms if triangle wave, according to Wikipedia)
The linked document suggests polypropylene film capacitors, but they are huge and expensive, and the selection is very small at <= 100 VDC. I'm having trouble finding ripple current specifications for other types of film capacitors, but I found one: KEMET R60. Maybe something like an R60DR54705050K would work in the VESC? It would be fun to try it sometime! (To test it I guess one would need a testbench where one could run a motor at 50 % duty cycle at a fixed/limited rpm?)
Re: Current limit of motor
If your voltage ripple formula does not ask for how much current you're drawing, it cannot be right. A fixed capacitor size will ripple proportional to the current you're drawing.
Re: Current limit of motor
Hmm... so what they do (in the paper) is to look at the phase leg ripple current (which should be dependent on load inductance and switching frequency, but not load current, right?) and then assert that "the ripple current in the bus link capacitor is essentially the same as the ripple current in the phase leg" (page 2, sect IV).
I think that might be true if you use a switching scheme where one leg of the bridge is always on  I'm not sure what this is called, "pushpull" might be a fitting name? I.e. you switch the load between +Vbus and Vbus, but never to 0 V.
If the switching scheme involves "all off" states where the load current does not pass through the DC link cap, then I see your point, and of course this is the case with SVM and the VESC.
Too bad, I was hoping to maybe avoid the need for a soft start circuit (as discussed in some other thread here) altogether by using a smallervalued bus capacitor.
I think that might be true if you use a switching scheme where one leg of the bridge is always on  I'm not sure what this is called, "pushpull" might be a fitting name? I.e. you switch the load between +Vbus and Vbus, but never to 0 V.
If the switching scheme involves "all off" states where the load current does not pass through the DC link cap, then I see your point, and of course this is the case with SVM and the VESC.
Too bad, I was hoping to maybe avoid the need for a soft start circuit (as discussed in some other thread here) altogether by using a smallervalued bus capacitor.
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