Testing the efficiency of the buck converter at different step down ratios. Muppet 2 project.
By Julian
Youtuber, shed dweller, solar charge controller aficionado
Good morning all…
Youtuber, shed dweller, solar charge controller aficionado
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My question is if i have a source 10v and i put two stepp down to 5v in series and on the other end again in parallel so i converting 5v to 5v on each steppdown so a have a eficiency of 100%??
It is the opinion of some excellent viewers who feel that you have let us all down (a wee bit). Capacitors?
No output capacitor!?
It would have been helpful at the start to mention something about the load. If one is needed, if the value is important etc
it does not increase the current, must be the inductor and freq used. The L used is commonly 150khz on most converters that increase the current when bucking
Did you ever do the vid with the output cap? I can't see one.
Maybe a True RMS Multimeter could help out. Maybe someone could lend you two Gossen Metrahit Energy Multimeters?
Did the capacitor video ever get posted? I just watched the buck converter series but don't see it… Maybe I'm just blind.
How did you decide on a "reasonable" switching frequency?
Do you actually like that sound CRT TVs used to make, or is it rather meant as a joke to annoy someone else?
Could the bump (similar efficiency reading) in the output be because that is near the natural frequency of the circuit?
You stepped power down with voltage, how bout you try to keep power level constant while stepping voltage down (one step at a time), changing two values at the same time does not paint the same picture…
Can you please run the same tests at much lower switching frequency, such as 100 Hz? Semiconductor switching losses will be affecting measured efficiency. And what percentage of the losses are also the result of inductor core losses? Your meters will also be introducing measurement errors, because of the non-sinusoidal waveforms. Finally, can you check the currents by using the same meter on input, then output? That would eliminate the differences in readings between two meters, neither of whose errors are exactly linear wrt current levels. Thanks.
As the voltage output goes down, the current wants to go up, so you should match the impedance to conserve the power input/output.
Think of class D amplifiers.