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Application: solar USB charger.

I have a 9V solar panel. I use a buck converter circuit to bring it down to 5V before going into the USB battery charging circuit.

Exact components:

http://www.amazon.co.uk/gp/product/B00I0MOVBO http://www.amazon.co.uk/gp/product/B00EYT1DWW

This link has more information about the buck converter circuit (I think it's the same one I bought from amazon above):

http://www.minikits.com.au/LM2596-PSU-01

I was hoping that the buck converter would also increase the output current (say, for example, from 333mA to 450 mA, for a ~75% efficiency), but after measuring it, it seems to track exactly the input current (Ioutput = Iinput). That's why I didn't simply use a 7805 regulator, which I understand dissipates extra power as heat: I was hoping that the buck converter circuit would convert more efficiently, and translate in increased output current.

Are there any alternative buck converters (PCB or a design that I could build myself) which would reduce from 9V to 5V while also increasing the current somewhat? If yes, in practice, would their efficiency be worth it? (I think that anything below ~75% efficiency might not be worth it for such low power).

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    \$\begingroup\$ Any buck converter should be able to produce more output current than it draws as input current. Show your circuit and how you measured the currents so we can start to guess why yours does not. \$\endgroup\$ – The Photon Apr 12 '15 at 13:58
  • \$\begingroup\$ Are you sure you have enough solar input to generate the solar cell's maximum 333 mA output? \$\endgroup\$ – The Photon Apr 12 '15 at 14:03
  • \$\begingroup\$ Do you simultaneaously measure the input and output current and voltage? \$\endgroup\$ – Wouter van Ooijen Apr 12 '15 at 14:30
  • \$\begingroup\$ Yes, at maximum sun exposure, the output current of the solar panel is 333mA. If sun intensity/angle is not optimal, I get lower currents, but the input and output currents are still always the same. I measure the input and output current with a multimeter. I don't measure them simultaneously, but with a 2-3 second delay (the time it takes me to move the contacts from the input connections to the output ones. \$\endgroup\$ – Mayec Apr 12 '15 at 14:58
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    \$\begingroup\$ The meter connections you describe are fine for measuring voltage, but you must not measure current that way. To measure current, you must break the circuit, and insert the meter in that break. The current you want to measure must flow through the meter. \$\endgroup\$ – Peter Bennett Apr 12 '15 at 17:03
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I measure by shorting the converter with the multimeter. So to measure input current/voltage, with the whole circuit connected:I place multimeter black on in-, multimeter red on in+. For output current/voltage: multimeter black on out-, multimeter red on in+ (shorting the battery)

This means you are not measuring the current that the solar cell is delivering to the switching converter. You are short circuiting the cell with the ammeter, and measuring what the cell can provide into a short circuit.

Then you short circuit the converter output and measure what it can provide into a short circuit. The way a buck converter works, when you short its output, it will essentially just short its input to its output. And again you will measure what the cell can drive into a short circuit.

Like Peter Bennett says in comments, in order to know what the currents in and out of the converter are in the operating circuit, you must break the circuit and insert the ammeter between the cell and the converter input; then insert the meter between the converter and the load.

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  • \$\begingroup\$ Ok, then my measurements for the currents were definitely wrong. I will measure again correctly and see if and by how much the converter is boosting my current. \$\endgroup\$ – Mayec Apr 13 '15 at 10:06
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A buck converter doesn't draw a continuous input current. The problem is that your solar cell is specified at the maximum current it can supply continuously, not the average of a pulsed current. In order to draw a more continuous current you have to add filters at the input of the converter. Once the output draws a little too much current the input drops dramatically(remember a solar cell acts almost like a current limited voltage source). Even better than a filter would be to get the maximum amount of power out of the cell by using a maximum power point tracking charger. The battery will act as a huge filter making it possible for your circuit to draw higher peak currents and to operate when there is less or no sunlight.

Example: In case of an ideal buck converter at high switching frequencies if you convert 10 volts to 5 volts 1A you will draw 1A current pulses at a 50% duty cycle, so 500mA average current. If there is no input filter the input source has to deliver the full output current. Beyond the maximum power point the voltage of the cell will rapidly drop. The converter will try to draw a larger current to compensate for this causing an even greater drop.

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  • \$\begingroup\$ Do you mean that the buck converter is actually outputting a higher peak current, but due to the buck's inherent switching/pulse, the average DC current is brought back down to roughly the same current found in the input? Not sure that's what you're saying, but that would explain what I'm seeing in my measuring. \$\endgroup\$ – Mayec Apr 12 '15 at 15:26
  • \$\begingroup\$ A buck converter is simply switching the input rapidly on and of and filters the resulting pulses using an inductor and a capacitor. The input current is therefore pulsed, not continuous. Half the time the solar input is not used. \$\endgroup\$ – elechris Apr 12 '15 at 16:44
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    \$\begingroup\$ It sounds like the OP was using the ammeter function incorrectly. This answer may be on-track, but the OP should get an accurate measurement first. In any event, bucks should always have some kind of input cap close to the high current input point. \$\endgroup\$ – mkeith Apr 12 '15 at 17:16
  • \$\begingroup\$ @elechris the buck converter in question has an input capacitor, so the story about it only using the panel for half the time does not hold. There will be ripple, but not significant in this discussion. A multimeter will read correctly, the average of the current. \$\endgroup\$ – tomnexus Apr 12 '15 at 18:45
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    \$\begingroup\$ @Mayec what load did you use? You say you expect a buck converter to provide more current compared to a linear voltage regulator. But current is determined by the load and the voltage, not topology. If you overload the output causing a drop in voltage the output current will be equal to the input current since the buck converter will get to a 100% duty cycle. \$\endgroup\$ – elechris Apr 12 '15 at 19:17

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