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I am looking at prototyping a buck based MPPT solar cell system but am mainly confused by how the solar cell I-V characteristics are manipulated by the buck converter to give the optimum power to the load. I believe it is essentially an impedance matching but am struggling to see it mathematically.

Lets say I have a 12V nominal 130W solar cell, that at max power is rated @ 7A39 @ 17V6 and I am trying to charge a 12V battery. Being a 12V battery it requires 13V2 to 14V4 to charge.

Now looking at the model below, the load could be modelled by a 12V voltage source, but how does one apply the standard buck converter ideal equations (ignoring losses as just trying to keep it simple).

Is there a mathematical equation/explanation of how the buck converter presents a 17V6 7A39 load, i.e. 2R38, to the solar cell, which is actually a 12V battery but requies 14V4?

enter image description here

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    \$\begingroup\$ Please use decimal point! \$\endgroup\$
    – winny
    Commented Mar 31, 2018 at 19:35
  • \$\begingroup\$ No. In the constant current charge region, you manipulate the buck (duty cycle) measuring the power at each step to maximise the latter. When you reach the CV stage you don't exceed that output voltage; at that point you've left MPPT. \$\endgroup\$
    – user16324
    Commented Mar 31, 2018 at 19:36

1 Answer 1

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As you've mentioned, for the maximum power transfer the impedance seen by the solar cell should be equal to the impedance of the solar cell, which at maximum power level is 17.6V/7.39A=2.38ohm.

With no losses, the full power will be transferred to the load, i.e., Pout=Pin

Let's assume, for the moment, that your load requires 12V and has the impedance of 3ohm. The buck converter will convert the input 17.6V to the output 12V, buy using a duty cycle of D=12V/17.6V=0.68.

The current in the load will be Iout=12V/4ohm=3A, which, due to the preservation of power, will draw from the solar cell Iin = P/Vin=12V*3A/17.6V=2.05A, which, of course, is 0.68 of the output current. So, the apparent load impedance for the solar cell will be 17.6V/2.05A=8.59ohm.

So, using one basic duty cycle equation of the buck converter, we can see how the solar cell sees a 4ohm load as a 8.59ohm load.

Of course, in this case the transferred power is only 36W (12V*3A=17.6V*2.05A), while the maximum power of the solar cell is 17.6V*7.39A=130W.

To maximize the power for the 12V output voltage, the load has to be R=V^2/P=12V*12V/130W=1.11ohm, in which case the output current would be 12V/1.11ohm=10.81A. The input current will be 130W/17.6V=7.38A and the apparent load impedance will be 2.38ohm.

When we charge a battery, we can control the current and that will determine the voltage on the battery or we can control the voltage and that will determine the charge current, but we cannot control both.

So, if we wish to charge the battery at 3A and, at that moment, this battery happens to develop 12V (at that charge rate), the maximum power we'll be able to transfer from the solar cell to the battery would be 36W.

If, on the other hand, we don't care about the charge current (probably, because we believe that, given the power source, it cannot exceed the maximum safe charge current) and we just want to maximize the power transfer from the solar cell (basic MPPT premise), we should keep increasing the voltage on the battery (by increasing the duty cycle) until the power on the battery reaches 130W.

The equivalent or effective resistance on the battery at that moment will depend on the type and state of the battery, but, as long as the power on the battery is 130W, the battery will be seen by the solar cell as 2.38ohm load.

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