Capacitor For MM 4S

[BrianG]

I guess I still don't know what you are disagreeing with, but I'll assume you disagree with a previous statement I made, so I'll try to explain:

Current is going to increase when you increase the voltage, but that's because there is the same "load" on the electromagnetic nail (zero in this case).

Please correct me if I am wrong here; but in a motor, loading (among other things) plays a big role.

So, take an actual motor of whatever wind.

Apply 1v, measure current. Apply 2v, measure current. Current will increase by about a factor of 2 (probably not exactly double because rpms are higher and is inducing more opposing voltage in the adjacent coils). Is this correct?

Now, apply 1v to it and add some physical resistance to the shaft. Measure the current. Now, apply 2v, but remove the physical resistance. Measure the current. The current will not double from 1v to 2v because the load was reduced even though the voltage doubled.

So, if all other factors being equal (gearing, motor wind, etc), when going with higher voltage, current will increase. But if you gear down (on the same motor), which reduces the mechanical load, current will be less even though you increased the voltage.

[GriffinRU]

Originally Posted by suicideneil

Im lost...

What are you trying to get at exactly? This all rather beyond me.

That is exactly what I am trying to say, to make it simple current will go always UP with voltage in our simplified world of electronics.

When you use your throttle control, you are changing average voltage applied to the motor. So if you start from 4S by throttle you apply, using PWM, voltage from near 0 volts to max of 14.8V, which corresponds to RPM change from 0 to Kv*Volts.
And in 5S case you do the same from 0V up to 18.5V...

[GriffinRU]

Brian do not change my nail to the motor, right away, you might think that you can jump to the conclusion so quick, but we are not here yet.

In simple motor example applying DC voltage to phase will blow it or overload. Voltage and Current in the motor are out of phase, but load defines that. But we are not here yet.

So, how much change in current would be when you move magnet around nail tip, on the same scale as going from 1V to 2V?


My main problem here is that there is a believe that BL motor will regulate power and that is what I am not agreeing with.

Second, it would be impossible to come to conclusion here if we not agree on system parameters, Like motor is fixed (9XL) and load is fixed. Otherwise we not just changing voltage applied to the motor, right?

[BrianG]

OK, I see now. Yeah, current is limited by the DC resistance of the coils which is really low generating really high potential currents. When the motor starts moving, back-EMF becomes the major factor in limiting current.

Maybe I'm not explaining it right, but I don't think a motor will regulate power either. All I was trying to say was that current is not dictated solely by the voltage because there ARE other factors.

[GriffinRU]

Now we can easy explain operation voltage for given motor, right? Meaning you will never apply 10S to 6XL motor :)

From here each motor has:
- load limit, not related to voltage. Motor need to spin.
- Maximum operational voltage, when power regulation can be managed by gearing change, within safety envelope of maximum load. RPM limit.

[BrianG]

I agree with both those points exactly...

[GriffinRU]

Ok, then from here 9XL will draw more amps from battery with higher voltage.
All the rest is proper setup!

[BrianG]

Yes, unless you reduce the gearing for lighter load...

[GriffinRU]

Originally Posted by BrianG

Yes, unless you reduce the gearing for lighter load...

Just to make sure...

After you done with gearing, measure your current draw at 5S and then switch to 4S, current draw should drop, right?

In simple world you can treat a motor as a light bulb:
- low voltage low light/poor efficiency/long life
- Normal voltage good light/good efficiency/expected life span
- high voltage sligtly more light than at norm/poor efficiency/short life
- too high voltage brief light(flash) -> broken

[BrianG]

Originally Posted by GriffinRU

After you done with gearing, measure your current draw at 5S and then switch to 4S, current draw should drop, right?

Yes.

[rschoi_75]

dang ... after reading this thread, I feel like the biggest electronics noob of all time. Just wanted to say thanks for this informative discussion. I learn something new every time I'm on here.

[Maciolus]

Will Panasonic FC 3300uf 25V be also suitable? I am using 4S2P A123 cells.

[GriffinRU]

Originally Posted by Maciolus

Will Panasonic FC 3300uf 25V be also suitable? I am using 4S2P A123 cells.

Yes,
18X25/2750mA/0.020Ohms - looks alright for 4S A123

[Maciolus]

Originally Posted by GriffinRU

Yes,
18X25/2750mA/0.020Ohms - looks alright for 4S A123

Thanks!

[Bernie Wolfard]

Hmm, there have been a couple of things discussed in this thread I feel I need to comment on. I will also try to provide some information about the use of capacitors, the original reason for this thread.

First, the motor’s Kv affect on current.
Kv means RPM per volt. It has nothing to do with kilovolt or other electrical measurement. This is a constant all motors have. Part of Kv is determined by the number of turns of copper on the motor but turns affect different motor types differently so are not a constant and hence not really useful in determining a motors performance. When a motor spins up it becomes either a generator (brushed) or alternator (three phase brushless). As the motor goes faster, the amount of voltage it generates goes up. When the voltage of the motor matches the voltage of the battery pack it cannot go any faster, hence it has reached its Kv.
Given two similar motors with different Kvs and the same voltage, the higher Kv motor will pull more current (AMPS) than one with lower Kv. This is simply because the higher Kv is trying to get a higher RPM and it takes more current to go faster. There are other things that will cause the higher Kv motor pull more current such a lower copper to magnet ratio, less backplane, perhaps wider airgap etc. but you get the point. To get the efficiency advantages of getting watts from volts instead of AMPS you need to lower the Kv of the motor as your volts go up or you wind up pulling a lot more current as your voltage goes up.

