Here I design a boost converter with 5 V input and 19.37 V output and it is working very well when the source is DC battery but when the source is a solar panel with also 5 V output I have no boost for the voltage did anyone know why this happens??

Solar panel

Here is the solar panel description:

  • Brand: GH Solar
  • Solar panel 10 W
  • Solar cell poly technology
  • Dimensions: 25.5 X 34.5 cm
  • Voltage at Pmax: 17.8 V
  • Current at Pmax: 0.57 A


Here the boost converter's input parameters

  • Vin: 5 V
  • RL: 100 Ω
  • Cout: 470 µF
  • Cin: 470 µF
  • L: 560 µH
  • \$\begingroup\$ Can you clarify the basic setup? The output of the solar panel is not 5V. It is much higher. When you run the boost converter with 5V input, what is the output voltage of the boost converter? \$\endgroup\$
    – user57037
    Commented Jul 13, 2019 at 16:44
  • \$\begingroup\$ my basic set up is 5V/19.37V and for the solar cell i keep it indoor to give me 5 volt \$\endgroup\$
    – HAYA
    Commented Jul 13, 2019 at 22:58
  • \$\begingroup\$ Solar cells need strong illumination to reach their target output power. If the solar cell output voltage is 5V, the cell is not putting out hardly any power at all. \$\endgroup\$
    – user57037
    Commented Jul 14, 2019 at 1:12

2 Answers 2


The huge difference between a battery which is a voltage source and a PV panel which a current source is impedance.

A charge needs a low impedance based on the need for a low voltage drop to rise in current ratio.

Any time voltage is transformed up by N, impedance is also transformed down by N\$^2\$ thus stressing worse your design problem of impedance mismatch.

Your goal , should you wish to learn, is to compute the required PV source impedance to drive your effective load impedance using : Load ESR/N^2=Voc/Isc=Zsource=V/I @ Pmax.

  • for OC=open circuit and SC= short circuit

Z(L)=2pi*f * L


The problem is a solar panel has very different output characteristics to a battery. Usually we would talk about 'max power point tracking' (MPPT) where you control the impedance to extract the most power for the conditions.

At the very least you need to re-think your values. Sunnyskyguy has posted information on that.

Though I think maybe a redesign with an MPPT before your boost converter will give you better results. I've put PVs directly into bucks, but I have never tried going directly to a boost..

  • \$\begingroup\$ A boost increases load current and drops Zin by N^2 so imagine the problem with voltage ratios is N^4 power worse for boost regulators than buck !!! So don’t try it unless you understand Zsource and match the load ALWAYs. So bad idea \$\endgroup\$ Commented Jul 13, 2019 at 6:42
  • \$\begingroup\$ @SunnyskyguyEE75 Yeah, my feeling was that it would be pretty difficult to get good results. But is using MPPT really the best answer? If it's a small/cheap device with well known load characteristics, MPPT adds a lot of complexity (and cost) to what is otherwise a simple problem. \$\endgroup\$
    – hekete
    Commented Jul 16, 2019 at 7:52
  • \$\begingroup\$ You are getting 5V when you should be getting 17.8V so 2/3 of your power is lost .. so run at 30% duty cycle and high enough f to make output current COntinuous \$\endgroup\$ Commented Jul 16, 2019 at 7:56

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