Voltage drop musings [RE-wrenches]

[Drake Chamberlin - Electrical Energy]

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Hi William,

I think we both agree that the MPPT tracking takes the excess power, 
represented by the voltage difference between the batteries and the array, 
and turns it into usable power.  We seem to have different ideas on the 
mechanism.

In my college chemistry class (30 years ago, forgive me if I'm a bit hazy 
on the details) they demonstrated how a chemical reaction was reversed in 
batteries by forcing back valance electrons.  It was demonstrated that it 
was the force of the chemical reaction that created battery current, and 
the forcing of electrons back to their original atoms that reversed the 
reaction.  It was a matter of replacing electrons.

To do this, a voltage was needed that was great enough to overcome the 
potential of the battery and the resistance of the system.  In general, if 
a higher voltage were used, more current would flow through the batteries, 
and result in a greater charge rate.  It is the equation:

  Amperage = Voltage / Resistance   I = V/R

The higher the voltage, the faster the charge is the general rule.  With 
solar modules, this is moderated by the fact that they have a very finite 
amount of current available.   But, the extra voltage does result in extra 
available power, which can be translated to more charging current for the 
batteries  with maximum power point tracking.  Without MPPT, the extra 
voltage doesn't help.

That ties in with the following question you raised:

>I disagree with the first sentence of the above paragraph.  A voltage drop
>in a series circuit will not reduce amperage one bit (Kirchhof's law).

Kirchhoff's Voltage Law only states that the sums of all voltage drops 
across all resistance in a series circuit will be equal to the EMF, or 
battery voltage.  Voltage drops occur across all resistive elements when 
current is present.  Ohm's law states the relationship between amperage, 
voltage and resistance.

                              V = IR  or  I = V/R

In other words, the higher the resistance, the lower the amperage.   The 
greater the voltage drop, the less the amperage.

I like your diagram.

>DC----------------wire resistance-----------------------------Load
>   Source                                                                |
>        |                                                                    |
>        |_______________________________________|

  I would add more elements of resistance to it.


DC source  ---- series fuse resistance -- array disconnect resistance--- 
wire resistance--solar room disconnect resistance --- Charge control and / 
or diode resistance--- metering shunt resistance--main DC battery 
disconnect resistance---Battery bank resistance---Load

The load in this scenario would be the chemical reaction in the batteries.

Drake


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