Timeline for Is there a voltage buck converter which also boosts output current?
Current License: CC BY-SA 3.0
10 events
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| Apr 12, 2015 at 19:17 | comment | added | elechris | @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. | |
| Apr 12, 2015 at 19:12 | comment | added | elechris | @tomnexus I didn't see the link of the buck converter because it was not on the next line. The third link was not there when I responded. You are right. That buck converter has an input capacitor. It may be incorrect measurement. | |
| Apr 12, 2015 at 18:45 | comment | added | tomnexus | @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. | |
| Apr 12, 2015 at 17:16 | comment | added | user57037 | 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. | |
| Apr 12, 2015 at 16:45 | history | edited | elechris | CC BY-SA 3.0 |
spelling error
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| Apr 12, 2015 at 16:44 | comment | added | elechris | 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. | |
| Apr 12, 2015 at 16:37 | history | edited | elechris | CC BY-SA 3.0 |
added example
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| Apr 12, 2015 at 15:26 | comment | added | Mayec | 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. | |
| Apr 12, 2015 at 14:58 | review | First posts | |||
| Apr 12, 2015 at 15:26 | |||||
| Apr 12, 2015 at 14:57 | history | answered | elechris | CC BY-SA 3.0 |