Re: (ET) Does battery voltage affect amp gauge?

["David Roden" <etpost drmm net>]

On 16 Oct 2021 at 15:40, Chris Zach via Elec-trak wrote:

> If the batteries really had a lower voltage then the tractor would need 
> to draw more amps to produce the same amount of power.

I've seen this stated several times on this list and elsewhere, and it 
always gnaws at me a bit.  

I'm not an engineer, so I'll defer to someone who is.  But the statement 
doesn't usually match what I see happening in practice.

I can understand how a motor and PWM controller combination that's trying 
to maintain a consistent motor power output has to draw more current from 
a 
lower voltage battery.  

That's the normal condition of a road EV, a Leaf or Bolt or Tesla or 
whatever.  When you're driving at a constant speed, the controller is in 
PWM, limiting power.  It has to be, because the EV is holding a bunch of 
power in reserve, for acceleration and higher speeds.

But I don't use my ETs that way.  (Do y'all?)  

I use the E15's PWM controller for smooth acceleration and for moving it 
in 
tight areas.  (Love it for that.)  But most of the time, I have the 
controller wide open for the best efficiency.  Also, most of the time the 
transaxle is in D1, the gear I use for moving snow.  

Meanwhile, the mower, lift, snow thrower, and tiller motors all run flat 
out with no speed control.

Aside from the action of back EMF, when a motor is running flat out from a 
battery, Ohm's law reigns.  That is, P = E * I or P = E^2 / R.  

To give an example, when I'm mowing, as my R36's battery voltage falls, 
its 
mower motors slow down.  So does its traction motor.  They're producing 
less power because there's less voltage, AND because there's also less 
current flowing as a result of lower voltage.  I = E/R.

I don't upshift to maintain forward speed.  What would be the point?  With 
slower motor mowers, the machine would just do a worse job of mowing. 

So the R36's battery voltage AND battery current are BOTH lower than when 
I 
started mowing.  (And pretty soon it'll be time to head back to the garage 
to give the battery a 30 amp whack for 45 minutes or so, but I digress.)

In case this still isn't clear, let's try a couple of water analogies.

My house has a well pump.  When I'm washing the car, I'm triggering the 
hose nozzle wide open to blast the grime off.  As the water tank's 
pressure 
falls (before the pump turns on), the dirt-blaster stream gets weaker.  

Not only is the water pressure (voltage) lower, the water volume (current) 
is lower.  Both contribute to the fact that I'm not getting much mud off 
until the pump turns on and pushes the water pressure back up.  

My water system is behaving just like electricity does - power = pressure 
* 
volume.

On the other hand, if I'm rinsing the car and using less hose power - the 
hose nozzle only, say, 1/3 open - when the tank pressure falls, I'll 
usually squeeze the nozzle trigger harder.  

This is comparable to running with the speed lever reduced, or the 
transaxle shifted to a lower gear - and nudging the speed lever up, or 
upshifting, as the ET battery drains.

Now my water system DOES make up for falling pressure (voltage) with 
increasing volume (current).  

And the same is happening with the ET's drive motor IF you try to make it 
go faster on a dying battery - though not with its mower motors.

But, again, how many ET owners actually DO raise the speed lever or 
upshift 
the trans when they're using the tractor and the battery starts going flat?

Thus it seems to me that for most practical purposes, "the tractor would 
need to draw more amps to produce the same amount of power" might be true, 
but it doesn't really apply, because we don't usually demand that the 
tractor produce the same amount of power.  At least I don't.

So, electrical engineers - is this analysis sensible, or am I missing 
something?  Or did I just confuse everyone trying to read this?


David Roden - Akron, Ohio, USA

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