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enter image description hereDoes the schottky diode current rating affect the performance of a dc to dc buck converter when loaded? Im using a 1N5822 shottky type instead of a 1N5825 on a National Semiconductor LM2596s-Adj based buck converter design. 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. My design is not yet on a PCB, im testing it on a breadboard.enter image description here

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    \$\begingroup\$ Actually, such level of drop indicates that either Cout is too low or DC resistance of L1 is too high. \$\endgroup\$ – Rohat Kılıç Jun 4 '17 at 15:23
  • \$\begingroup\$ how can resistance for L1 be measured? \$\endgroup\$ – Trust Chihowa Jun 4 '17 at 21:15
  • \$\begingroup\$ Datasheet shows actually. Or simply, measure the voltage across it while the circuit is loaded then divide that voltage by the load current. \$\endgroup\$ – Rohat Kılıç Jun 5 '17 at 0:05
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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\$.

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  • \$\begingroup\$ ok thank you. I was testing the circuit with that same 1N5822 didoe and all of a sudden the voltage at the output is higher than that on the input(higher than voltage supplied). Could the Lm2596 be dead. \$\endgroup\$ – Trust Chihowa Jun 4 '17 at 20:55
  • \$\begingroup\$ It's strange: if the IC is dead, the series-pass switch is likely to be shorted. In that case, Vout and Vin are almost equal. If Vout is larger than Vin, either the measurement is wrong or you have unintentionally built a boost which I doubt with that circuit. \$\endgroup\$ – Verbal Kint Jun 4 '17 at 21:31
  • \$\begingroup\$ how do i check if ic is dead. any actual menthod you know \$\endgroup\$ – Trust Chihowa Jun 4 '17 at 22:59
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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.

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