Cogs in mesh

This is an interesting question and I don’t believe the answer is as intuitive as it first seems. Whenever an electrical system is designed to have multiple simultaneous power sources, load management is engineered into the system somewhere. All but one of the sources has to have some give, or an ability to adapt to what the other source(s) is/are doing.

But lets take a few steps back and have a look at a simpler system. Consider the following circuit:

B1 and B2 are batteries, both supplying a resistive load (Load). All batteries have some internal resistance and this is represented by the resistors inside each battery block. Voltage points a and b are conceptual, they don’t actually exist in a real battery but give us a convenient point to make a distinction between the supply voltage and the internal resistance to use in calculations. Voltage point c is the voltage supplied to the load.

With this circuit configuration you would intuitively expect that both batteries would share the duty of supplying current to the load. Indeed if both batteries were of the same voltage it would be a trivial task to calculate, given the known quantities of the internal resistance and load resistance, exactly how the supply is shared using Ohm’s law and parallel/series resistance equations.

But what if the battery voltages are different? You might think that one battery would supply the load entirely and simultaneously attempt to charge the other battery, but this may not necessarily be the case. Let’s break it down mathematically. Using Ohm’s law we can infer the following:

V_a − V_c = I_B1R_B1
V_b − V_c = I_B2R_B2
V_c = (I_B1 + I_B2)R_Load

Where:
V_a−c are voltages at points a, b, and c,
I_B1,B2 are battery currents,
R_B1,B2 are battery internal resistances, and
R_Load is the load resistance.

Combining these equations and solving for I_B1 we arrive at:

I_B1 = (R_B2V_a + R_Load(V_a − V_b)) / (R_B2R_B1 + R_Load(R_B1 + R_B2))

If we plug in the following hypothetical values with somewhat different battery voltages:

V_a = 12
V_b = 15
R_B1 = 10
R_B2 = 12
R_Load = 5

We arrive at I_B1 = 0.561A

From this we can calculate the voltage at point c, the load current and battery current I_B2:

V_c = V_a − (I_B1R_B1) = 6.391V
I_Load = V_c / R_Load = 1.278A
I_B2 = I_Load − I_B1 = 0.717A

QED, both batteries are sharing the job of supplying the current to the load despite their different voltages. Lets now plug in some different values for the resistances, leaving the battery voltages the same as the previous example:

V_a = 12
V_b = 15
R_B1 = 2
R_B2 = 3
R_Load = 20

The formulas then return:

I_B1 = -0.226A
V_c = 12.453V
I_Load = 0.623A
I_B2 = 0.849A

Notice now that B₁ current is negative, and B₂ current is higher than the load current. B₂ is now the sole power supply and B₁ current is flowing in the opposite direction. If it’s a rechargeable battery it’s charging, if not it’ll probably leak or explode!

So how does this relate to PV systems? The principles are exactly the same but the engineering to make it work is very different. The mains supply is a scarily low impedance source so we can consider it the supply with no “give”, ie. the value of its series resistance is so low as to be negligible. Whether we draw 30A from it or supply 30A to it the voltage will not change much at all. The inverter of the PV system however is an incredibly versatile power supply. It can accept DC input(s) of anything up to several hundred volts, and produce output that precisely follows the AC cycle of the mains supply, while at the same time continuously varying the load on the PV array to maintain peak efficiency under wildly varying sunlight conditions.

Ultimately, the job of the PV system’s inverter is to provide as much current as it can, in the right direction during each half of each cycle of the mains supply. It doesn’t supply just a little more voltage as one answer suggested. Although that is what it’s attempting to do, the low impedance of the mains supply prevents it and this is part of the reason why grid connected inverters cannot supply power during a blackout. In understanding how a grid connected PV system works it’s important to think of it as a current source that adapts to the prevailing voltage conditions.

The supply of current to loads is determined both by the requirements of the load, and the instantaneous capability of the PV system at any given moment. Current flow from that point on follows convention like the example circuit. If PV system supply is insufficient, current to the load is shared between the grid and PV system as in the first example above. If generation exceeds requirement the system effectively tries to “charge” the grid, and all load current flows from the PV system - this is a similar scenario to the second example with the grid and metering represented by B1.

I hope this sheds some light on the topic!