This open source product from Sparkfun (named Sunny Buddy) features Linear Instruments LT3652 MPPT li-ion/li-poly battery charger IC that can support up to 2A charging current.

Unfortunately, the board has been designed to support by default only 450mA charging current rate, but by replacing R1 sense resistor the current can be raised up.

Anyway, on this board if we lower R1 too much, the current will be too high and the inductor will actually saturate, limiting the output current artificially. If I'm correct, I calculated that the maximum achievable current should be ≈0,84A (inductor DC current = 1,05A - 20%).

But this way, if I understood how this IC works, the most of my solar panel power will be wasted.

Solar panel specifications

Voc = 9,94V
Vp = 8,54V
Vmp = 7,95V
Isc = 0,90A
Wp = 7,15W

Low power

Discharged battery voltage: 3,2V
Max current: 0,84A
Max power: 3,36W

If I could change the electronic components involved in the switching process, to reach a current of 2,0A, I'll probably get much better results:

Discharged battery voltage: 3,2V
Max current: 2,0A
Max power: 6,40W

What components should be replaced in the circuit and what board design rules should I keep in mind to let the IC work at 2.0A max?

Board schematics
Board Eagle files

Thank you in advance.

  • \$\begingroup\$ Sense resistor would be 0.088Ohm, if there is such a beast. \$\endgroup\$
    – JRE
    Aug 26, 2016 at 18:40
  • \$\begingroup\$ How nice it would be if you always could increase current capability by switching out some components... \$\endgroup\$
    – PlasmaHH
    Aug 26, 2016 at 19:35

1 Answer 1


I will answer but I want to point out that your panel can only supply 7.15W according to the information in your question. In order to charge a single LiPo cell at 2A, the maximum output power would be 2A * 4.2V = 8.4W. It is physically impossible for output power to be higher than input power. But it is still a good idea to increase the charge rate to 2A. This will allow the charger to extract as much power as it can from the panel.

Here is a link to the datasheet for the part: http://cds.linear.com/docs/en/datasheet/3652fe.pdf

Here is what you should do:

1) Change R1 to 50 mOhms. This sets the charge current.

2) Change D1 to MBRS340 (available from several vendors). The diode in the Sunny Buddy may get over-loaded if you go to 2A charge.

3) Change L1 to 10uH capable of 2.5 to 3A. E.g., Bourns p/n SDE0805A-100M. It is tricky to calculate ripple current for a solar panel, because Vin max does not occur during Iout max. But 10 uH should give you low ripple current.

I did not study the whole design. I am assuming other things are set up in a reasonable fashion (such as the charge timeout and float voltage). I definitely recommend you review the datasheet, and, if Linear has a reference design for this part, review the reference design schematic and any notes or datasheet available for it.

Good luck!

  • \$\begingroup\$ Hi, my intention is not to achieve an higher output, perhaps I didn't explain well this detail in the question. My intention is to set up the circuit for LT3652 in order to best manage the electrical specifications of the solar panel. And given that the IC can manage up to 2A, this would seem the case to use its maximum current capabilities. I agree with you that in the early stages an higher current is required. Bat charging should start at VBAT=3,2V so the current may jump up to 2,23A. My concerns are more about inductor saturation and the components that are involved in the switching. \$\endgroup\$
    – Nic1337
    Aug 27, 2016 at 12:13
  • \$\begingroup\$ Well, it is a DC-DC converter, so the lower the battery voltage, the less input power is needed. As the battery approaches full charge, more input power will be needed. I thought you just wanted to know how to get 2A charge current. That is the question I answered. Maybe you should edit your question to make it more clear. \$\endgroup\$
    – mkeith
    Aug 27, 2016 at 15:40
  • \$\begingroup\$ Dear mkeith, I edited the question, please revise the answer accordingly. \$\endgroup\$
    – Nic1337
    Aug 27, 2016 at 19:18

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