# how the battery behaves in this diagram

My question is this. I'm using this ic mppt SM72445 along with a 250 watt solar panel, mppt output configured to 50.4 or 51 volts, 12-cell lithium-ion battery pack and bms, if the load is less than the output of mppt the battery will be charged, but what happens when the load is greater than the mppt output and there is a voltage difference between the mppt output and the battery? example ; mppt output = 50.4V 5A / battery pack = 46.8V 3.9Vcell * 12 / load = 10A my doubts are for the difference in voltage between the two sources, mppt and battery in parallel

• For me it is not really clear what the SM72445 does when the set maximum current is exceeded, will it drop the voltage to the level that only the maximum current is flowing or will it shut down completely. The datasheet (put a link to it in the question in the future) doesn't tell that (or I missed it). I'd try to contact TI and ask them how the chip behaves. May 24 '19 at 17:32
• I think the MPPT will supply its maximum and the battery will make up the difference. But I don't think the battery will ever charge in that circuit unless the load is switched off. The MPPT voltage will never be high enough to charge it... May 25 '19 at 6:19

A MPPT converter is a power source (as opposed to a current source or a voltage source). The output voltage and current settings of the SM72445 are maximums. The switches are driven to maximise the power output, as long as the resulting output voltage and current are below the maximum settings. So think of it this way: the output will be whatever voltage and current is necessary to deliver the power, unless the maximums are reached.

Let's apply that logic to the scenario at hand.

The presence of the battery and load is complicated, so consider one at a time. Imagine there is no battery. Suppose then the MPPT has 100W available from the PV. In a 5Ω load, the MPPT will produce 4.5A @ 22.4V. No qualms.

But if 200W becomes available from the PV, the current will hit the 5A limit and the load will cap the voltage at 25V. Only 125W will be delivered.

Now swap the battery for the load (and ignore the BMS for now). You can think of the battery as a voltage source in series with a resistor. The battery absorbs power when we apply more than its terminal voltage, limited by the series resistor. Says it's a 48V source with 1Ω of series resistance.

Again, suppose the PV has 100W available. The MPPT will produce 50V @ 2A, no qualms. Now suppose 200W becomes available. The voltage will hit the 50.4V limit, capping the current at 2.4A and only 121W will be delivered. In both cases the battery will be charging.

Okay, with all those fundamentals out of the way, on to your question, which I interpret to be: what if the load is greater than the MPPT can supply, but there is a battery present as well.

simulate this circuit – Schematic created using CircuitLab

Let's consider the numbers already used: 48V battery with 1Ω series resistance in parallel with 5Ω load. Now the battery is supplying power, so the series resistance causes its apparent voltage to drop. 48V across 1Ω + 5Ω means 8A will be flowing, so the apparent voltage is 48V - 8A * 1Ω = 40V. The load is absorbing 320W, all supplied by the battery.

Suppose first that the PV has 100W available. It will supply just enough current to deliver that power. That occurs at 2.4A and 42V. The battery now only needs to supply 6A. The increase in voltage from 40V to 42V means the load is now absorbing 353W, but the MPPT is supplying 100W of it so the battery is only supplying 253W.

Suppose now that the PV has 200W available. The happy equilibrium is now when it is delivering 4.56A @ 43.8V. The battery now only needs to supply 4.2A. The load is absorbing 384W but the battery's share has dropped to 184W.

In both cases, the load is able to absorb more than the MPPT can supply, so the battery ends up supplying part of it. The voltage will end up lower than is necessary to charge the battery.