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I want to power a bulb from solar panel(17V/.9A) but due to low resistace (2 Ohm)of load i have to reduce the input voltage to 5.5V @ 2.75A for max power. I am using generic buck converter (not any mppt module)to lower voltage but as i studied buck converter has not more output current than input current. So is there any way to increase output current?

Some days ago i found that increasing input capacitance at buck converter can let you increase o/p current. Is it true?

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    \$\begingroup\$ In an ideal Buck power is conserved, so output current increases as much as voltage output decreases.. \$\endgroup\$ – Andrés May 5 '18 at 10:30
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    \$\begingroup\$ You should bear in mind that if the solar panel is rated for 17V/0.9A then it will only produce that in ideal conditions. Most of the time it will produce less. Not enough power for the bulb. If you've measured 17V/0.9A, both at the same time, then that is better. \$\endgroup\$ – Jack B May 5 '18 at 11:00
  • \$\begingroup\$ What is V and W rating of bulb and do you have storage cells? \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 May 5 '18 at 11:24
  • \$\begingroup\$ And why do you want to light a bulb when the sun is shining? \$\endgroup\$ – Transistor May 5 '18 at 11:25
  • \$\begingroup\$ It's a psychological Engineering design challenge because if you can master nonlinear PTC bulb loads then you learn how to regulate MPPT \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 May 5 '18 at 11:26
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Yes it can have a higher output current then input current.
But let's have a look at your numbers:
- Input: 17V 0.9A that is 17*0.9 = 15.3W
- Output: you want 5.5V and 2.75A that is 5.5*2.75 = 15.125 W.
That is lower then at the input so at least we are not violating the law conversion of energy.

Your buck converter then needs an efficiency of 15.125/15.3 = 98.8%.
That however is a challenge, and as far as I know++ not possible for converters of about 15W.

++I have not got much experience with buck converters so I am happy to be corrected.

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    \$\begingroup\$ What?? I know that. \$\endgroup\$ – user3785133 May 5 '18 at 15:15
  • \$\begingroup\$ I just want to know if buck converter can produce more current then input current. Becuase buck converter have dis-continous input current. So o/p current<= i/p current \$\endgroup\$ – user3785133 May 5 '18 at 15:17
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"17V /0.9A" might imply the PV has a source impedance of 17/0.9 = 18.9 Ω
...
so guess what happens with a 2Ω load transformed to a higher impedance.

Due to the law of energy conservation , so Zin (Ω ) is just like transformers so \$P=\dfrac{V_{in}²}{Z_{in}}=\dfrac{V_{out}²}{Z_{load}}\$ (neglecting losses) so \$Z_{in}=(\dfrac{V_{in}}{V_{out}})² * Z_{load}\$

In your case load is 2Ω and the other variables are unknown with this load. So the design objective is to regulate the voltage rate for maximum power transfer (MPPT) to match the PV source impedance \$Z_{min}\text~ \dfrac{V_{oc}}{I_{sc}}\$ at mppt voltage which starts around 82%~85% \$V_{oc}\$ and drops >10% with useable solar power input, under varying solar input which means Z rises with useable lower power input.

If you understand maximum power transfer occurs at match impedances then you will understand one method used to regulate the above Buck impedance ratio. Your Buck uses a series switched inductor so its effective impedance rises with lower duty cycle, d, such that \$Z_L \text{~} 2\pi f*L/d~\$ is an approximation.

I have over-simplified this analysis for you to understand the fundamentals, so that you can examine more detailed design calculations not included in this answer.

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