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I have constructed and tested a circuit that achieves MPPT of solar panels. The MPPT circuits involves current sensors that requires 3.3V and -3.3V and also the MOSFET and MOSFET drivers that require 8V. To obtain these voltages, I am using an auxiliary power circuit that involves low power COTS chip converters. It begins by feeding the primary input to a LMR14006 based Buck converter. The output 3.3V is then used to feed the current sensors, a TPS6040X inverter that provides the required -3.3V and the LMR62421 boost converter that provides the required 8V. Please see the attached image which shows the auxiliary circuit component of my MPPT schematic.

enter image description here

As a side note, the MPPT circuit is intended for handling higher powers than that of the aforementioned auxiliary circuits. Its details are not important here as the source of my problem is not from them.

The issue I am getting is that the LMR14006 and (especially) the LMR62421 are not providing me a consistent 3.3V and 8V. When I connect an input voltage of 6V (and short circuit the input kill-switch in my schematic) and don't run any power through my MPPT circuit (easily achieved by turning off the MOSFETs), the LMR14006 and LMR62421 generates the required 3.3V and 8V. The resulting idle current from the 6V input is measured at 0.035A. However, once I begin running power through the circuit which means the MOSFET drivers will become a heavier load, I noticed that the LMR14006 will now drop its output to 3V and the LMR62421 drops its output to only 5V!

To verify that this problem exists on the auxiliary circuit side, alternatively to running power through the MPPT circuit, I connected an external 120 ohm resistor to the supposed output 8V of the LMR62421. Once again, the LMR14006 outputs 3V only and the LMR62421 outputs 5V instead of the originally desired 3.3V and 8V. Why is this happening?

As an additional note, I also noticed that the feedback voltage of the LMR62421 drops from the datasheet specified value of 1.225V to around 790mV. By the way, the TPS6040X voltage inverter is working as intended and I have no issues with it.

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Better, but not good enough:

enter image description here

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  • \$\begingroup\$ diagram unreadable visually. Pse make clear - blk white best - bold and adequate contrast. | Power in capability must be >= power out or Vin will collapse. \$\endgroup\$
    – Russell McMahon
    Mar 3, 2017 at 2:18
  • \$\begingroup\$ FYI: You can click the image to view a larger version of it. \$\endgroup\$
    – Trobby
    Mar 3, 2017 at 2:46
  • \$\begingroup\$ sadly, no I can't. Clicking the image gives a SMALLer version - and just as unreadable, or worse. I originally right click3ed image and tried "open image in new tab" with the same even worse result. If you want help here from the numerous very competent people you MUST use better images. Some screens work better than others and some eyes work better than others and the image is sub sub standard on many screens and with many eyes. It is easy for you to provide an image that is reasonable. If you expect all helpers to spend the effort required to do what you can do more easily your \$\endgroup\$
    – Russell McMahon
    Mar 3, 2017 at 7:38
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    \$\begingroup\$ Are you surprised at the lack of attention that your question is getting? It needs (at least) a diagram that people can see. Not a small unreadable one that is no better when you click it. \$\endgroup\$
    – Russell McMahon
    Mar 3, 2017 at 14:19
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    \$\begingroup\$ - you are right! - something is rotten in the state of Windows!. When I click on the image or open it in a new tab the resultant image is smaller than before. BUT if I save the image to disk and then open it it is usefully larger. Apologies for my blaming you for what appears to be one of Bill Gates problems :-). \$\endgroup\$
    – Russell McMahon
    Mar 4, 2017 at 1:23

1 Answer 1

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R9, R10, the voltage feedback divider for the LMR14006 buck converter, should be on net P$2 (right hand side of the 1st 3.3 uH) and NOT on net P$1 (lhs of 3.3 uH).

enter image description here

  • At very low loads IC1 will be off most of the time and the voltage across L1 will be low. When you load the output the IC1 feedback pin sees the divided output of the IC1 switch pin - either full Vin or a D4 diode drop above ground. IC1 either goes insane or if there is enough capacitance on IC1 pin 3 (FB pin) it may see an average DC value.

For a single PV panel you can produce a good approximation to MPPT by holding the voltage at IC1-5 at the optimum voltage for the panels (Vmp). This is typically 80 - 85% of Voc depending on the cells fill factor and can easily be determined in practice or (if you must) from theory. Increasing this voltage slightly with increasing load current produces an even better match to MPPT. For multiple panels this approach is "problematic" except when all panels are identical and equally illuminated.

I'm not sure what the input diodes are intended to do. They would serve to isolate the effects of shading on some of the panels. Efficiency losses will be incurred - how much depends on panel voltage. You do not say what Vmp is. AT 12V a diode at the input will lose you around 1V/12V ~= 8%. At 30V that's 1V/30V =~ 3%.

Adding a small capacitor across R9 and R12 will improve transient response of the two converters. Value tbd but probably in the 100's of pF range.

Components MUST have names on diagrams - L1, L2 etc.

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  • \$\begingroup\$ You're correct about the R9 and R10 part! This may actually be the main source of my problem that the feedback resistor rail was incorrectly connected (a silly mistake too that went long un-noticed). I'll try to modify my constructed PCB to reflect on this error next time I get to my lab. Thank you for pointing out the error. \$\endgroup\$
    – Trobby
    Mar 4, 2017 at 2:37
  • \$\begingroup\$ PS: The multiple input diodes is because I have multiple DC-DC converter inputs where each one does MPPT independently on a different set of solar panels. The multiple input diode connections allows me to power it via any of the input connections. It may not be the most efficient way to achieve this but it is a simple connection and also the auxiliary power circuit shown in my OP doesn't require much power. \$\endgroup\$
    – Trobby
    Mar 4, 2017 at 2:42

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