GS450H MG1 and MG2 Torque Ratio and oddity in inverter protocol [SOLVED]

[catphish]

The algorithm people are using to control the GS450H inverter sets mg1 torque to 125% of mg2 torque. Where did this ratio come from? Similarly, MG1 is often not used in reverse. Is there a discussion of the reasoning behind these decisions?

Also, does someone know why the control code takes the sum of the mg1 speed (negative) and mg2 speed (positive), divide the sum by 113, and send it back to the inverter. I'm interested to know what is really happening there.

Thanks!

[sfk]

I suspect the formula ratios outlined above are to do with the fact that once the input shaft from the combustion engine is locked (in relation to the transmission housing) MG1 will be turning at a higher RPM and in the opposite direction to MG2. Due to the nature of the planetary gearset.

[konstantin8818]

So that means, it would be impossible to use current firmware when planetary carrier is welded? =\ That makes things much harder for me.

[catphish]

So that means, it would be impossible to use current firmware when planetary carrier is welded? =\ That makes things much harder for me.

I really wouldn't worry about this. It's easily adjusted.

[xp677]

The algorithm people are using to control the GS450H inverter sets mg1 torque to 125% of mg2 torque. Where did this ratio come from? Similarly, MG1 is often not used in reverse. Is there a discussion of the reasoning behind these decisions?

Also, does someone know why the control code takes the sum of the mg1 speed (negative) and mg2 speed (positive), divide the sum by 113, and send it back to the inverter. I'm interested to know what is really happening there.

Thanks!

MG1 and MG2 torque difference is due to the gear ratio in the PSD. If the PSD gets welded then yes this would need to change.

No reason why you can't run MG1 in reverse. But at the same time, the additional torque isn't needed for a low speed reverse function. The same argument could be made that MG1 is not required for forward travel (at a loss of power output).

The inverter expects the first htm byte to be as you described. This is likely a sanity check / checkback to make sure the HVECU has correctly interpreted the MG1, engine, and MG2 speed feedback. Since MG1 rotates in the opposite direction to MG2 with zero engine speed, this is simplified to the speed difference between the two motors for our application. With the engine running, MG1 speed will be dependant on engine RPM for a given vehicle speed. Vehicle speed is a function of MG2 speed and gear ratio. From the sum of the motor speeds, engine RPM can be calculated as well and compared to the RPM input from the GI pin for further checks.

The divide by 113 is there most likely so that it fits within one byte for all given motor and engine speed combinations. And could be determined via experimentation (noting that MG1+MG2 speed in a running GS450h is always 113x HTM byte 0).

[DkubusEV]

This is very interesting and as I'm only beginning my gs450h zombie project I'm especially keen on the idea of the planetary lock rather than the external input shaft lock. Would be nice to therefore utilise the internal oil pressure? And circulation pump rather than running the external oil pump continuously?? Plus more speed range and possibly better overall efficiency due to less gear losses and fluid sheer/windage losses?? Just theoretical of course but I'm seriously keen on hearing more about what I'd need to alter within the "code" so that both motors would run in unison and offer the best total power at peak but best case efficiency at cruise speeds too... cheers for all your hard work and good information everyone involved.

Cheers from Australia

[xp677]

You'd just remove the *5/4 ratio form the MG1 torque, and I guess flip it's direction of rotation as well. Since it would be a 1:1 ratio, rather than a -1.25:1

[DkubusEV]

With the PSD locked the speed ratio between MG1 and MG2 would depend on what gear MG2 has activated. MG1 would only be at 1:1 with the tail shaft output speed. Im not entirely sure if this means the required torque split request should be "MG1 at 1 vs MG2 at 1.9" (0.5263:1) in top gear and "1:3.9" (0.2564:1) in first gear? or do we just feed them both the same current?