If I directly connect a 10v 500mA solar cell to a 5v-out (switching) buck converter, will I be able to draw out 5v at 1A?

It seems from the theoretical buck diagrams it is not possible, since at any point the PWM is switched off you are losing potential photons that could be harvested, however it seems a simple solution is to just use a capacitor to buffer the solar cell output.

In practice, do typical buck switching regulators already buffer input sufficiently to drive their maximum rated output current? If not, how would I calculate the buffer capacitance needed to achieve maximum power transfer from the solar cell?

 

Theoretically yes, practically no, expect 80-90% efficiency. The input caps on any regular switchmode supply should be adequate to filter the incoming voltage.

Also are you sure that those 10V and 500mA are at the same time? It might also be 10V open circuit voltage and 500mA short circuit current, which will give you much less than 5W at the maximum power point.

 

Not quite. What you're asking for would required 100% efficiency, which isn't possible. 80-90% is the best you can typically hope for.

 

You can come pretty close. With a fully synchronous circuit, oversized fets, and oversized inductor, you can hit 90% without much effort. Add in a limited output current range and things get even better. Linear Technology has demo circuits that hit 95% and above. All of that oversizing isn't cheap, which is why commercial designs sit in the 80's.

ak

 

Sorry yes I was not counting for the buck regulator efficiency, it was more of a theoretical example.

So if I understand correctly I should not need a cap to buffer the low current input in order for the buck to have a higher current output. Thank you all!

Also thanks for pointing that out kubeek, I will have to do some reading on open circuit vs short vs pmp ratings before I buy a solar cell.

Edit: to clarify I'm going to use an IC buck similar to: http://www.digikey.com/scripts/DkSearch/dksus.dll?Detail&itemSeq=149240926&uq=635347432301281242

 

Input capacitance is not directly proportional to efficiency. It is there for either energy storage or filtering. Solar converters typically use MPPT (maximum power point tracking) to increase their efficiency, but that can only be done if you have further power conversions stages down stream (usually a grid tied inverter).

Most of the power loss in a buck converter is due to switching loss in the transistor, but also conduction and switching losses in both the transistor and the diode, as well as I^2R loss in the inductor. When the PWM is off, you don't lose any energy, it just gets transferred to the capacitor. It is only in the act of switching that the transistor dissipates power in the form of heat, which is why switching loss is proportional to frequency. That is one way that capacitance can help you improve efficiency - higher capacitance allows for slower switching frequency which decreases switching loss in the transistor.

 

You can come pretty close. With a fully synchronous circuit, oversized fets, and oversized inductor, you can hit 90% without much effort. Add in a limited output current range and things get even better. Linear Technology has demo circuits that hit 95% and above. All of that oversizing isn't cheap, which is why commercial designs sit in the 80's.
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10v->5v conversion is pretty easy to get efficiency >90%.

You don't need synchronous rectification because the duty is about 55% and the 0.3v-0.4v schottky diode drop is only about a 4.4% energy loss.

The mains things needed are a low Rds-on FET and a low DC ohms inductor like a toroid type.

Most commercial SMPS these days are well above 90% with 95% becoming quite typical.

Most of those cheap ebay SMPS buck converter modules will run better than 90% with 10v->5v conversion.