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Discussion Starter #1
Comments please on this idea. It's intriguing.

I don't know enough about 3 phase BLDC and full bridge IGBT driver theory/maths to make a call on it.
I'm probably missing an obvious reason as to why it would not work in both assist and regen modes.

Please ignore safety/fusing/current matching/monitoring or practical concerns and concentrate on the basic idea.

Could/would it work?

The proposal.

Insert extra voltage/batteries directly into the motor phase wires.

That's it in a nutshell..

Imagine our system with three extra batteries say 50V inserted into each phase wire.

U IGBT ----->>>> Extra Battery ---->>>>> Motor Phase U
V IGBT ----->>>> Extra Battery ---->>>>> Motor Phase V
W IGBT ----->>>> Extra Battery ---->>>>> Motor Phase W

What is going to happen?

1) When system is off? Leakage through IGBT body diodes?
2) When system is assisting?
3) When system is regenning?

Over to the IC forum geniuses. :unsure:
 

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Southern California
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Oddly enough, I believe this would make absolutely no difference at all!

For reference, here is how a three-phase brushless DC motor looks electrically - three inductors with one end connected to a common node, and the other three ends brought out to be driven.
88482


When there is current flowing through a series loop of components (such as battery positive -> IGBT -> motor winding A -> motor winding common -> motor winding B -> IGBT -> battery negative), every component in that loop experiences the same current. This is a simplified explanation of Kirchhoff's current law.
Additionally, a voltage is a difference in electrical potential between two nodes. If you connect one end of a light to a 50V source and the other end to a 50V source, that light sees 0V and won't do anything.

The electrical potential between the three phases of the motor depends on how the IGBTs are switched. Let's say all three phases are pulled to ground, so at the drive module, the three phases are at 0V. The batteries in the three lines add 50V, so at the motor, each phase is at 50V. However, from the point of the motor, each phase might as well be at 0V, or at any other voltage. There is no voltage difference between any of the phases, so no work will be done.

Let's say one phase is at ground (0V), one is at battery voltage (150V), and one is floating (so no current flowing into or out of that phase). This is how three-phase brushless DC motors like the one in the Insight are driven. As the motor rotates, each phase is switched accordingly, but you'll always find that one phase is high, one low, and one floating.
So let's go around the loop starting at battery positive. We have 150V coming out of the IGBT, going into the 50V battery, so the voltage at the motor winding is 200V. It goes through one winding, into the common node, and out of the other winding (the one that is switched to ground). We encounter the 50V battery on the other winding, but since current is flowing into the positive side of the battery and out of the negative, we subtract 50V. This means there's 150V across the motor winding, which is the same as if there were no batteries in the windings at all. And since the current is still the same, the power (voltage * current) in the motor is exactly the same.

This explanation can be hard to visualize so I made a couple circuits to illustrate.
You can see here that the voltage across the load resistor and the current going through it is the same in either case! The voltage does get "boosted" as it goes through the first battery, but gets subtracted when it goes through the second.

88480


And here's a representation of a brushless DC motor being driven with and without voltage sources in the phases, and if you check the graphs, the current and voltage across each winding is the same in either case.

88481



Both circuits are here if you want to play with them:

https://www.falstad.com/circuit/circuitjs.html?ctz=CQAgjCAMB0l3BWcAOaAmA7MgnGB-kA2MAFgwTRAGYSQkFIAoAG3DSspO0PCrE+5QosJAFoYCDCQSFuhDNjTIqFDC178QXHmCqFIWjkJiQxEqTOwYwOMJGS6E63Zu299dDwZP1Y2ZGgk+sgkOAy6AKaiyIwAZiAYenRKCSHJyEJg0EgmaIwA7qm0oQZYxYJMhWVagtUImpVFNTrsAjxMAEpNJKHdYO1CJAZ2xsJOAOZNCD1TXoyTwVq9i0G0nSCLKpSbaN6Dw3ve2QUb9lpw7gZD3icrFyv9ULdnVGl3N-EyV0YrRsPZwkgeS6X2ohAyFAyegyBkoIxGh2ghEICGwaPRGPR8xA3AMhCGOI8KPajC6uLoM3J9UoewJCOMx0KVMUlzoLMazMo5MWHI8kh4nKe8Vw20ChLxYv+OVgeUKug811ZJD+J3lBlRcL06rmws0oKpHggWWlQLingM0NOUPB4HAANyZsSPFW3XxtuNgOB5sMlEhPqEcIOoxgVA4mPDaMIpISST9GEgtD9sPAQdhfmV2HqQKo2D4maowlDaAj4ajXSdWjd8eKbuT8L2aHTOaz7FzYHzwmRqJLGOjIo2YuQyAlaxAgZTg2gNmQCiBCawhBzpkLYZ76LLOL1HiH2obE-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

Good question, it took me a little while to come to this conclusion!
 