Second: Capacitors, why?
Sole purpose of adding Caps to a controller is to control ripple current. Ripple current is a side affect of pulling pulsed current from a DC source where nothing in the system is electrically perfect. In this case this means the components of the system, particularly the battery, have electrical resistance*. A speed controller works by switching full throttle current on and off really fast to the motor (in our case about 11,000 times per second). This is called pulse width modulation or PWM. The motor averages these pulses out. If 50% of each pulse if off, 50% on the motor sees 50% power. This means the ESC is also pulling pulsed current at the same rate from the battery. In a ideal world these pulses would form a square sine wave where the on part of the pulse went straight up to full throttle current, straight over to the cut off point, straight downs to off current, straight over to on then repeat over and over, then rinse. Unfortunately, as some of you may have found out by now, we don’t live in a ideal world, electrical or otherwise. Because of the batteries internal resistance each pulse it puts out starts a little late, slopes up, overshoots, flattens out, shuts off a little late and tapers down and undershoots. All of this lateness, sloping and overshooting is called ripple current. The caps simply help smooth this out. If the batteries are inadequate, there is high resistance wire or plugs between the pack and the ESC or a high resistance plug the caps will quickly become drained trying to mitigate ripple current. If this happens the caps overheat. Worst case they blow up. As they are overheating they allow more ripple current to reach the FETS which in turn heat up. All of this is bad and can lead to catastrophic failure of the ESC. However, it is important to remember that capacitors cannot make up of inadequate batteries or poor wiring, they provide no increase in power and if they are the wrong ones make matters worst by increasing resistance. A properly designed ESC with good batteries does not need more capacitance. With inadequate batteries no amount of capacitance will help. To ensure long life in an electric power system just make sure you use the best (lowest internal resistance) batteries available with large enough capacity. Always ensure are least 20% more battery discharge capability than you think you will need and everything is OK an power will actually go up as more amperage equals more power, more voltage equals more motor RPM.

Bernie

* A batteries internal resistance is the one and only number that determines how fast the cell can discharge. The lower the resistance, the quicker the cell can discharge. Wouldn’t it be nice if battery companies published this statistic so we could easily judge which batteries are the best?

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For the more technically minded here is a responce from our chief engineer on why ESCS have caps.

You might notice that every brushless controller on the market has an input capacitor that goes across the battery leads - the plus of the capacitor goes to positive battery and the negative of the capacitor across the negative of the battery. The reason for this capacitor is to smooth the ripple current from the battery, so that the battery sees a smoother current demand.

At partial throttle, the controller is turning the motor on and off at a high rate (for our controllers it is typically 13khz.) During the "on"
cycle, there is a fairly high current demand on the batteries. During the "off" cycle, the motor current is recirculated through the controller, and the battery has a near zero current demand. The capacitor recharges during the controller "off" cycle, and discharges during the controller "on" cycle.

The apparent ESR of the battery is reduced, and some of the strain on the battery is transferred to the capacitor.

In most cases this works well -- the battery efficiency is increased because it sees a lower peak current demand. Also the strain on other components (MOSFETs, back-emf detection circuits etc.) is reduced because of a reduction in the ripple voltage on the battery rail.
Without the capacitor, some setups would see ripple voltages exceeding 50% of the battery voltage. So you have to think of a battery as you would any other device - - there is parasitic resistance (ESR) and inductance (ESL.) Batteries are pretty good DC sources when the load is constant, but when the load makes large step changes, their output voltage also makes large step changes.

If, for example, I tried to run a motor which would draw 10 amps on a "perfect" DC supply, on a battery that had .1 ohm of resistance and an output voltage of 10V at no load, we would see a ripple voltage of just under 1V, with a loaded voltage of just above 9V, and an open voltage of 10V. Placing a capacitor across the battery would average the current demand on the battery, and smooth the output to a constant 9.5V.

Ok, at this point I can assume you understand the role of the capacitor. The reason electrolytic capacitors are used is because they are inrush current surge resistant (tantalums are not -- so they are dangerous to use in this type of application) they have a high Q so they tend to suppress ripple well at lower frequencies, and they have good bulk capacitance at moderate to high voltages (where ceramics do not.) As you pointed out, lower ESR would be desirable, especially in marginal systems where the batteries have very large voltage ripple (high ESR.) Tantalums cannot be used because the inrush currents would destroy them (they burn up) so electrolytics are used instead. With electrolytics, ESR and bulk go hand in hand -- so we selected a capacitor with a low enough ESR for the majority of applications, that won't be too large or too expensive. However, in some marginal systems, the capacitor is forced to do a lot of work, and the ESR of the capacitor creates heat. With an electrolytic, there is a failure mode where the temperature of the capacitor exceeds the boiling point of the electrolyte, causing a catastrophic failure of the capacitor. Usually, this will only happen on systems where the current demands of the motor are significantly higher than the capability of the battery to supply current.

I hope this helps clear it up!