# Can the output current of a buck converter be more than the input current?

I want to power a bulb from solar panel (17 V, 0.9 A) but due to low resistance (2 Ω) of the load I have to reduce the input voltage to 5.5 V @ 2.75 A for maximum power.

I am using a generic buck converter (not an MPPT module) to lower the voltage. I studied the buck converter, and it doesn't seem to have more output current than input current. Is there any way to increase the output current?

Some days ago I found that increasing the input capacitance of a buck converter can let you increase the output current. Is this true?

• In an ideal Buck power is conserved, so output current increases as much as voltage output decreases.. Commented May 5, 2018 at 10:30
• 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. Commented May 5, 2018 at 11:00
• What is V and W rating of bulb and do you have storage cells? Commented May 5, 2018 at 11:24
• And why do you want to light a bulb when the sun is shining? Commented May 5, 2018 at 11:25
• It's a psychological Engineering design challenge because if you can master nonlinear PTC bulb loads then you learn how to regulate MPPT Commented May 5, 2018 at 11:26

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.

• What?? I know that. Commented May 5, 2018 at 15:15
• 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 Commented May 5, 2018 at 15:17

"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.

The average (and RMS) current will be increased with a buck convertor. Power out = Power in * efficiency (say 90%). Post the source that states otherwise. This answers your question. But doesn't answer your problem...

The specs that you provided for your solar panel are likely the open-circuit voltage (17V) and the short-circuit current (0.9A) @ solar noon (best-case on a bright sunny day). Ironically, neither of these operating points are useful in practice because no power can be delivered operating at these boundaries.

Starting with the open-circuit case, as you start to draw current, you increase power delivered to the load; the voltage will fall. Keep on drawing current, the voltage will fall some more until you get to a peak power point. Draw additional current, voltage will continue to fall, power will still be delivered at falling rate. Then finally, draw all the current possible (short), and you no longer deliver power; the energy gets burned in the PN junctions.

When you factor in efficiency of the buck convertor, the operating points, and calculations based on perfect solar conditions, you don't have enough juice to illuminate your 2 ohm bulb.

An LED would be much more appropriate for solar/battery applications even in less than ideal sun conditions.

Can the output current of a buck current be more than the input current?

The output current of a (standard) buck converter (without an input capacitor) cannot be more than peak input current. However, as other answers have pointed out, the output current of a buck converter is almost always greater than the average input current. In fact, for an ideal buck converter, the average output current is larger than the average input current by the same ratio as the (average) input voltage to the (average) output voltage.

$$\frac{I_{out(avg)}}{I_{in(avg)}} = \frac{V_{in(avg)}}{V_{out(avg)}}$$

The difference in the behavior of the peak current vs the average current is possible because, in a buck converter, the input only supplies current to the inductor for part of a switching cycle, so the average input current is only a fraction of the peak input current.

Peak input current can be an important issue when dealing with a power source that is current limited, such as a solar panel.

Some days ago I found that increasing the input capacitance of a buck converter can let you increase the output current. Is this true?

Adding an input capacitor, or increasing its size, can reduce the peak current drawn from the external supply. The input side capacitor can store charge while the buck switch is off, and then release some of that charge when the buck switch is on, thus allowing the peak current seen by the switch to be larger than the peak current supplied by the source. This in turn can allow the output current to be greater than would be the case if the supply had limited current (such as a solar panel) and the converter had no input capacitor or only a small one.

There are alternative topologies that will also allow buck conversion with less peak input current than output current. One is a flyback converter.

A buck converter's output can, and usually does, have a higher output current than input current. At 100% efficiency, Pin = Pin, so Vin·Iin = Vout·Iout. When you go from a higher voltage to a lower voltage, output current will be larger than input current.

What you are seeing, however, is the buck converter's input impedance pulling the solar panel's voltage down to about 5.5 V. As the input and output voltages of the buck converter are now the same, so will the input and output currents be.

Why does this happen? The solar panel is roughly a current source over most of its characteristics, and the voltage it delivers is determined by the impedance at its output.

Maximum output power is only delivered at the solar panel's MPP, and the input impedance of your buck converter is setting a solar panel voltage that is nowhere near VMPP or the 17 V from the specs, but very near the output voltage of the buck converter.

Now the buck converter's Vin ≈ Vout, so its Iin ≈ Iout.

If you want to increase the output current you will have to force the solar panel to sit closer to its MPP voltage by controlling the buck converter's input impedance; you would be building a MPPT.