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Let's say one phase is at ground (0V), one is at battery voltage (150V), and one is floating (so no current flowing into or out of that phase). This is how three-phase brushless DC motors like the one in the Insight are driven. As the motor rotates, each phase is switched accordingly, but you'll always find that one phase is high, one low, and one floating.
So let's go around the loop starting at battery positive. We have 150V coming out of the IGBT, going into the 50V battery, so the voltage at the motor winding is 200V. It goes through one winding, into the common node, and out of the other winding (the one that is switched to ground). We encounter the 50V battery on the other winding, but since current is flowing into the positive side of the battery and out of the negative, we subtract 50V. This means there's 150V across the motor winding, which is the same as if there were no batteries in the windings at all. And since the current is still the same, the power (voltage * current) in the motor is exactly the same.
So, in the audio world, we used to bridge amplifiers together by running one amplifier out of phase with the other. The loudspeaker is fitted with the (+) on the output of the amplifier running in phase and the (-) on the output of the amplifier running out of phase. You get double the voltage between the peaks, therefore power is increased by a theoretical factor of 4 because of V=IR, minus losses.

Would it theoretically be possible to add a second IMA battery so that the car runs a +/- high voltage system, with two IGBTs running in opposite phase? The in-phase IGBT runs as normal, and the out-of phase IGBT applies a negative voltage to the opposite pole instead of 0V.

I'm trying to get my head round how you would do this. I think you'd need to build a second motor controller with all the transistors mirrored, so PNP where there are NPN and NPN where there are PNP, so that it switches a negative voltage relative to 0V. Flip the signal with an op amp and run it 'dumb' so that the monitoring on the in-phase IGBT doesn't see it?
 

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Discussion Starter #7
You would eventually saturate the rotor/stator and start to demagnetise it well before your broke splines.
As I understand it current would then go up for no gain in torque/power.

I managed to stuff 30kw through it and have been hacking for years without ruining an IMA motor yet.

But how much of that 30kw in my test was turned it mechanical effort,
and how much went into heating and melting things who knows. :eek:

The opposite phase idea is clever. Good luck LOL...
 

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So, in the audio world, we used to bridge amplifiers together by running one amplifier out of phase with the other. The loudspeaker is fitted with the (+) on the output of the amplifier running in phase and the (-) on the output of the amplifier running out of phase. You get double the voltage between the peaks, therefore power is increased by a theoretical factor of 4 because of V=IR, minus losses.

Would it theoretically be possible to add a second IMA battery so that the car runs a +/- high voltage system, with two IGBTs running in opposite phase? The in-phase IGBT runs as normal, and the out-of phase IGBT applies a negative voltage to the opposite pole instead of 0V.

I'm trying to get my head round how you would do this. I think you'd need to build a second motor controller with all the transistors mirrored, so PNP where there are NPN and NPN where there are PNP, so that it switches a negative voltage relative to 0V. Flip the signal with an op amp and run it 'dumb' so that the monitoring on the in-phase IGBT doesn't see it?
IMO, it would be much easier to just double the battery voltage and make a higher-voltage motor driver.
 

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Great question/concept Peter and even greater response Mario!

I'm not in the Insight business anymore but love this forum compared to to my "new" forum of GM-Volt owners....

Your idea seems like a good application for most any LTO conversion... I have my 360V 16KW used pack from my Volt (bought it at end-of-battery-life and swapped in a lower mileage used pack) which surely has plenty of cycles left if managed more thoughtfully... I'm using them for PV buffering now, but see that 3 48V sections might provide power for your concept. Seems like some clever SOC management could come into play, but if you add grid-charging it seems like a daily charge/balance while on a charger, this could work well!

The full battery is about 450lbs so this 144v subset would be about 1/2 that... maybe not too high of a weight budget with/out rear suspension boost? The replaced NiMH pack is probably a good 100lbs by itself?
 
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