Archived: the OG noob question thread! 😀

Love the analogy,
Though to be a pain in the ass and out of interest, what would be the mathematical proof of it?
Considering that voltage in the end is a form of pressure caused by the amount of electrons moving. There would be something in the conversion from DC -> AC that can use the increased pressure in order to draw less current from the battery.

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I don’t know of others, but there could have been.

A lot of these analogies won’t hold up to a proof. One of the reasons EE is considered difficult is because electricity and how it interacts within a circuit is pretty difficult to visualize.

The analogies help explain some core concepts but the most common one of electricity vs water/pipes/pressure doesn’t hold up particularly well beyond sound bytes.

On the topic of higher voltage/less amps, that’s a lot of the core theory behind why full size EVs trend upwards in voltage. At higher voltage you can deliver the same watts at less amps, reduce operational ceilings, increase robustness of the system.

The takeaway is that it’s generally easier to just think of batteries in Watt-hr and motor output in Watts once the system has already been specified.

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It is the maximum difference between the p-groups.

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I have had basic courses in EE, but they don’t really go into anything practical. But they do show how much difficult it can be. Especially once AC is introduced.

The analogy I usually use for U = RI is that it’s a system of water flowing through a tube where I = how much water is going through, R = inverse to the width of the inner diameter of the tubing and U = the pressure the amount of water puts on the tubing. But once you start adding parts it becomes harder and harder to explain.

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@Krindor increasing the inductance in a circuit means increasing the time it takes for current to increase or decrease with applied or removed voltage.

so for example at 50% duty cycle, the battery could be supplying pulses of 60a, half the time, which averages to 30a. the battery side of the circuit has low inductance and so the current can ramp up to 60a and down to 0a very quickly.

on the motor side, the inductance is very high, so when the current from the battery shuts off, the circuit is rewired/switched so that the leads coming from the motor forms a loop, which allows the current to flow inductively (without applied voltage) for a short time.

the time scale of the on-off pulses from the battery are so short in duration that during the off times when there is no battery current, the current continues to flow in the motor entirely from the inductance.

so 60a from the battery half the time (30a average) can result in 60a in the motor the whole time, because the current still flows in the motor inductively while it is not flowing from the battery, but we say the 60a the motor sees the whole time is half the voltage of the battery if the duty cycle is 50%.

hope this helps.

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if you happen to be wondering what “powers” the current that continues to flow in the motor when you remove the battery current… it’s the collapsing magnetic field of the stator which stores energy in the magnetic field itself.

the inductance is directly proportional to the magnetic field produced by a given amount of current, and since the motor produces a large magnetic field for a given current, it has high inductance, and the current still flows for a while when you remove the battery current if you quickly short the motor to itself.

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That does help,

Correct me if I’m wrong,
But if I were to make an analogy with a spring, if the inductor were to be a spring, the voltage the power used to push the spring back and the current the power/second when the spring retracts back into equilibrium. The retraction meter/second is constant. This would mean that the higher the voltage is, the more amps it can push back. Meaning that in the end, it would result in less amp usage with more voltage used instead.

Does this sound kind of sane?

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when you send current through the motor, because of the high inductance (created by the coil shape of the winding combined with the iron in the stator, which forms an elecromagnet) it takes extra time to reach a given amount of current, compared to the time it would take if you used a straight wire with no iron stator.

during this “extra time” you are building a magnetic field or “compressing a spring” or “storing energy”… when you remove the battery voltage, the spring releases and powers the current that continues to flow for a short time without any battery voltage.

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That does make sense, thanks a lot for the help!

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Can you use a Boosted charger for a 10s battery? The nominal output for a Boosted charger is 42.24V 2A

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For that one maybe, the og charger was 50v, you sure that yours isn’t?

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I don’t have one on hand but I read it of somewhere online

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If I find one that is 42.24V would it be safe to use?

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The 42 and change charger should work though, .24v isn’t alot of overcharge. Just don’t let it sit like that for too long, you know the drill

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It’s about that thing.

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I somehow lost my reply.

I would say that drift doesn’t look great, but it’s also not indicative of something like a cell failure.

You’re using an active balance charger already? Do they balance okay at full charge?

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I swapped it for a new bms, that Baja sended to a friend of mine, before I picked it up four weeks ago. Now it runs on the new bms. The battery had a huge drift then.

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The fact that this is a fairly new battery + you’re using an active balance charger is somewhat concerning.

Is the drift there every single time you discharge? And doing full discharge/charge cycles doesn’t lower the delta?

Feedback from APS on this subject is that brushless motors, by and large, are waterproof.

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