Does the Schottky diode current rating affect the performance of a DC to DV buck converter when loaded? I'm using a 1N5822 Schottky diode instead of a 1N5825 on a National Semiconductor LM2596S-ADJ based buck converter design.
My output voltage drops when I load the circuit with a 100 Ω or 50 Ω resistor. The voltage drops by nearly 0.25 V and 0.40 V respectively.
My design is not yet on a PCB, I'm testing it on a breadboard.
Yes, the diode voltage drop \$V_f\$ in a buck appears in the input-to-output transfer function \$\frac{V_{out}}{V_{in}}\$ weighted by \$(1-D)\$, \$D\$ being the duty ratio. So if the \$V_f\$ is too big, you will see a drop in the output but the loop should compensate by an increase of the duty ratio. Nevertheless, efficiency will drop as \$V_f\$ grows especially in CCM when \$V_{out}\$ is much smaller than \$V_{in}\$. As the diode voltage drop at a given current is directly related to its die size, a higher current rating type will imply a lower \$V_f\$.
My output voltage drops when i load the circuit by a 100 ohm resistor or 50 ohm resistor. Drops by nearly 0.25V and 0.40V respectively.
Then either your input supply voltage can't supply the current needed to operate the chip correctly under load or your breadboarding and ground techniques are poor. These, and naivety in taking accurate measurements can easily account for the discrepencies you see.
