PHEV Inverter Question

Three reasons that I can think of off the top of my head:

1) Batteries are far more efficient with less current transfer (regen or draw), thus the higher the voltage the higher the efficiency (depending on total power output of course). Remember that the area under the voltage drop curve represents lost energy (burned as heat) and the voltage drop increases with current based on internal resistance.

While many of the batteries available today claim to support high C rate discharges, you end up sacrificing significant portions of the battery to heat. High amperage draw can also cause premature lifespan degradation depending on the battery (most cells are rated for lifespan at 1C discharge). Additionally, not all batteries are created equal: many times cells will come from the factory with vastly different internal resistances from each other--thus they will deplete at different levels and heat up differently.

Also, think about cold weather performance. Internal resistance /drastically/ increases in extreme cold conditions thus severely limiting your total output if you have a low voltage connection. Battery voltage also drops in the cold (depending on the chemistry used) so you have to increase current further to maintain a given power output.

2) The higher the current draw the more expensive the core components generally are (though this is of course proportional to the selected voltage range)--and the more heat they tend to generate. The more heat generated requires more cooling thus more cost. Higher current requires larger gauge cabling which also increases cost.

3) It's arguably more safe and reliable to use a higher voltage system because of the lower current draw but there are pros and cons to that argument.

Really the only real benefit to a lower voltage system is lower cost for monitoring hardware (BMS) and possibly lower costs on a charger (depending on implementation of course).

Andrew

 

The data I have seen is that the Enginer converter is about 90% efficient when cool and 85% when hot and the fans all come on.

It draws about 60a at 52v with charged batteries to start. Then I have seen more like 75a at 45.5v just before it shuts down from the cells being low.

I expect you will get more like 80% from a converter set to run 1/2 the voltage form the cells. Then you would pull about 160a. You would have to run heavier cabling and components through out the system to handle that much current. Also I expect that the added heat would give you more variation in the cells and therefore more balancing issues. Also your Balancer would need to transfer more like 5a than the 2a in the current balancer. So the balancer wiring would all need to be heavier.

Thanks,
Dan Lander

 

All good points and interesting. But is cold really a factor for higher vs. lower voltage, beyond different batteries and chemistry of course. If cold drops battery output to say 80%, that would be the same factor no matter the number of batteries. No the input voltage range of the CONVERTER, not Inverter BTW, comes into play.

 

To recap my previous point, the cold does 2 important things to batteries:

1) It lowers the sitting battery voltage.
2) It increases the internal resistance, requiring the application to draw more current to get the same amount of power than if the batteries were warm.

If the total system voltage was higher (say 200v versus 50v), the impact of point #2 is far less significant because you don't have as much additional voltage drop from further increased current (V=IR). A higher voltage system sees a small increase in current / decrease in voltage as a result of cold weather operation whereas a low voltage system sees a rather significant increase / decrease in said values. Many times the lower voltage system total power output has to be scaled back rather aggressively to prevent damaging the batteries.

A higher voltage system will pretty much always be more efficient than a lower voltage system of equal capacity if implemented properly simply because it reduces energy lost to heat (EG: it's more efficient to draw 100a at 200v than it is to draw 430a at 48v even though they end up producing roughly the same power output).

Andrew

 

In cold weather, perhaps just saying only 80% capacity (a simple factor that would impact all batteries of that type) is just too simplistic. I would be interesting to see some actual tests that were done on this point.

 

It is much more complex than simply saying 80% capacity in the cold--I don't think I ever said anything to that effect.

Take Thunder-sky cells for example in the datasheet I link to below--note how the voltage sag is different at colder temperatures in the discharge graph at the bottom? Lets just say that the graphs in this datasheet are...more ideal than you might find in real life. In the real world the voltage sag would be more defined than in these graphs--but that should give you at least a starting point. Note also that the voltage sag increases proportionally to the current as the internal resistance has increased (the curve is at 0.5C which would be 20A, a realistic draw for an application like this would be 100+ amps which would be ~3C and would produce a much more significant voltage drop).

Link to datasheet: http://www.thunder-sky.com/pdf/TS-LFP40.pdf

Andrew

 

There is no question cold temp has an adverse impact. (But of course the same warm and cold temp would apply to both high and low voltage which is the variable/question.) Temp would obviously be a constant or consistent when comparing lowand high voltage. Right? I assume you are referring to the bottom left chart (thanks for providing some data), but that single chart does not specify a variable current draw, correct? If not, then where does this come from as you say above? "Note also that the voltage sag increases proportionally to the current" Should I be combining graphs? If so, specifically why do you think that is the case? Where is the specified graph shows a change in "c" and in temp? Please provide the key

 

I'm not really sure what your getting at so I'm going to provide a real-world example:

Take thundersky again (I chose thundersky because I figured most people here would be familiar with it and it exhibits my point quite well).

At room temperature, a 40Ahr cell has an internal resistance of ~7 mOhm. That translates to roughly a 0.7v drop when drawing 100 amps. At 32F the resistance is roughly ~28 mOhm which is significantly higher. 28 mOhm translates to roughly a 2.8v drop when drawing 100 amps which would /destroy/ the battery considering the open (sitting) voltage is ~3.2v to start with. You'd have to scale back your current to around 30a to prevent damaging your batteries thus seriously hampering your power output.

Lets assume for a moment that you could actually draw 100a at those temperatures: your losses would be astronomical (~87% energy loss from voltage drop).

I calculated the above voltage drops by using ohms law as mentioned before (v is voltage drop, I is current and R is resistance):

Vdrop = I * R

so, 0.7 = 100 * 0.007 for the first example.

Does this help at all? The impact of temperature is /not/ the same between high and low voltage because a higher voltage pack does not require as much current to be drawn to achieve the same power output thus less voltage drop due to the higher resistance...I'm not sure how many other ways I can say this .

Oh, and this isn't just for thundersky--pretty much all batteries exhibit this behavior in the cold (some far more than others of course). You can try googling data sheets for various batteries and see if they provide discharge data in the cold.

Andrew